INFORMS
Vol. 00, No. 0, Xxxxx 0000, pp. 000–000
issn 0000-0000 | eissn 0000-0000 | 00 | 0000 | 0001
doi 10.1287/xxxx.0000.0000
c 0000 INFORMS
The Design of Experiential Services
with Acclimation and Memory Decay:
Optimal Sequence and Duration
Aparupa Das Gupta, Uday S. Karmarkar, Guillaume Roels
UCLA Anderson School of Management, 110 Westwood Plaza, Los Angeles, CA 90095, USA
For many consumer-intensive (B2C) services, delivering memorable customer experiences is a source of
competitive advantage. Yet, there are few guidelines available for designing service encounters with a focus
on customer satisfaction. In this paper, we show how experiential services should be sequenced and timed to
maximize satisfaction of customers who are subject to memory decay and acclimation. We find that memory
decay favors positioning the highest service level near the end, whereas acclimation favors maximizing the
gradient of service level. Together, they maximize the gradient of service level near the end. Although memory
decay and acclimation lead to the same design individually, they can act as opposing forces when considered
jointly. Overall, our analysis suggests that short experiences should have activities scheduled as a crescendo
and duration allocated primarily to the activities with the highest service levels, whereas long experiences
should have activities scheduled in a U-shaped fashion and duration allocated primarily to activities with
the lowest service level so as to ensure a steep gradient at the end.
Key words : service design, experience, scheduling, social psychology, behavioral operations
1.
Introduction
As epitomized by Walt Disney’s entertainment parks and Benihana’s restaurants, firms that deliver
outstanding customer experiences achieve greater customer satisfaction, and therefore greater customer loyalty (Voss and Zomerdijk 2007, DeVine and Gilson 2010, DeVine et al. 2012), which
ultimately drives future sales (Heskett et al. 1997).
Because service experiences are dynamic processes, the utility customers derive from an experience evolves dynamically, in response to varying stimuli (Bitran et al. 2008, Baucells and Sarin
2012). For instance, Figure 1 depicts a map, over time, of the customer experience at Starbucks.
The total experienced utility of a customer from the service will therefore integrate all these instantaneous utilities experienced throughout the service encounter (Edgeworth 1881).
The customers’ remembered utility, which drives their future purchase decisions, may however
differ from their total experienced utility (Kahneman et al. 1997). In particular, their remembered
1
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Figure 1
Map of Starbucks in-store customer experience.
Note. Source: Lee (2008). The blue stars represent subprocesses that are both important to the customer and differentiate theEmotion
brand.Curve–Credit
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are Multiple
either Touch-Points
not important
to the customer or do not differentiate the brand.
Experience
What do your customers remember? The memories of Starbucks customers are filled with her core brand values
utility from
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son 2000).
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effect) due to attention decrement (Crano 1977) or assimilation (Asch 1946). Behavioral biases also
affect how much (instantaneous) utility is derived from an outside stimulus. For instance, acclimation makes people adapt to states but react to changes (Hsee and Tsai 2008), satiation reduces
the marginal utility obtained from accumulated consumption (Baucells and Sarin 2013), and loss
aversion weights losses more than gains (Tversky and Kahneman 1991). Savoring and dread may
furthermore affect the utility experienced before the service delivery (Loewenstein 1987).
As a result, maximizing customer satisfaction in services requires a deep understanding of customer behavior (Chase and Dasu 2008), both during and between service encounters (Bitran et al.
2008, Watkinson 2013). In fact, Cook et al. (2002) argue that, by drawing on concepts from psychology and sociology, service encounters can be designed with the same depth and rigor as manufacturing processes. However, there exist to date few guidelines for designing service experiences,
despite their ubiquity in today’s economy (Pine and Gilmore 1998).
In this paper, we show how to sequence and allocate duration to activities in a service encounter
so as to maximize the satisfaction of a customer who is subject to acclimation and memory decay.
Acclimation makes people adapt to states but react to changes (Hsee and Tsai 2008); because
of acclimation, the first bites of a dish are usually the most enjoyable (Miller 2014). Memory
decay, on the other hand, makes people forget what happened in the past; accordingly, the last
moments of an encounter are the most memorable. We focus on acclimation and memory decay for
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the following reasons. First, they have a long tradition in psychology and have therefore received
strong empirical support. Second, these seemingly simple biases are sufficiently rich to explain
classical empirical findings on remembered utility, such as that “more pain can be preferred to
less” (Kahneman et al. 1993) or that interrupting a pleasant experience with an annoying break
can result in greater satisfaction (Nelson and Meyvis 2008). Third, they appear mathematically
symmetrically in our models, and we therefore offer a joint treatment. Nevertheless, we acknowledge
that other behavioral biases may also be salient in service experiences, and we leave it for future
research to investigate how our service design results are affected by these other biases.
To answer our research question, we build a stylized model of a service encounter with the
following characteristics:
• The encounter consists of a given set of activities that have a common acclimation parameter
and has a fixed total duration.
• Each activity is associated with a given unidimensional service level, independent of its place-
ment in the sequence and of its duration.
• Customers are captive, i.e., customers cannot arrive after the beginning of the encounter or
leave before its end.
• The provider, and not the customer, has control over the sequencing of activities and/or the
allocation of duration to them.
Although stylized, our model could apply, as a first-order approximation, to an executive education program on a specific topic (e.g., “Current trends in supply chain management”). The different
activities would correspond there to the different sessions offered (e.g., sustainability, offshoring),
and their respective service levels could be measured by their historical teaching ratings. In that
context, both memory decay and acclimation are salient (Hoogerheide and Paas 2012). In practice,
however, student satisfaction may be based on multiple attributes (Marsh 1984). Moreover, the
program director may have additional levers than just the scheduling and duration of each class
to influence student satisfaction (such as company visits and the timing of the main reception).
Although our stylized model ignores those dimensions, we expect that our managerial insights
would remain applicable to those more complex settings.
Other examples of service encounters where sequence and duration of events affect the overall
experience are concerts, magic shows, museum exhibitions, fireworks, city walking or bus tours,
cruises, spa treatments, fitness classes, or medical procedures like colonoscopy and dental visits.
We find that a service design that is optimal for memory decay alone is also optimal for acclimation alone, even though memory decay and acclimation are fundamentally different mechanisms. In
particular without precedence constraints, it is optimal to sequence activities in increasing order
of service levels so as to finish at a peak and, with variable durations, to lengthen the duration
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of those activities with the highest service level. Hence, memory decay and acclimation act in the
same way when considered individually.
However, when memory decay and acclimation are considered jointly, we find that they may
act in opposite directions. Specifically, when both phenomena are present, it may be optimal to
sequence activities in a U-shaped fashion, beginning and finishing with high service levels, and to
lengthen the duration of activities with the lowest service level.
Because memory decay favors a high service level near the end and acclimation favors a steep
rise in service levels, the combination of these effects favors a steep gradient of service level near
the end of the encounter. Hence, the often-heard “finish strong” recommendation (Chase and Dasu
2001) should be understood as “finish with a steep ascent” in the presence of acclimation and
memory decay. A grand finale becomes even more memorable when it comes just after a respite.
Overall, our analysis suggests that short experiences should have activities scheduled as a
crescendo and duration allocated primarily to the activities with the highest service levels, whereas
long experiences should have activities scheduled in a U-shaped fashion and duration allocated
primarily to activities with the lowest service level so as to ensure a steep gradient at the end.
The remainder of this paper is organized as follows. In §2, we review the related literature. In §3,
we introduce the model for customer satisfaction. In §4, we characterize the optimal solution for
the sequencing and duration allocation problem in three different settings, depending on whether
the sequence and durations are fixed or variable. In §5 we numerically compare different sequencing
heuristics when customers are heterogenous in terms of their rates of acclimation and memory
decay. We present our concluding remarks in §6. The proofs appear in appendix. The mathematical programming formulation of the problem and heuristic algorithms appear in an electronic
companion.
2.
Literature Review
This paper is related to three bodies of research: the experimental study of human behavior in
psychology and marketing, the analytical models of experienced utility in decision theory, and the
prescriptive analytical models of service design in operations management. We next review these
three streams of literature and position our paper accordingly.
2.1.
Behavioral Aspects
Research in psychology and marketing has identified many factors that influence the retrospective
evaluation of experiences, including the trend of an experience (Loewenstein and Prelec 1993), its
rate of change (Hsee and Abelson 1991), and its maximum and final intensities (Frederickson and
Kahneman 1993). Underlying these factors are psychological phenomena such as acclimation, loss
aversion, or memory decay. We contribute to this literature by adopting a design perspective on
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two such phenomena, namely, memory decay and acclimation. We then relate these two phenomena
to the celebrated “peak-end” rule in retrospective evaluation.
2.1.1.
Memory Decay. People naturally tend to forget past events more than recent ones.
The most prevalent models of memory decay are an exponential and a power models; see Anderson
(1995) for a review. The exponential model of memory decay was constructed by Ebbinghaus
(1913) from his study on how much memory of monosyllabic words was retained after several days.
Bahrick (1984) also report an exponential decay of memory of Spanish words over a time span of
50 years, though memory ultimately does not tend to zero, but remains constant after 3-6 years.
Wickelgren (1974) derives a combined exponential-power model of memory decay and argues that
the exponential term provides a better fit for short-term memory data and the power term is a
better fit for longer-term memory data. Nevertheless, the difference between the exponential and
the power models of memory decay is often marginal. For instance, Wickelgren (1974) report that
both models account for 99% of the variance of the original data collected by Ebbinghaus.
Moreover, exponential models are widely used. For instance, Naik (1999) uses an exponential
decay model to estimate the half-life of advertising campaigns and Watt et al. (1993) apply it to
measure the salience of news. Surveying marketing models of awareness of a new product introduction, Mahajan et al. (1984) report that almost all models explicitly capture forgetting, and the
effect is in general exponential. In this paper, we adopt Ebbinghaus’ exponential model of memory
decay due to its simplicity and relevance.
The rate of memory decay is, in general, context-dependent (Anderson 1995): While some experiments show memories over a few seconds (Anderson 1995), data from Ebbinghaus (1913) is over
several days, Naik (1999) reports that the half life of an advertising campaign can be as long as
three months, and Bahrick (1984) reports a half-life span of multiple years.
2.1.2.
Acclimation. Acclimation makes people adapt to states but react to changes (Hsee
and Tsai 2008). This phenomenon is ubiquitous in physiological processes, going as far back as
Newton’s law of cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (Incropera
et al. 2011). Acclimation is often referred to as adaptation, though life sciences tend to distinguish the two phenomena by associating the former to changes within an organism’s lifetime and
the latter to changes across several generations of a particular species. Besides physiological processes, Helson (1964) suggests that adaptation also governs cognitive processes and Brickman and
Campbell (1971) propose that adaptation also governs emotions, giving rise to a hedonic treadmill.
For instance in waiting lines, customers have been reported to become gradually demoralized as
they wait, but to positively respond to each advance of the queue (Carmon and Kahneman 1995).
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In learning experiences, moderately difficult material generates more utility if it follows difficult
material than if it follows easy material (Hoogerheide and Paas 2012). Tversky and Griffin (1991)
highlight the duality between endowment and contrast: Positive (negative) experiences make us
happy (unhappy), but also render similar experiences less exciting (less bad). Hsee and Abelson
(1991) emphasize the importance of the rate of change, as opposed to just the contrast, in satisfaction. Because of acclimation, the satisfaction from a pleasant experience may actually be higher
when the experience is interrupted by an annoying break than with no interruption (Nelson and
Meyvis 2008, Nelson et al. 2009).
Similar to memory decay, the rate of acclimation is highly context dependent. For physiological
processes, complete adaptation is often in the order of minutes (Gent and McBurney 1978, Overbosch 1986, Dalton and Wysocki 1996) or even seconds (Dawes and Watanabe 1987). On the other
hand, the experiments on commercial interruptions by Nelson et al. (2009) span several minutes.
Yet, other research fails to find contrast in hedonic consumption (Novemsky and Ratner 2003).
2.1.3.
Peak-End Rule. Remembered utility and experienced utility are different constructs
and require different modeling approaches (Kahneman et al. 1997). On the one hand, the total
experienced utility, which has normative appeal, integrates all (undiscounted) instantaneous utilities (Kahneman et al. 1997), and therefore weights them by their respective durations, in the same
spirit as Edgeworth (1881); see Kahneman et al. (2004) for practical measurement methods.
On the other hand, the remembered utility from an experience, which governs future decisions
(Kahneman et al. 1993), is highly correlated with its maximum and final intensities, a phenomenon
called “the peak-end rule;” see Varey and Kahneman (1992), Frederickson and Kahneman (1993)
and Fredrickson (2000). As a consequence, remembered utility is typically not significantly affected
by the duration of the experience and may violate temporal monotonicity, in the sense that adding
more pain may result in greater remembered utility (Kahneman et al. 1997). In particular, in waiting lines, a long queue that ends with a very rapid advance may elicit a more favorable retrospective
evaluation than a shorter queue (Carmon and Kahneman 1995). In learning experiences, adding
moderately difficult material to a difficult lesson may make the overall lesson perceived as easier
to learn (Hoogerheide and Paas 2012).
Although the peak-end rule is a well accepted paradigm, it is only a “good first approximation”
(Kahneman 2000a). In particular, the peak-end rule suggests that the remembered utility from an
experience is independent of the sequence of all but the last activity, i.e., irrespective of whether
there are multiple peaks, valleys, breaks and intermissions, etc. However, subsequent research has
shown the importance of both the valence of trends (Loewenstein and Prelec 1993, Ariely 1998,
Rozin et al. 2004, Dixon and Verma 2013) and their rates (Hsee and Abelson 1991, Baumgartner
et al. 1997), as well as, in some cases, of duration (Ariely 1998, Schreiber and Kahneman 2000).
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Instead of enhancing the peak-end rule to account for trend and/or duration, as is done in some
studies (e.g., Dixon and Thompson 2013b), we propose here an alternate model of remembered
utility. Specifically, we assume that the remembered utility is the discounted sum of instantaneous
utilities, where discounting applies backwards in time, consistent with the memory decay models
reviewed in §2.1.1. In this stylized model of remembered utility, the end utility carries greater
weight, because it is the least discounted; and the peak utility has more impact than the other
utilities by definition of it being a peak. Moreover, temporal monotonicity may also be violated in
our model, as we illustrate in §3.1.1, and trends may affect remembered utility. Hence, our model
captures the key features of the peak-end rule, as well as trends and (if the discount rate is not too
high) durations. Moreover, our model is rooted on a fundamental behavioral bias, namely memory
decay, unlike the peak-end rule, which captures only its symptoms.
2.2.
Decision Theory
Kahneman et al. (1997) propose that there exist two kinds of experienced utility, the one reported
in real time (instantaneous) and the other based on retrospective evaluation (remembered). Using
this idea, our model represents the remembered utility as a function of the instantaneous utility.
Following Constantinides (1990), Wathieu (1997) proposes a model of habit formation, which
adapts the reference level to the current consumption level. Baucells and Bellezza (2012) extend
that model to account for anticipation and recall and Baucells and Sarin (2013) embed it into a
general framework for happiness. Consistent with Wathieu’s habit formation model, our acclimation
model adapts the reference level to the current service level.
We contribute to this literature by adopting a design perspective and by focusing on maximizing
(ex-post) remembered utility as opposed to (ex-ante) total experienced utility. There is indeed a lot
of evidence that people make decisions based on their memories of experiences (Kahneman et al.
1997). Accordingly, services should be designed to enhance future memories (Watkinson 2013).
Because of our different objective, our results differ. For instance, Wathieu (1997) shows that in
the presence of acclimation the optimal consumption pattern that maximizes ex-ante discounted
utility should be decreasing, increasing, or U-shaped, so as to trade off time discounting (favoring
immediate consumption) with acclimation (favoring a positive gradient of consumption). In contrast, we show that, with retrospective evaluation, the optimal sequence could be either increasing
or U-shaped, but never decreasing. Baucells and Sarin (2007) and Baucells and Sarin (2010) show
that U-shaped sequences maximize total experienced utility in the presence of satiation, respectively with time discounting and with acclimation. We corroborate their result with memory decay
and acclimation by taking the perspective of remembered utility.
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2.3.
Service Operations
Analytical models of service operations typically focus on managing capacity (Gans and Zhou
2003), scheduling resources (Pinedo 2005, Sampson and Weiss 1995), and pricing (Talluri and Van
Ryzin 2004). In the context of service design, Bellos and Kavadias (2014) study how to allocate
tasks between the service provider and the customer. In contrast to their model, which considers
customer experience as random, we adopt here a behavioral model of customer experience.
Lately, some papers have studied how operational decisions are affected by customer behavior.
The literature in this area is diverse in terms of modeling paradigms and behavioral regularities.
In particular, Nasiry and Popescu (2011) study pricing and Dixon and Verma (2013), Dixon and
Thompson (2013a,b) study event scheduling decisions, and Verhoef et al. (2004) study call center
satisfaction in the light of peak-end rule. Aflaki and Popescu (2013), Popescu and Wu (2007), and
Nasiry and Popescu (2011) study how loss aversion affects pricing and service level policies. Caro
and Martı́nez-de-Albéniz (2012) show how satiation affects assortment decisions. Ely et al. (2013)
characterize the optimal revelation of information when customers have preference for suspense or
surprise. Adelman and Mersereau (2013) study how customer memory effects impact a supplier’s
profits when the supplier dynamically allocates limited capacity among a portfolio of customers.
Focusing on queueing experiences, Plambeck and Wang (2013) consider the effect of hyperbolic
discounting on unpleasant services and Carmon et al. (1995) show how customer dissatisfaction
can be reduced by spreading the service across the waiting period in queues.
Our work is most closely related to Aflaki and Popescu (2013) and Dixon and Thompson
(2013a,b). Aflaki and Popescu (2013) characterize the optimal service level between service encounters so as to maximize long-term customer retention and profitability. By contrast, we focus on
activities within a service encounter, assuming that customers are captive, and we control the
duration and sequence of activities with fixed service levels. Hence, our approach can be viewed as
complementary to theirs. Dixon and Thompson (2013a,b) study an event scheduling problem by
creating bundles of events in the presence of peak-end, trend, and other outcomes of psychological
phenomena. In contrast, we directly model the psychological phenomena, and not their outcomes.
Moreover we adopt an analytical approach, whereas theirs is computational.
3.
Model
We consider a service provider who seeks to maximize ex-post customer satisfaction from a service
encounter. The service encounter consists of a sequence of n activities. Each activity i ∈ {1, .., n}
Pn
is designed to offer a fixed service level xi for a duration ti , with i=1 ti = T . We assume that
the service level is unidimensional and is independent of the activity’s placement in the sequence
or its duration. For instance, the executive education program in supply chain management may
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9
Acclimation affects instantaneous utility and memory decay determines its relative weight in the overall
remembered utility.
Acclima2on!
!Service!
!!!Level!
!!!!!x(t)%
Memory!Decay!
Instantaneous!
!!!!!!!U2lity!
!!!!!!!!!u(t)%
Overall!
customer!
sa2sfac2on!!S%
assessed!at!
!!2me!T%
consist of different sessions (e.g., sustainability, offshoring), each with a specific duration (e.g., one
or two hours) and a specific service level (e.g., based on historical teaching ratings). Our model is
applicable to both positive and negative values of the service level xi .
A service design is characterized by the following two design variables: 1) the sequence of activities
(i1 , .., in ) where ik ∈ {1, .., n} is the kth activity in the sequence and 2) the duration for each activity
Pk
ti , ∀i. For any sequence (i1 , .., in ), let Tk = j=1 tij be the time passed by the end of the kth activity
Pn
and T k = j=k tij be the time remaining from the beginning of the kth activity to the end of the
encounter. We next build the customer satisfaction utility model and then formulate the service
provider’s design problem.
3.1.
Customer Satisfaction Utility
We consider a customer who is subject to memory decay and acclimation. We denote by uik (t) the
customer’s utility at time t while experiencing the kth activity in the sequence, with Tk−1 < t ≤ Tk .
The customer satisfaction from the whole experience, evaluated at the end of the encounter, will
then integrate all such instantaneous utilities. Acclimation affects how much instantaneous utility
uik (t) is obtained from each activity, whereas memory decay determines the relative weight of the
contribution of uik (t) in the overall remembered utility S((i1 , . . . , in ), t), as illustrated in Figure 2.
3.1.1.
Customer Memory Process. We assume that, in the satisfaction, i.e., remembered
utility, from a service encounter, the instantaneous utilities experienced throughout the encounter
are weighted by their recency. In contrast, the peak-end rule gives positive weight only to the
peak and the end utilities, with no consideration of their duration (Fredrickson 2000). For the
memory decay process, we use the exponential forgetting model proposed by Ebbinghaus (1913).
According to this model, the memory from an encounter weights activities at the end of the
encounter more heavily than the ones at the beginning. Accordingly, memory decay operates like a
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Figure 3
Because of memory decay, adding more pain at the end can result in greater remembered utility.
9
8
8
7
7
6
Discomfort
-u(t)
9
Patient A
Remembered
discomfort
S = -6.9
6
Remembered
discomfort
S = -12.1
5
Patient B
Discomfort
-u(t)
4
5
4
3
3
2
2
1
1
0
0
0
5
10
15
20
25
0
time
5
10
15
20
25
time
Note. To measure overall satisfactions (S), we assume w = 0.3 per minute.
backward discounting process. When the customer discounts past experiences exponentially with
rate w ∈ [0, ∞), her remembered utility or satisfaction from the service experience is given by
n Z Tk
X
S((i1 , .., in ), t) =
uik (t)e−w(T −t) dt.
(1)
k=1
Tk−1
As an illustration, we apply Equation (1) to the data collected during the colonoscopy experiment
by Redelmeier and Kahneman (1996). Figure 3 plots the instantaneous (dis)utility reported by two
patients experiencing two different colonoscopy procedures, one short (Patient A) and one long
(Patient B). Assuming a memory decay rate of w = 0.3 per minute (which corresponds to a half
life of about 2.3 minutes), we find that Patient A experiences a total discomfort of S = −12.1,
whereas Patient B experiences a total discomfort of S = −6.9.1 Hence, our memory decay model
explains why, in that experiment, the patient subject to the longest procedure reported overall
less discomfort than the patient subject to the shortest procedure, similar to what Redelmeier
et al. (2003) reported. Although there exist of course other interpretations of that result (including
acclimation), this illustrates that memory decay alone is rich enough to explain the surprising
outcome that “more pain can be preferred to less” (Kahneman et al. 1993).
3.1.2.
Customer Acclimation Process. We next introduce our acclimation model, follow-
ing the adaptation model of Wathieu (1997) and Baucells and Sarin (2013). As illustrated in Figure
2, the acclimation process affects how much instantaneous utility is derived from a service level.
Let b(t) be the customer’s reference level at time t. The instantaneous utility experienced at time
t is a function of the difference between the current service level and the reference point:
uik (t) = U (xik − b(t))
1
We assume here that patients evaluate their satisfaction S just at the end of the procedure. In the experiment, they
were asked to report it within an hour after the procedure. Although some memory decay could also happen during
that relapse period, potentially at a different rate than the one prevailing during the operation (Baucells and Bellezza
2012), our result would continue to hold in that case, provided that both patients had the same relapse period.
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Hence, the instantaneous utility, uik (t), captures both the valence (good or bad) and the intensity
(mild to extreme) of the instantaneous experience (Kahneman 2000b).
The rate of change of the reference point is proportional to the difference between the service
level and the reference point, akin to Newton’s law of cooling and other adaptation models in life
sciences (e.g., Overbosch 1986):
db(t)
= α(x(t) − b(t)),
dt
where α > 0 is the rate of acclimation. A high value of α implies that the customer is less influenced
by past service levels. Solving this differential equation for activity ik yields,
b(t) = xik − (xik − b(Tk−1 ))e−α(t−Tk−1 ) , Tk−1 ≤ t ≤ Tk ,
(2)
where the initial reference level b(0) captures the history of past experiences.
We assume that the activities in a service encounter have a common acclimation parameter,
such as different sessions in a one-day executive program or different art galleries in a guided
museum tour. Accordingly, the reference point at the end of an activity carry over to the beginning
of the next activity, i.e., lim b(t) = lim b(t). If activities potentially have different parameters for
t↑Tk
t↓Tk
acclimation (e.g., waiting for a table and eating dinner), then the reference point could be discontinuously reset once a new activity is started. Baucells and Sarin (2010) make a similar assumption
to generate insights.
In addition, we assume that the utility function is linear, U (x − b) = u0 + (x − b), where u0 is the
intrinsic utility from the experience, normalized to zero in the sequel since it does not affect the
design decisions. In practice, utility could be nonlinear, potentially exhibiting different behaviors
for gains and losses (Kahneman 2000b). Moreover, it could be non-monotone. For instance, with
temperature, it is likely that one’s utility would peak when the temperature is close to 70◦ F . In
those cases, a linear model can be viewed as a first-order approximation of the utility function
around x, if the variations in service levels tend to be within the same order of magnitude.
Figure 4 illustrates the acclimation process under those assumptions for a service encounter
consisting of three activities such that x2 > x3 > x1 . The acclimation level b(t) always trails the
service level. Therefore, for a constant service level x(t), the instantaneous utility decreases over
time and eventually becomes zero (Ariely 1998). In contrast, an upward or downward jump in
service level across consecutive activities leads to a rapid increase or decrease in the instantaneous
utility level, respectively. Hence, our simple acclimation model captures the observation that people
adapt to states but react to changes (Hsee and Tsai 2008).
Under the assumption of same rate of acclimation for all activities, we can expand (2) to obtain
a closed-form expression for the reference level:
k
X
b(t) = xik − (xi1 − b(0)) +
(xij − xij−1 )eαTj−1
j=2
!
e−αt , Tk−1 ≤ t ≤ Tk .
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Figure 4
The acclimation process leads to a decay in utility over time for a fixed service level and yields discontinuous jumps in utility after a change in service level.
Figure 5
Interrupting a positive experience disrupts acclimation, making the experience more enjoyable.
Without Interruption
90
80
70
u(t)
60
50
Intensity of Enjoyment
Intensity of Enjoyment
70
x(t)
40
30
20
b(t)
10
u(t)
60
50
x(t)
40
30
20
b(t)
10
0
0
0
-10
With Interruption
90
80
10
20
30
40
50
60
0
Time (seconds)
-10
10
20
30
40
50
60
70
Time (seconds)
Note. In this experiment, we assume α = 1.8 per minute.
Therefore, because uik (t) = xik − b(t), we obtain
uik (t) = (xi1 − b(0))e
−αt
k
X
+
(xij − xij−1 )e−α(t−Tj−1 ) , Tk−1 ≤ t ≤ Tk .
(3)
j=2
The instantaneous utility, uik (t), can thus be expressed as a discounted sum of the past changes
in service levels, consistent with the intuition that acclimation depends on the gradient of service
levels.
As an illustration, we apply the acclimation model to replicate the results from the break insertion
experiment by Nelson and Meyvis (2008, p. 660). In their experiment, the inclusion of a break
(e.g. an irritating guitar feedback) in an enjoyable experience (e.g., a playlist of songs) disrupts the
acclimation process. Since in their experiment the break is of the same nature as the experience,
we consider the same acclimation rate throughout, and we assume it to be set to α = 1.8 per
minute (which corresponds to a half-life of 0.4 minutes). As shown in Figure 5, the acclimation
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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model indicates that if there is no break (left figure), the instantaneous utility gradually decreases
over time. Whereas if there is an unpleasant break (right figure), the instantaneous utility first
gradually declines, then falls sharply during the break, and then goes back up at the end of the
break. Moreover, at the end of the experiment, the utility of the customers who experienced the
unpleasant break (right) dominates the utility of the customers who didn’t experience the break
(left), consistent with Figure 3 in Nelson and Meyvis (2008). Hence, our simple acclimation model
can explain why inserting unpleasant breaks may result in higher utility following the break.
Moreover, we can show that, with both acclimation and memory decay, the overall satisfaction
from the experience may be higher with the unpleasant break than without, as was empirically
observed by Nelson and Meyvis (2008).
3.1.3.
Customer Satisfaction Model. Combining (1) and (3) leads to the following model
of cumulative remembered utility in the presence of acclimation and memory decay:
n Z Tk
n
X
X
(e−αT k − e−wT k )
−w(T −t)
.
S((i1 , .., in ), t) =
uik (t)e
dt =
(xik − xik−1 )
w−α
k=1 Tk−1
k=1
(4)
Equation (4) reveals that, under our modeling assumptions, memory decay and acclimation, despite
being fundamentally different mechanisms, play a symmetric role mathematically on ex-post satisfaction. Specifically, in (4), the customer weights a change in service level from the (k − 1)th to
the kth activity by
. e−αT k − e−wT k
Φ(α, w, T k ) =
=
w−α
Z
Tk
e−αt e−w(T k −t) dt.
(5)
xik (Φ(α, w, T k ) − Φ(α, w, T k+1 ))
(6)
0
Rewriting (4) as follows:
S=
n
X
k=1
(xik − xik−1 )Φ(α, w, T k ) =
n
X
k=1
indicates that memory decay, which is fundamentally a weighting of the activity service levels xik ,
can also be interpreted as a weighting of the activity increments (xik − xik−1 ), and, conversely,
that acclimation, which is fundamentally a weighting of the service level increments, can also be
interpreted as a weighting of the service levels themselves.
3.2.
Service Provider’s Design Problem: Sequence and Duration
We consider a service provider who seeks to optimize the design of a service encounter to maximize
ex-post customer satisfaction. We assume that the population of customers is homogeneous, i.e.,
they have the same values of the parameters α and w, and we test the robustness of our results to
that assumption in §5. We consider two design variables, namely the sequencing and the duration
of activities.
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Due to resource limitations, activity durations are typically constrained. In particular, we assume
that activity durations must lie within certain bounds, i.e., τ i ≤ ti ≤ τ i such that τ i ≥ 0, τ i > 0, ∀i.
When τ i = 0, we are effectively allowing for activity i not to be included in the encounter. Moreover,
we consider a situation in which the total duration of the service encounter is fixed to T , i.e.,
P
2
i ti = T , such as a one-day executive program.
In addition to (or instead of) controlling the duration of activities, a service provider may also
have some control over the sequence of activities. The admissible sequence of activities can be
restricted by precedence constraints; for instance, a fundamental course in operations management
must be taken before an elective. Let P be the set of all possible feasible sequences of activities.
Then, using (4), the service provider’s design problem (SPDP) is given by
max S((i1 , ., in ), t) =
(i1 ,..,in ),t
subject to
Pn
k=1 (xik
− xik−1 )Φ(α, w, T k )
τ i ≤ ti ≤ τ i ,
Pn
i=1 ti = T
(7)
∀i
(i1 , .., in ) ∈ P .
(8)
(9)
(10)
In the sequel, we will denote the optimal sequence by (i?1 , .., i?n ) and the optimal duration by t? . In
the electronic companion, we propose a mathematical programming formulation of SPDP.
Different service industries may face different degrees of flexibility regarding the sequencing
and the allocation of duration to activities. In certain industries (e.g., music performances), the
sequence of activities is variable, but their duration is fixed (VSFD). In SPDP, this can be modeled
by setting τ i = τ i , ∀i. In other industries (e.g., spa treatment), the sequence of activities is fixed,
but their duration is variable (FSVD). In SPDP, this can be modeled by having only one element
in P . Finally, other industries (e.g., fitness classes) are characterized by both variable sequences
and variable durations (VSVD). Overall, this leads to three different design problems, see Table 1.
To obtain a general characterization, we will assume that, whenever the sequence is variable, i.e.,
in VSFD and in VSVD, there are no precedence constraints, i.e., P is the set of all permutations.3
2
With variable total duration, one would also need to explicitly model the pricing decision and how it relates to the
customer’s expectations, since different durations (e.g., a 1/2-day or 1-day executive program) are typically associated
with different prices and therefore with different customer’s expectations. In contrast, the durations of the activities
within the encounter (e.g., sessions on specific topics) are typically not specified contractually (within certain bounds)
and are typically not reflected in the price of the service. We leave it for future research to explore the interrelationship
between total duration, pricing, and customer expectations. In case there is no such interrelationship, such as in
colonoscopy procedures (Redelmeier et al. 2003), our model can easily be extended to accommodate variable durations.
3
Because our design insights remain identical between VSFD (no precedence constraints) and FSVD (fixed sequence),
we expect them to carry over to cases of variable sequences with precedence constraints.
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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15
Three stylized models of a service encounter.
Sequence
Duration
Fixed
Variable
Variable Sequence-Fixed Duration
(VSFD)
Fixed
Variable
4.
session chair at a conference, music performance
Fixed Sequence-Variable Duration
(FSVD)
Variable Sequence-Variable Duration
(VSVD)
spa treatment, dental procedure
personal fitness class, museum tours, fireworks
Analysis
In this section, we characterize the optimal sequence and duration allocation solutions. After some
preliminaries, we sequentially consider VSFD, FSVD, and VSVD. Underlying the optimal design
in each scenario are the following two properties of memory decay and acclimation:
• Memory decay favors a high service level at the end of the encounter.
• Acclimation favors large positive gradients for successive service levels.
Together, they maximize the gradient of the service level near the end of the service encounter.
4.1.
Preliminaries
We first characterize three properties of the weighting function Φ(α, w, t), defined in (5), which will
allow us to characterize next the optimal service designs.
First, the weighting function is symmetric in α and w, i.e., Φ(α, w, t) = Φ(w, α, t). Consequently,
a design that is optimal with pure acclimation α and no memory decay is also optimal with pure
memory decay, with w = α, and no acclimation. That is, acclimation and memory decay, despite
being fundamentally different phenomena, lead to same designs when considered individually.
In fact, this mathematical symmetry allows us to reduce the problem dimensionality. Instead of
characterizing a problem instance by a pair of acclimation and memory decay rates (α, w), it turns
out that we can characterize it by a single number, namely the logarithmic mean (Carlson 1972)
of the acclimation rate and the memory decay rate:

0
.
m(α, w) = α
 w−α
ln w−ln α
if α = 0 or w = 0,
if α = w,
otherwise.
Accordingly, all our design results will be expressed in terms of m(α, w). As depicted in Figure
6, m(α, w) is symmetric and increasing in both α and w. Moreover, it lies between the geometric
√
and the arithmetic means, i.e., α · w ≤ m(α, w) ≤ (α + w)/2, for any α, w ≥ 0.
Second, Φ(α, w, t) is pseudo-concave in t and peaks when t = 1/m(α, w); see Lemma A-1 in
appendix. Accordingly if the service encounter consists of only one activity, i.e., n = 1, such that S =
(x1 − b(0))Φ(α, w, t1 ), and x1 > b(0), satisfaction peaks when the duration is 1/m(α, w). Intuitively,
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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Figure 6
The logarithmic mean m(α, w) is symmetric and tends to zero when either α or w tends to zero.
2
1.8
1.6
1.4
m(α,w)
1.2
1
0.8
0.6
0.4
0.2
0
2
w
1
0
0
0.2
0.4
0.6
0.8
α
1
1.2
1.4
1.6
1.8
2
the duration of the activity must be long enough to give customers an opportunity to enjoy it, but
short enough to avoid having them acclimating to it.
Third, Φ(α, w, t) is concave-convex in t, and its inflection point occurs at t = 2/m(α, w); see
Lemma A-3 in appendix. Based on (6), this shows that the weights on service levels, Φ(α, w, T k ) −
Φ(α, w, T k+1 ), are decreasing up to time T k = 2/m(α, w) and then increasing.
Using these three properties, we next characterize the optimal design in terms of the maximum
of Φ(α, w, T k ) (i.e., 1/m(α, w)) and of its inflection point (i.e., 2/m(α, w)).
4.2.
Variable Sequence and Fixed Duration (VSFD)
In VSFD, the service provider can freely change the sequence of activities, but their durations are
fixed. The next proposition shows that, individually, both memory decay and acclimation favor
designs with increasing service levels. However, when they act together, the optimal sequence is in
general U-shaped. Figure 7 illustrates Proposition 1 for different values of m(α, w).
Proposition 1. In VSFD,
1. When m(α, w) ≤ 2/T , it is optimal to sequence activities in increasing order of service levels,
i.e., xi?1 < ... < xi?n .
2. When m(α, w) > 2/T , it is optimal to sequence activities in a U-shaped fashion of service
levels, i.e., there exists a k such that xi?1 > ... > xi?k−1 > xi?k < xi?k+1 < · · · < xi?n . In particular, either
this kth activity (ik ), or its direct predecessor (ik−1 ), or its direct successor (ik+1 ) should take place
when there remains 2/m(α, w) time units until the end of the encounter. Furthermore, k < n, i.e.,
the last two activities are always sequenced in increasing order of service levels.
Hence, when m(α, w) ≤ 2/T , i.e., when either α or w is small, a crescendo is optimal; see Figure
7 (left). This is the case, in particular, with either pure acclimation (w = 0) or pure memory decay
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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17
The optimal sequence of activities is either increasing or U-shaped, depending on the magnitude of
m(α, w). The last two activities should always be in increasing order.
x(t)$
x(t)$
Moderate$m(α,w)$
10$
7$
7$
7$
5$
5$
5$
2$
2$
2$
2
m( , w)
10$
15$ 18$
24$ 4me$ 0$
5$
15$ 18$
T
!!!!!m(α,%w)=0.0215!
(e.g.,%w=0.1,%α=0.001)!
2
m( , w)
!!!!!m(α,%w)=0.2050!
(e.g.,%w=1.5,%α=0.001)!
High$m(α,w)$
x(t)$
10$
0$
T
Low$m(α,w)$$
10$
24$ 4me$ 0$
3$
8$
18$
T
24$ 4me$
2
m( , w)
!!!!m(α,%w)=0.5869!
(e.g.,%w=5,%α=0.001)!
(α = 0). We next discuss the intuition, both for the case where w is small and for the case where α is
small. First, when w is small, the customer gives equal weight to past and recent utilities. Because
of acclimation, any drop in service level creates a negative instantaneous utility (as illustrated in
Figure 4), which will be remembered for a long time since w is small. Hence with small w, it is
optimal to have a monotonically increasing sequence of service levels. Second, when α is small,
the customer acclimates to the service level of an activity very slowly. Therefore, the higher the
service level, the larger the magnitude of instantaneous utility. Because of memory decay, these
high utilities are more likely to be remembered if they happen at the end of the encounter. Hence,
with small α, it is also optimal to have a monotonically increasing sequence of service levels. The
optimality of crescendo is consistent with people’s preferences for increasing outcomes over a time
frame (Loewenstein and Prelec 1993), with loss aversion (Tversky and Kahneman 1991), and also
with what a basic application of the peak-end rule (Fredrickson 2000) to design would imply.
By contrast when m(α, w) > 2/T , i.e., when both α and w are large, Proposition 1 shows that the
optimal sequence of activities is U-shaped: The service level should fall gradually in the beginning
and rise steeply at the end. See Figure 7 (middle and right). In particular, the last two service
levels should always be increasing. The intuition is that, with both acclimation and memory decay,
it is desirable to maximize the gradient of service level towards the end of the encounter. Because
high acclimation α generates little utility from slow, gradual increases in service level and high
memory decay w makes the customer heavily discount past experiences, the service provider can
gradually decrease the service level in the early part of the sequence, which will be forgotten due
to high memory decay, so as to ensure a steep positive gradient towards the end.
Hence, even though it is sometimes recommended to “finish strong” (Chase and Dasu 2001),
Proposition 1 indicates that it is not merely the intensity of the last activity that matters, but the
steepness of the gradient leading to it, i.e., one should “finish with a steep gradient.” For instance, in
18
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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00(0), pp. 000–000, Tchaikovski’s 1812 Overture, the finale, which celebrates the Russian victory over Napoleon’s army
with cannon fire, ringing chimes, and brass fanfare, is especially climatic because it is contrasted
with the preceding score, which represents the French invasion of Moskow. When designing song
playlists, people derive the highest utility when they not only schedule their favorite song at the
end, but also precede it with a less preferred song (Kahn et al. 1997).
U-shaped sequences of activities appear in many experiences, such as concerts (Baucells et al.
2013), concertos (Rozin et al. 2004), opera season scheduling (Dixon and Thompson 2013a), movies
and theme parks (Lawrence 2006). U-shaped sequences turn out to be also optimal in consumption
planning in the presence of either acclimation (Wathieu 1997) or satiation (Baucells and Sarin
2010). We thus reach similar conclusions to those consumption planning models (albeit slightly
different since decreasing sequences are never optimal here) from a different perspective, by making
evaluation retrospective instead of prospective. We also note that a U-shaped sequence could
emerge for other reasons, such as because of the primacy and recency effects, and we thus offer a
complementary, yet more subtle, justification of the optimality of such sequences.
Moreover, what happens within the U-shape may differ. In particular, the requirement that one
should finish with a steep gradient may differ from these other models. For instance, Dixon and
Verma (2013) speculate that, for a performing arts season, it may be optimal to start with the
peak activity, immediately followed by a steep drop in service levels, and then gradually increase
the service levels until the end of the season. Although this is probably motivated by other reasons
than acclimation and memory decay (e.g., engagement), we find here that it is optimal to adopt
the opposite pattern, i.e., a gradual drop followed by a steep increase.
Also, to maximize the end gradient, it may not always be optimal to place the activity with
the highest service level at the end, contrary to the crescendo sequence. In particular, when the
activity with the highest service level has a long duration, the customer becomes acclimated to
that high service level and her utility diminishes over time. In that case, it may be optimal to move
that activity at the beginning of the encounter so as to ensure a steep gradient near the end.4
Although Proposition 1 characterizes the optimal sequence, it offers little indication on how to
construct the optimal U-shape. This problem turns out to be exponentially complex and commercial
solvers may fail to return a solution when the number of activities becomes large.5 In the electronic
companion, we propose a heuristic to construct a U-shape sequence so as to ensure a steep gradient
near the end. Numerical experiments suggest that the heuristic performs well.
4
For instance, when (x1 , x2 , x3 , x4 ) = (2, 5, 7, 10) and (t1 , t2 , t3 , t4 ) = (5, 4, 3, 8), the optimal sequence when w = 1 and
α = 0.7 is (i1 , i2 , i3 , i4 ) = (4, 2, 1, 3), i.e., the activity with the highest service level is placed at the beginning.
5
Because VSFD can be
P formulated as a single machine job scheduling problem with the objective of maximizing the
total weighted profit k xik f (Tik ), where f (Tik ) is pseudoconvex (see Lemma A-2 in appendix), the computational
complexity of such problems, even without precedence constraints, is an open problem (Höhn and Jacobs 2012).
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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19
Optimal duration allocation for an increasing sequence. The shaded activities are allocated duration
above their lower bound.
Moderate m(D,w)
Low m(D,w)
4.3.
High m(D,w)
Fixed Sequence and Variable Duration (FSVD)
In FSVD, the service provider controls the duration of activities, but their sequence is fixed.
The following proposition characterizes the optimal duration allocation across a subsequence of
activities.
Proposition 2. In FSVD,
1. When m(α, w) < 1/T , if it is optimal to allocate more than the minimum duration to activity
q, i.e., if t?q > τ q , then it is optimal to allocate maximum duration to all subsequent activities if
they have increasing service levels, i.e., t?j = τ j , j = q + 1, . . . , r if xq < · · · < xr , and to all preceding
activities if they have decreasing service levels, i.e., t?j = τ j , j = l, . . . , q − 1 if xl > · · · > xq .
Pn
2. When 1/T ≤ m(α, w) ≤ 1/ ( i=r τ i ), for some r, it is never optimal to allocate more than
the minimum duration only in the middle of an increasing sequence, i.e., if xl < · · · < xr and
t?l = τ l , t?r = τ r , then t?i = τ i , ∀i, l < i < r; and it is never optimal to allocate more than the minimum
duration only in the extremities of a decreasing subsequence, i.e., if xl > · · · > xr and t?l > τ l , t?r > τ r ,
then t?i > τ i , ∀i, l < i < r.
3. When m(α, w) > 1/ (
Pn
i=r
τ i ) for some r, if it is optimal to allocate more than the minimum
duration to activity q, q < r, i.e., if t?q > τ q , then it is optimal to allocate maximum duration to all
preceding activities if they have increasing service levels, i.e., t?j = τ j , j = l, . . . , q − 1 if xl < · · · < xq ,
and to all subsequent activities, up to activity r, if they have decreasing service levels, i.e., t?j =
τ j , j = q + 1, . . . , r if xq > · · · > xr .
Figures 8 and 9 illustrate Proposition 2 with two service encounters: (i) an encounter with all
increasing service levels (Figure 8) and (ii) an encounter with all decreasing service levels (Figure
9). In the figures, the shaded activities are allocated duration above their lower bound, i.e., t?i > τ i .
When m(α, w) < 1/T , i.e., either α or w is low, Proposition 2 shows that duration should be
allocated to the activities in decreasing order of service levels (see (i) in Figures 8 and 9). We next
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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Figure 9
Optimal duration allocation for a decreasing sequence. The shaded activities are allocated duration
above their lower bound.
Low m(D,w)
Moderate m(D,w)
High m(D,w)
discuss the intuition, both for the case where w is small and for the case where α is small. First,
when w is small, this design strategy naturally ensures that the maximum time is spent at high
service levels. Second, when α is small, this design strategy brings closer to the end the activities
associated with a high service level, so that the customer does not heavily discount them.
By contrast when m(α, w) > 1/τ n , i.e., when both α and w are large, then duration should be
allocated in increasing order of service levels so as to push the steep rise and gradual fall in service
level towards the end of the encounter (see (iii) in Figures 8 and 9). Large memory decay implies
that the customer forgets the initial experience of the encounter. Therefore with an increasing
sequence, it is optimal to allocate duration to the activities in the beginning so as to bring closer
to the end the steep rise in service levels. Conversely in a decreasing sequence, it is optimal to
allocate duration at the end so as to push the steep drop in service levels as far as possible from
the end of the sequence and make the gradient of service level near the end tend to zero.
In the intermediate cases, when 1/T ≤ m(α, w) ≤ 1/τ in , duration can be allocated in any of
the three possible ways shown in Figures 8 and 9, respectively for an increasing and a decreasing
sequence. However, duration should never be allocated only to the activities in the middle of an
increasing sequence or in the extremities of a decreasing sequence.
Figure 8 (iii) indicates that, in the presence of both acclimation and memory decay, the optimal
design should induce a steep gradient at the end of the encounter, even if that involves spending
less time on high-service level activities. A grand finale is desirable, but it needs to be short
and contrasting with the preceding activities. This corroborates the optimality of the U-shaped
sequence in VSFD, which introduced some respite before the final ascent in service levels.
Similarly, Figure 9 (iii) indicates that, in the presence of both acclimation and memory decay,
it is preferable to spend more time at low service levels than to provide great service initially and
then create disappointment just before the end of the encounter. Pushing that logic further, if
τ i = 0, ∀i, it would then be optimal to allocate zero duration to the (initial) activities associated
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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with the highest service level, effectively removing them from the encounter. Hence, an anticlimax
is never optimal when both acclimation and memory decay are salient.
Together, Figures 8 (iii) and 9 (iii) indicate that, instead of offering a constant service level, a
service provider may actually create more satisfaction by interrupting the experience with a less
pleasant break, if the customer is subject to both memory decay and acclimation, consistent with
what was empirically observed by Nelson and Meyvis (2008). Hence, deteriorating an experience
may be optimal, provided that it induces a steep gradient near the end. In fact, the “service recovery
paradox” stipulates that, in some situations, the satisfaction of customers who experienced a service
failure and who were recovered by the firm may actually exceed the satisfaction of customers who
have not encountered any problems (McCollough and Bharadwaj 1992).
Given the design scope of our model, we are of course not suggesting that the service provider
should plan for a service failure and recovery (since customers would be able to internalize the whole
process in their ex-ante expectations). Nevertheless, the service provider could time a less enjoyable
activity in the next-to-last position so as to induce a steep gradient near the end. For instance, a
coffee shop may actually take advantage of the disagreeable checkout process by ensuing it with
an enjoyable activity (e.g., offering a free cookie). Similarly, an instructor should not relinquish
at covering material that is difficult to learn provided that the lesson ends with easier material
(Hoogerheide and Paas 2012). In music, some scores purposely feature a ritardando before finishing
with an explosive finale.
We close this section by noting that Proposition 2 offers only a partial characterization of the
optimal duration allocation. In general, FSVD is ill-behaved, with multiple local optima, unless the
sequence of activities is increasing in service levels, in which case the objective function is concave
(see Lemma A-5 in appendix). Yet, we show in the electronic companion that a simple gradient
coordinate algorithm performs reasonably well.
4.4.
Variable Sequence and Variable Duration (VSVD)
In VSVD, the service provider controls both the sequence and the duration of activities. Combining
Propositions 1 and 2 shows that the optimal solution for VSVD is obtained by constructing the
steepest positive gradient near the end.
Corollary 1. In VSVD,
1. When m(α, w) < 2/T , a sequence with increasing service levels is optimal. Moreover,
(a) When m(α, w) < 1/T , there exists an index k such that minimum duration is allocated
to the activities preceding the kth activity, i.e., tij = τ ij , 1 ≤ j ≤ k − 1 and maximum duration is
allocated to the activities subsequent to the kth activity, i.e., tij = τ ij , k < j ≤ n.
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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Figure 10
Optimal sequence and duration allocation as m(α, w) increases. The shaded activities are allocated
duration above their lower bound.
x(t)
1
< ݉ ߙ, ‫ < ݓ‬2/T
σ௜ ߬௜
time
(iii)
x(t)
x(t)
݉ ߙ, ‫ < ݓ‬1/ܶ
(i)
time
x(t)
1/ܶ < ݉ ߙ, ‫ < ݓ‬min 2/ܶ,
(ii)
1 time
σ௜ ߬௜
x(t)
݉ ߙ, ‫ > ݓ‬max 2/ܶ,
(v)
2/ܶ < ݉ ߙ, ‫< ݓ‬
(iv)
1
σ௜ ߬௜
time
1
σ௜ ߬௜
time
(b) When 1/T < m(α, w) ≤ 1/τ in , the set of activities with minimum duration is contiguous,
i.e., for any l, r, if t?il = τ il and t?ir = τ ir , then t?ij = τ ij , ∀j, l < j < r.
(c) When m(α, w) > 1/τ in , there exists an index k such that maximum duration is allocated
to the activities preceding the kth activity, i.e., tij = τ ij , 1 ≤ j ≤ k − 1 and minimum duration is
allocated to the activities subsequent to the kth activity, i.e., tij = τ ij , k < j ≤ n.
2. If m(α, w) > 2/T , a U-shaped sequence bottoming out at the kth activity is optimal such that:
• In the increasing part, the set of activities with minimum duration is contiguous, i.e., for any
k ≤ l < r ≤ n such that t?il = τ il and t?ir = τ ir then t?ij = τ ij , ∀j, l < j < r;
• In the decreasing part, the set of activities with strictly more than the minimum duration is
contiguous, i.e., for any 1 ≤ l < r ≤ k such that t?il > τ il and t?ir > τ ir then t?ij > τ ij , ∀j, l < j < r.
Figure 10 illustrates Corollary 1 as m(α, w) increases, i.e., as either acclimation or memory decay
or both become stronger. Initially, when m(α, w) is low, crescendo is the optimal sequence, with
duration initially allocated to the activities at the end of the sequence, as in (i), and then, as
m(α, w) increases, also at the beginning of the sequence, as in (ii) and (iii), so as to induce a
steeper gradient near the end of the encounter. As m(α, w) keeps increasing, the optimal sequence
may switch to a U-shaped sequence, as in (iv) and (v), and more duration is then allocated to the
activities in the middle of the encounter, so as to induce more gradual transitions at low levels of
service and steeper transitions at higher service levels.
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Alternatively, one may interpret Corollary 1 in terms of the total service duration T , since most
structural results can be expressed in terms of the product m(α, w) · T . Keeping m(α, w) fixed, the
optimal service design changes from (i) to (v) as T increases. Accordingly, short experiences (e.g.,
one-day executive courses) should have activities scheduled as a crescendo and duration allocated
primarily to the activities with the highest service levels, whereas long experiences (e.g., one-week
executive courses) should have activities scheduled in a U-shaped fashion and duration allocated
primarily to activities with the lowest service level so as to ensure a steep gradient near the end.
In terms of complexity, VSVD is at least as difficult as FSVD or VSFD. In the electronic
companion, we propose a heuristic to first construct a U-shaped sequence when m(α, w) > 2/T
or a crescendo sequence otherwise, and then allocating duration to activities by following the
gradient ascent algorithm used for FSVD. Comparing the performance of the heuristic to the
optimal solution (for small n) or to an upper bound (for large n) reveals that this heuristic performs
reasonably well.
5.
Service Design for a Heterogeneous Population of Customers
In practice, a service provider may not be able to adjust the service design to different customers,
characterized by different rates of acclimation (α) and memory decay (w), or to precisely assess
those rates. In this section, we provide guidelines regarding the best sequence of activities in the
absence of complete knowledge of α and w for every individual customer. Based on our structural results in §4.2, we propose three different sequencing heuristics and numerically test their
performance for a randomly generated population of customers.
In our experiments, we consider a service encounter consisting of 7 activities and generate 150
instances of service levels and durations. For each instance, both the service level and the duration
of each activity are randomly drawn from independent Gamma distributions with scale 2 and shape
2, i.e., xi ∼ Gamma(2, 2) and ti ∼ Gamma(2, 2) for i = 1, . . . , 7.
For each instance of service encounter, keeping fixed the service levels and durations, we randomly
generate a population of 100 customers (α, w), where the distribution α and w are independent
Gammas, with means and standard deviations respectively equal to (µα , σα ) and (µw , σw ). For each
customer, we compare their satisfaction generated with the following sequences:
• The optimal sequence for that particular customer (α, w), denoted as S ∗ ;
• A crescendo sequence, which orders activities in increasing order of service levels, i.e., xi1 ≤
· · · ≤ xin , denoted as S cresc ;
• A sequence that maximizes the gradient at the end, by ordering the (n − 1) smallest service
level activities in decreasing order of service levels and placing the activity with the highest service
level at the end, i.e., xi1 ≥ · · · ≥ xin−1 ≤ xin = max xi , denoted as S steep ;
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• The optimal sequence if the provider ignored customer heterogeneity, i.e., based on the means
(µα , µw ), denoted as S mean .
Note that the crescendo and steep gradient sequences require no information about the rates of
acclimation and memory decay. The last heuristic, by contrast, requires that the provider know
the mean rates of acclimation and memory decay.
Table 2 displays the average optimality gaps, averaged across all 150 instances of service encounters, and, for each encounter, across all 100 customers, for the crescendo sequence, the steep
gradient sequence, and the sequence based on the mean rates, when µα = µw ∈ {0.2, 0.5, 0.8} and
σα = σw ∈ {0.1, 0.3, 0.5}.
Table 2
Average suboptimality gap for the sequence based on the mean rates, i.e., S mean , the steep gradient
sequence, i.e., S steep , and the crescendo sequence, i.e., S cresc , when µα = µw and σα = σw .
1 − S mean /S ∗
1 − S steep /S ∗
1 − S cresc /S ∗
µ α = µw
0.2
0.5
0.8
0.2
0.5
0.8
0.2
0.5
0.8
0.1
4.5%
1.5%
0.3%
28.9%
13.1%
32.9%
16.4%
53.2%
72.1%
σα = σ w
0.3
9.9%
9.7%
3.9%
20.7%
13.6%
27.5%
8.3%
50.6%
68.8%
0.5
6.1%
16.9%
8.4%
11.6%
16.9%
21.6%
4.1%
28.2%
64.0%
From Table 2, we make the following observations:
• The performance of the sequence based on the mean rates naturally improves when their
standard deviation is low (since it is optimal when σα = σw = 0), but also improves when the mean
rates increase, indicating that its performance mostly depends on the coefficient of variation.
• By contrast, the performance of the crescendo improves when the mean rates are low (since, by
Proposition 1, crescendo is optimal when either α or w is small) or when their standard deviation
increases. With a large standard deviation, the gamma distribution indeed tends to have a large
probability mass near zero, similar to the exponential distribution.
• The sequence based on the mean rates overall exhibits a very good performance, with an
optimality gap mostly in the single digits, especially in comparison to the other two heuristics. This
shows that (i) estimating the mean acclimation and memory decay rates and using those means
to guide the service design is very beneficial, but that (ii) knowing more than the mean rates, e.g.,
by attempting to elicit the actual rates for each individual customer and reconfiguring the service
design accordingly, tends to have only a marginal impact.
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• The steep gradient sequence appears to generally outperform the crescendo, except when the
mean rates are low. Moreover, the steep gradient sequence appears to have a relatively stable performance, in contrast to the crescendo, which performs very poorly at high rates of acclimation
and memory decay. This suggests that, in absence of information about the mean rates of acclimation and memory decay, a service provider may be better off finishing the encounter with a steep
gradient, even if that involves lowering the service levels initially, rather than gradually increasing
the service level, as could be potentially inferred from a “finish strong” recommendation.
As a robustness check, we next consider a situation when µα 6= µw . We moreover assume that
σw = 0.001 so as to concentrate all random variations on α. (Because of the symmetry between
α and w, the same results could have been obtained by setting σα = 0 and varying w.) Table 3
displays the optimality gaps when µα , µw ∈ {0.2, 0.5, 0.8} and σα ∈ {0.1, 0.3, 0.5}.
Table 3
Average suboptimality gap for the sequence based on the mean rates, i.e., S mean , the steep gradient
sequence, i.e., S steep , and the crescendo sequence, i.e., S cresc , when µα 6= µw and σw = 0.001.
1 − S mean /S ∗
1 − S steep /S ∗
1 − S cresc /S ∗
µα
0.2
0.5
0.8
0.2
0.5
0.8
0.2
0.5
0.8
µw = 0.2
σα = 0.1 σα = 0.3 σα = 0.5
2.9%
8.3%
9.3%
0.4%
3.8%
8.1%
0.2%
1.8%
3.3%
34.9%
23.3%
23.0%
17.8%
24.7%
20.1%
11.4%
11.7%
7.9%
18.0%
12.7%
8.4%
37.2%
27.5%
26.5%
44.4%
40.9%
41.3%
µw = 0.5
σα = 0.1 σα = 0.3 σα = 0.5
2.5%
9.3%
10.3%
0.6%
4.8%
10.4%
0.1%
1.3%
3.5%
16.7%
17.2%
15.1%
8.6%
16.2%
17.9%
29.3%
18.5%
27.3%
35.3%
25.2%
17.4%
54.5%
48.8%
44.3%
71.6%
63.7%
65.5%
µw = 0.8
σα = 0.1 σα = 0.3 σα = 0.5
2.6%
5.9%
9.1%
0.5%
3.4%
8.5%
0.4%
2.5%
5.4%
8.5%
13.0%
14.0%
21.5%
18.6%
17.5%
35.1%
30.2%
25.4%
36.6%
27.8%
18.2%
66.8%
61.8%
55.6%
72.2%
69.3%
64.5%
In addition to confirming our earlier observations, Table 3 reveals that the performance of the
sequence based on the mean rates remains very stable with respect to asymmetric changes in
means. Hence, that heuristic does not fit a particular mean pattern. By contrast, the performance
of crescendo improves when at least one of the means becomes smaller since, by Proposition 1, it
becomes near-optimal when either rate is zero. In that case, crescendo obviously outperforms the
steep gradient sequence, which starts with a diminuendo. On the other hand, when both means
are large, crescendo does not perform well and is significantly outperformed by the steep gradient
sequence. Hence, the steep gradient and the crescendo sequences are in some way complementary,
although the former generally outperforms the latter, and sometimes by a significant amount.
6.
Conclusion
In this paper, we analytically determine how to sequence and allocate duration to activities in
a service encounter so as to maximize ex-post customer satisfaction in the presence of memory
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decay and acclimation. Memory decay favors high service levels at the end of the encounter and
acclimation favors steep positive gradients of service level.
We show that, individually, the two behavioral biases yield the same service design. However,
when considered jointly, they can act as opposing forces. In particular, if either memory decay or
acclimation is low, it is optimal to sequence activities in increasing order of service levels and to
lengthen the duration of activities with the highest service levels. In contrast, when both memory
decay and acclimation are high, it is optimal to sequence activities in a U-shaped fashion and
to lengthen the duration of activities with the lowest service levels, so as to accentuate the final
gradient.
Our results can alternatively be interpreted in terms of the total duration of the encounter: short
experiences should have activities scheduled as a crescendo and duration allocated primarily to the
activities with the highest service levels, whereas long experiences should have activities scheduled
in a U-shaped fashion and duration allocated primarily to activities with the lowest service level
so as to ensure a steep gradient near the end.
Overall, our results suggest that even though “finishing strong” is optimal in the presence of
either memory decay or acclimation, the optimal design when both effects are present is more
subtle as one should aim at maximizing the gradient near the end. While an anti-climax is never
desirable, a never-ending grand finale should also be avoided. Moreover, it may be a good idea to
introduce some respite just before the grand finale so as to accentuate its strong character. Going
one step beyond the already controversial idea of adding light pain at the end to induce customers
to forget a more painful event (Kahneman et al. 1993), our results suggest it may be desirable
to voluntarily degrade the service level of the middle activities to accentuate the strong character
of the ending activities as empirically observed by Nelson and Meyvis (2008). In fact, our model
suggests it may be desirable to even get rid of some high-intensity activities in order to either avoid
a decline in service level (if they are sequenced at the beginning) or accentuate the final gradient in
service levels (if they are sequenced towards the end). This suggests that properly sequencing and
allocating duration to a service encounter can more than compensate for inferior service levels.
This paper takes a first step toward building an analytical framework for designing service experiences. Natural research extensions could relax the assumptions outlined in the introduction, such
as (i) characterizing the optimal encounter duration and the optimal choice of service levels (subject to a budget constraint); (ii) characterizing the optimal design in the presence other behavioral
biases (e.g., satiation or loss aversion); (iii) considering services where customers are not captive,
i.e., can join after the beginning of the encounter or leave before the end, and/or where they selfcontrol the sequence of activities (e.g., amusement parks); (iv) investigating the effect of precedence
constraints and/or sequence-dependent service levels. We hope that this paper will induce further
interest in engineering service experiences to maximize customer satisfaction.
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
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Acknowledgments
The authors would like to thank Serguei Netessine, the associate editor, and three anonymous
referees for insightful comments, which have significantly improved this paper.
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Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
c 0000 INFORMS
00(0), pp. 000–000, 32
Appendix. Proofs
In the proofs, we will use, when there is no ambiguity, Φ(t) instead of Φ(α, w, t) .
Lemma A-1. The function Φ(α, w, t) is strictly pseudoconcave in t with a stationary point at 1/m(α, w).
Proof.
Suppose α < w. We have:
Φ0 (t) =
e−αt (−α) − e−wt (−w) e−wt (w − αet(w−α) )
ln w − ln α
=
> 0 ⇔ w > αet(w−α) ⇔
> t.
w−α
w−α
w−α
Since Φ0 (t) > 0 when t < 1/m(α, w), Φ0 (t) = 0 when t = 1/m(α, w), and Φ0 (t) < 0 when t > 1/m(α, w), Φ(t)
is strictly pseudoconcave. Note that the result is symmetric for w < α.
Lemma A-2. The function f (Tik ) = Φ(α, w, T − Tik + tik ) − Φ(α, w, T − Tik ) is strictly pseudoconvex in
Tik .
Proof.
Suppose α < w. We have:
f 0 (Tik ) =
αe−α(T −Tik +tik ) − we−w(T −Tik +tik ) αe−α(T −Tik ) − we−w(T −Tik )
−
.
w−α
w−α
Consider t0 = T − 1/(w − α) · ln (w(1 − e−wtik )/ (α(1 − e−αtik ))). Since f 0 (Tik ) < 0 when Tik < t0 , f 0 (Tik ) =
0 when Tik = t0 , and f 0 (Tik ) > 0 when Tik > t0 , f (Tik ) is strictly pseudoconvex. Note that the result is
symmetric for w < α.
Lemma A-3. Φ(α, w, t) is concave in t ∀t ∈ [0, 2/m(α, w)] and convex in t ∀t ∈ [2/m(α, w), T ].
Proof.
Since Φ00 (t) = (α2 e−αt − w2 e−wt ) / (w − α), Φ00 (t) ≤ 0, ∀t ∈ [0, 2/m(α, w)] and Φ00 (t) ≥ 0, ∀t ∈
[2/m(α, w), T ]. Hence, Φ(t) is concave ∀t ∈ [0, 2/m(α, w)] and convex ∀t ∈ [2/m(α, w), T ].
Lemma A-4. In VSFD, the optimal sequence (i?1 , .., i?n ) is such that for any two consecutive activities xi?k
and xi?k+1 , the following applies:
(i) If both activities start and finish within [0, T − 2/m(α, w)], then xi?k > xi?k+1 .
(ii) If both activities start and finish within [T − 2/m(α, w), T ] or if k = n − 1, then xi?k < xi?k+1 .
Proof.
The proof uses an interchange argument. Let S ? be the satisfaction obtained from the optimal
sequence (i?1 , . . . , i?n ). For any j, let Sj? be the satisfaction obtained by interchanging i?j−1 and i?j in the optimal
sequence. Therefore, by (6),
S ? − Sj? = xi?j−1 − xi?j · (−Φ(T j ) + Φ(T j +1 )) + (Φ(T j−1 ) − Φ(ti?j−1 + T j +1 )) .
We next consider three scenarios: (i) when the two activities start and end within [0, T − 2/m(α, w)], (ii)
when the two activities start and end within [T − 2/m(α, w), T ], and (iii) when j = n.
(i) If i?j−1 and i?j start and finish within [0, T − 2/m(α, w)], then by Lemma A-3, the corresponding Φ0 (t)
is increasing ∀t ∈ [2/m(α, w), T ]. Hence,
Φ(T j ) − Φ(T j +1 ) < Φ(ti?j−1 + T j ) − Φ(ti?j−1 + T j +1 ).
Therefore, by optimality of (i?1 , . . . , i?n ), we have xi?j < xi?j−1 .
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
c 0000 INFORMS
00(0), pp. 000–000, 33
(ii) If i?j−1 and i?j start and finish within [T − 2/m(α, w), T ], then by Lemma A-3, the corresponding Φ0 (t)
is decreasing ∀t ∈ [0, 2/m(α, w)]. Hence,
Φ(T j ) − Φ(T j +1 ) > Φ(ti?j−1 + T j ) − Φ(ti?j−1 + T j +1 ).
Therefore, by optimality of (i?1 , . . . , i?n ), we have xi?j > xi?j−1 .
(iii) By interchanging activities i?n−1 and i?n , we obtain a suboptimal sequence with corresponding satisfaction Sn? . Therefore,
S ? − Sn? = xi?n − xi?n−1 · Φ(ti?n ) + Φ(ti?n−1 ) − Φ(ti?n−1 + ti?n )
xi?n − xi?n−1 n
o
−wt ?
−αt ?
=
· 1 − e in−1 1 − e−wti?n − 1 − e in−1 1 − e−αti?n .
w−α
Hence, by optimality of (i?1 , . . . , i?n ), we have xi?n > xi?n−1 .
Proof of Proposition 1.
The proof follows from Lemma A-4 in appendix.
Lemma A-5. Suppose that xi is the service level of the activity at the ith position in the sequence. When
x1 < · · · < xn , S(t) is strictly pseudoconcave, whereas, when x1 > · · · > xn , S(t) is strictly pseudoconvex.
Proof.
By equation (6), S(t) =
to ti , ∀i we obtain
Pn
i=1
(xi − xi−1 )Φ(T i ). Taking the partial derivatives of S(t) with respect
i
∂S(t) X
=
(xj − xj−1 )Φ0 (T j ), ∀i.
∂ti
j =1
Because Φ0 (t) = 0 if and only if t = 1/m(α, w), any stationary point has to satisfy the following set of
equations:
n
X
j =i
tj =
1
, ∀i.
m(α, w)
(A-1)
Therefore, t? = (0, . . . , 0, 1/m(α, w)) is the unique stationary point satisfying equations (A-1).
We next evaluate the Hessian of S(t) at t? , denoted H, to obtain for any y ∈ <n


!2
!2
n
n
α
α w−α
X
X
−(x2 − x1 )
yT Hy = α
yi − (x3 − x2 )
yi · · · − (xn − xn−1 )yn2  .
w
i=2
i=3
Clearly, for x1 < · · · < xn , yT Hy < 0, and for x1 > · · · > xn , yT Hy > 0 ∀y ∈ <n . Hence, S(t) is strictly
pseudoconcave when x1 < · · · < xn and strictly pseudoconvex when x1 > · · · > xn .
Lemma A-6. In FSVD the necessary conditions for optimality are given by
∂S(t)
= λi + µ, ∀i ∈ IU
∂ti
∂S(t)
= µ, ∀i ∈ IM
∂ti
∂S(t)
= µ − λi , ∀i ∈ IL ,
∂ti
(A-2)
together with constraints (8) and (9), in which µ ∈ <, λi ≥ 0 and λi ≥ 0; and IU , IM , IL are three disjoint sets
such that I = IU ∪ IM ∪ IL and ti = τ i , ∀i ∈ IU , ti = τ, τ i < τ < τ i , ∀i ∈ IM and ti = τ i , ∀i ∈ IL .
Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
c 0000 INFORMS
00(0), pp. 000–000, 34
Proof.
The necessary conditions for optimality of t? are given by the following Karush-Kuhn-Tucker
conditions (Boyd and Vandenberghe 2004):
1. The stationarity condition gives ∂S(t)/∂ti − µ + λi − λi = 0, ∀i at t? .
Pn
2. Complementary slackness gives µ(T − i=1 t?i ) = 0, λi (τ i − t?i ) = 0, ∀i and λi (t?i − τ i ) = 0, ∀i.
3. Primal feasibility implies t? satisfies the constraints (8) and (9).
4. Dual feasibility implies µ ∈ < and λi , λi ≥ 0, ∀i.
From the complementary slackness condition, λi = 0, ∀i ∈ IU and λi = 0, ∀i ∈ IL and both λi = 0, λi =
0, ∀i ∈ IM .
Proof of Proposition 2.
We use Lemmas A-1 and A-6 for this proof. By (6), we have
i
∂S(t? ) X
=
(xj − xj−1 )Φ0 (T j ), ∀i,
∂ti
j =1
where x0 = b(0).
1. Suppose that xq < · · · < xr . Since T < 1/m(α, w), Φ0 (t) > 0, ∀t ∈ [0, T ) by Lemma A-1, and therefore
∂S(t)/∂tq < · · · < ∂S(t)/∂tr , ∀t ≥ (τ 1 , . . . , τ n ). Hence, from condition (A-2), if tq > τ q , ∂S(t? )/∂tq = µ or
∂S(t? )/∂tq = µ+λq . Therefore, ∂S(t? )/∂tj = µ+λj , j = q +1, . . . , r. From Lemma A-6, t?j = τ j , j = q +1, . . . , r.
The proof for a decreasing subsequence is similar.
2. We show the result by contradiction. Suppose that xl < · · · < xr and that tl = τ l and tr = τ r . There-
fore, condition (A-2) implies that ∂S(t? )/∂tl = µ − λl and ∂S(t? )/∂tr = µ − λr . If for any i, l < i < r, we
Pn
have t?i > τ i , then ∂S(t? )/∂ti = µ or ∂S(t? )/∂ti = µ + λi , by condition (A-2). If h=i t?h ≤ 1/m(α, w), then
Pn
Φ0 (t) > 0 for t ≤ h=i+1 t?h < 1/m(α, w) from Lemma A-1 and therefore ∂S(t? )/∂tr > ∂S(t? )/∂ti , thereby
Pn ?
Pn
a contradiction. If
t > 1/m(α, w), then Φ0 (t) < 0 for t ≥ h=i t?h from Lemma A-1 and therefore
h=i h
∂S(t? )/∂tl > ∂S(t? )/∂ti , thereby a contradiction. Therefore, from Lemma A-6, t?i = τ i , ∀i, l < i < r. The
proof for a decreasing subsequence is similar.
Pn
Pn
3. Suppose that xl < · · · < xq . When i=r τ i > 1/m(α, w), we have Φ0 (t) < 0, ∀t ∈ ( i=r τ i , T ] by Lemma
A-1, and therefore ∂S(t)/∂tl > · · · > ∂S(t)/∂tq , ∀t ≥ (τ 1 , . . . , τ n ). From condition (A-2), since tq > τ q ,
∂S(t? )/∂tq = µ or ∂S(t? )/∂tq = µ + λq . Therefore, ∂S(t? )/∂tj = µ + λj , j = l, . . . , q − 1. Therefore, from
Lemma A-6, t?j = τ j , j = l, . . . , q − 1. The proof for a decreasing subsequence is similar.
e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
This page is intentionally blank. Proper e-companion title
page, with INFORMS branding and exact metadata of the
main paper, will be produced by the INFORMS office when
the issue is being assembled.
ec1
ec2
e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
Computational Study
In this electronic companion, we first present a mathematical programming formulation of SPDP, then
present heuristics for VSFD, FSVD, and VSVD, and finally derive a computationally-tractable upper bound
on SPDP.
EC.1.
Mathematical Programming Formulations of SPDP
Let us define zij as a binary variable indicating if activity i directly or indirectly precedes j and define ci as
the completion time of activity i. We keep constraints (8)-(9) and express constraint (10) in SPDP in terms
of zij ’s and ci ’s. Constraints (EC.1)-(EC.2) ensure that, for any pair of activities, one has to precede the
other. Let Q be the set of precedence constraints between pairs of activities. Then constraint (EC.3) ensures
that activity j precedes activity k for all such pairs of precedence constraints in Q. Constraint (EC.4) ensures
that the completion time of an activity must be greater than its duration and smaller than the total duration
of the service encounter. Finally, constraint (EC.5) ensures that for any pair of activities, the completion
time of the preceding activity is less than that of the succeeding activity.
max S(z, t, c) =
z,t,c
subject to
Pn
i=1
xi (Φ(T − ci + ti ) − Φ(T − ci ))
constraints (8)-(9)
∀i, j, i 6= j
(EC.1)
zii = 0,
∀i
(EC.2)
zjk = 1,
∀(j, k) ∈ Q
(EC.3)
∀i
(EC.4)
∀i, j, i 6= j
(EC.5)
∀i, j.
(EC.6)
zij + zji = 1,
ti ≤ ci ≤ T,
ci ≤ cj − tj + (1 − zij )T,
zij ∈ {0, 1},
EC.2.
Heuristics and Computational Study
EC.2.1.
VSFD
We solved VSFD with BARON and LINDOGlobal available on NEOS with a maximum run time of 500,000
seconds for n = 6, . . . , 10. Because commercial solvers failed to return a feasible solution within the set time
limit for n ≥ 9 we propose a heuristic based on Proposition 1.
The heuristic (formally stated as Algorithm 1) assigns activities sequentially, going backwards in time. It
starts allocating activities in decreasing order of service levels, starting from the activity with the highest
service level, until time T − 2/m(α, w) is reached. At that point, it allocates the remaining activities in
increasing order of service level, starting from the activity with the lowest service level. Doing so results in
a U-shaped sequence, or in an only increasing sequence if T − 2/m(α, w) < 0.
We tested the performance of the heuristic on small problem instances that can be solved optimally by
exhaustive enumeration. We generated the problem instances with α and w uniformly drawn between 0 and
1 such that the condition T > 2/m(α, w) is satisfied since an increasing sequence is known to be optimal
otherwise. The values of ti ’s were uniformly drawn from the interval [1, 100], and the service levels (xi ’s)
ec3
e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
Algorithm 1 Heuristic for VSFD
Sort activities such that x1 > .. > xn .
Comment: opt is the optimal sequence.
Initialize: time= 0, i = n, j = 1, k = n.
1
do
while time≤ 2 m(α,w)
opt[i]=x[j]
i=i-1
time=time+t[j]
j=j+1
end while
while i > 0 do
opt[i]=x[k]
i=i-1; k=k-1;
end while
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
Table EC.1
n
6
7
8
9
10
Mean
0.77%
1.17%
1.43%
1.88%
1.92%
xi ∼ U nif orm[1, 100]
Median Std Dev
Min.
0.44%
0.95%
0.00%
0.86%
1.03%
0.00%
0.93%
1.36%
0.00%
1.44%
1.69%
0.00%
1.39%
1.75%
0.00%
Max.
4.60%
4.29%
5.68%
7.59%
7.68%
Optimality Gap of VSFD Heuristic
Mean
0.53%
1.15%
1.23%
1.52%
1.61%
xi ∼ Gamma(1, 2)
Median Std Dev
Min.
0.15%
0.84%
0.00%
0.61%
1.34%
0.00%
0.89%
1.34%
0.00%
1.19%
1.39%
0.00%
1.22%
1.45%
0.00%
Max.
5.97%
5.92%
7.47%
6.14%
6.89%
Mean
0.45%
0.75%
0.99%
1.26%
1.36%
xi ∼ Gamma(9, 0.5)
Median Std Dev
Min.
0.18%
0.61%
0.00%
0.49%
0.82%
0.00%
0.68%
1.03%
0.00%
1.16%
1.18%
0.00%
0.87%
1.43%
0.00%
Max.
3.68%
3.57%
5.59%
5.71%
6.02%
were drawn from a Uniform [1,100] or from a Gamma distribution with shape (k) and scale (θ) parameters,
with either k = 1 and θ = 2 or k = 9 and θ = 0.5. Table EC.1 reports the mean, median, standard deviation,
minimum, and maximum suboptimality gaps of our heuristic relative to the optimal solution found through
complete enumeration, over 100 randomly generated instances for each value of n. Overall, Table EC.1 shows
that the average suboptimality gap for this heuristic is within 2%, and the maximum gap is within 8%.
The suboptimality gaps naturally increase with n, but the deterioration in performance remains moderate
relative to the exponential increase in run time of the enumeration method. Moreover, the gaps appear to
be robust to the distribution of service levels.
EC.2.2.
FSVD
We solved FSVD with LINDOGlobal using the NEOS optimization solver. The commercial solver returned
the optimal or a near optimal solution (with a 1% optimality gap) for small problem instances, but it failed
to solve larger problem instances (e.g., n ≥ 24) with decreasing subsequences of service levels within one
week. Since the objective function of FSVD with a decreasing sequence is pseudoconvex (Lemma A-5), the
maximization problem has potentially multiple local optima when τ in < 1/m(α, w) < T , and there is no
guarantee that a solution can be obtained in a reasonable amount of time.
We therefore propose a modified coordinate ascent heuristic to solve FSVD in a smaller computational
time on average than LINDOGlobal. The heuristic starts with t0 = τ and iteratively allocates time increments
∆t to the activity with the highest marginal return, i.e., arg max ∂S(t)/∂ti until the total duration is equal
i
to T . The formal algorithm is presented as Algorithm 2.
To compare the heuristic solution against the solution obtained from the solver, we consider four patterns
of sequences of service levels, namely, increasing, decreasing, U-shaped, and Λ-shaped. For each pattern
of sequence, we consider sequences of different lengths, i.e., n = 8, 16, 24, 32. We randomly generate 100
instances for each n. All instances are generated with α and w uniformly drawn from (0, 1) and with xi ,
ec4
e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
Algorithm 2 Heuristic for FSVD
Comment: ∆t is the step size. I is the set of activities such that ti = τ i , ∀i ∈ I.
0
Initialize: tP
= τ ; m = 0; I = ∅
while T − n
tm > 0 do
i=1 i
∂S(tm )
i = arg max
k,k∈I
/
∂tk P
∆ti = min{∆t, T − n
tm , τ i − tm
i }
j=1 j
if ∆ti = 0 then
I = I ∪ {i}
end if
tm+1
= tm
i + ∆ti
i
m=m+1
end while
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
Λ-shaped U-shaped Decreasing Increasing
Table EC.2
n
8
16
24
32
8
16
24
32
8
16
24
32
8
16
24
32
Heuristic Solution vs LINDOGlobal Solution for FSVD
?
(S (t?
solver )−S (theuristic ))×100
|S (t?
)|
solver
Mean Median Std Dev Minimum Maximum
0.00% 0.00%
0.00%
0.00%
0.00%
0.00% 0.00%
0.00%
0.00%
0.00%
0.00% 0.00%
0.00%
0.00%
0.00%
0.00% 0.00%
0.00%
0.00%
0.00%
0.00% 0.00%
0.00%
0.00%
0.01%
0.01% 0.00%
0.29%
0.00%
1.25%
0.02% 0.00%
0.10%
0.00%
1.01%
0.11% 0.00%
0.30%
0.00%
1.04%
0.00% 0.00%
0.00%
0.00%
0.02%
0.00% 0.00%
0.00%
0.00%
0.04%
0.00% 0.00%
0.00%
0.00%
0.07%
0.00% 0.00%
0.00%
0.00%
0.07%
0.00% 0.00%
0.00%
0.00%
0.00%
0.00% 0.00%
0.00%
0.00%
0.00%
0.00% 0.00%
0.00%
0.00%
0.04%
0.00% 0.00%
0.00%
0.00%
0.06%
Mean Run Time
Heuristic
Solver
0.07 sec
0.10 sec
0.08 sec
0.29 sec
0.09 sec
1.53 sec
0.11 sec
3.15 sec
0.08 sec
0.49 sec
0.10 sec 1092.55 sec
0.11 sec 1575.66 sec
0.12 sec 1888.51 sec
0.08 sec
1.32 sec
0.11 sec
1.51 sec
0.14 sec 100.01 sec
0.15 sec 210.03 sec
0.09 sec
2.10 sec
0.10 sec
2.31 sec
0.11 sec 139.11 sec
0.14 sec 256.02 sec
No. of Aborted
Instances by Solver
0 cases
0 cases
0 cases
0 cases
0 cases
0 cases
13 cases
19 cases
0 cases
0 cases
5 cases
13 cases
0 cases
0 cases
6 cases
12 cases
P
τ i , τ i , ∀i uniformly drawn from [1, 100]. The total duration of the service encounter is fixed to T = i τ i +
P
(τ i − τ i )/n. For each n, the U-shape sequence is created by forming a decreasing subsequence with n/2
i
randomly selected activities and an increasing subsequence with the remaining n/2 activities. The Λ-shape
is constructed similarly. The time increment for the heuristic is fixed as ∆t = 0.01.
Table EC.2 benchmarks the solution as well as the running times from the heuristic to the best solution
found by LINDOGlobal, within a 1% optimality gap. All codes were implemented in MATLAB 7.12.0. and
GAMS. The computational tests on MATLAB were run on a workstation with a 2.26 GHz Intel Core 2 Duo
processor, 8 GB of RAM, and Mac OS X 10.6.8 as the operating system.
Apart from the sequences of activities with increasing service levels, which yield a concave maximization
problem (Lemma A-5), all other sequences had some cases aborted by NEOS when the solver failed to find
an optimal or a near-optimal solution with a 1% optimality gap in 500,000 seconds (reported in the last
column in Table EC.2). Only the cases for which the solver returned an optimal or near-optimal solution
were included when computing the performance gaps.
Overall, the FSVD heuristic gives solutions that closely match the solver solutions (within 1.5%) in a
significantly smaller run time (i.e., in less than 1 second). Moreover, the computation time of the coordinate
ascent heuristic appears less sensitive to the size or specific shape of the sequence.
e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
EC.2.3.
ec5
VSVD
VSVD is at least as complex as VSFD since an instance of VSFD can be constructed from VSVD by setting
τ i = τ i , ∀i. When T < 2/m(α, w), the sequence of service levels is increasing, and therefore the duration
allocation problem is pseudoconcave (see Lemma A-5). When T > 2/m(α, w), Corollary 1 shows that a Ushaped sequence is optimal and the duration allocation problem may therefore have multiple local optima.
In general, commercial solvers like BARON and LINDOGlobal on NEOS failed to return a feasible solution
in one week for many instances with n ≥ 10. We therefore propose a heuristic for SPDP when T > 2/m(α, w)
and we present in §EC.2.5 an upper bound on SPDP against which we test the performance of our heuristic.
The heuristic first finds a U-shaped sequence and then allocates duration for the corresponding FSVD
problem so as to maximize the gradient of service levels at the end of the encounter. We next detail the two
steps of the heuristic.
Step 1: Sequencing. The heuristic constructs a U-shaped sequence. Without loss of generality, we assume
that x1 ≥ · · · ≥ xn . The following optimization problem freely allocates duration among all activities, so as
to maximize customer satisfaction by generating a steep gradient in service level.
max S(t) =
t
subject to
Pn−1
i=1
(xi − xi+1 )Φ(
Pi
0 ≤ ti ≤ τ i ,
Pn
t ≤ T.
i=1 i
t)
j =1 j
∀i
Compared to SPDP, the lower bounds on activity duration (8) are set to zero and the constraint on the
encounter’s total duration (9) is relaxed to an inequality. Moreover, the sequence is fixed. An allocation
of zero duration to an activity suggests that that activity does not contribute to creating a steep gradient
and should therefore be moved at the beginning of the encounter. Accordingly, let L be the set of activities
that are allocated zero duration, i.e., if t? denotes the optimal solution then L = {j|t?j = 0}. Therefore, for
all j ∈ L, we order activities in decreasing order of xj ’s and for all j ∈
/ L, we order activities in increasing
order of xj ’s. The combination of the decreasing subsequence followed by the increasing subsequence forms
a U-shaped sequence.
Step 2: Duration Allocation. For the U-shaped sequence obtained in Step 1, we apply the coordinate
ascent method described in Algorithm 2 to allocate duration.
To test the performance of the heuristic against the optimal solution, when computable, and an upper
bound (§EC.2.5) otherwise, we randomly generated problem instances with α and w uniformly drawn from
[0, 1], and xi , τ i , τ i , ∀i uniformly drawn from [1, 100]. The total duration of the service encounter was set as
P
P
T = i τ i + i (τ i − τ i )/n and the condition T > 2/m(α, w) was verified.
Table EC.3 compares the solution obtained from the heuristic with the optimal solution and the upper
bound when n ≤ 9. For every instance of size n, the gaps were evaluated over 100 instances, and the corresponding sample was used for evaluating the mean, median, standard deviation, minimum, and maximum
optimality gap. We observe that the gaps against the optimal solution are within 3% on an average, whereas
against the upper bound the average gaps are within 6%.
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e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
Table EC.3
Optimality Gaps for the VSVD heuristic with T > 2/m(α, w)
n
Solver
Mean Median Std Dev Minimum Maximum
Upper Bound
Mean Median Std Dev Minimum Maximum
5
6
7
8
9
0.72%
1.09%
1.42%
2.19%
2.67%
3.11%
3.77%
3.94%
4.02%
5.11%
0.75%
0.71%
1.30%
2.01%
2.32%
0.35%
0.81%
1.33%
1.47%
1.41%
Table EC.4
0.00%
0.00%
0.08%
0.22%
0.52%
1.35%
2.46%
4.89%
5.56%
5.76%
3.92%
3.01%
3.31%
3.80%
4.10%
2.51%
2.95%
3.10%
3.68%
3.18%
0.11%
0.25%
0.22%
0.30%
0.73%
7.13%
8.31%
8.91%
10.01%
10.19%
Optimality Gaps for the VSVD heuristic with T > 2/m(α, w)
n
Upper Bound
Mean Median Std Dev Minimum Maximum
10
15
20
25
30
5.99%
6.91%
7.09%
8.01%
9.81%
4.91%
5.95%
6.81%
7.77%
8.97%
3.18%
2.97%
3.01%
3.86%
3.55%
0.75%
0.83%
1.03%
1.99%
2.08%
10.53%
11.79%
13.86%
14.01%
14.19%
Table EC.4 compares the performance of the heuristic against the upper bound for larger instances, over
a sample of 100 instances for each n. The optimality gap of the heuristic is on average less than 10% of the
upper bound, although it increases with n. Given that the upper bound appears to be less tight for larger
values of n from Table EC.3, the performance of the heuristic could be within a few percent of the optimal
solution, even for large problem instances.
EC.2.4.
SPDP-U
Based on Proposition 1, the optimal sequence is either U-shaped or crescendo. Based on that result we
introduce an alternate formulation of SPDP that allows the sequence to be either U-shaped or monotonically
increasing in service levels. Without loss of generality, we assume that x1 ≥ · · · ≥ xn . A U-shaped sequence
consists of a decreasing subsequence followed by an increasing one. Let yis ∈ {0, 1}, ∀i, ∀s be a variable
indicating if activity i belongs to subsequence s, where s = l for the decreasing subsequence on the left and
s = r for the increasing subsequence on the right of the U-shape. The variables tsi , ∀i, ∀s denote the duration
allocated to activity i when it is in subsequence s. With these decision variables, SPDP-U is formulated as:
! n−1
!
n
i−1
i
X
X
X
X
l
r
l
r
l l
r r
max S(y , y , t , t ) =
(xi − xi−1 )Φ T −
t j yj +
(xi − xi+1 )Φ
tj yj
(EC.7)
yl ,yr ,tl ,tr
i=1
subject to
j =1
i=1
yil + yir = 1,
Pn−1
1 ≤ i=1 yir
j =1
∀i
(EC.8)
(EC.9)
yis τ i ≤ tsi ≤ yis τ i , s = l, r,
Pn
(yil tli + yir tri ) = T
i=1
∀i (EC.10)
yil , yir ∈ {0, 1},
∀i. (EC.12)
(EC.11)
where x0 = b(0). The objective function consists of a decreasing and an increasing subsequences, such that
each subsequence is composed of all the activities. When tli = 0, the coefficient of xi in the decreasing
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e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
subsequence is equal to zero, i.e., no weight is associated with activity i in the decreasing subsequence.
Similarly if tri = 0, no weight is associated with activity i in the increasing subsequence. Constraints (EC.8)
allow an activity to be either a part of the decreasing or a part of the increasing subsequence. Constraint
(EC.9) eliminates the possibility of having only a decreasing sequence of activities. Therefore, constraints
(EC.8)-(EC.9) ensure a U-shaped or a monotonically increasing sequence. The remaining constraints (EC.10)(EC.11) correspond to the duration constraints. In particular, constraints (EC.10) ensure that if yis = 0, then
the corresponding duration tsi = 0. This rules out the possibility of allocating duration to an activity when
it does not belong to the corresponding subsequence.
Lemma EC.1. When T > 2/m(α, w), SPDP is equivalent to SPDP-U.
Proof.
We establish the equivalence result in two steps by showing that (i) any optimal solution for SPDP
is feasible in SPDP-U and has the same objective value, and (ii) any optimal solution for SPDP-U is feasible
in SPDP and has the same objective value.
(i) From Proposition 1 we know that any optimal solution of SPDP for T > 2/m(α, w) will either be
monotonically increasing or U-shaped in service levels. Let ((i?1 , . . . , i?n ), t? ) be an optimal solution of SPDP
such that xi?1 ≥ · · · ≥ xi?k ≥ xi?k+1 ≤ xi?k+2 ≤ · · · ≤ xi?n , for 0 ≤ k ≤ n − 2. We construct a corresponding feasible
solution for SPDP-U by assigning yirj = 0, trij = 0, yilj = 1, tlij = t?ij , ∀j = 1, . . . , k and yilj = 0, tlij = 0, yirj =
1, trij = t?ij , ∀j = k + 1, . . . , n.
Pn
t = T and τ ij ≤ t?ij ≤ τ ij , ∀j,
?
j =1 ij
By construction, (EC.8), (EC.9), and (EC.12) are satisfied. Since
(EC.10)-(EC.11) are also satisfied. Hence, such a solution is feasible in SPDP-U.
We next show that the objective values are equal. The objective value for SPDP at the optimal point is
given by
S SP DP ((i?1 , . . . , i?n ), t? ) =
n
X
j =1
=
k+1
X
j =1
(xi?j − xi?j−1 )Φ(t?ij + · · · + t?in )
(xi?j − xi?j−1 )Φ(
n
X
p=j
X
j =1
=
k+1
X
j =1
=
k+1
X
j =1
=
n
X
j =1
=
n
X
j =1
(xi?j − xi?j−1 )Φ(
j =k+2
j−1
k+1
=
n
X
t?ip ) +
(xi?j − xi?j−1 )Φ(T −
(xi?j − xi?j−1 )Φ(T −
X
X
t?ip ) +
j =k+2
n
X
tlip ) +
p=1
j =k+2
ij −1
(xi?j − xi?j−1 )Φ(T −
(xj − xj−1 )Φ(T −
(xj − xj−1 )Φ(T −
X
p=1
j−1
X
p=1
= S SP DP −U (yl , yr , tl , tr ).
n
X
l
p
t )+
p=1
j−1
X
t?ip )
p=j
n
p=1
j−1
X
n
X
tlp ) +
(xi?j − xi?j−1 )Φ(
(xi?j − xi?j−1 )Φ(
(xi?j − xi?j−1 )Φ(
j =k+2
n−1
X
j =1
tlp ypl ) +
(xj − xj +1 )Φ(
n−1
X
j =1
j
X
n
X
t?ip )
(EC.13)
trip )
(EC.14)
trp )
(EC.15)
p=j
n
X
p=j
ij
X
p=1
trp )
(EC.16)
p=1
(xj − xj +1 )Φ(
j
X
p=1
trp ypr )
(EC.17)
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e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
Equality (EC.13) is because
Pn
t = T and (EC.14) is because tlij = t?ij , ∀j ≤ k, trij = t?ij , ∀j ≥ k + 1. Equal-
?
j =1 ij
ities (EC.15)-(EC.16) are due to tlij = 0, ∀j ≥ k + 1, trij = 0, ∀j ≤ k, and x1 ≥ · · · ≥ xn by assumption. Finally,
equality (EC.17) is due to tsp > 0 if and only if yps = 1, ∀p, and for s = l, r by construction.
(ii) Let (yl? , yr? , tl? , tr? ) be an optimal solution of SPDP-U. We construct a sequence of activities
(i1 , . . . , in ) from (yl? , yr? ) and the corresponding duration (ti1 , . . . , tin ) as follows.
Pn−1
We define i0 = 0. Let k = j =1 yjl? . For j = 1, . . . , k, let ij = min{p > ij−1 such that ypl? = 1} and set tij =
l? l?
r? r?
tl?
p . We then assign ik+1 = n and set tik+1 = yn tn + yn tn . Finally, for j = k + 2, . . . , n, let ij = max{p <
ij−1 such that ypr? = 1} and set tij = tr?
p .
By equation (EC.8), each activity is such that either yjl? = 1 or yjr? = 1. Hence, by construction, all activities
are allocated and no activity is allocated twice, i.e., (i1 , . . . , in ) represents a sequence of activities. Since,
x1 ≥ · · · ≥ xn , we have xi1 ≥ · · · ≥ xik+1 ≤ xik+2 ≤ · · · ≤ xin . Because of constraints (EC.8) and (EC.10),
Pn
τ ij ≤ tij ≤ τ ij , ∀j. Moreover by (EC.11), j =1 tij = T . Hence, this constructed solution is feasible for SPDP.
We next show that the objective value of SPDP-U is the same as SPDP at (yl? , yr? , tl? , tr? ). The objective
value for SPDP-U at optimality is given by
S SP DP −U (yl? , yr? , tl? , tr? )
j−1
j
n
n−1
X
X
X
X
l?
r?
=
(xj − xj−1 )Φ(T −
tl?
(xj − xj +1 )Φ(
tr?
p yp ) +
p yp )
j =1
=
n
X
j =1
p=1
(xj − xj−1 )Φ(
n
X
=
X
j =1
=
n
X
j =1
l?
tl?
p yp +
p=j
k+1
(xij − xij−1 )Φ(
EC.2.5.
X
p=j
n
X
r?
tr?
p yp ) +
p=1
n
(xij − xij−1 )Φ(
j =1
n
X
p=1
n−1
X
j =1
(xj − xj +1 )Φ(
n
t ip ) +
X
j
X
r?
tr?
p yp ), since
p=1
n
X
r? r?
yjl? tl?
j + yj t j = T
j =1
n
(xij − xij−1 )Φ(
X
j =k+2
s?
tip ), since ts?
i > 0 if and only if yi = 1, s = l, r, by (EC.10)
p=j
and by construction, xi1 ≥ · · · ≥ xik+1 ≤ · · · ≤ xin
tip ) = S SP DP ((i1 , . . . , in ), t). p=j
Upper Bound Formulation on SPDP
We obtain an upper bound on SPDP-U (§EC.2.4) by dropping constraints (EC.8)-(EC.9), relaxing constraint
(EC.10) by replacing the lower bounds τ i by zero ∀i, and relaxing (EC.11) by converting the equality sign to
Pn
an inequality sign, i.e., i=1 (yil tli + yir tri ) ≤ T . We furthermore relax constraint (EC.12) by allowing yi to be
s
a continuous variable between 0 and 1, and then introduce a new set of variables, namely, ti = tsi yis , s = l, r.
After dualizing the relaxed constraint (EC.11), we obtain the upper bound U B = min U L(λ) + U R(λ) + λT ,
λ≥0
in which
U L(λ) = max U L(λ, t) =
t
n
X
i=1
subject to
U R(λ) = max U R(λ, t) =
t
(xi − xi−1 )Φ(T −
j =1
tj ) − λ
n
X
n−1
X
(xi − xi+1 )Φ(
0 ≤ ti ≤ τ i , ∀i,
ti
i=1
0 ≤ ti ≤ τ i , ∀i
i=1
subject to
i−1
X
(D-1)
i
X
j =1
tj ) − λ
n
X
ti
i=1
(D-2)
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l
r
in which we renamed the duration variables ti = ti , ∀i in U L(λ) and ti = ti , ∀i in U R(λ), for simplicity of
exposition. The Lagrangian relaxation U B = min U L(λ) + U R(λ) + λT is then solved by the subgradient
λ≥0
method (Fisher 2004).
Next we show that U R(λ) and U L(λ) can be solved in polynomial time. Without loss of generality, we
assume that the service levels of all activities are unique, i.e., x1 > · · · > xn . If we have xi = xi−1 , for some i,
we can delete activity i − 1 and increase the upper bound on ti to τ i + τ i−1 since
U L(λ, t) =
i−1
X
k=1
=
i−1
X
k=1
(xk − xk−1 )Φ(T −
(xk − xk−1 )Φ(T −
k−1
X
j =1
k−1
X
tj ) + (xi − xi−1 )Φ(T −
tj ) + 0 +
j =1
n
X
i−1
X
n
X
tj ) +
j =1
(xk − xk−1 )Φ(T −
k=i+1
(xk − xk−1 )Φ(T −
k=i+1
k−1
X
j =1
j6=i,i−1
k−1
X
j =1
tj − (ti−1 + ti )) − λ
tj ) − λ
n
X
n
X
ti
i=1
ti
i=1
= U L(λ, (t1 , . . . , ti−2 , 0, ti−1 + ti , ti+1 , . . . , tn )).
and similarly, for U R(λ, t).
Lemma EC.2. Without loss of generality, suppose that x1 > · · · > xn and T > 2/m(α, w). For any λ ≥ 0
only O(n2 ) points need to be considered to find an optimal solution for U L(λ).
Proof.
Let t? be the optimal solution for U L(λ). Hence, t? should satisfy the first order optimality
conditions:
1. The stationarity condition gives
∂U L(λ,t)
∂ti
+ µi − µi = 0, ∀i.
?
i
2. Complementary slackness gives µi (τ i − t ) = 0, ∀i and µi (t?i ) = 0, ∀i.
3. Primal feasibility implies t? satisfies the constraints (D-1).
4. Dual feasibility implies µi , µi ≥ 0, ∀i.
These conditions reduce to the following:
∂U L(λ, t? )
= µi ≥ 0, ∀i such that t?i = τ i
∂ti
∂U L(λ, t? )
= 0, ∀i such that 0 < t?i < τ i
∂ti
∂U L(λ, t? )
= −µi ≤ 0, ∀i such that t?i = 0.
∂ti
(EC.18)
We observe that for any two consecutive activities the following holds
i−1
X
∂U L(λ, t? ) ∂U L(λ, t? )
−
= −(xi − xi−1 )Φ0 (T −
tp ), ∀i = 2, . . . , n.
∂ti−1
∂ti
p=1
∂U L(λ,t? )
∂tn
= −λ ≤ 0, we may take t?n = 0.
Pn
1
Suppose that T − p=1 t?p > m(α,w
. Then we have
)
Since
(EC.19)
∂U L(λ,t? )
∂t1
<
∂U L(λ,t? )
∂t2
< ··· <
∂U L(λ,t? )
∂tn−1
< 0 by Lemma
A-1 and condition (EC.19). By (EC.18), t? = (0, . . . , 0) is an optimal solution.
Suppose next that there exists a k = min{i : i ∈ {2, . . . , n + 1} such that T −
by (EC.19),
∂U L(λ, t? ) ∂U L(λ, t? )
<
, ∀i < k
∂ti−1
∂ti
∂U L(λ, t? ) ∂U L(λ, t? )
≥
, ∀i ≥ k,
∂ti−1
∂ti
Pi−1
?
p=1 p
t ≤
1
m(α,w)
}. Therefore
(EC.20)
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e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
1
since x1 > · · · > xn and Φ(t) is strictly pseudoconcave and maximized at t = m(α,w
(Lemma A-1), and the
)
Pi−1 ?
1
latter inequalities are strict unless there exists an l ≥ k such that T − p=1 tp = m(α,w
for i = {k, . . . , l}.
)
Therefore, in case of strict inequalities there exists at most one i < k − 1 and at most one i > l such that
∂U L(λ,t? )
∂ti
= 0. By (EC.18) the optimal solution is of the form t? = (0, . . . , 0, t?l , τ l+1 , . . . , τ r−1 , t?r , 0, . . . , 0).
Pi−1
1
Suppose next that there exists an index j ≥ k such that T − p=1 t?p = m(α,w
for i ∈ {k, . . . , j} and
)
Pj
1
. Then, t?p = 0, p = k, . . . , j − 1. We consider three cases.
T − p=1 t?p < m(α,w
)
(i) Suppose that
∂U L(λ,t? )
∂tk−1
> 0. Therefore,
∂U L(λ,t? )
∂ti
> 0 for i = k − 1, . . . , j by (EC.19). By (EC.18), t?p =
τ p , p = k − 1, . . . , j, a contradiction since t?p = 0, p = k, . . . , j − 1.
(ii) Suppose that
∂U L(λ,t? )
∂tk−1
= 0. Therefore,
∂U L(λ,t? )
∂ti
t?p = 0 for p = k, . . . , j − 1. Therefore by (EC.20),
= 0 for i = k − 1, . . . , j by (EC.19). By assumption
∂U L
∂U L
< 0 for i < k − 1;
< 0 for i > j.
∂ti−1
∂ti
By (EC.18), the optimal solution is therefore of the form t? = (0, . . . , 0, tk−1 , 0, . . . , 0, tj , 0, . . . , 0).
(iii) Suppose that
∂U L(λ,t? )
∂tk−1
< 0. Therefore,
∂U L(λ,t? )
∂ti
(EC.20),
< 0 for i = k − 1, . . . , j by (EC.19). This implies by
∂U L
∂U L
< 0 for i ≤ k − 1;
< 0 for i > k − 1.
∂ti−1
∂ti
By (EC.18), the optimal solution is therefore of the form t? = (0, . . . , 0).
In summary, the optimal solution of U (λ, t) can be characterized by two indices (l, r) such that
t? = (0, . . . , 0, t?l , τ l+1 , . . . , τ r−1 , t?r , 0, . . . , 0), or t? = (0, . . . , 0, t?l , 0, . . . , 0, t?r , 0, . . . , 0),
where t?l and t?r , are such that 0 ≤ t?l ≤ τ l , 0 ≤ t?r ≤ τ r and T −
Pl
(EC.21)
?
p=1 p
t ≤ m(α, w) with r ≥ l. We next claim
that, for any such l and r if there exists a solution such that 0 < t < τ l and 0 < t?r < τ r , then there exists
?
l
only one such solution. In particular, t?l and t?r have to satisfy (EC.18), which reduces to:
f1 (tl ) = (xl+1 − xl )Φ0 (T − tl ) + · · · + (xr − xr−1 )Φ0 (T − tl − τ l+1 − · · · − τ r−1 ) = 0
f2 (tl , tr ) = (xn − xr )Φ0 (T − tl − τ l+1 − · · · − τ r−1 − tr ) − λ = 0.
Solving for f (tl ) = 0 gives the closed form solution
"
#
Pr−1
1
(xl+1 − xl ) + · · · + (xr − xr−1 )ew( p=l+1 τ p )
t̂l = T −
ln
.
Pr−1
w−α
(xl+1 − xl ) + · · · + (xr − xr−1 )eα( p=l+1 τ p )
Pr
Hence, there exists at most one tl , 0 < tl < τ l solving f1 (tl ) = 0. Moreover, because r ≥ l, T − p=1 t?p ≤
Pl
Pr
1
T − p=1 t?p ≤ m(α,w
, and therefore Φ00 (T − p=1 t?p ) < 0 by Lemma A-3. Hence, f2 (t?l , tr ) is a strictly
)
increasing function of tr given that xn < xr and there can be therefore at most one tr , 0 < tr < τ r solving
f2 (t?l , tr ) = 0.
Hence, for any l, r, there exists only a constant number of potential optimal solutions of type (EC.21)
Pl
1
and r ≥ l. Since there are O(n2 ) different ways of selecting l, r, the number
such that T − p=1 t?p ≤ m(α,w
)
of candidate solutions is therefore of the order O(n2 ).
1
Lemma EC.3. Without loss of generality, suppose that x1 > · · · > xn and T > 2 m(α,w
. For any λ ≥ 0,
)
only O(n) points need to be considered to find the optimal solution for U R(λ).
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e-companion to Das Gupta, Karmarkar, and Roels: The Design of Experiential Services
Proof.
Let t? be the optimal solution for U R(λ). Hence, t? should satisfy the first order optimality
conditions:
1. The stationarity condition gives
∂U R(λ,t)
∂ti
+ µi − µi = 0, ∀i.
?
i
2. Complementary slackness gives µi (τ i − t ) = 0, ∀i and µi (t?i ) = 0, ∀i.
3. Primal feasibility implies t? satisfies the constraints (D-2).
4. Dual feasibility implies µi , µi ≥ 0, ∀i. These conditions reduce to the following:
∂U R(λ, t? )
= µi ≥ 0, ∀i such that t?i = τ i
∂ti
∂U R(λ, t? )
= 0, ∀i such that 0 < t?i < τ i
∂ti
∂U R(λ, t? )
= −µi ≤ 0, ∀i such that t?i = 0.
∂ti
Observe that for any two consecutive activities the following holds
i−1
X
∂U R(λ, t? ) ∂U R(λ, t? )
−
= (xi−1 − xi )Φ0 (
tp ), ∀i = 2, . . . , n.
∂ti−1
∂ti
p=1
Since
∂U L(λ,t? )
∂tn
= −λ ≤ 0, we may set t?n = 0.
Suppose next that there exists a k = max{i : i ∈ {2, . . . , n + 1} such that
?
?
Pi−1
?
p=1 p
?
?
t ≤
(EC.22)
(EC.23)
1
m(α,w)
}. Therefore
∂U R(λ, t ) ∂U R(λ, t )
∂U R(λ, t ) ∂U R(λ, t )
≥
, ∀i < k and
<
, ∀i ≥ k,
(EC.24)
∂ti−1
∂ti
∂ti−1
∂ti
Pi−1
1
and the former inequalities are strict unless there exists an l ≤ k such that p=1 t?p = m(α,w
for i = l + 1, . . . , k
)
Pr
1
0
since x1 > · · · > xn and by Lemma A-1. If no such k exists, i.e., if t1 > m(α,w) , then Φ ( p=1 )t?p < 0, ∀r =
1, . . . , n. Therefore,
∂U R(λ,t? )
∂t1
< ··· <
∂U R(λ,t? )
∂tn−1
< 0 by (EC.22) and since
∂U R(λ,t? )
∂tn
?
≤ 0. Hence, the first-order
conditions (EC.22) imply that the optimal solution is t = (0, . . . , 0).
Pi−1
Pl−1
1
Suppose that there exists an l ≤ k such that p=1 tp = m(α,w
for i = l + 1, . . . , k and p=1 tp <
)
By (EC.23),
∂U R(λ,t? )
∂tk
?
l
∂U R(λ, t? ) ∂U R(λ, t? )
<
, ∀i = k + 1, . . . , n.
∂ti−1
∂ti
?
?
)
)
< · · · < ∂U R∂t(λ,t
≤ 0 and therefore, ∂U R∂t(λ,t
= ··· =
n
l
?
n
implies that t = · · · = t = 0. Since
∂U R(λ,t? )
∂ti
.
∂U R(λ, t? ) ∂U R(λ, t? )
>
, ∀i = 1, . . . , l
∂ti−1
∂ti
and
In particular,
1
m(α,w)
∂U R(λ,t? )
∂ti−1
>
∂U R(λ,t? )
∂ti
= 0.
∂U R(λ,t? )
∂tk
< 0. By (EC.22) this
, ∀i = 1, . . . , l, there is at most one i such that
Therefore, by (EC.22), when there exists an index k such that
Pk−1
?
p=1 p
t ≤
1
m(α,w)
, the optimal solution has
the form t? = (τ 1 , . . . , τ r−1 , t?r , 0, . . . , 0), with at most one activity t?r assigned a duration 0 < t?r < τ r , in which
Pr
Pr
1
1
r ≤ k, i.e., i=1 t?i ≤ m(α,w
. We next claim that for any r such that i=1 t?i ≤ m(α,w
, there exists at most
)
)
one interior solution tr , 0 < tr < τ r , such that
∂U R(λ,t? )
∂tr
= 0. In particular, t?r has to satisfy the first-order
optimality conditions (EC.22):
f (tr ) = (xr − xn )Φ0 (τ i1 + · · · + τ r−1 + tr ) − λ = 0.
(EC.25)
Pr
P
P
k−1
r
1
Because p=1 t?p ≤ p=1 t?p ≤ m(α,w
, Φ00 ( p=1 t?p ) < 0 by Lemma A-3. Moreover, xr > xn by assumption.
)
Hence, f (tr ) is a strictly decreasing function of tr . Therefore, there can be at most one solution for f (tr ) = 0.
Since there are O(n) different ways of selecting r, the number of candidate solutions is of the order O(n).