The University of Chicago
Booth School of Business
Harold’s Hot Dog Stand∗
Part I: Deterministic Process Flows
December 28, 2011
Harold runs a hot dog stand in downtown Chicago. After years of consulting and a stint on wall street, Harold decided to simplify his life. While
the hot dog business is good, he wonders if he could improve his operation.
He knows that customers will sometimes not stop if the line is too long —
could he improve his service?
1
System I
Harold has an analytical mind. He divided up the service process into 3 main
tasks. A) Retrieving a warm steamed bun and placing a hot dog inside. B)
Putting on the toppings. C) Getting the beverage, and collecting payment.
He had his brother Bernie measure how long it took to perform each task.
The flow diagram in Figure 1 shows the time required for each task. The
total time required for all three tasks is 90 sec.
Harold performs all three tasks. He notes that the rate at which he can
make money, his revenues, cannot exceed his maximum speed or capacity.
He denotes his service capacity for System I by µI , so that
Capacity: µI = 1 cust/90 sec.
∗
This case was prepared by Donald D. Eisenstein, Booth School of Business, University
of Chicago, Chicago, IL 60637.
E-mail: [email protected]
c
Copyright 2010
Donald D. Eisenstein
1
Harold’s Hot Dog Stand
System I
(A)
Dog & Bun
15 secs
(Harold)
(B)
Toppings
45 secs
(Harold)
(C)
Drink&Pay
30 secs
(Harold)
Figure 1: The original 3 task process that Harold performs to serve hot dogs
to customers.
Harold opens his hot dog stand between 11:00am–2:00pm — we consider
a day for Harold to be 3 hrs. No day or time is the same, queues fluctuate
sometimes forming long lines, and at other times the queue is empty. Harold
collects receipts for three weeks, and calculates that the average customer
spends $5.00. Thus Harold’s capacity can be expressed in terms of the
maximum rate he could generate revenues:
µI = 1 cust/90 sec
= 120 cust/day (using 10800 Sec in a 3 hour day.)
= $600/day.
In total, Harold actually averages $500.00 per day of revenues. And thus
$500/day
he calculates that he averages $5/cust = 100 cust/day.
He now lets λ represent his throughput — the average rate at which
customers move through his system:
Throughput: λ =
$500/day
$5/cust
= 100 cust/day
= 1 cust/108 sec.
Harold realizes that on most days there are times when he is idle, even
though there are other times when he has a number waiting in line. He
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2
Harold’s Hot Dog Stand
measures his Capacity Utilization, denoting it by ρ, to be
Capacity Utilization: ρI = λ/µ =
100 cust/day
1 cust/108 sec
=
= 5/6.
120 cust/day
1 cust/90 sec
Harold realizes that his Capacity Utilization must always be less than 1,
since it is impossible for him to serve more customers in a day than he has
capacity. He jots down:
λ ≤ µ,
and
0 ≤ ρ ≤ 1.
He also notes that he might calculate his Slack Capacity to be the difference between his service rate capacity and the rate customers move through
his system:
Slack Capacity : µI − λI = 120 cust/day − 100 cust/day = 20 cust/day.
2
System II
Harold wondered what would be the effect of adding some more workers to
his hot dog stand. He thinks his niece Sarah would help out, and of course
his brother Bernie. They each would handle one of the tasks, such as the
following:
System II
(A)
Dog & Bun
15 secs
(Harold)
(B)
Toppings
45 secs
(Bernie)
(C)
Drink&Pay
30 secs
(Sarah)
Figure 2: Harold now shares the tasks with Bernie and Sarah.
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3
Harold’s Hot Dog Stand
Now how fast can they together process customers — that is, what would
be the new system capacity? Harold realized that the Topping task was the
Bottleneck, the task that took the longest time (45 secs). It is the task
that would “throttle” the process. That is, even though Dogs&Buns is
faster (15 secs), it did not do much good — piling up naked dogs in front
of the Topping process would only serve to get them cold. And similarly,
Drink&Pay was faster (30 secs), but again, that extra speed will not help
the process — Drink&Pay would be idle 15 secs for every customer. The
capacity of System II, µII , would then be determined by the bottleneck
process, Toppings, and would therefore be
µII =
1
cust/sec.
45
Or as Harold explained, each step, A, B, and C for System II had a
capacity as follows:
1
cust/sec,
15
1
cust/sec, and
µB =
45
1
µC =
cust/sec.
30
µA =
Each capacity is expressed as a rate of flow in units of customers per second.
Since each customer must process through all three tasks, the bottleneck is
the task with the smallest capacity. So the system capacity is determined
by the task with the smallest capacity:
µII = min{µA , µB , µC }
= min{1/15, 1/45, 1/30} cust/sec
= 1/45 cust/sec.
Harold tried to explain this to Bernie, but Bernie failed to grasp the
concept. Harold then drew a diagram that maps out the maximum rate of
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4
Harold’s Hot Dog Stand
flow of customers through his system over time and showed it to Bernie.†
This finally convinced Bernie.
This system, in which all three combined efforts for each customer did
improve the capacity of the system; from µI = 1/90, to µII = 1/45 cust/sec.
The capacity doubled — but the labor tripled! Both Task A (Harold) and
Task B (Sarah) have slack in System II whenever the system is operating at
capacity (λ = 1/45 cust/sec).
Slack capacities for System II when flow of customers is exactly
λ = 1/45 cust/sec:
SlackA = 1/15 − 1/45 = 2/45 cust/sec.
SlackB = 1/45 − 1/45 = 0 cust/sec.
SlackC = 1/30 − 1/45 = 1/90 cust/sec.
The capacity utilization of each worker is:
Capacity Utilization for System II when flow of customers is exactly λ = 1/45 cust/sec:
1/45
= 1/3
1/15
1/45
ρB =
=1
1/45
1/45
ρC =
= 2/3
1/30
ρA =
†
The diagram Harold drew has been found and is more generally known as a Gantt
Chart. Your professor may be good enough to share this diagram with you during class.
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Harold’s Hot Dog Stand
Harold was not pleased that so much of his labor would be idle when
the system was so busy. If he could somehow keep all his workers busy then
they could equally share the total 90 sec time required to produce each dog.
The system capacity would increase to 3/90 sec. Another way to view this,
is that his system has some Cost of Imbalance:
For System II:
Cost of Imbalance =
Theoretical Capacity of
a Balanced System
−
Actual
Capacity
= 3/90 − 1/45 cust/sec
= 1/90 cust/sec.
Harold could regain this capacity if he could re-engineer the tasks somehow to evenly spread the load among the workers. Or, perhaps set up a line
like the local Chipotle, where workers tend to stay busy by multi-tasking,
moving to wherever the work is. But such a re-engineering task would have
to wait for Harold.
3
System III
Harold wondered if instead he could operate some processes in parallel. That
is, couldn’t a customer’s hot dog be assembled at the same time their drink
and payment is done? But that would require them to write out the entire
order first, send the hot dog to assembly and simultaneously the drink and
payment could be made. This is in contrast to the current system where
the customer places his order as the hot dog is assembled — “A bit more
mustard, but no onions!”. Harold devised the following process shown in
Figure 3 — he estimated the processing times from some experiments he
ran with Sarah and Bernie.
Now, what is the system capacity? Each customer order must pass
through all 3 Tasks. Again, the system bottleneck is the task with the
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Harold’s Hot Dog Stand
System III
Dog, Bun,
and Toppings
35 secs
(Bernie)
Place Order
30 secs
(Harold)
EXIT
Drink&Pay
25 secs
(Sarah)
Figure 3: The processes are re-engineered with two of them performed in
parallel.
largest processing time; Dog, Bun, and Toppings at 35 secs. So, the capacity
of System III is:
µIII = 1/35 cust/sec
He was able to increase his capacity by a bit in System III, but not too
much. He realized that even though he parallelized some tasks, his capacity
was still at the mercy of the bottleneck process. He was not able to smooth
the load evenly over all three of his workers.
But something else did seem significantly better about the new process
— indeed, there was a measurement he has yet to take! How long does it
take a customer to go through his system? He called this the Cycle Time.
Now certainly customers often suffered some time to wait in a queue, but he
decided he would tackle that later.‡ So for now, Harold will think about the
cycle time just through the production process. And furthermore, Harold
‡
In fact, your professor has documented this exploration of queues in a Part II com-
panion to this write-up.
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Harold’s Hot Dog Stand
assumes that once a customer enters service, he is not delayed by another
customer.
So for the original serial system, shown in Figure 1, the cycle time of a
customer is 90 secs. For the new process, shown in Figure 3 the cycle time
of a customer is reduced to 65 secs. It is the top path through the network,
30 secs to place an order plus 35 secs to assemble the hot dog. This top path
is the longest path, it is known as the critical path through the network.
So Harold realized that placing tasks in parallel may allow one to reengineer some of the tasks so as to balance the system better and thus
improve its capacity. But parallelizing tasks is primarily a tool to reduce
the critical path through the network and thereby decrease the cycle time
of a customer through the process.
4
Little’s Law
Bernie and Harold then went to the library and picked up a book on Process
Flows. They found the following
Little’s Law:
INV = λ CT.
They soon understood that Little’s Law equates the Average Inventory
(INV) in a system to the product of the Average Throughput (λ) and average Cycle Time (CT). The Inventory in the system is the number of “jobs”
or customers in the system. INV is the time-average number of customers
in the system.
Harold was not sure he really understood the law, so he collected some
data from four customers who entered and exited his system. When the
first customer entered, the system was empty; and when the last customer
exited, there were no other customers waiting. He produced the following
table of each customers arrival and departure times (all units of time are in
minutes).
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Harold’s Hot Dog Stand
Cust
Arrival
Departure
Cycle Time
1
0.0
1.5
1.5
2
1.0
3.0
2.0
3
1.0
4.5
3.5
4
3.0
6.0
3.0
In the last column he computed the Cycle Time of each customer —
this is the time each customer spent in the queue plus service. And so, the
average cycle time, CT is easily computed as:
1.5 + 2.0 + 3.5 + 3.0
= 5/2 min.
4
The average throughput can also be calculated. If one considers the 6.0
CT =
min as the time frame of interest, then the average throughput, λ, is simply:
4 cust
2
= cust/min.
6 min
3
So, by Little’s Law, he would compute the average number of customers
λ=
in the system, INV to be:
INV = λ CT = (2/3 cust/min) (5/2 min) = 5/3 cust
.
Almost 2 customers in the system on average. Harold produced the
following plot shown in Figure 4 of Customers (Inventory) over time to help
him understand Little’s Law.
From the plot in Figure 4, and a recall of a lecture in Calculus by his
high school teacher, Harold realized he could calculate INV directly from
the plot as follows:
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Harold’s Hot Dog Stand
Customers
(Inventory)
3
3
2
2
3
4
2
3
1
1
0
1
2
3
4
4
5
6
Minutes
Figure 4: A plot of customers in the system over time. Each customer is a
series of blocks, as they occupy a place in service over time. The number in
each block represents the customer number.
Area Under the Curve
Time
(Area of Bottom Row) + (Area of Middle Row) + (Area of Top Row)
=
Time
((1 × 6) + (1 × 3.5) + (1 × 0.5)) cust-min
=
6 min
10
5
= = cust
6
3
INV =
Harold was impressed with himself. He indeed has verified Little’s Law
on this small example.
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Harold’s Hot Dog Stand
Harold and Bernie decided to make a list of what they have learned so
far:
µ : The average service capacity of the system (customers/time).
λ : The average throughput rate of customers in the system (customers/time).
λ ≤ µ:
The average rate of throughput of customers cannot exceed the
average capacity of the system.
CT: The average time a customer spends in the system, queueing plus
service (time).
Little’s Law:
INV = λ CT: The average number of customers in the
system (INV) equals the average Throughput (λ) times the average
Cycle Time (CT).
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Harold’s Hot Dog Stand
Harold and Bernie’s Practice Exercises
1. After a busy week under System I, Harold determines that he averaged
$550 per day in revenues. Estimate Harold’s slack capacity for this
week.
2. Harold is considering adding a new cash register which would change
System II by decreasing Sarah’s task time from 30 secs to 20 secs.
What is the improvement to system capacity?
3. Back to the original System II. Harold is now considering adding a
station in which customers will self-serve mustard and relish after purchase. As a result Harold estimates that Bernie can now complete the
Toppings task in 20 seconds. What is the improvement to system capacity? Suppose customers arrive at a rate of 1 cust every 35 sec, what
is the new capacity utilization for Task A, B, and C? What is the new
Cost of Imbalance?
4. For a one hour period under System II, Harold asks his son Jim to
walk by the stand every 5 minutes and write down how many people
he sees in the system. At the end of the period, Jim reports data of
0, 1, 2, 2, 1, 3, 4, 5, 3, 2, 1, and 0 customers observed.
In addition, over the one hour time period, Harold’s receipts indicated
that he served 20 customers. Estimate of the average cycle time for
customers during this period.
5. Suppose in Harold’s experiment (described in the Section 4) one more
customer was observed, arriving at minute 2.5 and departing at minute
5. What is this customer’s cycle time? What is the system’s new
average cycle time? What is the new average number of customers in
the system? Verify that Little’s Law holds.
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Harold’s Hot Dog Stand
SOLUTIONS
1. Harold’s service capacity under System I is 120 customers per day. At
$5 per dog, we estimate that Harold averaged a throughput of 110
customers per day. Thus his average slack capacity is estimated to be
120 - 110 = 10 cust/day.
2. There is no improvement. Bernie is still the bottleneck with a processing time of 45 secs.
3. Task (C) is now the bottleneck with 30 secs. So system capacity
improves to 1/30 cust/sec. The capacity utilizations are:
1/35
= 15/35
1/15
1/35
ρB =
= 20/35
1/20
1/35
= 30/35
ρC =
1/30
ρA =
The updated Cost of Imbalance is 3/65 - 1/30 = 0.0128 cust/sec.
4. Within the fidelity of an observation every 5 min, we estimate:
INV =
0+1+2+2+1+3+4+5+3+2+1+0
= 24/12 = 2 cust,
12
and we have:
λ = 20/60 = 1/3 cust/min.
And thus by Little’s Law we have:
CT = INV/λ = 2 cust/(1/3 cust/min) = 6 min
NOTE: In calculating INV above, the units were sloppily left
out. Be sure you can verify the calculation.
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Harold’s Hot Dog Stand
5. The new customer’s cycle time is 2.5 min. So we now have:
CT =
1.5 + 2.0 + 3.5 + 3.0 + 2.5
= 25/10 min.
5
and
5
cust/min.
6
And, after updating the graph of inventory over time we have:
λ=
((1 × 6) + (1 × 4.0) + (1 × 2.5)) cust-min
6 min
25
= cust.
12
INV =
So we can verify Little’s Law with:
INV = λ CT
25
5
25
cust = cust/min ×
min
12
6
10
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