Exam Inventory Management
Part 1: Theory (1 hour)
• Closed book: Inspection of book/lecture notes/personal notes is not allowed;
computer (screen) is off.
• Maximum number of points for this part: 30.
• Answers can be given in Dutch or in English; formulate concisely.
• Make clear assumptions for results that you need to solve a problem if you were
not able to obtain these results from an earlier part of the problem.
(8) 1. (a) Summarize at least 5 different reasons for holding stock.
Solution:
• To provide a buffer between different operations.
• To allow for mistakes between supply and demand rates.
• To allow for demands that are larger than expected or at unexpected
times.
• To allow for deliveries that are delayed or too small.
• To take advantage of price discounts on large orders.
• The possibility to buy items when the price is low and expected to rise.
• To create the possible usage or sale of special articles (e.g. spare parts)
for the future although they are no longer produced.
• To allow for full loads (reducing transport costs).
• To provide cover for emergencies.
(b) There is a number of performance measures that is frequently used in inventory
models. Most can be considered individually as well as aggregate. Summarize
at least 5 of these KPIs.
Solution:
• Inventory turnover: (annual sales) / (average inventory value)
• Number of back-orders per year
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• Total / pct value short per year
• Number / pct of missed order lines per year
• Number of stockout occasions per year
• Number of replenishments per year
• Average holding cost per year
• Total safety stock value
(c) Which possibilities do you have in practice to estimate the lead time demand
(LTD) variance?
Solution:
• Collect LTD-forecast errors in non-overlapping intervals, and compute
the variance.
• Use an empirical relation, such as σ̂L (t) = Lc σ̂1 (t). See SPP 4.6.2.
• Derive an approximating formula (LN 3.6).
(d) Give the expression of a time supply, based on the EOQ formula.
Solution: EOQ expressed as time supply
r
TEOQ = EOQ/D =
2A
.
Dvr
(e) Define the service criteria P2 and P3 , and explain why P2 ≈ P3 .
Solution:
P2 : Fraction of demand to be satisfied routinely from shelf
P3 : Fraction of time during which net stock is positive
If the demand pattern is stationary and smooth, then the demand during
the interval in which net stock is positive (negative) is proportional to the
length of this interval.
(10) 2. (a) What is the general effect of using estimates instead of the unknown true values
when determining the order-up-to level in a (R, S)-inventory control model based
on the minimization of total costs or on a service criterion?
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Solution: Higher costs/lower service.
Consider a retailer selling a certain commodity. Assume that he uses an (R, S)inventory control model with zero lead-time. Net demand during review can be
negative due to goods returned by customers. Assume that this net demand during
review, DR , can be described well by a stationary normal distribution with a relatively
high coefficient of variation (νR = σR /µR ); further assume that subsequent review
demands are independent, and, for now, that µR and σR are known with certainty.
(b) Give an expression for P (DR < 0) using Φ(.), the CDF of a standard normal
distribution, and νR .
Solution:
P (DR < 0) = P (Z < (0 − µ)/σR ) = Φ(−1/νR )
(c) What happens to the net stock and the inventory position when DR < 0 in a
certain order cycle? What should the retailer (strictly speaking) do when this
occurs? Does this influence the attained service level?
Solution: Net stock and inventory position are raised beyond S, the excess
demand should be returned to the supplier; not returning the excess demand
will increase the service level.
Now assume that is µR unknown, and that σR can be very well estimated based on
historical demand (so, assume σ̂R ≈ σR , and E(σ̂R ) = σR ). Assume further that the
retailer wants to attain a cycle service level of α = 0.95 and that he determines after
each review moment, t, a new order-up-to level, Ŝ(t) = µ̂(t) + cα σR using the average
of the last n demands during review for µ̂R (t).
(d) Define cα such that Ŝ(t) is unbiased (that is, the long run average of Ŝ(t) equals
the value of S that the retailer should take when µR would be known exactly).
Show that this expression for Ŝ(t), however, leads to a lower attained cycle
service than α on average.
Solution: Take cα = Φ−1 (α). Then
E[µ̂R (t)] = µR
⇒
E[Ŝ(t)] = µR + cα σR = S.
Note: DR and Ŝ(t) are independent and that both are normally distributed.
Furthermore:
P (DR < Ŝ(t)) = P (DR − Ŝ(t) < 0)
p
p
= Φ(cα σR /(σR 1 + 1/n)) = Φ(cα / 1 + 1/n) < α.
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(12) 3. Consider the (s, Q) inventory control system.
(a) Define the so-called “two-bin”-system, which is a special case of the (s, Q)system. Which condition has to be fulfilled to call a system a “two-bin”-system?
Solution: No more than one replenishment order can be outstanding at any
point in time.
(b) Describe precisely when (s, Q)-systems coincide with (s, S)-systems. Give a
realistic example when this occurs.
Solution: No undershoot; unit-size demand.
(c) Even (R, S) and (s, Q)-systems can coincide. Under which (extraordinary) condition does this occur? What is the relation between s, S, and Q in that case?
Solution: DR = Q; s + Q = S
(d) Summarize all conditions necessary to assure that Gu (k) = Q(1 − P2 )/σL is a
valid fill rate service model, where k is the safety factor and
R ∞Gu (k) is the loss
function for the standard normal distribution, i.e., Gu (k) = k (x − k)φ(x)dx.
Solution: Q has to be predetermined and must be expressed in the same
units as σL ; P2 and/or Q/σL has to be relatively large; lead time demand
is stationary and normally distributed; demands for non-overlapping time
intervals are independent; no undershoots.
(e) A high value of D × v (demand × value) for an item, and, at the same time, a
small lead time demand (e.g. < 10) asks for a special (s, Q)-system. Describe an
(s, Q) model based on a B2 -cost criterion that will often be appropriate in this
case. How would you (pre)determine the order quantity, Q, in that case? Give
an expression for the expected total relevant costs (ETRC) as a function of s.
Solution: Cf. slides “(s, Q)-control systems for A-items”.
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Exam Inventory Management
Part 2: Practice (2 hours)
• Open book: Inspection of book and lecture/personal notes is allowed.
• Maximum number of points for this part: 50.
• Excel and MATLAB may be used.
• E-mail and Internet are not allowed with the exception of Blackboard.
• Motivate all your answers. If necessary indicate how you have obtained your
results using the software.
• Make clear assumptions for results that you need to solve a problem if you were
not able to obtain these results from an earlier part of the problem.
(20) 1. The manager of a company is unhappy about the order sizes that are being used for
some important items. He wants his support staff to give a report on the use of the
economic order quantity (EOQ) instead of the company’s current ordering method.
He wants the computations to be carried out with a fixed replenishment costs of
e 30 and a carrying charge of 10% per year. Table 1 gives an overview of the yearly
demand and value of the six most important items.
Item
Yearly demand (units)
Value (e)
400
500
900
1600
1600
1600
98
89
23
71
48
74
1
2
3
4
5
6
Table 1: Item overview.
(a) Compute the EOQ and corresponding average number of yearly replenishments
for all items.
Solution: Apply the formula Qi =
(unrounded) results:
Exam Inventory Management
p
2ADi /(vi r). This yields the following
i
Qi
1
2
3
4
5
6
49.4872
58.0585
153.2262
116.2804
141.4214
113.8990
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(b) Give the general formula for the total relevant costs (TRC) of an item as a
function of the order size Q under the assumptions of the EOQ-model. Compute
the (aggregate) TRC for all items.
Solution:
i
+ 12 Qvi r
TRCi (Q) = AD
Q
P
TRC = i TRCi (Qi ) = 3701.41
(c) When the manager is confronted with the results he insists that the total number of replenishments is too large and should be halved. He argues that his
initial estimate of the fixed replenishment costs is too low. What value of fixed
replenishment costs should be used to obtain the desired result?
Solution: For the model above we find REP = 61.7. We use the model for
the EOQ-exchange curve:
2
1 Xp
Di vi ≡ c
TACS × REP =
2 i
TACS
A
=
REP
r
We want REP0 = 30.8, hence the first equation yields TACS0 = c/REP0 .
From the second: A0 = r × TACS0 /REP0 = 120.
Alternatively, since we are changing A for all items we can simply look at
the EOQ-formula. The number of replenishments is halved if Q is doubled,
which happens when A is increased by a factor 4.
(d) Suppose that the newly computed value of the fixed replenishment costs is indeed
correct, but that actually the initial estimate is used to compute the EOQ.
Compute the percentage cost penalty of using the wrong fixed replenishment
costs.
Solution: Let Q0i denote the EOQ evaluated with A0 . Then
X
TRC0 (Q1 , . . . , Q6 ) =
A0 Di /Qi + Qi vi r/2
i
0
TRC (Q) − TRC0 (Q0 )
PCP =
× 100% = 25%
TRC0 (Q0 )
(e) The manager still is not satisfied because of the increased total average cycle
stock (TACS) of the new approach. Give a formula that relates the total number
of replenishments to the TACS and compute the number of replenishments that
corresponds to a TACS of e 30.000.
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Solution: The EOQ-exchange curve gives the relationship between TACS
and REP:
2
P √
1
Di vi
i
2
REP =
TACS
For TACS = 30000, this yields REP = 38.1.
(30) 2. A company uses a (R, S)-system to replenish one of its items. The review period R
is 4 weeks. The company has a contract with the item’s supplier for a lead time (L)
of 3 weeks.
A forecasting analysis on the weekly demand for this item has resulted in an average
weekly demand of 30 items and a corresponding forecast error standard deviation of
15 items. It is assumed that a normal distribution is a good approximation of the
weekly demand.
The company uses a P1 -criterion with a target service level of 95% to set the orderup-to level S.
(a) Compute the order-up-to level S that the company should use for this service
criterion.
Solution: With µ = 30 and σ = 15 we find µDR+L = (R + L)µ and σDR+L =
√
R + Lσ. Furthermore
S − µDR+L
P1 = Pr(DR+L < S) = Φ
σDR+L
Taking the inverse of the normal cdf yields S = 275.3.
(b) Give the service equation for the P2 -criterion and the (R, S)-system. Compute
the induced P2 service for the previously computed order-up-to level.
Solution:
E[(DR+L − S)+ ] − E[(DL − S)+ ] = (1 − P2 )E[DR ]
σDR+L Gu ((S − µDR+L )/σDR+L ) − σL Gu ((S − µL )/σL ) = (1 − P2 )µR
with
Z
k
(x − k)φ(x)dx = φ(k) − kΦ(−k)
Gu (k) =
−∞
If we substitute S into the service equation, then we obtain P2 = 0.9931.
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Analysis of the supplier lead time performance has indicated that the lead time is
not always equal to the promised 3 weeks. The following list shows the supplier lead
times for the last 12 deliveries.
2 3 4 4 4 2 4 4 2 3 3 3
(c) Compute a more accurate estimate for both the mean and standard deviation
of the demand during lead time plus review period (R + L) based on this information. You may assume that the lead time distribution is stationary.
Solution: The table leads to the following estimates for the lead time: µL =
3.1667 and σL = 0.8348. Now, if we assume independence between L and
D, then result follows from
µ̂DR+L = E[R + L]E[D1 ] = (R + µL )µ = 215
p
σ̂DR+L = E[R + L] VAR[D1 ] + (E[D1 ])2 VAR[R + L]
q
= (R + µL )σ 2 + µ2 σL2
= 47.3
(d) Recompute the order-up-to level that should be used to obtain the desired P1
service level target.
Solution: S 0 = 292.8
Instead of implementing this new order-up-to level the manager has been guaranteed
by the supplier that future lead times will be exactly 3 weeks. Hence, from now on
we may assume L = 3 weeks.
It turns out that the normality assumption is not really satisfied and that demand is
better described by a gamma distribution.
(e) Compute the P1 service level for the order-up-to level from part (a) assuming
that demand follows a gamma distribution.
2
Solution: We assume that DR+L ∼ Gamma(ρ, λ) with ρ = µ2DR+L /σD
R+L
2
and λ = µDR+L /σD
.
Then
P
=
F
(S)
=
gamcdf(S,
rho,
1/lambda)
=
1
ρ,λ
R+L
0.9408.
(f) Give an approximation of the average on hand stock using the same assumptions
as in (e).
Solution:
E[OH] = SS + 1/2µR = S − µDR+L + 1/2µR = 125.3
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