ISyE 3104: Introduction to Supply Chain Modeling:
Manufacturing and Warehousing
Instructor : Spyros Reveliotis
Summer 2006
Solutions for Homework #3
ISYE 3104 Spring 2006
Homework 7 Solution
Section 5.2
2. Deterministic variation is systematic variation in demand that can be predicted in
advance. This variation is typical in the demand of seasonal products, i.e., products that
tend to present the main volume of their demand at certain periods of a regular cycle.
Examples of such seasonal products are products pertaining to summer recreational
activities or toys that are consumed primarily during the Christmas season. Aggregate
and production planning problems to be discussed in the next part of the course seek to
characterize and accommodate this systematic variation of the demand in the most
efficient / cost-effective manner.
On the other hand, random variation is the unpredictable component in the variation of
the experienced demand, and it is typically absorbed through safety stock.
Section 5.3
7. If he only keeps track of the number of sales, he has no way to accurately estimate the
demand since demand = sales + lost sales. He would need some way to gauge the lost
sales. One method would be to increase his supply for a period of time so that he would
be able to meet all demand. An alternative approach would be to consult some published
statistics on this demand.
8. a)
c0 = 0.08 - 0.03 = 0.05
cu = 0.35 - 0.08 = 0.27
0.27
 0.84375
Critical ratio =
0.05  0.27
From the given distribution, we have:
Q
0
5
10
15
20
f(Q)
0.05
0.1
0.1
0.2
0.25
F(Q)
0.05
0.15
0.25
0.45
0.7
25
30
35
0.15
0.1
0.05
0.85
0.95
1
0.8438
Since the critical ratio falls between 20 and 25, the optimal is Q = 25 bagels.
b) The answers should be close since the given distribution appears to be close to the
normal. The quality of approximation could be assessed more systematically by
using the relevant tests from your statistics (or simulation) classes.
c)    xf ( x)  (0)(0.05)  (5)(0.10)  ...  (35)(0.05)  18
 2   x 2 f ( x)   2  402.5  (18) 2  78.5
  78.5  8.86
The z value corresponding to a critical ratio of 0.84375 is 1.01.
Hence, Q* = z +  = (8.86)(1.01) + 18 = 26.95 ~ 27
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ISYE 3104 Spring 2006
Homework 7 Solution
10. a) A period is three months. Holding cost per year is $500, which means that the
cost for a 3-month period is:
500/4 = $125 = c0
cu = 250 (emergency shipment cost)
250
 0.667  z  0.44
Critical ratio =
250  125
Hence, Q* = z +  = (6)(0.44) + 60 = 62.64 ~ 63 cars.
b) cu = 150
c0 = 125
150
 0.5454  z  0.11
150  125
Hence, Q* = z +  = (6)(0.11) + 60 = 60.66 ~ 61 cars.
Critical ratio =
c) This corresponds to an infinite horizon problem with lost sales. In this case,
cu = lost profit = $3,500
c0 = holding cost = 125
3500
 0.9655  z  1.82
Critical ratio =
3500  125
which gives Q* = (6)(1.82) + 60 = 70.92 ~ 71 cars.
Section 5.5
14. a) Note: We assume 4 weeks/month and 48 weeks/year.
Monthly demand is normal (= 28= 8)
 = 14 weeks = 3.5 months => Lead time demand is normally distributed with =
28(3.5) = 98, = (8) 3.5  15
h = Ic = (0.3)(6) = $1.8
= (28)(12) = 336 units per year
p = 10
K = 15
= 98
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ISYE 3104 Spring 2006
Homework 7 Solution
2(336)(15)
 75
1.8
(75)(1.8)
F ( R1 ) 
 0.04
(10)(336)
 z  1.75
 R1  z    124, n( R1 )  L( z )  0.2426
Q0 
2(336)
(15  (10)(0.2426))  81
1.8
(81)(1.8)
F ( R2 ) 
 0.0434
(10)(336)
 z  1.71
 R2  z    124
Since R2 = R1, we stop. Conclude that (Q,R) = (81,124)
Q1 
b) S = R -  = 124 - 98 = 26 units
15. a) EOQ = 75 (from 14(a))
Type 1 service of 90% => F(R) = 0.10, z = 1.28
R = z = (15)(1.28) + 98 = 117
(Q,R) = (75,117)
b) Find Type II service level achieved in part (1)
n( R )
L( z ) (15)(0.0475)
 1  

 0.0095
Q
Q
75
   0.9905 (99.05% service level)
So it is actually much higher than the 90% fill rate that he wanted.
Section 5.6
22. a) F(Q) = 0.95. The corresponding z value is z = 1.645, giving Qz=
(6)(1.645) + 60 = 69.87 ≈ 70.
b) n(Q) = (1-) = (0.05)(60) = 3.
L(z) = n(Q)/ = 3/6 = 0.5
z = -0.19
Qz= 58.86 ≈ 59.
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