ISyE 3104: Introduction to Supply Chain Modeling:
Manufacturing and Warehousing
Instructor : Spyros Reveliotis
Summer 2006
Solutions for Homework #2
ISYE 3104 Summer 2006
Homework 2 Solution
Chapter 4
1. The questions that inventory control addresses are when to order and how much to
order.
2. Too much:
a) Money that is tied up in inventories could be invested elsewhere.
b) Costs of supporting inventory could be high, including cost of storage, taxes,
insurances, pilferage, etc.
c) Production/distribution inefficiencies could arise.
d) Stock can become obsolete.
Too little:
a) Insufficient stock to meet customer demand, i.e., opportunity cost of lost sales
b) Loss of customer goodwill
c) Expediting costs for filling potential backorders
d) Production disruptions in case of raw material or work-in-process (WIP)
inventory; for example, in a sequential production process, a lack of sufficient
buffer inventory could cause production to come to a halt.
5. Total cost in 5 years = 12,500 + 5(2,000) = $22,500. Let s = yearly savings as fraction
of sales. The s solves
(80,000)(s)(5) = $22,500
s = 0.05625
10. a. The optimal order quantity
Q* 
2 K

h
2(45)( 280)
 229
.2(2.40)
b. The time between placement of orders
Q * 229
T

 0.8179 yrs  9.81 months

280
c. The average annual cost of holding and setup
K
Q*
h
 2 Kh  2(45)( 280)(.2)( 2.40)  $109.98
Q*
2
d. The reorder level r =  = 280(3/52) = 16.15
r = 16 units
12.
 = 1250(0.18) = 225
c = $18.50
I = 0.25
h = Ic =0.25 (18.50) = 4.625
K = 28
a. The optimal order quantity Q* 
2 K

h
2(28)( 225)
 52
4.625
2
ISYE 3104 Summer 2006
The time between placement of orders T 
Homework 2 Solution
Q*


52
 0.2311 yrs  12.02 weeks.
225
b. The reorder point r =  = 255 (6/52) = 25.96 ≈ 26 units
c. If Q = 225, the average inventory level is Q/2 = 225/2 = 112.5. The annual holding
cost is (112.5)(4.625) = $520.31. At the optimal solution, the annual holding cost is
(52/2)(4.625) = $120.25. The excess holding cost is $400.06 annually.
The annual holding and set-up cost incurred by this policy is $520.31 + 28 =
$548.31 since there is only one set-up annually.
The average annual holding and set-up cost at the optimal policy is
2Kh  2(28)(225)(4.625)  $241.40
Therefore, the annual difference = $306.91.
2(15)( 280)
 132
.2(2.40)
The actual annual holding plus setup cost, when ordering at Q=229, is equal to
K
Q 15 * 280
229
h 
 .2 * 2.40 *
 73.3
Q
2
229
2
Hence, the resulting error is equal to 109.98-73.3=36.5$.
13. If the set up cost is $15, the true optimal Q* 
17. a. Notice that an annual demand of 0.6 million pounds with 250 working days per
year is equivalent to 2400 pounds per day. Hence,
2 K
2(1500)( 2400)( 250)
Q* 

 44,612 lbs
Ic (1   / P)
(0.22  0.12)(3.50 )(1 - 2400/10000 )
b. Up time = Q/P = 44,612 / 10,000 = 4.46 days
Cycle time = Q/ = 44,612 / 2,400 = 18.59 days
Proportion of uptime in a cycle = 4.46/18.59 = 0.24
Proportion of downtime in a cycle = 1-0.24 = 0.76
c. Annual holding and set-up cost
2KIc(1   / P)  2(1500)(600000)(0.34)(3.50)(0.76)  $40347.5
If the compound sells for $3.9 per lb, then the annual profit is
(3.9-3.5)(600,000) – 40347.5 = $199,652.5
22.
 = 20,000
K = 100
I = 0.20
c0= $2.50
c1= $2.40
c2= $2.30
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ISYE 3104 Summer 2006
Homework 2 Solution
Q ( 0) 
2 K

Ic 0
Q (1) 
2 K
2(100)( 20000)

 2887
Ic1
(0.20)(2.40)
Q ( 2) 
2 K

Ic 2
2(100)( 20000)
 2828
(0.20)(2.50)
2(100)( 20000)
 2949
(0.20)(2.30)
Only Q(0) is realizable.
Hence, the minimum achievable cost for source C is at Q = 4000 and it is equal to:
0.2(2.30)( 4,000) (100)( 20,000)
(20,000)(2 .30) 

 $47,420
2
4,000
The minimum achievable cost for source B is at Q = 3000 and it is equal to:
0.2(2.40)(3,000) (100)( 20,000)
(20,000)(2 .40) 

 $49,386.67
2
3,000
Finally, the minimum achievable cost for source A is at Q = Q(0) = 2828 and it is equal to:
(20,000)(2.50)  2 KIc0  50,000
It follows that the optimal order size is Q = 4,000.
b. Holding cost and set up cost =
0.2(2.30)( 4,000) (100)( 20,000)

 $1,420
2
4,000
c. T = Q/ = 4000/20000 = 0.2 years = 2.4 months. Hence  >T. This implies that we
need to place the replenishment order more than one cycle ahead from the cycle in
which it will be consumed. More specifically, setting the reorder point at r =  () =
20,000 [(3-2.4)/12] = 1,000, we shall receiving replenishment orders according to the
pattern indicated below:
IP
Q*
ROP
Order 0
t
Order 1
Order 2
Order 3
4
ISYE 3104 Summer 2006
Homework 2 Solution
24. Since Q(2) is realizable, it must be optimal. This can be argued as follows: Every other
point on the “total annual cost (TAC)” curve corresponding to price c2 will be above to
the point of this curve corresponding to Q(2) (since this is the corresponding EOQ value).
Furthermore, for any given Q, the corresponding points of the TAC curves obtained for
prices c1 and c0 will be above the point of the TAC curve obtained for price c2 (we
showed that in class). Hence, it is not possible to have a better total annual cost than that
obtained with Q(2).
800 875
750
925
900
5