ISyE 3104: Introduction to Supply Chain Modeling: Manufacturing and Warehousing
Instructor: Spyros Reveliotis
Spring 2002
Homework #3
Due Date: Wednesday, 2/20/02
Reading Assignment:
 From your textbook: Sections 2.2, 2.3, 2.4.3 to 2.4.8, 4.1 (except for 4.1.2.4), 6.1.1 and 6.1.2.
 The material on computing the EOQ under finite production rate, posted on the library reserves.
Problem Set:
2.1 List the major types of inventory and the key factors that determine the optimal quantity of each type
Solution:
 Raw materials – the level of this inventory is determined by the acquisition lead time distribution and
predictability of production starts;
 Finished goods – the level for this inventory is determined by the setup/ordering and holding costs and the
average demand rate (c.f., the EOQ formula); also the EOQ quantity might be inflated by the required safety
stock in order to guard against any variability in the demand;
 Work-in-process – this inventory should be set in such a level that the target production rate is met in spite
of any operational variability experienced in the system (e.g., non-deterministic processing times, equipment
and/or labor unavailability, etc);
 Pipeline – the level of this inventory is determined by the transportation time between facilities.
2.3 Give two examples of how information can be used to improve service and reduce costs in a production
system.
Solution:
 More accurate information on demand can allow reduction of safety stock without increasing the risk of
experiencing any shortages. It also can prevent the production of unnecessary products.
 Accurate, advanced information on production starts allows the just-in-time delivery of raw materials.
 Information on worker and machine availability can improve plant scheduling and reduce planning lead
times.
2.13 A popular buzzword in recent years has been “zero inventory.” It is possible to achieve a level of zero
inventory and still meet demand? Why or why not?
Solution:
 In general, zero inventory is impossible. At a minimum, pipeline and in-process inventory (to match
operation times) is essential. To see this, notice that under the assumption of finite processing and transfer
times, Little’s law implies that operating with zero WIP will result in zero throughput/production rate.
Furthermore, additional inventory may be economical or necessary due to setup times and various sources of
inherent variability.
 More specifically, the following treasons can justify the need for carrying raw material inventory:
 Batching due to quantity discount or economy of scale in deliveries
 Randomness caused by quality problems, late deliveries, etc.
 Similarly, the following five reasons can justify the need for carrying finished goods inventory:
 Customer responsiveness – when customer quoted lead time is shorter than production lead time.
 Batch production – when production occurs in pre-specified quantities.
 Forecast error – the actual order profile will not match completely the forecasted demand.
 Production variability – when there are variations in production timing, quality, etc.
 Seasonality – when demand varies with season, we might build inventory during off-season to meet
peak demand.
2.15 In a stable system inventory averages $10,000,000. The company sells $35,000,000 per year. What is the
average time product spends in the system? How many inventory turns are there per year?
Solution:
 Little’s Law: T = I/X = 10,000,000/35,000,000 = 0.286 yrs.
 Inventory Turns = Annual sales/Average Inv. = 3.5
2.22 Due to the complicated process required to set up a machine, a company has decided to produce a particular
product only once a month. Demand occurs at a continuous and constant rate of 1,000 units per month. The
company also keeps 50 units available as safety stock, just in case production is delayed. Sketch the level of
cycle inventory and safety stock over the month. What is the average inventory level?
Solution:
Inventory level
1,050
Average inv. = 50 + 1000/2 = 550
Safety Stock = 50
50
Month

1
2
Inventory level follows a “saw-tooth” curve varying between 1,050 following the completion of a batch
and a low of 50. Average inventory level is 550 units.
4.3 What are the competitive advantages of vertical integration?
Solution:
 The company can control the quality, price, and availability of raw materials
4.4 List the different strategies available for achieving flexibility in manufacturing operations.
Solution:
 Purchasing excess capacity – allow more flexibility to adapt to demand changes.
 Cross-training workers and purchasing flexible operation machines.
 Engineering product designs that (re-)use a standard set of components, and can be produced through a
standardized set of operations.
 Flexible material handling systems and policies, and good monitoring of the production floor status, so that
production activity can be re-routed, if necessary, to less loaded production units.
4.6 A company can purchase a component for $0.35. However, the facility has extra capacity, and if they invest
$5,000 in new tooling, the item can be made for and estimated $0.23 per unit. Find the break-even point for
determining whether the component should be made or bought.
Solution:
 Breakeven point analysis: C = $5,000, C1 = $0.35, C2 = $0.23.
 Breakeven Point: = X = C/(C1-C2) = 5,000/(0.35-0.23) = 41,667 units.
4.7 A common bracket can be purchased in large quantities for $0.67. The company can make the bracket at a
variable cost of $0.41 by buying a machine for $33,000. Demand over the next two years (the anticipated life of
the product and machine) is estimated to be normally distributed with mean 100,000 units and a standard
deviation of 10,000 units. Would you recommend making or buying this bracket?
Solution:
 Make VS Buy Decision: we will produce if: X > C/(C1-C2)
 C = $33,000, C1 = $0.67, C2 = $0.41
 Breakeven Point: = X = C/(C1-C2) = 33,000/(0.67-0.41) = 126,923 units.
 Prob{Demand > Breakeven point} = 1-[(126,923-100,000/10,000)] = 1- (2.69) = 0.36%, where function
() denotes the cumulative distribution function (cdf) of the standard normal distribution N(0,1), and its
values can be read from statistical tables.
 Since the probability that we can economically make the items is very low, we recommend to buy.
4.8 The purchasing manager for a job shop must decide whether to stock a particular material or just order it as
needed for custom jobs. To stock the item, the inventory status would have to be checked once a week at a cost
of $500 annually. The cost to place an order is estimated at $125. Irregular orders placed for specific jobs would
require an additional $85 delivery charge. The company would have to place approximately twenty such orders
per year. Demand for the $8 item is 2,300 units per year. The annual holding cost rate is 40%. Should the
manager decide to stock the item?
Solution:
 Make-to Stock VS Make-to-Order condition: we will stock the item if:
2 ADh  Cs 
(A  ) D
Qns
2 ADh  Cs  2(125)(2,300)(0.4  8)  500  1,856.46
( A   )  D (125  85)(2300)

 4, 200
Qns
(2300 / 20)


Thus, we recommend to stock the item
Using EOQ, the order quantity will be Q = SQRT(2*125*2300/3.2) = 424 units.
4.9 A plant is being built to manufacture two product families – A and B. The facility must produce 100,000
units of each family per year. Three technologies are available. Technology one costs $125,000 per system per
year, and a system can produce 50,000 units of product A, or 25,000 Bs (or any proportional combination)
annually. Technology two costs $65,000 per year and can produce 45,000 of either product per year. A third
technology exists that can only produce product B. The system could produce all 100,000 units for a cost of
$180,000 per year.
a.) Develop a mathematical programming formulation for the problem of selecting technologies.
c. ) In addition to expected cost, what other factors should be considered?
Solution:
 DA =DB = 100,000/yr. C1 = $125,000; C2 = $65,000; C3 = $180,000.
a. Let Yi = the number of units of system i purchased;
xij = number of units of technology i used to make family j.
Minimize Cost  125,000  Y1  65,000  Y2  180,000  Y3
subject to:
50,000  x1A  45,000  x2 A  100,000
25,000  x1B  45,000  x2 B  100,000  x3 B  100,000
x1A  x1B  Y1
x2 A  x2 B  Y2
x3 B  Y3
Yi integer, x ij  0
c. In addition to expected cost, one might consider the flexibility to cope with uncertainty in demand. For
instance, we might prefer flexible machines to machines that can produce only one product. Also, the extent to
which these three technologies follow developing industry trends and might be able to support future, more
stringent specifications should be taken into consideration.
6.1 Describe the difference in assumptions between the EOQ and EMQ models. How does this affect the
development of the cost function in the two models?
Solution:
 While the derivation of the EOQ formula assumes instantaneous replenishment, the derivation of the EMQ
formula in section 6.1.2 in your textbook assumes that the replenishment quantity is produced in-house, over
a finite time interval. Thus, we need to account for the WIP inventory cost incurred between release of
material to the shop and completion of the batch.
6.2 An automotive repair shop stocks many sizes of tires. One particular size and model is purchased for $15 and
sold for $30. The manager estimates the cost to order at $75, including the delivery charge and paperwork. Using
the cost of rent, interest, and utilities, the manager estimates the cost of carrying inventory at approximately 50%
per year based on average inventory value. The shop sells approximately 2,000 of these tires per year. Determine
the optimal reorder quantity and reorder point. Orders are received two weeks after placement.
Solution:
 EOQ model: C = $15, p = $30, A = $75, i = 0.50/yr, D = 2,000/yr,  = 2 weeks
 h = iC = (0.5)(15) = $7.5 /unit/yr
 Q = SQRT(2AD/h) = SQRT(2*75*2,000/7.5) = 200 units
 r = D = (2 weeks/ 52 week per yr)(2,000 tires/yr) = 77 tires
6.3 An office furniture warehouse sells 5,500 chairs of a popular style per year. Chairs are purchased from the
factory at a cost of $45. The comptroller has estimated the inventory carrying cost rate at 0.4 per year. Labor,
mail, and receiving costs per order are estimated at $125. Determine the optimal order quantity for the
warehouse.
Solution:
 EOQ model: D = 5,500/yr, C = $45, i = 0.4/yr, A = $125
Q
2 AD

h
2(125)(5,500)
 276
(0.4)( 45)
6.4 Suppose that in the previous problem, the warehouse and factory were jointly owned and the objective was to
minimize combined cost. For each order, the factory sets up the machines and produces the requested amount. A
setup cost $235 and takes two hours. Each unit requires one hour of processing time. The $45 cost comprises $20
of purchased material and $25 of value added. Throughput time in the plant is approximately five times total
processing time. Determine the optimal order quantity and total system cost per year. Assume the plant operates
2,500 hours per year.
Solution:
 EMQ model: D = 5,500/yr, i = 0.4/yr, A = 235+125 = $360, s = 2 hrs, p = 1 hr, M = $20, v = $25, w = 5,
Hrs/yr = 2,500.
Q
2 AD

i[( M  v)  2 pDw( M  v / 2)]
2(360)(5,500)
 114
1
0.4[( 20  25)  2(
)(5500)(5)( 20  25 / 2)]
2500
Note that the WIP cost is substantial compared to finished goods cycle inventory.
2. From the handout on the EOQ computation under finite production rates, problems 17 and 18.
17. The Wod Chemical Company produces a chemical compound that is used as a lawn fertilizer. The compound
can be produced at a rate of 10,000 pounds per day. Annual demand for the compound is 0.6 million pounds per
year. The fixed cost of setting up for a production run of the chemical is $1,500, and the variable cost of
production is $3.50 per pound. The company uses an interest rate of 22 percent to account for the cost of capital,
and the costs of storage and handling of the chemical amount to 12 percent of the value. Assume that there are
250 working days in a year.
a. What is the optimal size of the production run for this particular compound?
b. What proportion of each production cycle consists of uptime and what proportion consists of downtime?
c. What is the average annual cost of holding and setup attributed to this item? If the compound sells for $3,90
per pound, what is the annual profit the company is realizing from the item?
Solution:
 First, we compute h = (0.22+0.12)(3.5) = (0.34)(3.5) = $1.19/unit/year. The modified holding cost is



h’ = h(1-D/P) = (1.19)(1-600,000/(250*10,000)) = (1.19)(1-6/25) = $0.9044/unit/year. Substituting into the
EOQ formula and using h’ for h, we obtain
Q* = SQRT(2AD/h’) = SQRT(2*1,500*600,000/0.9044) = 44,612.44 = 44,612 units
Time between production runs is T = Q/D = 44,612/600,000 = 0.07435 year. The uptime = T1 = Q/P =
44,612/2,500,000 = 0.01784 year, and the down time each cycle is T2 = T-T1 = 0.07435-0.01784 = 0.05651
year. Thus proportion of uptime is 0.01784/0.07435 = 24% and downtime is 76%
The average annual cost of holding and setup is AD/Q + h’Q/2 = (1,500)(600,000)/(44,612) +
0.9044(44,612)/2 = 20,174 + 19,970 = $40,144.
Profit per unit is 3.90 – 3.50 = $0.40/pound. Annual profit is (0.40)(600,000)-40,144 = $199,856.
18. Determine the batch size that would result in Problem 17 if you assumed that the production rate was infinite.
What is the additional average annual cost that would be incurred using this batch size rather than the one you
found in Problem 17?
Solution:
 Notice that as P tends toward infinity, D/P tends toward zero, and this EPQ formula reduces to the basic
EOQ equation. This is as it should be, since setting P to be infinity is equivalent to ensuring instantaneous
production as assumed in the EOQ model.
 EOQ = SQRT(2AD/h) = SQRT(2*1,500*600,000/1.19) = 38,892.2 = 38,892 units. The average annual cost
of holding and setup is AD/Q + hQ/2 = (1,500)(600,000)/(38,892) + (1.19)(38,892)/2 = $46,282.
 Therefore, the average annual cost will be increased by 46,282-40,144 = $6,138.
Extra credit
1. (15%) Using your favorite solver, solve the IP formulation developed for problem 4.9, and also, do the part
(b) (essentially solve the Linear Program obtained by relaxing/removing the integrality requirement for the y
variables, and convert the resulting solution to a "meaningful" one by rounding up the non-integral y values
to some close integer in a way that feasibility is maintained.) Compare the results obtained through the two
solution approaches.
Solution:
In put file for LINGO:
MODEL:
1]SETS:
2]tech/1..3/:C,y;
3]prod/1..2/:D;
4]links(tech,prod):v,a,x;
5]ENDSETS
6]
7]min = @sum(tech:C*y)+@sum(links:v*x);
8]
9]@for(prod(j): @sum(tech(i): a(i,j)*x(i,j)) > D(j););
10]
11]@for(tech(i): @sum(prod(j): x(i,j)) < y(i););
12]
13]@for(tech: @GIN(y););
14]
15]
16]DATA:
17]C = 125000, 65000, 180000;
18]D = 100000, 100000;
19]v = 0, 0,
20] 0, 0,
21] 0, 0;
22]a = 50000, 25000,
23] 45000, 45000,
24] 0, 100000;
25]ENDDATA
END
The output file will be:
LP OPTIMUM FOUND AT STEP
6
OBJECTIVE VALUE = 288888.900
FIX ALL VARS.( 2) WITH RC > 35555.6
SET Y( 2)
TO >=
5 AT 1
BND= -325000.00 TWIN= -.10000000E+31
NEW INTEGER SOLUTION AT STEP
12
OBJECTIVE: 325000. BOUND ON OPTIMUM: 324444.
DELETE Y( 2)
AT LEVEL 1
RELEASE FIXED VARIABLES
FIX ALL VARS.( 1) WITH RC > 17222.2
SET Y( 2)
TO >=
3 AT 1
BND= -375000.00 TWIN= -.10000000E+31
DELETE Y( 2)
AT LEVEL 1
ENUMERATION COMPLETE. BRANCHES=
2 PIVOTS= 22
LAST INTEGER SOLUTION IS THE BEST FOUND
RE-INSTALLING BEST SOLUTION...
VARIABLE
VALUE
C( 1)
125000.0
C( 2)
65000.00
C( 3)
180000.0
Y( 1)
.0000000
Y( 2)
5.000000
Y( 3)
.0000000
D( 1)
100000.0
D( 2)
100000.0
V( 1, 1)
.0000000
V( 1, 2)
.0000000
V( 2, 1)
.0000000
V( 2, 2)
.0000000
V( 3, 1)
.0000000
V( 3, 2)
.0000000
A( 1, 1)
50000.00
A( 1, 2)
25000.00
A( 2, 1)
45000.00
A( 2, 2)
45000.00
A( 3, 1)
.0000000
A( 3, 2)
100000.0
X( 1, 1)
.0000000
X( 1, 2)
.0000000
X( 2, 1)
2.222222
X( 2, 2)
2.222222
X( 3, 1)
.0000000
X( 3, 2)
.0000000
ROW
1
2
3
4
REDUCED COST
.0000000
.0000000
.0000000
125000.0
65000.00
180000.0
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
SLACK OR SURPLUS
DUAL PRICE
325000.0
1.000000
.0000000
.0000000
.0000000
.0000000
.0000000
.0000000
5
6
.5555556
.0000000
.0000000
.0000000
This yields the actual purchase cost of 65,000*5 = 325,000
b. Find the least expensive alternative for each product family ignoring integer restrictions.
Costs are approximated by
j=A
j=B
Ci  Dj
aij
. These values are
i=1
250,000
500,000
i=2
260,000
144,444.
The cheapest method for family A is technology 1. Set x1 A 
The cheapest method for family B is technology 2. Set
x2 B 
i=3
180,000
DA
 2 . and Y1 = 2.
a1a
100,000
 2.22 and Y2 = 3.
45,000
This yields an actual purchase cost of 125,000(2) + 65,000(3) = $445,000.
Result: Using software package obtains better purchase cost.
2.
(10%) Read Section 6.1.4 from your textbook and use the relevant theory to solve problem 6.21.
6.21 A plant purchases 150,000 units of an electrical component per year. The fixed ordering cost is $60. The
unit cost is $0.25 if ordered in quantities less than 1000, $0.24 if order size is between 1,000 and 10,000, and
$0.23 for orders of more than 10,000. The holding cost rate is 0.4 per year. Find the optimal order quantity and
annual cost.
 D = 150,000, A = $60, i = 0.4/yr.

$0.25, if Q  1000


C  $0.24, if 1000  Q  10, 000
 $0.23, if Q  10, 000


Qi* 

Since Q* > 10,000, this option is feasible.

Cost / Year 
2(60)(150, 000)
2 AD
*
 13,988.
For C = $0.23, Q 
0.4(0.23)
iCi
(60)(150, 000)
(13,988)
 (0.23)(150, 000)  (0.4)(0.23) 
 $35, 787.
13,988
2