6.1 Objective
This chapter is the first on inventory. We use it as an introduction to “supply chain management” focusing on “inventory” and its role in business. In a class of 100 minutes without break, we talk about the general cost and benefits of inventory, the different types of inventory and why they exist.
The focus of the remainder of the chapter/class is then on economies of scale as inventory’s first reason of existence. This should be an easy class to teach.
6.2 Additional Suggested Readings
We assign a short case as supplemental reading for the economies of scale. The case is used to show simple EOQ calculations and the benefits of centralization. It will also be useful for Chapter 7 to talk about the benefits of postponement.
 “Hewlett-Packard: DeskJet Printer Supply Chain (A)”.
Stanford Case 1993. Authors: Laura Kopczak and Hau L. Lee.
Suggested assignment questions:
1.
What factors should be taken into consideration when deciding the batch size of a particular type of printer shipped to Europe (or any other destination)? What do you think of the fact that “inventory growth had tracked sales growth closely?”
2.
Assume that each printer costs HP $100 and the fixed cost incurred in each production run (lost production due to setup) is $1,000. Consider the Option A printers for world wide demand. How many of these printers should Vancouver produce in each batch?
Problem 6.1
The data in the question is: flow rate R = 50,000 parts/yr, fixed ordering cost S = $800, purchasing cost C
= $4/part, and cost of capital r = 20%/yr. Thus, the annual unit holding cost is H = rC = $0.8/yr. The economic order quantity tells us to purchase each time a) Q =
2 RS
H

2

50 , 000

800
= 10,000 units.
0 .
8 b) Order R/Q = 5 times per year.
Problem 6.2
BIM Computers: Assume 8 working hours per day.
49

Chapter 6

We know Q = 4 wks supply = 1,600 units; R = 400 units/wk = 20,000 units/yr; purchase cost per unit C = $1250*80% = $1,000. Thus, holding cost H = rC = 20%/year × $1,000 =
$200/yr. Switch over or setup cost S = $2,000 + (1/2hr×$1,500/day×1day/8hr)= $2,093.75.
Thus, # of setups per year = R/Q = 20,000 units/yr / 1600 units/setup = 12.5 setups/yr. Thus,

Annual setup cost = ( R/Q ) * S = 12.5 setups/yr × $2,093.75/setup = $26,172/yr.

Annual Purchasing Cost = R*C = 20,000 units/yr × $1,000/unit = $ 20 M/yr.

Annual Holding Cost = ( Q/2 ) ×H = 800 × $200/yr = $160,000/yr.

Thus, total annual production and inventory cost = $20,186,172.


EOQ =
2 RS

H
= 647 units.
200
 number of setups = R/Q = 20,000 /647 = 30.91. Thus, annual setup cost =
30.91setups/yr × $2,093.75/setup = $64,718/yr.
 annual holding cost = ( Q /2) × H = 323.5 × $200/yr = $64,700/yr (notice that at optimal EOQ annual holding cost equal setup costs)
 annual purchasing cost remains $20M/yr

The resulting annual savings equals $20,186,172 - $20,129,418 = $56,754.
Problem 6.3
Victor's data: flow unit = one dress, flow rate R = 30 units/wk, purchase cost C = $150/unit, order lead time L = 2 weeks, fixed order cost S = $225, cost of capital r = 20%/yr. Victor currently orders ten weeks supply at a time, hence Q = 10wks × 30 units/wk = 300 units. a. Costs for Victor's current inventory management:

Annual variable ordering (purchasing) cost = RC = $150/unit × 30 units/wk × 52 wks/yr
= $234,000/yr.

Annual fixed ordering (setups) cost = (# of orders/yr) S = ( R/Q ) S = (30×52/yr/300) $225
= $1,170/yr.

Annual holding cost = H (Q/2) = ( rC ) ( Q/2 ) = $30/yr × 150 = $4,500/yr.

Total annual costs = $239,670. b. To minimize costs, Victor should order in batches of
Q* = EOQ =
2 RS
H

2

30

52

225
= 153 units.
30

Thus, he should place an order for 153 units two weeks before he expects to run out. That is, whenever current inventory drops to R×L = 30 units/wk × 2 wks = 60 units, which is the reorder point.

His annual cost will be
RC + 2 RSH

2

30

52

225

30 + $234,000 = $4,589 + $234,000 = $238,589.

Chapter 6 c. Inventory turns = R/I, where average inventory I = Q/2 with cycle stock only.

Current policy: turns = R/(Q/2)=2 × R/Q = 2 × 30units/wk / 300units = 2/10week = 52 × 2/10 per year = 10.4 times per year.

Proposed policy: Q is roughly halved, so turns roughly double to 20.4 times per year.
Problem 6.4
The retailer: Current fixed costs, S
1
= $1000. Current optimal lot size Q
1
= 400. New, desired lot size Q
2
= 50. We must find the fixed cost S
2
at which Q
2
is optimal. Since Q
1
is optimal for S
1
, we have
Q
1
= 400 =
2 RS
1
H

2

R

1000
. So, R/H = 160000/2000 = 80.
H
Now,
Q
2
= 50 =
2 RS
2
H
, or S
2
= 50 2 /(2×80) = 15.625. So the retailer should try to reduce her fixed costs to $15.625.
Problem 6.5
Major Airlines: This question illustrates the basic tradeoff between fixed and variable costs in a service industry; thus the concepts of EOQ discussed in class in the context of inventory management are much more generic.
The process view here is illuminating and it goes as follows: flow unit = one flight attendant (FA). The process transforms an input (= "un-trained" FA) into an output (= "quitted" FA). The sequence of activities is: undergo training for 6 weeks, go on vacation for one week, wait in a buffer of "trained, but not assigned FA" until being assigned, serve as a FA on flights, and finally quit the job.
Untrained
FA
Training Vacation Pool of trained FAs
? wks
Serve on flights
R
Quitted
FA
T = 6 wks 1 wks 2 years
The question asks for the tradeoff between training costs (higher class size is preferred) versus 'holding costs' in the buffer (smaller class size -> fewer attendants waiting in buffer is preferred).

(a)

Flow rate R = 1000 every two years = 500 attendants per year = 10 per week.

Fixed costs of training involves hiring ten instructors and support personnel for 6 weeks.
Thus, fixed costs of training S = 10 × ($220+$80) × 6 weeks = $18,000 per training session.

Annual holding cost is the cost incurred to hold one flow unit (FA) in the buffer for one year:
H = $500 per month × 12 = $6,000 / person / year.

Chapter 6

Thus, Economic Class Size (EOQ) = 54.77 or 55 per class. Thus, we should run R/Q = 500 /
55 = 9.09 classes per year

Per person variable cost of training is the stipend paid for 6 weeks of training + stipend for a week of vacation = $500/mo. perperson × 7 wk × 12 mo/yr / 50 wk/yr = $840 per person.
Notice that the annual variable cost is constant $840/person × 500 person/yr = $420,000/yr regardless of the class size .

Total Annual Cost = Fixed Costs of Training + Variable Costs of Training + Holding Costs =
($18,000 × 9.09) + ($840 × 500) + (55/2)($6000) = $748,636.36 per year.

Time Between starting consecutive classes (say, T) = Q/R = 5.5 weeks. Thus, we will have two classes overlap for a 1/2 week (and thus we need two sets of trainers and training class rooms). The inventory-time diagram looks as follows (assuming for simplicity that we start the training process at time 0):
I (in training)
110
55
Class 1 Class 2 Class 3
55
0
5.5
6
I (on vacation)
Class 1 Class 2 Class 3 t (weeks)
0 6 7
55
I (in buffer)
Class 1 Class 2 Class 3 t (weeks)
0 6 7 t (weeks)

(b): This part of the question illustrates the following: Often, in reality, people wish to adopt policies that are simple (e.g., starting training every 6 weeks is simpler than trying to track the exact days to start training when subsequent trainings start every 5.5 weeks. But what is the implication of deviation from the optimal? In this case, quite small. This is because the optimal cost structure near the (optimal) EOQ is quite flat. Thus any solution close to optimality will suffice.

Chapter 6

If time between classes ( T = Q/R ) has to be 6 weeks, then Q = TR = 6 wks × 10 attendants/wk = 60 attendants.

Total Cost of this policy = ($18,000)(500/60) + ($840)(500) + (60/2)($6000) = $750,000 per year.
Problem 6.6
Fixed cost of filling an ATM m / c , S = $100.
To estimate demand, observe that the average size of each transaction = $80. With 150 transactions per week, annual demand R is estimated to be = 150×52×80 = 624,000.
With cost of money of 10%, unit holding cost, H = $0.10 / year
Then, the economic quantity to place in the ATM machine is given by the EOQ formula:
Q =
2 R S
=
2 ґ
624000 ґ
100
=
H 0.1
35, 327
The number of times the ATM needs to be filled = R / Q = 624000/35327 = 17.66 per year.
Problem 6.7
The annual demand, R = 150,000 lbs/yr. The purchase price per lb is $1.50. However the shipping cost exhibits a quantity discount model. The holding cost per year is then 15% of the sum of the purchase and shipping cost. The administrative costs of placing an order = $50/order.
(a) In addition, rental cost of the forklift truck adds to the fixed cost giving a total fixed cost, S =
50+350 = $400/order. We can use a spreadsheet model as shown in Table TN 6.1. The optimal order quantity = 22,000 lbs with an annual cost of $249,916.77.
(b) If GC buys a forklift and builds a new ramp, then the per-transaction fixed cost will simply be the administrative cost of $50 per order. The economic order quantity and annual operating costs of this option is shown in Table TN 6.2. The economic order quantity is 15000 lbs. with an annual operating cost = $246,833.75. The annual savings = 249,916.77 - 246,833.75 = $3,083.02. The net present value of cost savings (over 5 years) with cost of capital of 15% = $10,334.76.
Assuming a useful life of 5 years for the forklift and ramp, an investment of less than $10,334 generates a positive NPV.
Problem 6.8
Changeover time = 4hrs resulting in a fixed cost, S = 4 × 250 = $1,000.
Annual demand, R = 1000/mo × 12 = 12,000 units / yr.
Unit cost, C = 100
Holding cost = $25 / unit / yr. a) The optimal production batch size is

Chapter 6
Q

2 RS

H

980
25 b) To reduce batch size by a factor of 4, the setup cost needs to be reduced by a factor of (4) 2 = 16.
That is, S should reduce to 1000/16 = $62.5. This reduction can be achieved by reducing the changeover time or the cost per unit time during changeovers.
Problem 6.9 a) From the EOQ formula, observe that the order quantity is proportional to the square root of annual demand ( R ). Since cycle inventory is half of the order quantity, it too is proportional to R .
Since HP motors has a higher R , the cycle inventory for HP motors is also higher. b) Average time spent by a motor T = I cycle
/ R . Since I cycle
is proportional to to 1/
R , T is proportional
R . Therefore time spent by a HP motor is less than the time spent by an LP motor.
Problem 6.10
Each retail outlet faces an annual demand, R = 4000/wk × 50 = 200,000 per year. The unit cost of the item, C = $200 / unit. The fixed order cost, S = $900. The unit holding cost per year, H = 20 % × 200 =
$40 / unit / year. a) The optimal order quantity for each outlet
Q

2 RS
H

40

3000 with a cycle inventory of 1500 units. The total cycle inventory across all four outlets equals 6000 units. b) With centralization of purchasing the fixed order cost, S = $1800. The centralized order quantity is then,
Q

2 RS

H and a cycle inventory of 4242.5 units.
40

8485
Usually, tests will combine concepts from Chapters 6 and 7. Therefore, refer to questions for Chapter
7 that include concepts from Chapter 6.We list here a few questions that exclusively rely on concepts in Chapter 6.
[1] Complete Computer (CC) is a retailer of computer equipment in Minneapolis with four retail outlets. Currently each outlet manages its ordering independently. Demand at each retail outlet averages 4,000 units per week. Each unit costs $200 and CC has a holding cost of 20%. The fixed cost of each order (administrative + transportation) is $900. Assume 50 weeks in a year.

Chapter 6
(a) Given that each outlet orders independently and gets its own delivery, the optimal order size at each outlet is (5 points) [SHOW WORK]
(i) 424
(ii) 3,000; EOQ = sqrt (2RS/H) = sqrt (2*4000*50*900/40) = 3000
(iii) 6,000
(iv) 42,426
(v) None of the above
(b) CC is thinking of centralizing purchasing (for all four outlets). In this setting, CC will place a single order (for all outlets) with the supplier. The supplier will deliver the order on a common truck to a transit point. Since individual requirements are identical across outlets, the total order is split equally and shipped to the retailers from this transit point. This entire operation has increased the fixed cost of placing an order to $1,800. If CC manages ordering optimally in the new setting, average inventory in the CC system (across all four outlets) can be expected to (5 points)
(i) Increase
(ii) Decrease
(iii) Remain unchanged
Give arguments for your answer:
In the decentralized system, cycle stock per outlet = 3000/2 = 1500. Total cycle stock = 1500*4 =
6000. In the centralized system, Demand increases by a factor of 4 over a single outlet. Fixed costs increase by a factor of 2. So EOQ will increase by a factor of sqrt (4×2) = sqrt (8) = 2.83. So cycle stock = 2.83 × (3000/2) = 4245 < 6000.
Students may work out the EOQ all over again by the original formula and do the comparison. Either approach is fine.

16000
17000
18000
19000
20000
20500
21000
21500
22000
8000
9000
10000
11000
12000
13000
14000
15000
22500
23000
23500
24000
24500
25000
25500
Chapter 6
Table TN 6.1: Problem 6.7 (with use of rental forklift)
S
R
( h + r ) purchase price
400 per order
150000 lbs / yr
15.00% % per year
1.5 per lb.
Order size ( Q )
Shipping cost per pound
Total variable cost = purchase cost + shipping
( C ) holding cost pound per year
( H )
Number of orders
( R/Q )
Annual order cost
( S × R/Q )
Average
Cycle
Inventory
( Q/2 )
Annual holding cost
( H × Q/2 )
Annual procurement cost
( C × R )
Total
Annual costs TC
0.13
0.13
0.13
0.13
0.13
0.13
0.13
0.13
0.13
0.17
0.17
0.15
0.15
0.15
0.15
0.15
0.13
0.13
0.13
0.13
0.13
0.13
0.13
0.13
1.67
0.2505
1.67
0.2505
1.65
0.2475
1.65
0.2475
1.65
0.2475
1.65
0.2475
1.65
0.2475
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
1.63
0.2445
18.75
7500.00
16.67
6666.67
15.00
6000.00
13.64
5454.55
12.50
5000.00
11.54
4615.38
10.71
4285.71
10.00
4000.00
9.38
3750.00
8.82
3529.41
8.33
3333.33
7.89
3157.89
7.50
3000.00
7.32
2926.83
7.14
2857.14
6.98
2790.70
6.82
2727.27
6.67
2666.67
6.52
2608.70
6.38
2553.19
6.25
2500.00
6.12
2448.98
6.00
2400.00
5.88
2352.94
4000
4500
5000
5500
6000
6500
7000
7500
1002
1127.25
1237.5
1361.25
1485
1608.75
1732.5
1833.75
8000
8500
9000
9500
1956
2078.25
2200.5
2322.75
10000 2445
10250 2506.125
10500 2567.25
10750 2628.375
11000 2689.5
11250 2750.625
11500 2811.75
11750 2872.875
12000 2934
12250 2995.125
12500 3056.25
12750 3117.375
250500 259,002.00
250500 258,293.92
247500 254,737.50
247500 254,315.80
247500 253,985.00
247500 253,724.13
247500 253,518.21
244500 250,333.75
244500 250,206.00
244500 250,107.66
244500 250,033.83
244500 249,980.64
244500 249,945.00
244500 249,932.95
244500 249,924.39
244500 249,919.07
244500 249,916.77
244500 249,917.29
244500 249,920.45
244500 249,926.07
244500 249,934.00
244500 249,944.10
244500 249,956.25
244500 249,970.32

Chapter 6
Table TN 6.2: Problem 6.7 (without use of rental forklift)
11500
12000
12500
13000
13500
14000
14500
15000
7500
8000
8500
9000
9500
10000
10500
11000
15500
16000
16500
17000
17500
S
R
( h + r ) purchase price
50 per order
150000 lbs / yr
15.00% % per year
1.5 per lb.
Order size ( Q )
Shipping cost per pound
Total variable cost = purchase cost + shipping
( C )
6000
6500
7000
0.17
0.17
0.17 holding cost pound per year
( H )
1.67 0.2505
1.67 0.2505
1.67 0.2505
Number Annual Average of orders order cost
( R / Q ) ( S x R / Q )
Cycle
Inventory
( Q /2)
25.00 1250.00
23.08 1153.85
21.43 1071.43
Annual holding cost
( H x Q /2)
3000 751.5
3250 814.125
3500 876.75
Annual Total procurement Annual costs cost ( C x R ) TC
250500 252,501.50
250500 252,467.97
250500 252,448.18
0.15
0.15
0.15
0.15
0.15
0.15
0.15
0.13
0.17
0.17
0.17
0.17
0.17
0.15
0.15
0.15
0.13
0.13
0.13
0.13
0.13
1.67 0.2505
1.67 0.2505
1.67 0.2505
1.67 0.2505
1.67 0.2505
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.65 0.2475
1.63 0.2445
1.63 0.2445
1.63 0.2445
1.63 0.2445
1.63 0.2445
1.63 0.2445
20.00 1000.00
18.75 937.50
17.65 882.35
16.67 833.33
15.79 789.47
15.00 750.00
14.29 714.29
13.64 681.82
13.04 652.17
12.50 625.00
12.00 600.00
11.54 576.92
11.11 555.56
10.71 535.71
10.34 517.24
10.00 500.00
9.68 483.87
9.38 468.75
9.09 454.55
8.82 441.18
8.57 428.57
3750 939.375
4000 1002
4250 1064.625
4500 1127.25
4750 1189.875
5000 1237.5
5250 1299.375
5500 1361.25
5750 1423.125
6000 1485
6250 1546.875
6500 1608.75
6750 1670.625
7000 1732.5
7250 1794.375
7500 1833.75
7750 1894.875
8000 1956
8250 2017.125
8500 2078.25
8750 2139.375
250500 252,439.38
250500 252,439.50
250500 252,446.98
250500 252,460.58
250500 252,479.35
247500 249,487.50
247500 249,513.66
247500 249,543.07
247500 249,575.30
247500 249,610.00
247500 249,646.88
247500 249,685.67
247500 249,726.18
247500 249,768.21
247500 249,811.62
244500 246,833.75
244500 246,878.75
244500 246,924.75
244500 246,971.67
244500 247,019.43
244500 247,067.95