CHAPTER 12:
INVENTORY MANAGEMENT
Teaching Notes
This is a fairly long quantitative chapter, and hence not all the topics can be covered in even two sessions.
The models are for independent demand items such as finished goods, spare parts and supplies (the
dependent demand items use MRP or JIT, and are covered in later chapters). Typical operations that use
inventory control models are warehouses and storerooms. The basic models are the EOQ/ROP and Fixed
Interval models. If time allows, the variations of EOQ, economic production quality and quantity
discount may be covered. Another fairly important model is the single period (newsboy) model.
Answers to Discussion and Review Questions
1.
Inventories are held:
1. while items are being transported
2. to protect against stock-outs
3. to take advantage of economic lot sizes and quantity discount
4. to smooth seasonal demand or production
5. to decouple operations
6. to hedge against price increases
2.
A company can reduce its needs for inventories by addressing the reasons (purpose) of
inventories, as given in the previous question:
a. reducing supplier delivery lead times (e.g., locate closer) and increasing delivery reliability
b. cross-docking or eliminating the DC
c. using faster and more reliable transportation mode
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
3.
Establishing close long-term relationship with suppliers and customers
Ensuring high quality material
improving forecasting of demand and reducing forecast error
reducing ordering time and cost by using tools such as EDI
performing faster set ups
smoothing out demand by demand management so need for seasonal production is reduced
using preventive maintenance on equipment and machines
training workers and accepting idle time
smoothing out parts of production so need for decoupling inventory is reduced
using Futures Markets to hedge the requirements, so that there is no need to hold inventory in
anticipation of price hikes
It is inappropriate to compare the inventory turnover ratios of companies in different industries
because:
 structure of industries are different. In some industries, suppliers hold inventories for the
customers.
 nature of products are different, resulting in different mode and time of transportation and
storage. Grocery stores have high turnover ratio bec. groceries, especially fresh produce, are
perishable.
 lead times are different depending on the production process, length of production run, etc.
 demand variabilities are different resulting in different levels of safety stock.
 demand seasonal patterns are different.
Instructor’s Manual, Chapter 12
12-1
4.
Effective inventory management requires
 safe storage and use of inventory
 a tracking system and replenishment models
 forecasts of demand and lead time (amounts and variabilities)
 inventory costs estimation
 a priority system (i.e., A-B-C classification).
5.
Purchase lead time is the time interval between ordering and receiving the order. It is made of
supplier order processing time, any production or purchase of the item by the supplier if not on
hand, picking, packing, and shipment.
6.
Bar codes allow quick and accurate identification of items and storage locations.
7.
Holding (carrying) cost is the cost to keep an item in inventory, consisting of cost of money tied
up in inventory and storage cost. Ordering cost is the cost of the actual placement of an order
(not including the purchase cost), including the time of buyer, receiver, and accounts payable.
Shortage cost is the cost of demand exceeding supply of inventory on hand; includes unrealized
profit per unit, loss of goodwill, etc.
8.
The A-B-C classification is sorting the inventory items according to annual dollar value (unit cost
× annual volume), and calling the top 15-20% of stock keeping units (SKUs) accounting for
about 70-80% of annual dollar value as A items, and bottom 50-60% of SKUs accounting for
about 5-10% of annual dollar value as C items. The rest are called B items as they have controls
that lie between the two extremes.
9.
The total inventory-control cost-curve is relatively flat in the vicinity of the EOQ, so there is a
range of values of order quantity for which the total cost is close to its minimum. The fact that the
EOQ calculation involves taking a square root lessens the impact of data estimation errors. Also,
errors may cancel each other out.
10.
As holding inventory becomes more expensive, the company will order in smaller quantities
(note H in the formula for EOQ). This will result in more frequent orders.
11.
As machine setup time decreases, set up cost decreases, resulting in lower run size (note S in the
EPQ), which results in lower average amount of WIP inventory. Overall inventory costs would
decrease. Also lower WIP results in less confusion and better quality.
12.
The EPQ model is similar to the EOQ model, except that instead of orders being received in a
single delivery, units are received incrementally. This is because the production rate is finite.
This results in some of the units being consumed before the maximum inventory level is reached.
𝑝
To compensate for this, EPQ is larger than EOQ by the proportion √𝑝−𝑑 where p = production
rate and d = demand rate.
13.
12-2
Planned shortage would lead to lower holding cost. For example, many large retailers such as
Sears Canada, Staples, and Home Depot do not stock big-ticket items. Or auto-part suppliers
such as Canadian Tire would not stock low-demand items. As demand occurs, orders are placed.
This is a form of planned shortage.
Operations Management, 5/C/E
14.
Safety stock is inventory held in excess of average/expected demand. Its purpose is to reduce the
risk of stock-out due to variability in lead time and/or demand.
15.
Safety stock should be large when large variations in lead time and/or usage/sales are present.
Conversely, small variations in usage/sales and lead time require small safety stock. Safety stock
should be zero when usage and lead time are constant, or when the service level is 50 percent
(and hence, z = 0).
16.
Lead time service level is the probability that demand will not exceed supply during a lead time.
Lead time Service level determines the safety factor z, and safety stock is z × standard deviation
of demand during a lead time (for the EOQ/ROP model). Therefore, increasing the LT service
level requires increasing the amount of safety stock.
17.
Annual service level (SLannual) is the percentage of annual demand filled (directly from
inventories). It is related to z through the following two steps:
1. Calculate E ( z ) 
Q(1  SLannual )
 dLT
2. Take the E(z) to Table 12-3 to determine the associated z value.
18.
EOQ/ROP model orders a fixed quantity when inventory drops to or below the ROP. Fixed
Interval model orders only after a fixed interval of time. EOQ/ROP is used for individual SKUs,
whereas Fixed Interval model is used for a group of SKUs bought from the same supplier.
19.
The Fixed Interval model orders each SKU at every order time, even if the order quantity is very
small. However, it might cost less to order some items every two, three, or more order intervals
than ordering them at every interval, because their holding cost may be less than the line-item
ordering cost. The Coordinated Periodic Review model accommodates this by determining an
order interval OI for reviewing every SKU and a multiple mi of OI for ordering SKUi.
20.
The single period model is a model for ordering of perishables and other items with limited useful
lives. It is appropriate for any SKU that is not carried over from one period to the next (i.e.,
unused or unsold items are salvaged).
21.
Yes. When excess cost Ce is higher than the shortage cost Cs, SL = Cs / (Cs + Ce) will be less than
0.5 which means that Q = a value of demand less than its median.
22.
Multi-echelon control uses echelon inventory (all inventory at a Distribution Center (DC) and
retailers) for a DC, echelon demand (all POS customers’ demands), and echelon lead time (sum
of LT of DC from its supplier and LT of retailers from DC) to make replenishment decisions.
Distribution Requirement Planning plans distribution of shipments from a DC to retailers, given
forecasts of customer demands at retailers and lead times from DC to retailers.
Inventory optimization determines the location and size of inventory and promised lead times in a
supply chain in order to meet customer-promised lead time at minimum total inventory holding
cost.
Instructor’s Manual, Chapter 12
12-3
Answers to Taking Stock Questions
1.
a. If we buy additional amounts of a particular good to take advantage of quantity discounts,
then we will save money on a per unit purchasing cost of the item. We will also save on
ordering cost because we will order this item less frequently. However, as a result of ordering
larger quantity, we will have to hold larger inventory, resulting in an increase in inventory
holding cost, including the need for a larger storage space.
b. Conducting inventory count once a quarter instead of once a year will result in an increase in
labour cost. However, the more frequent counting would also lead to more timely detection of
errors and improved inventory accuracy which in turn would lead to better customer service
and lower cost.
2.
Buyer, planner/expeditor, and inventory clerk would be directly involved in day-to-day inventory
decisions. The materials manager, production planning and control manager, purchasing manager,
and inventory manager would be involved in critical decisions, large contracts, setting goals, etc.
Top management will be involved in deciding total inventory investment and customer service
levels.
3.
Technology has aided inventory management by providing:
 bar codes and scanners and RFID for identification, electronic communication for order
picking, replenishment, and tracking trucks/shipments
 inventory software for storing data, tracking inventory position, forecasting, and making
inventory replenishment decisions.
4.
Use RFID, cameras, secure storerooms, employ a security guard to check employees leaving.
Answers to Critical Thinking Exercises
1.
Expansion of menu offerings has created problems for inventory management in a fast food
restaurant because there are more ingredients and therefore more inventory items to keep, track,
and order. The small storage spaces around the production lines in the compact kitchen (in order
to be more efficient) may not have enough room for more items.
2.
I’d look at the demand per day, the item’s profit margin, and degree of substitutability (would
customers buy a substitute which is available?).
Answers to Experiential Learning Exercises
1.
(a)
(b)
(c)
2.
(a)
gas:
EOQ/ROP model
(b)
groceries:
Fixed Interval model
(c)
cash:
EOQ/ROP model.
Running out of gas has the highest shortage cost.
3.
Generally, items are available on the shelves, display fridges, etc. Once in a while, the store runs
out of a SKU. Some shelf locations don’t have price label. Sometimes, items get knocked down
on the floor and are left there. Sometimes items are misplaced. Sometimes the shelf label doesn’t
match the store computer price. The store can monitor these more closely and fix them faster.
12-4
groceries:
gas:
pocket money:
shop once a week
fill up as it is running out
go to bank as it is running out
Operations Management, 5/C/E
Answers to Internet Exercises
1.
`
Retail Pro’s Inventory Control & Management enables the following:
 Define an item, including several data fields, including special fields for serial no.,
whether it is a kit (package), a service or fee (non-stock) item, etc.
 Keep a memo for each transaction involving quantity, cost, or price, e.g., quantity
adjustment
 Keep a history of transactions for each item, including the ability to drill down by store
 Update inventory balance
 Show commitments, open orders, layaways, transfers
 Issue point-of-sale customer receipt, check stock level, including across stores
 Show stats such as days of supply, inventory turns, etc.
2.
Answers will vary.
3.
a.
b.
c.
4.
Conveyables can be moved using conveyers. E.g., a case of oil filters.
Non-conveyable cannot be moved by conveyers. E.G., a gas grill.
Paper and RF order picking systems
Popup wheel, sliding shoe
The automated library system of UBC stores about half (2 million) of its collection of books. The
floor space is 9,600 ft2 with 4 aisles, each 61 pairs of bays long and 39 tiers high, for a total of
19,032 book bins. Each bin is 2 feet wide and 4 feet deep, with 4 heights: 6, 10, 12, and 16
inches. Each bin has dividers that make compartments. A max. of 500 pounds of books can be
stored in each bin. Each aisle has one storage and retrieval machine that moves the bins between
the front-of-aisle workstations and bays. Placing in and retrieving a bin out of a bay uses handles
at each side of the bin and lasers for accurate positioning. The Dematic management software
works with the library circulation software by tracking books in the bins.
Instructor’s Manual, Chapter 12
12-5
5.
SmartOps is an inventory optimization software that was bought by SAP. It determines the
location and size of safety stocks (and reorder point) in a supply chain in order to minimize total
inventory holding cost subject to meeting end-customer desired lead time and service level.
SmartOps was purchased in 2007 to be used as a tool internally (Celestica has over 20 facilities
worldwide) and for the supply chain management service that Celestica started to offer to its
customers. SmartOps starts with end-customer demand forecasting, trying to increase forecast
accuracy. Celestica has applied SmartOps to over 20 supply chains so far. Results show more than
10% inventory reduction (including excesses and obsoletes), visibility (and reduction in info.
latency—e.g., two SAP’s in North America and Asia not communicating well) in the supply chain,
lead time reduction (leading to more flexibility), and better customer service.
Solutions to Problems
1.
a.
Step 1:
Compute the Annual Dollar Value (= Unit Cost * Annual Volume) for each item:
Item
Unit
Cost
1
2
3
4
5
6
7
$100
80
15
50
11
60
10
Annual
Volume
(00)
25
30
60
10
70
85
60
Annual
Dollar
Value
$2,500
$2,400
$900
$500
$770
$5,100
$600
Step 2:
Sort the data in descending order of Annual Dollar Value (ADV). Sum the ADV and compute
the % ADV for each item. In this case, there should be only 1 A (1/7 = 14%) and 2 Bs (2/7 =
28%).
12-6
Item
Unit
Cost
6
1
2
3
5
7
4
60
$100
80
15
11
10
50
Annual
Volume
(00)
85
25
30
60
70
60
10
Annual
Dollar
%ADV
Value
$5,100 39.94%
$2,500 19.58%
$2,400 18.79%
$900
7.05%
$770
6.03%
$600
4.70%
$500
3.92%
$12,770
A
B
B
C
C
C
C
Operations Management, 5/C/E
b.
Q0 
2 DS
2(4,500)36

 180 units
H
10
c.
2𝐷𝑆
𝑝
2(18,000)(100)
120
√
𝑄𝑝 = √
=√
= 300(2) = 600 𝑢𝑛𝑖𝑡𝑠
√
𝐻 𝑝−𝑑
40
120 − 90
2.
PROBLEM NO. 12 - 2
Item
4021
9402
4066
6500
9280
4050
6850
3010
4400
2307
Monthly Usage Unit Cost
50
$1,400
300
$12
40
$700
150
$20
10
$1,020
80
$140
2,000
$15
400
$20
7,000
$25
1958
$14
Sort by Monthly $ v alue:
Item
4400
4021
6850
4066
2307
4050
9280
3010
9402
6500
Monthly Usage Unit Cost
7,000
$25
50
$1,400
2,000
$15
40
$700
1958
$14
80
$140
10
$1,020
400
$20
300
$12
150
$20
% of cumul
Monthly
Monthly
$ value
$ value
Class
$17 5,000
47 .8%
A
$7 0,000
19.1%
A
$30,000
8.2%
B
$28,000
7 .7 %
B
$27 ,412
7 .5%
B
$11,200
3.1%
C
$10,200
2.8%
C
$8,000
2.2%
C
$3,600
1.0%
C
$3,000
0.8%
C
$366,412
A reasonable classification is given above. Its properties are:
Class
% cumul $ value
% of SKUs
A
B
C
66.9%
23.4%
9.9%
20%
30%
50%
Instructor’s Manual, Chapter 12
12-7
3.
a.
PROBLEM NO. 12 - 3
Item
K34
K35
K36
M10
M20
Z45
F14
F95
F99
D45
D48
D52
D57
N08
P05
Unit Cost Monthly Usage
$10
200
$25
600
$36
150
$16
25
$20
80
$80
700
$20
300
$30
800
$20
60
$10
550
$12
90
$15
110
$40
120
$30
40
$16
500
Sort by Monthly $ v alue:
Item
Z45
F95
K35
P05
F14
D45
K36
D57
K34
D52
M20
F99
N08
D48
M10
Unit Cost Monthly Usage
$80
700
$30
800
$25
600
$16
500
$20
300
$10
550
$36
150
$40
120
$10
200
$15
110
$20
80
$20
60
$30
40
$12
90
$16
25
Monthly
$ value
$56,000
$24,000
$15,000
$8,000
$6,000
$5,500
$5,400
$4,800
$2,000
$1,650
$1,600
$1,200
$1,200
$1,080
$400
$133,830
% of cumul
Monthly
$ value
41.9%
17 .9%
11.2%
6.0%
4.5%
4.1%
4.0%
3.6%
1.5%
1.2%
1.2%
0.9%
0.9%
0.8%
0.3%
Class
A
A
A
B
B
B
B
B
C
C
C
C
C
C
C
A reasonable classification is given above. Its properties are:
b.
c.
12-8
Class
% cumul $ value
% of SKUs
A
B
C
71.0%
22.2%
6.8%
33.3%
46.7%
20%
He/she could exert most control on A items and least on C items. For example, A items
could be ordered daily, Bs weekly, and Cs monthly.
It might be important for some reason other than dollar value, such as cost of a stock-out,
sole sourcing, becoming obsolete, etc.
Operations Management, 5/C/E
4.
D = 4,860 bags/yr.
S = $10
H = $5
2DS
H
a. EOQ =
=
2(4,860)10
= 139.4 bags, round to 139
5
b. Q / 2 = 139 / 2 = 69.5 bags
c.
D
=
Q
4,860 bags
139 bags/order
d. TC = Q/2 H +
=
D
Q
139
2
= 34.96 orders ≈ 35 orders
S
(5) +
4,860
(10) = 347.5 + 349.64 = $697.14
139
e. Using S = $15, EOQ =
EOQ increases by
Instructor’s Manual, Chapter 12
171 – 139
139
2(4,860)15
5
= 170.8 ≈ 171
= 23%
12-9
5.
D = 10/day × 260 days/year = 2,600 packages/year
S = $10, H = $1
a. EOQ =
b. TC =
TC228 =
Q
2
228
(1) +
2
2DS
H
=
H+
D
S
Q
2,600
228
2(2,600)10
= 228 packages
1
(10) = 114 + 114.04 = $228.04
c. Yes
d. TC100 =
100
(1) +
2
2,600
(10)
100
= 50 + 260 = $310
Yes, the firm will save $310 - $228.04 = $81.96 per year
12-10
Operations Management, 5/C/E
6.
D = (250 pots/month) × (12 months/year) = 3,000 pots / year
Price = $2/pot, H = ($2)(0.30) = $0.60/unit/year
S = $20, Current Q = 250 pots
a. EOQ =
2DS
=
H
b. TC447 =
447
2
2(3,000)20
= 447.2  447 pots
0.60
(.60) +
3,000
(20)
447
= 134.10 + 134.23 = $268.33
c.
250
3,000
(.6) 
(20)
2
250
= 75 + 240 = $315
TC 250 
Therefore, the additional cost of using order size of 250 is:
$315 – $268.33 = $46.67
Instructor’s Manual, Chapter 12
12-11
7.
800/month, so D = 12(800) = 9,600 pallets/year
Price = $5 each
H = 0.25 Price = 0.25($5) = $1.25/pallet/year
S = $28 per order
Current Q = 800 pallets
TC800 =
EOQ =
TC658 =
800
9600
(1.25) +
(28) = $500 + 336 = $836
2
800
2DS
=
H
2(9,600)$28
= 655.8 [round to 656]
$1.25
656
9,600
(1.25) +
(28) = 410 + 409.76 = $819.76
2
656
Savings of only $836 - $819.76 = $16.24 per year.
12-12
Operations Management, 5/C/E
8.
D = 100/month, H = $2/unit/month, S = $55
a. EOQ =
2DS
H
EOQ =
2(100)55
= 74.16
2
round to 74 units
Note: You can use monthly demand in EOQ, so long as H is also monthly.
b. Discount of $10/order is equivalent to S - 10 = $45 (revised ordering cost)
TC74 =
74
100
(2) +
(55) = 74 + 74.32 = $148.32 per month
2
74
TC50 =
50
100
(2) +
(45) = 50 + 90 = $140* per month
2
50
100
100
(2) +
(45) = 100 + 45 = $145 per month
2
100
Yes, use order size of 50.
TC100 =
c.
B = $5/unit/month
Q
2 DS  H  B 


H  B 
2(100)(55)  2  5 

 = 87.75 round to 88 units
2
 5 
 H 
 2 
Qb  Q 
  88
 = 25.1 round to 25 units
H B
 25
Instructor’s Manual, Chapter 12
12-13
9.
D = 27,000 jars/month
Current Q = 4,000 jars
H = $.01/unit/month
S = $20 per order
a.
TC = (Q/2) H + (D/Q) S
TC(4,000) = (4,000/2)($.01) + (27,000/4,000)($20) = 20 + 135 = $155 per month
EOQ 
2 DS
2(27,000)( 20)

= 10,392.3 round to 10,392 jars
H
.01
TC(10,392) = (10,392/2)($.01) + (27,000/10,392)($20) = 51.96 + 51.96 = $103.92
Difference = $155 – $103.92 = $51.08 per month.
b.
EOQ 
2 DS
2(27,000) S

 4,000
H
.01
Solving, S = $2.96.
12-14
Operations Management, 5/C/E
10. p = 500 kg/day
d = 100 kg/day
300 days per year
S = $12
H = $4/kg/year
a. Qo =
2DS
H
D = 100 kg/day × 300 days/year = 30,000 kg/year
p
=
p -d
2(30,000)12
4
500
= 474.34 [round to 474 kg]
500 -100
b. D / Qo = 30,000 / 474 = 63.29 or 63 runs/year (every approx. 5 days)
c. Run length: Qo / p = 474 / 500 = 0.95 day (approx. 1 day)
Instructor’s Manual, Chapter 12
12-15
11. p = 50 tonnes/day
d = 20 tonnes/day
200 days/year
S = $400
H = $200/tonne per year
Bag = 100 kg
D = 20 tonnes/day × 200 days/year = 4,000 tonnes/year
a. Qo =
2DS
H
b. Imax =
Q
163.30
(p - d) =
(50 - 20) = 97.98 tonnes
p
50
Avg inven:
p
=
p-d
2(4,000)400
200
50
= 163.30 tonnes = 1,633 bags
50 - 20
Imax
97.98
=
= 48.99 tonnes = 490 bags
2
2
c. Run length =
d. Runs per year=
Q
P
=
163.30
= 3.27 days
50
D
4,000
=
= 24.5
Q
163.30
e. If S = $200,
Q' 0 
2 DS
H
p

pd
2(4,000)200
50
=115.47 tonnes
200
50  20
Imax = (Q/p)(p - d) = (115.47/50)(50 - 20) = 69.28 tonnes
Avg inventory = Imax/2 = 69.28/2 = 34.64 tonnes
TC = (Imax/2)H + (D/Q)S
TC(S = $400) = 48.99($200) + (4,000 /163.3)$400 = $9,798 + $9,797.92 = $19,595.92
TC(S = $200) = 34.64($200) + (4,000 /115.47)$200 = $6,928 + $6,928.21 = $13,856.21
Saving = $19,595.92 - $13,856.21 = $5,739.71
12-16
Operations Management, 5/C/E
12. p = 200/day
d = 80/day
Operating 5 days a week, 50 weeks a year
S = $300
H = $10/unit/year
D = 20,000 (250 days × 80 = 20,000)
a. Qo =
2DS
H
p
=
p-d
2(20,000)300
10
200
200-80
= (1,095.44) ( 1.2910) = 1,414.2 or 1,414 units
b. Run length =
Q
P
=
1,414
= 7.07 days
200
c. p – d = 200 - 80 = 120 units per day
d. Cycle length = Q0 / d = 1414 / 80 = 17.7 days
Yes, after making this component for 7 days, we can set up for day and then make the other job
for 8 days, then set up for a day and make this component again, and so on.
Instructor’s Manual, Chapter 12
12-17
13. p = 800 units per day
d = 300 units per day for 250 days; Therefore, D = 300 × 250 = 75,000/year
Current Q = 2,000 units
a. D/Q = 75,000/2,000 = 37.5 batches per year
b. The number of units produced in two days = (2 days)(800 units/day) = 1,600 units
The number of units used in two days = (2 days) (300 units/day) = 600 units
Inventory buildup after the first two days of production = 1,600 - 600 = 1,000 units
c.
Imax = (Q/p)(p - d) = (2000/800)(800 - 300) = 1250 units
Avg. inventory = Imax/2 = 1250/2 = 625 units
d. Another job requires 4 days per cycle of hairdryers.
Q 2,000

 2.5 days
P 800
Setup time per batch = 1/2 day
Total time per batch = 2.5 + 0.5 = 3 days
Cycle length = Q / d = 2,000 / 300 = 6.7 days
No, after setup for half a day and making the heating element for 2.5 days, we only have 6.7 – 3
= 3.7 days to make the other component (3.7 < 4 days required).
Production time per batch =
12-18
Operations Management, 5/C/E
14.
D = 18,000 boxes/year
i = 20% of unit cost (R) per year
S = $32
a. EOQR=$1.10 =
2 DS

iR
2(18,000)32
 2,288.3
.20(1.10)
boxes
Bec. 2,288.3 < 10,000, this EOQ is not feasible. Try the next higher price $1.15:
EOQR=$1.15 =
2 DS

iR
2(18,000)32
 2,238
.20(1.15)
boxes
Bec. 2,238 < 5,000, this EOQ is not feasible. Try the next higher price $1.20:
EOQR=$1.20 =
2 DS

iR
2(18,000)32
 2,191
.20(1.20)
boxes
Bec. 2,000 < 2,191 < 4,999, this EOQ is feasible. Next, we need to compare the total cost of Q
= 2,191 units with those of Q = 5,000 and Q = 10,000 units.
2,191
18,000
TC2,191 =
(.20)(1.20) 
($32)  $1.20(18,000) =
2
2,191
 262.92  262.89  21,600  $22,125.81
TC5,000 =
5,000
18,000
(.20)(1.15) 
($32)  $1.15(18,000) 
2
5,000
= 575 + 115.2 + 20,700 = $21,390.20
TC10,000 = 10,000 (.20)(1.10)  18,000 ($32)  $1.10(18,000) 
2
10,000
= 1,100 + 57.6 + 19,800 = $20,957.60
[lowest]
Hence, the best order quantity would be 10,000 boxes.
Total
Annual
cost

2,191
5,000
10,000
Quantity
b.
D 18,000

 1.8 orders per year
Q 10,000
Instructor’s Manual, Chapter 12
12-19
15.
Quantity
Unit Price
1 - 399
$10
400 - 599
9
600 +
8
Operates 200 days a year
D = 25 stones/day × 200 days/year = 5,000 stones/year
S = $48
a. H = .30R
EOQ$8 =
2(5,000)48
= 447.2
.30(8)
Bec. 447.2 < 600, this EOQ is not feasible. Try the next higher price $9:
EOQ$9 =
2(5,000)48
= 421.64 ≈ 422
.30(9)
Bec. 400 ≤ 422 ≤ 599, this EOQ is feasible. Next, we need to compare the total cost of Q = 422 units
with those of Q = 600:
TC =
Q
D
(.30R) +
(S) + RD
2
Q
TC422 =
422
[.30($9)] +
2
5,000
($48) + $9(5,000) = 569.7 + 568.72 + 45,000 = $46,138.42
422
TC600 =
600
[.30($8)] +
2
5,000
($48) + $8(5,000) = 720 + 400 + 40,000 = $41,120 [lower]
600
b. ROP = (25 stones/day) (6 days) = 150 stones.
16.
Range
D = 4,900 pulleys/ year
0 - 999
H = .2 R
1,000 - 3,999
S = $50
4,000 - 5,999
6,000 +
EOQR=$4.85 =
2 DS

iR
R
$5.00
4.95
4.90
4.85
2(4,900)50
 710.7
.20(4.85)
or 711 pulleys
Bec. 711 < 6,000, this EOQ is not feasible. Try the next higher price $4.90:
EOQR=$4.90 =
2 DS

iR
2(4,900)50
 707.1
.20(4.90)
or 707 pulleys
Bec. 707 < 4,000, this EOQ is not feasible. Try the next higher price $4.95:
12-20
Operations Management, 5/C/E
EOQR=$4.95 =
2 DS

iR
2(4,900)50
 703.53
.20(4.95)
or 704 pulleys
Bec. 704 < 1,000, this EOQ is not feasible. Try the next higher price $5.00:
EOQR=$5.00 =
2 DS

iR
2(4,900)50
 700
.20(5.00)
pulleys
Bec. 0 ≤ 700 ≤ 999, this EOQ is feasible. Next, we need to compare the total cost of Q = 700
units with those of Q = 1,000, Q = 4,000, and Q = 6,000 units.
700
4,900
TC700 =
(.2)($5.00) +
($50) + $5.00(4,900) = 350 + 350 + 24,500 = $25,200
2
700
1,000
4,900
TC1,000 =
(.2)($4.95) +
($50) + $4.95(4,900) = 495+245+24,255 = $24,995 lowest
2
1,000
4,000
4,900
TC4,000 =
(.2)($4.90) +
($50) + $4.90(4,900) = 1,960+61.25+24,010 = $26,031.25
2
4,000
6,000
4,900
TC6,000 =
(.2)($4.85) +
($50) + $4.85(4,900) = 2,910+40.83+23,765 = $26,715.83
2
6,000
Order 1,000 units at a time.
TC





700 704 707 711
1,000
4,000
6,000
Quantity
Instructor’s Manual, Chapter 12
12-21
17.
D = 800 units/month × 12 = 9,600 units/year
S = $40
H = .25 R
For Supplier A:
2(9600)(40)
Q13.6 
 475.27 (not feasible)
(.25)(13.6)
Q13.8 
2(9600)(40)
 471.81 (feasible, round to 472)
(.25)(13.8)
9600
 472 
(40)  
 (.25)(13.8)   (13.8)(9600) 
472
 2 
 813.56  814.2  132, 480
TC472 
$ 134,107.76
9600
 500 
(40)  
 (.25)(13.6)   (13.6)(9600) 
500
 2 
 768  850  130,560
TC500 
TC500
TC500  $132,178 *
For Supplier B:
Q13.7 
2(9600)(40)
 473.53 (feasible, round to 474)
(.25)(13.7)
9600
 474 
(40)  
 (.25)(13.7)   (13.7)(9600) 
474
 2 
 810.13  811.72  131,520
TC474 
 $133,141.85
Since $132,178 < $133,141.85, choose supplier A.
The optimal order quantity is 500 units.
12-22
Operations Management, 5/C/E
18.
S = $40 /order
H = .40R per year
D = 3600 boxes per year
Q = 800 boxes (recommended)
Range
1-199
200-800
801+
R
$1.20
$1.10
$1.00
If the company decides to order 800, the total cost is as follows:
TC = (Q/2)H + (D/Q)S + RD
TC(800) = (800/2)(.4)($1.10) + (3,600/800)$40 + 3,600($1.10)
= 176 + 180 + 3,960 = $4,316.
Even though the inventory total cost curve is fairly flat around its minimum, when there are
quantity discounts, there are multiple U-shaped total inventory cost curves, one for each unit price.
Therefore, when the quantity changes from 800 to 801, we shift to a different total cost curve.
If we take advantage of the quantity discount and order 801 units, the total cost is calculated as
follows:
TC(801) = (801/2)(.4)($1.00) + (3,600/801)$40 + 3,600($1.00)
= 160.20 + 179.78 + 3,600 = $3,939.98
The order quantity of 801 is better than 800 because $3,939.98 < $4,316.
However, the EOQ for R = $1.00 may have even lower cost:
EOQ=
2 DS

H
2(3,600)$40
 848.53
.4($1)
round to 849 boxes
TC(849) = (849/2)(.4)($1.00) + (3,600/849)$40 + 3600($1.00)
= 169.80 + 169.61 + 3,600 = $3,939.41
In fact, order size of 849 has a slightly lower total cost than 801 ($3,939.41 vs. $3,939.98).
Instructor’s Manual, Chapter 12
12-23
19.
Average daily usage = 800 m/day
Lead time = 6 workdays
Desired lead time service level = 95 percent.
Safety stock
Stock out prob.
1,500m
.1
1,800m
.05
2,100m
.02
2,400m
.01
ROP?
Stock out risk should be at most 1.00 - 0.95 = 0.05. Therefore, safety stock = 1,800 m.
ROP
= d (LT) + safety stock
= 800 (6) + 1,800 = 6,600 m
20. Expected demand during a LT = 300 units
dLT = 30 units
Demand during a LT ~ Normal
a.
SS? for 1% LT stock-out prob.
z = 2.33 from Appendix B, Table B
SS = z d LT = 2.33 (30) = 69.9 round to 70 units
b.
2% stock out risk  smaller z  less SS
21. Expected dLT = 600 kg
d LT = 52 kg
Stock out risk = 4%  z = 1.75 from App. B, Table B
a.
SS?
SS = z d LT = 1.75 (52) = 91 kg
b.
12-24
ROP = Average demand during a lead time + safety stock = 600 + 91 = 691 kg
Operations Management, 5/C/E
22.

d = 21 litres/wk
d = 3.5 litres/wk
Lead time SL = 90%
LT
= 2 days
Open 7 days a week
90%
0
6
a. SL = 90 %  z = 1.28 from App. B., Table B.

ROP = d (LT) + z (d)
LT = 21(2/7) + 1.28(3.5)
b. OI = 7 days
On hand = 8 litres

Imax = d (OI + LT) + zd
OI + LT
1.28 z-scale
8.39
litres
(2/7) = 6 + 2.39 = 8.39 litres
= 21(7 + 2)/7 + 1.28(3.5)
(7+2)/7
= 27 + 5.08 = 32.08 round to 32 liters
Q = Imax – on hand = 32 – 8 = 24 litres
c.
1 day after order;
from part a, ROP = 8.39
on hand now = ROP - 2 litres = 6.39
Prob (stock- out) in 2 days = ?
6.39 = 21 (2/7) + z (3.5) 2/7
0.39 = 1.87 z
z = .2085 ~ .21  SL = .5832 from App. B, Table B
Prob (stock out) = 1 - .5832 = .4168 or about 42%
23.
d = 30 litres/day
ROP = 170 litres
LT = 4 days
Stock out risk = 9%
SS? to make stock-out risk = 3%
First we need to determine 𝜎𝑑𝐿𝑇
Stock out risk = 9%  z = 1.34 from App. B., Table B
SS = ROP - d(LT) = 170 - 30 (4) = 50 litres
SS = z 𝜎𝑑𝐿𝑇  50 = 1.34 𝜎𝑑𝐿𝑇  𝜎𝑑𝐿𝑇 = 37.31 litres
Stock out risk = 3%  z = 1.88 from App.B, Table B
SS = z 𝜎𝑑𝐿𝑇 = 1.88(37.31) = 70.15 litres
Instructor’s Manual, Chapter 12
12-25
24.
𝑑̅ = 85 capacitors/day
𝑑 = 5 capacitors/day
ROP = 625 capacitors
LT = 6 days
LT = 1.1 days
Prob. of stock-out?
2
ROP  d ( LT )  z LT d2  d 2 LT
625 = 85 (6)  z 6(5) 2  (85) 2 (1.1) 2
115 = 94.3 z  z = 1.22
From App. B, Table B, SL = .8888
Therefore, prob. of stock-out = 1- .8888 = .1112
25.
approx. 11%
SL  96%

d = 12 units- /day
d = 2 units/day

LT = 4 days
LT = 1 day
ROP?
SL  96%  Z > 1.75 from App. B, Table B
2
ROP  d ( LT )  z LT d2  d 2 LT
 12(4)  1.75 4(2) 2  (12) 2 (1) 2
> 48 + 22.14 = 70.14 or 70 units
12-26
Operations Management, 5/C/E
26. New LT = 3 days
D = 4,500 litres per year
360 days/year
d = 2 litres per day
S = $10
H = .40R/litre/year
Qty.
1 - 399
400 - 799
800+
Unit Price
$2.00
1.80
1.60
a. Q?
EOQ 
2 DS

H
2(4,500)(10)
= 375 litres, not feasible.
.4($1.60)
EOQ 
2 DS

H
2(4,500)(10)
= 353.55 litres, not feasible.
.4($1.80)
EOQ 
2 DS

H
2(4,500)(10)
= 335.4 round to 335 litres. Feasible.
.4($2)
Need to calculate the TC of Q = 335, Q = 400, and Q = 800.
TC = (Q/2)H + (D/Q)S + RD
TC(335) = (335/2)(.4)($2.00) + (4,500/335)$10 + 4,500($2.00)
= 134 + 134.33 + 9,000 = $9,268.33
TC(400) = (400/2)(.4)($1.80) + (4,500/400)$10 + 4,500($1.80)
= 144 + 112.5 + 8,100 = $8,356.50
TC(800) = (800/2)(.4)($1.60) + (4,500/800)$10 + 4,500($1.60)
= 256 + 56.25 + 7,200 = $7,512.25 [lowest]
b. ROP? for risk of stock-out = 1.5%
Risk of stock-out = 1.5%  z = 2.17 from App. B, Table B
4,500

= 12.5 / day
d=
360

ROP = d LT + z LT d
= 12.5 (3) + 2.17 3 (2)
= 37.5 + 7.52
= 45.02 round to 45 litres
Instructor’s Manual, Chapter 12
12-27
27. 
d = 5 boxes/day
d = .5 box/day
LT = 2 days
S = $10
H = $10/box/year
a.
EOQ? Assuming 250 workdays per year
D = 5 boxes/day × 250 days/year = 1,250 boxes/year
EOQ =
2DS
=
H
b. ROP = 12 boxes
Risk of stock-out?

ROP = d (LT) + z
Solving for z:
2(1250)(10)
= 50 boxes
10
LT(d)

ROP - d (LT)
z=
12 - 5(2)
=
LT (d)
2 (.5)
= 2.83
 SL = .9977 from App. B, Table B, so risk of stock-out = 1 - .9977 = .0023 or 0.23%
c. OI = 7 days
Shortage risk? If Q = 36 boxes when on hand = 12 boxes
Imax = 36 + 12 = 48 boxes

Imax = d (OI + LT) + zd
OI + LT
48 = 5(7 + 2) + z(.5) 7 + 2
3 = 1.5 z  z = 2.0
 SL = .9772 from App. B, Table B
Therefore, shortage risk = 1 - .9772 = .0228 or 2.28%
12-28
Operations Management, 5/C/E
28.

d = 8 kg
d = 1 kg

LT = 8 days
LT = 1 day
ROP? Stock-out risk of 10%
SL = 100 – 10 = 90%  z = 1.28 from App. B, Table B

 

ROP = d (LT ) + z
LT 2d + d 22LT
= 8 (8) + 1.28
= 74.86
8(1)2 + 82(1)2
= 64 + 1.28(8.49)
[round to 75 kg]
29. Open 360 days per year

d = 10 rolls/day
d = 2 rolls/day
S = $10
H = $.40/roll per year
LT = 3 days
a.
D = 10 rolls/day × 360 days/year = 3,600 rolls/year
EOQ =
2DS
=
H
2(3,600)10
= 424.3 round to 424 rolls
.40
b. Lead time SL of 96%  z = 1.75 from App. B, Table B

ROP = d (LT) + z LT(d) = 10(3) + 1.75 3(2) = 36.06 [round to 36]
c. SLannual = 96%
𝜎𝑑𝐿𝑇 = 𝜎𝑑 √𝐿𝑇 = 2√3 = 3.46
E( z) 
Q(1  SLannual )
 dLT

424 (1  .96)
 4.90
3.46
 Table 12-3 to get z, but E(z) = 4.90 is not in the table. However, note that for values of E(z)
> 2.4, z = - E(z). Therefore, for E(z) = 4.90, z = -4.90.

ROP = d (LT) + z 𝜎𝑑𝐿𝑇 = 10(3) - 4.90(3.46) = 13.05 [round to 13]
Instructor’s Manual, Chapter 12
12-29
30.
D = 1,200 cases/year
S = $20
H = $3 per case/year
SLannual = 99%
2 DS
a.
EOQ 

H
b.
2(1,200)($20)
= 126.49 round to 126 cases
$3
Expected dLT = 80d LT = 5
ROP?
E( z) 
Q(1  SLannual )

126(1  .99) = .252
5
 dLT
From Table 12-3, z = 0.32
ROP = Expected dLT + zd LT = 80 + .32 (5) = 81.6 cases
31.
250 days a year

d = 250 litres/day
d = 14 litres/day
H = $0.3/litre/year
S = $10
LT = 1 day
a.
D = 250 litres/day× 250 days/year = 62,500 litres/year
EOQ 
b.
2 DS

H
2(62,500)($10) = 2,041.24 round to 2,041 litres
$0.3
ROP? if SLannual = 99.5%
E( z) 
Q(1  SLannual )
 dLT

2,041 (1  .995)
14 1
= .729
E(z) = .729 → z = -0.54 from Table 12-3 (midpoint of E(z) = .712 and .740)

ROP = d (LT) + z dLT = 250(1) - 0.54(14) = 242.44
12-30
Operations Management, 5/C/E
32.
ROP = 18 units
LT = 3 days
Demand during last 10 days:
1
3
Day
Units
2
4
3
7
4
5
5
5
6
6
7
4
8
3
9
4
10
5
SL?
Mean and standard deviation of daily demand:
𝑑̅ = 4.6
(3−4.6)2 +(4−4.6)2 +(7−4.6)2 +⋯(5−4.6)2
10−1
𝜎𝑑 = √
= 1.265
ROP  d (LT)  z ( d ) LT
18  4.6(3)  z (1.265) 3
18 = 13.8 + 2.191z
4.2 = 2.191z
z = 1.92
From Appendix B, Table B, SL = 0.9726 = 97.26%.
33.
Fixed Interval model
SL = .98
OI = 14 days
LT = 2 days
Q?
SL = .98  z = 2.055 from App. B, table B.
Q
= Imax – On hand
= d (OI  LT )  z d OI  LT - on hand
QK033 = 60(14  2)  2.055(5) 14  2 - 420 = 960 + 41.1 – 420 = 581.1 ≈ 581
QK144 = 50(14  2)  2.055(4) 14  2 - 375 = 800 + 32.88 – 375 = 457.88 ≈ 458
QL700 = 8(14  2)  2.055(2) 14  2 - 160 = 128 + 16.44 – 160 = -15.56 ≈ 0 (do not order)
Instructor’s Manual, Chapter 12
12-31
34.
50 week/year
P34
P35
4 weeks
OI

60 units/week
d
4 units/week
d
R
$15
H
(.30)(15) = $4.50
S
$70
LT
2 weeks
Stock-out Risk
= 2.5%
70 units/week
5 units/week
$20
(.30)(20) = $6.00
$30
2 weeks
2.5%
D
z
= 70(50) = 3,500
1.96
a.
= 60(50) = 3,000
1.96
From App. B, Table B.

ROPP34 = d × LT+ z LTd
ROPP34 = 60(2) + 1.96 2 (4) = 131.1 round to 131 units
2 DS

H
2(3,000)($70)
$4.50
b.
EOQ 
c.
Q? if on hand = 110 units

Q = d (OI + LT) + z OI + LT d - on hand
QP35 = 70 (4 + 2) + 1.96
=305.51
round to 306 units
4 + 2 (5) – 110
QP35 = 420 + 24 – 110
QP35 = 334 units
12-32
Operations Management, 5/C/E
35.
a.
PROBLEM NO. 12 - 35
Estimated
Annual
Demand
20,000
60,200
9,800
16,300
6,250
4,500
21,000
45,000
800
26,100
Item
H4-010
H5-201
P6-400
P6-401
P7-100
P9-103
TS-300
TS-400
TS-041
V1-001
Sort by Annual $ value
Estim ated
Annual
Item
Demand
TS-400
45,000
TS-300
21,000
P6-400
9,800
H5-201
60,200
P6-401
16,300
V1-001
26,100
P9-103
4,500
P7-100
6,250
H4-010
20,000
TS-041
800
Ordering Holding Cost
Cost
(%)
50
20
60
20
80
30
50
30
50
30
50
40
40
25
40
25
40
25
25
35
Annual
Unit Price
$ value
$40.00 $1,800,000
$45.00 $945,000
$28.50 $27 9,300
$4.00 $240,800
$12.00
$195,600
$4.00 $104,400
$22.00
$99,000
$9.00
$56,250
$2.50
$50,000
$20.00
$16,000
$3,786,350
Unit Price
$2.50
$4.00
$28.50
$12.00
$9.00
$22.00
$45.00
$40.00
$20.00
$4.00
% of cumul
Annual
$ value
48%
25%
7%
6%
5%
3%
3%
1%
1%
0%
Class
A
A
B
B
B
C
C
C
C
C
b.
Item
H4-010
H5-201
P6-400
P6-401
P7-100
P9-103
TS-300
TS-400
TS-041
V1-001
Estimated
Annual
Demand
20,000
60,200
9,800
16,300
6,250
4,500
21,000
45,000
800
26,100
Instructor’s Manual, Chapter 12
Ordering Holding Cost
Cost
(%)
50
20
60
20
80
30
50
30
50
30
50
40
40
25
40
25
40
25
25
35
Unit Price
$2.50
$4.00
$28.50
$12.00
$9.00
$22.00
$45.00
$40.00
$20.00
$4.00
EOQ
2000.0
3005.0
428.2
672.9
481.1
226.1
386.4
600.0
113.1
965.5
12-33
36.
Cost = $3.20 per dozen
Rev = $4.80 per dozen
Salvage = 4.80/2 = $2.40 per dozen
Cs = Rev - Cost = $4.80 - $3.20 = $1.60
Ce = Cost - Salvage = $3.20 - $2.40 = $.80
Cs
$1.60
1.6
SL =
=
=
= .67
Cs + Ce
$1.60 + $.80
2.4
Since this falls between the cumulative
probabilities of .63 (x = 24) and .73 (x = 25),
The larger should be chosen, i.e., make 25 dozen.
doughnuts.
The resulting service level = 73%
37.
x
Demand
19
20
21
22
23
24
25
26
27
28
29
P(x)
.01
.05
.12
.18
.13
.14
.10
.11
.10
.04
.02
Cum.
P(x)
.01
.06
.18
.36
.49
.63
.73
.84
.94
.98
1.00
Cost = $100 per unit
Holding cost = 145% of unit cost
Cs = $8,000 per unit
Demand ~ Poisson with mean 3.2 units
a. Ce = $100 + 1.45($100) = $245
Cs
$8,000
SL =
=
= .97
Cs + Ce
$8,000 + $245
Using the Poisson probabilities, the minimum
stocking level that will provide the desired service
is 7 spares (cumulative probability = .983).
[From Poisson Table with  = 3.2]
x
Cum. Prob.
0
.041
1
.171
2
.380
3
.603
4
.781
5
.895
6
.955
7
.983
8
.994
9
.998
.
.
b. Cs range for 6 spares?
𝐶
𝑠
SL = 𝐶 +𝐶
𝑠
𝑒
𝐶
𝑠
. 955 ≥ 𝐶 +245
 .955(Cs + 245) ≥ Cs  .045 Cs ≤ 233.975  Cs ≤ $5,199.44
𝑠
𝐶
𝑠
. 895 < 𝐶 +245
 .895Cs + 219.275 < Cs  .105 Cs > 219.275  Cs > $2,088.33
𝑠
Therefore, carrying 6 spare parts is best if the shortage cost > $2,088.33 and ≤ $5,199.44.
12-34
Operations Management, 5/C/E
38. Cost = $4.20/kg
Rev = $5.70/kg
Salvage = $2.40/kg
Normal demand

d = 80 kg/day
d = 10 kg/day
Optimal stocking level?
Cs = Rev - Cost = $5.70 - $4.20 = $1.50/kg
Ce = Cost - Salvage = $4.20 - $2.40 = $1.80/kg
$1.50
Cs
$1.50
SL =
=
=
= .4545
$3.30
Cs + Ce
$1.50 + $1.80
SL = .4545  z = -0.114 from Appendix B, Table B (interpolate).

Optimal stocking level = d + z d = 80 - 0.114(10) = 78.86 kg
39.
Demand ~ Normal

d = 40 litres/day
d = 6 litres/day
Ce = $0.35/litre
Order quantity Q = 49 litres
a. Cs = ?
A stocking level of 49 litres translates into:

z = Q - d = 49 - 40 = 1.5
d
6
z = 1.5  SL = 0.9332 from App. B, Table B.
Cs
Cs
SL =
Thus, 0.9332 =
Cs + Ce
Cs + $0.35
.9332
0
1.5
z-scale
40
49
quarts
0.9332(Cs + 0.35) = Cs
0.0668 Cs = 0.32662
Cs = $4.89 / litre
b. At first it may seem a little high for the loss of profit from selling a litre of strawberries. However,
customers may buy other items along with the strawberries (ice cream, whipped cream, etc.).
Instructor’s Manual, Chapter 12
12-35
40. Demand is Poisson with mean of 6 cakes per day
Cost = $9 per cake
Rev. = $12 per cake
Salvage value = $7 per cake
Optimum stocking level?
Cs = Rev - Cost = $12 - $9 = $3/cake
Ce = Cost - Salvage = $9 - $7 = $2/cake
SL =
Cs
=
Cs + Ce
$3
$3 + $2
= 0.6
Since 0.6 falls between the cumulative
probability for demand of 5 and 6, the
optimum stocking level is 6 cakes.
[From Poisson Table with  = 6.0]
Demand Cum. Prob.
0
.002
1
.017
2
.062
3
.151
4
.285
5
.446
6
.606
.
.
.
.
41. Cost = $3 per kg
Salvage value = $2 per kg
8 burgers per kg
Price of a burger = $2
Cost of a burger = $1
Demand ~ Normal (400kg, 50kg)
Optimal order quantity?
Cs = ($2 - $1)/burger × 8 burgers/kg = $8/kg
Ce = Cost - Salvage = $3.00 - $2.00 = $1/kg
Cs
$8
SL =
=
= .8889
Cs + Ce
$8 + $1
SL = .8889  z =1.22 (App. B, Table B)
Optimal order quantity =  + z  = 400 + 1.22(50) = 461 kg
.8889
0
400
12-36
1.22 z-scale
461
kg
Operations Management, 5/C/E
42.
Profit = $10 per machine per day
Store has 4 machines
a. Range for Ce = ?
Cs = lost profit = $10
Demand Freq.
0
.30
1
.20
2
.20
3
.15
4
.10
5
.05
1.00
For four machines to be optimal, the SL ratio must be
.85 <
Cum.
Freq.
.30
.50
.70
.85
.95
1.00
$10
 .95.
$10 + Ce
Re-writing the left inequality: 8.5 + .85 Ce < 10  .85 Ce < 1.5 
Ce < $1.76
Re-writing the right inequality: 9.5 + .95 Ce  10  .95 Ce  .5 
Ce  $.53
b. If Ce is too low, the number of machines should be decreased: the higher excess costs are, the lower SL
becomes, and hence, the lower the optimum stocking level.
43.
Cost = $200 per unit
Salvage value = $50 per unit
Shortage  2 days down time; down time cost = $400 per day
Q?
a. Ratio method:
# of spares
0
1
2
3
Probability of Demand Cumulative Probability
0.10
0.10
0.50
0.60
0.25
0.85
0.15
1.00
Cs = Cost of stock out = ($400 per day) (2 days) = $800
Ce = Cost of excess inventory = Unit cost - Salvage Value = $200 - $50 = $150
𝐶𝑠
800
SL = 𝐶 +𝐶
= 800+150 = .842
𝑠
𝑒
Bec. 0.842 is between cumulative probabilities of 0.85 and 0.60, we need to order 2 spares.
b. Tabular method:
Stocking
Level
0
1
2
3
Demand = 0
Prob. = 0.10
$0
.10(1)($150)=$15
.10(2)($150)=$30
.10(3)($150)=$45
Demand = 1
Prob. = 0.50
.50(1)($800)=$400
$0
.50(1)($150)=$75
.50(2)($150)=$150
Demand = 2
Prob. = 0.25
.25(2)($800)=$400
.25(1)($800)=$200
$0
.25(1)($150)=$37.50
Demand = 3
Prob. = 0.15
.15(3)($800)=$360
.15(2)($800)=$240
.15(1)($800)=$120
$0
Expected
Cost
$1,160
$455
$225*
$233.50
Same answer: order 2 spares because the expected cost is the least for 2.
Instructor’s Manual, Chapter 12
12-37
44.
Cost = $33 per cake
Rev. = $60
Salvage value = $30 for 1/3 of cakes; rest have 0 salvage value.
Demand
0
1
2
3
Probability of Demand
0.15
0.35
0.30
0.20
Cumulative Probability
0.15
0.50
0.80
1.00
Cs = Cost of stock-out = Selling Price - Unit Cost = $60 - $33 = $27
Ce = Cost of excess inventory = Unit Cost - Salvage Value = $33 - $30(1/3) = $23
SL 
Cs
27

 0.54
Cs  Ce 27  23
Bec. the service level of 0.54 falls between cumulative probabilities of 0.50 and 0.80, the bakery
should stock 2 wedding cakes.
45.
No shows ~ Normal(18, 4.55)
Profit per passenger = $99.
If a passenger is bumped, the company pays that passenger $200.
Number of tickets to overbook?
Cs = $99, Ce = $200 for overbooking
SL 
Cs
99

 .3311
Cs  Ce 99  200
Using Appendix B, Table B, .3311 falls closest to .3300  z = -0.44.
𝑆𝑜 = 𝜇 + 𝑧𝜎 = 18 + (−0.44)(4.55) = 15.998 = 16 overbooks
12-38
Operations Management, 5/C/E
46.
D = 210(12) = 2,520 vaccines
S = .5 ( $17/hr) = $8.5 per order
Holding cost rate i = 8 + 8 = 16% per unit per year
Price R = $16 per vaccine
a.
EOQ =
2 DS
2(2,520)($ 8.5)

 129.36, round to 129 vaccines
iR
(.16)($ 16)
b. LT = 2 days
d =7
d = 2
Service level = 98% 
ROP = d . LT + z
d
z = 2.055 from App. B. Table B
LT = 7(2) + 2.055(2)(
2 ) = 14 + 5.81 = 19.81 round to 20 vaccines.
c. OI = 2 weeks
47.
d
OI  LT
= 7(14 + 2) + 2.055(2) 14  2 = 112 + 16.44 = 128.44, round 128 vaccines.
i.
Imax = d . (OI + LT) + z
ii.
On hand = 34 units
Order quantity = Imax – on hand = 128 – 34 = 94 vaccines.
D = 500 bottles/year
S = $10 per order
H = $1 per bottle per year
B = $10 per bottle per year
2 DS  H  B 
2(500)(10)  1  10 



 = 104.88 round to 105 units
H  B 
1
 10 
a.
Q
b.
 H 
 1 
Qb  Q 
  105 
 = 9.55 round to 10 units
H B
 1  10 
Instructor’s Manual, Chapter 12
12-39
48.
Price = $12.99 per unit
Cost = $6 per unit
Demand (in 1000s)
Prob.
Cumul Prob.
90
.1
.1
100
.2
.3
110
.4
.7
120
.2
.9
130
.1
1.0
Salvage value = $5 per unit
Q?
Ce = cost – salvage value = $6 - $5 = $1
Cs = price - cost = $12.99 – $6 = $6.99
SL = Cs / (Cs + Ce) = $6.99 / ($6.99 + $1) = 0.875
Bec. 0.875 is just smaller than 0.9, the cumul prob. of 120(000) demand, choose Q =120(000).
49.
D = 6.7 units per month
R = $15 per unit
i = 20% of unit cost per year
S = $2 per order
LT = 20 days
30 days in a month
a.
D = 6.7 (12) = 80.4 units
Q
2 DS

iR
2(80.4)( 2) 10.35 round to 10 units

.2(15)
b.
10 / 6.7 = 1.5 months or 45 days
c.
TC(10) = (Q/2)iR + (D/Q)S = (10/2)$3 + (80.4/10)$2 = $15 + $16.08 = $31.08
d.
TC(7) = (7/2)$3 + (80.4/7)$2 = $10.50 + $22.97 = $33.47
TC(7) – TC(10) = $33.47 - $31.08 = $2.39 more expensive per year
e.
ROP = d (LT) = (6.7/30)20 = 4.47 round to 4 units
f.
95% lead time service level
 = 5.14 units per month
ROP?
95% service level  z =1.645 from App. B, Table B.
ROP  d ( LT )  z
LT  6.7(20 / 30)  1.645(5.14) (20 / 30)
= 4.47 + 6.90 = 11.37 round to 11 units
12-40
Operations Management, 5/C/E
50.
LT = 15 days
R1 = $3.85, D1 = 5,767 units per year
R2 = $7.54, D2 = 603 units per year
364 days a year
a.
S = $3.50, s = $0.50, i = 0.20
OI * 
b.
2 ( S  ns )
i DjRj


2(3.50  2  .50)
 0.041
.20(5,767  3.85  603  7.54)
year or 15 days
d1 = 138 per week
 1 = 37 units per week
On hand = 555 units
Service level = 98.5%  z = 2.17 from App B, Table B.
Imax  d1 (OI  LT )  z1 OI  LT  138(15  15) / 7  2.17(37) (15  15) / 7
= 591.43 + 166.22 = 757.67
round to 758 units
Q = Imax – on hand = 758 – 555 = 203 units
c.
Use the forecast demand for the next 30 days = first 4 weeks plus 2/7 of week 5
Imax  d1 (OI  LT )  z 1 OI  LT
= 144.2 + 144.2 + 133.1 + 133.1 + 122(2/7) + 2.17(37) (15  15) / 7
= 589.46 + 166.22 = 755.7 round to 756 units
Instructor’s Manual, Chapter 12
12-41
51.
Forecast D = 27.48 units per month
 = 18.84 units per month
LT = 14 days
R = $1.40 per unit
i = 20% of unit cost per year
S = $1 per order
Assume 30 days in a month
a.
D = 27.48(12) = 329.76
Q
2 DS
2(329.76)(1)

 48.53 round to 49 units
iR
.2(1.40)
b.
49/27.48 = 1.78 months or 53 days
c.
TC(49) = (Q/2)iR + (D/Q)S = (49 2)($.28) + (329.76 49)($1)
= $6.86 + $6.73 = $13.59
d.
Currently Q = 30 units
TC(30) = (30 / 2)($.28) + (329.76 / 30)($1) = $4.20 + $10.99 = $15.19
TC(30) – TC(49) = $15.19 - $13.59 = $1.60 more expensive per year
e.
97.5% service level  z =1.96 from App B, Table B.
ROP  d ( LT )  z LT  ( 27.48)(14 / 30)  1.96(18.84) 14 / 30
= 12.82 + 25.23 = 38.05 round to 38 units
f.
RP = 1 month or 30 days
97.5% service level  z =1.96
Min  d ( RP  LT )  z RP  LT
 27.48(30  14) / 30  1.96(18.84) (30  14) / 30
= 40.30 + 44.72 = 85.02 round to 85 units
Max = Min + EOQ = 85 + 49 = 134 units
12-42
Operations Management, 5/C/E
52.
Forecast D = 7.25 units per month
 = 3.13 units per month
R = $2.48 per unit
i = 20% of unit cost per year
LT = 14 days
Service level = 97.5%  z = 1.96 from App B, Table B.
Assume 30 day months
a.
S = $1 per order
Q
2 DS

iR
2(7.25)(12)(1) 18.73 round to 19 units

.2(2.48)
ROP  d ( LT )  z
b.
LT  (7.25)(14 / 30)  1.96(3.13) 14 / 30
= 3.38 + 4.19 = 7.57 round to 8 units
RP = 1 month or 30 days
Min  d ( RP  LT )  z
RP  LT
 7.25(30  14) / 30  1.96(3.13) (30  14) / 30
= 10.63 + 7.43 = 18.06 round to 18 units
Max = Min + EOQ = 18 + 19 = 37 units
c.
i.
Forecast for D2 = 2.23 per month
R2 = $5.06
S + s = $3
s = $0.5
OI * 
2 ( S  ns )

i DjRj

2(2.50  2  .50)
.20(7.25  12  2.48  2.23  12  5.06)
= 0.3157 year or 113.65 or 114 days
ii. On hand = 13 units
Imax  d (OI  LT )  z OI  LT
 7.25(114  14) / 30  1.96(3.13) (114  14) / 30
= 30.93 + 12.67 = 43.61 round to 44 units
Q = Imax – on hand = 44 – 13 = 31 units
Instructor’s Manual, Chapter 12
12-43
53.
Demand (in 1000s)
Prob.
Cumul Prob.
6
.1
.1
7
.2
.3
8
.4
.7
9
.2
.9
10
.1
1.0
60% hamburger, 40% hot dog
Burger : cost = $2.25, price = $5.00
Hot dog: cost = $1.34, price = $4.00
Salvage value = $0
Q for burgers and hot dogs?
Burgers:
Ce = cost – salvage value = $2.25 - $0 = $2.25
Cs = price - cost = $5.00 – $2.25 = $2.75
SL = Cs / (Cs + Ce) = $2.75 / ($2.75 + $2.25) = .55
Bec. .55 is just smaller than .7 (cumul prob.), choose Q = 8,000 (.6) = 4,800 burgers.
Hot dogs:
Ce = cost – salvage value = $1.34 - $0 = $1.34
Cs = price - cost = $4.00 – $1.34 = $2.66
SL = Cs / (Cs + Ce) = $2.66 / ($2.66 + $1.34) = .665
Bec. .665 is just smaller than .7 (cumul prob.), choose Q = 8,000 (.4) = 3,200 hot dogs.
12-44
Operations Management, 5/C/E
54.
S + s = $23
s = $3
i = .24
LT = 1 week
50-week year
7 days per week
SKU
Annual
demand
Unit cost
1
450
$8
2
2000
12.5
3
200
3.52
4
3000
33.3
demand ×
Cost
$3,600
$25,000
$704
$99,900
Largest Annual demand × Unit cost is for SKU4  m4 = 1
𝑚1 = √𝐷
𝑠
𝐷4 𝑅4
1 𝑅1
𝑚2 = √𝐷
𝑠
𝑆+𝑠
𝐷4 𝑅4
2 𝑅2
𝑚3 = √𝐷
𝑠
𝑆+𝑠
𝐷4 𝑅4
3 𝑅3 𝑆+𝑠
3
= √3,600
3
99,900
23
= √25,000
99,900
23
3 99,900
= √704
23
= 1.90 round to 2
= 0.72 round to 1
= 4.30 round to 4

n
1 

 1 1 1 1
2 S  s
2  20 3     
j 1 m 

56.5
j
 2 1 4 1


OI *   n

.24 23, 600125, 000 4704199, 900
32, 379.84
i
mjDjRj

 j 1
= .04177 years × 350 = 14.62
Instructor’s Manual, Chapter 12
round to 15 days
12-45
55.
Forecast D = 44 cases per month
Price = $120 per case
S = $15 per order
i = .15
LT = 2 days
2 DS

iR
2(44  12)($15)
 29.66
.15($120)
(a)
EOQ 
(b)
TC = (Q/2)(iR) + (D/Q)S = (30/2)(.15)($120) + (44×12/30)($15) = 270 + 264 = $534
(c)
SL = 96%  z = 1.75 from Table B, App B, and
ROP  d ( LT )  z d
(d)
round to 30 cases
d
(monthly) = 16.82 cases,
LT  44(2 / 30)  1.75(16.82) 2 / 30
 2.93  7.60  10.53 round to 11 cases
Fixed Interval model, OI = 2 weeks, SL = 94%  z = 1.555 from Table B. App B,
I max  d (OI  LT )  z d OI  LT  44(14  2) / 30  1.555(16.82) (14  2) / 30
 23.47 19.10  42.57 round to 43 cases
56. ROP for 1–litre bottles = 1 month demand
On hand 1 litre Vodka = 144 cases
D for Vodka = 312 cases per month
S = $73.23
Cost = $29.31 per case
i = .15
a.
D = 312(12) = 3,744
EOQ 
b.
2 DS

iR
2(3,744)($ 73.23)
 353.16
0.15($29.31)
round to 353 cases
d
(monthly) = 98 cases
Supervisor produces for 2 months
SL?
OI = 1 month
LT = negligible
Imax = 2 months’ demand = 2(312) = 624
I max  d (OI  LT )  z d OI  LT
624  312(1  0)  z (98) 1  0
z = 312 / 98 = 3.18  Service level = 0.9993 From Table B, App B.
12-46
Operations Management, 5/C/E
57.
Demand ~ Normal (2,150, 807)
Price = $173
Cost = $130
Salvage value = $115
Q?
Cs = $173 - $130 = $43
Ce = $130 - $115 = $15
Choose Q such that Prob (D ≤ Q) = Cs / (Cs + Ce) = 43 / (43 + 15) = .7414
 z = .65 from Table B, App. B.
Q = 2,150 + .65(807) = 2,674.55 round to 2675 units
58.
D = 10,000 units per year
S = $50 per order
i = .20 per unit per year
Order Quantity Range
1 - 500 units
501 - 750
751 - 1500
1501 - 2000
EOQ 
Unit Price
$29.50
26.87
24.77
23.93
2 DS
2(10,000)($50)

 457.1not in the range (≥ 1501)
iR
.20($23.93)
Try a more expensive price range. Bec. EOQ only decrease as price increases, only EOQ of
most expensive price will be in its range.
EOQ 
2(10,000)($50)
 411.7
.20($29.50)
Round to 412
TC = DP + (D/Q)S + (Q/2)iR
Q
412
501
751
1501
Purchasing cost
10,000($29.50)
= $295,000
10,000($26.87)
= $268,700
10,000($24.77)
= $247,700
10,000($23.93)
= $239,300
Ordering cost
(10,000/412)($50)=
1,213.59
(10,000/501)($50)=
998
(10,000/751)($50)=
665.78
(10,000/1501)($50)=
333.11
Holding cost
(412/2)(.20)($29.50)=
1,215.40
(501/2)(.20)($26.87)=
1,346.19
(751/2)(.20)($24.77)=
1,860.23
(1501/2)(.20)($23.93)=
3,591.89
Total cost
297,429
271,044
250,226
243,225*
Choose Q = 1501
Instructor’s Manual, Chapter 12
12-47
59.
Currently: Q = 100 cases, R = $65 + $4.50 = $69.50.
D ~ 100 cases per month = 1200 per year
LT = 6-8 weeks
Quantity discount: Q = 500, R = $50 + $2.50 = $52.50
Storage space cost = $400 per month
i = 20% per unit per year
S = $50 per order
Q?
TC(100) = (Q/2)iR + (D/Q)S + RD
= (100/2)(.2)($69.50) + (1200/100)($50) + $69.95(1200)
= $695 + $600 + $83,940 = $85,235
TC(500) = (Q/2)iR + (D/Q)S + RD + storage cost
= (500/2)(.2)($52.50) + (1200/500)($50) + $52.50(1200) + $400(12)
= $2,625 + $120 + $63,000 + $4,800 = $70,545
Yes, buy in full truckloads (Q = 500 cases) bec. $70,545 < $85,235.
60.
unit = 50 kg bag of herbicide
Price = $56.93 per unit
Cost = $45.54 per unit
Excess transportation cost for storage after season = $1.09 per unit
i = 10% per year (charge only for .5 year)
10% of shortage will be lost, 90% incur $2.19 per unit expediting cost
Demand
100
400
1500
Prob.
.1
.5
.4
Cumul Prob.
.1
.6
1.0
Q?
Ce = cost of excess transportation + holding cost per unit = $1.09 + (.10/2)($45.54)
= $1.09 + $2.28 = $3.37 per unit
Cs = .10 (price – cost) + .90 (expediting cost)
= .10 ($56.93 – $45.54) +.90 ($2.19) = $1.14 + $1.97 = $3.11
SL = Cs / (Cs + Ce) = $3.11 / ($3.11 + $3.37) = 0.48
Bec. 0.48 is just smaller than 0.6 (cumul prob.), choose Q = 400 units.
12-48
Operations Management, 5/C/E
Answers to CAMECO PROMOTIONAL ITEMS Mini-case:
LT = 2 weeks ~ 0.5 month
OI = 2 months
SL = 70%  z = 0.525 from Table B. App B
a
Executive Sports Bag
I max  d (OI  LT )  z d OI  LT  12(2  0.5)  0.525(4) 2  0.5
= 30 + 3.32 = 33.32 round to 33 units
Q = Imax – on hand = 33 – 0 = 33 units
33 > 25 = min  order 33 units
Retro Stainless Steel Mug
I max  d (OI  LT )  z d OI  LT  15(2  0.5)  0.525(2) 2  0.5
= 37.5 + 1.66 = 39.16 round to 39 units
Q = Imax – on hand = 39 – 4 = 35 units
35 < 60 = min  order 60 units
Lava Pen
I max  d (OI  LT )  z d OI  LT  35(2  0.5)  0.525(6) 2  0.5
= 87.5 + 4.98 = 92.48 round to 92 units
Q = Imax – on hand = 92 – 16 = 76 units
76 > 75 = min  order 76 units
Heavyweight Brushed Cotton Cap
I max  d (OI  LT )  z d OI  LT  60(2  0.5)  0.525(6) 2  0.5
= 150 + 4.98 = 154.98 round to 155 units
Q = Imax – on hand = 155 – 6 = 149 units
149 > 72 = min  order 149 units
b.
Heavyweight Brushed Cotton Cap
D = 60(12) = 720 units per year
i = 12%
S = $50
Quantity range
72-143
144-239
240-575
EOQ 
Unit price
$8.75
$7.95
$7.50
2 DS
2(720)($50)

 282.84
iR
.12($7.50)
round to 283, in range (≥ 240)
Because 283 is in the cheapest range, it follows that it is optimal.
Instructor’s Manual, Chapter 12
12-49
Answers to CAMECO MINE SUPPLIES Mini-case:
Code
20050008
20073791
20126073
20013742
Description
Cable Hanger CAB826
Grinding Wheel 5” ¼” 5/8” All Metal
Pyrolon Coveralls XL Blue w/ HD Elastic
Oil Pressure Gauge 2-1/2” 0-200 PSI
Price
$2.99
$4.57
$8.05
$44.75
Lead time
5 days
6
8
14
Current Min
151
31
4
1
Current Max
250
50
75
3
RP = 3 days
S = $10 per order
i = 20% per year
SL = 97.5%  z = 1.96 from Table B. App B
Cable
Hanger
Month
Grinding
Wheel
Coveralls
Oil
Pressure
Gauge
Jan
0
44
3
0
Feb
150
57
1
0
Mar
75
61
0
0
Apr
100
28
0
0
May
0
102
26
0
Jun
0
53
0
0
Jul
100
97
26
0
Aug
75
27
25
0
Sep
0
27
66
0
Oct
0
74
9
0
Nov
0
31
0
0
Dec
0
25
2
0
500
54.70
626
27.26
158
19.88
0
0.00
Sum =
std dev =
Oil Pressure Gauge 2-1/2” 0-200 PSI
Because this item was not demanded last year, it should not be reordered. In addition, after consultation
with users, it may be dropped as a stocked item.
Grinding Wheel 5” ¼” 5/8” All Metal
Q
Min
12-50
2 DS

iR
2(626)(10) 117.03 round to 117 units

.2( 4.57)
 d ( RP  LT )  z
RP  LT
Operations Management, 5/C/E
 626(3  6) / 360  1.96(27.26) (3  6) / 30
= 15.65 + 29.26 = 44.91 round to 45 units
Max = Min + EOQ = 45 + 117 = 162 units
Current Min (31) is a little too low and current Max (50) is very low.
Cable Hanger CAB826
Q
Min
2 DS

iR
2(500)(10) 129.3 round to 129 units

.2( 2.99)
 d ( RP  LT )  z
RP  LT
 500(3  5) / 360  1.96(54.70) (3  5) / 30
= 11.11 + 55.36 = 66.47 round to 66 units
However, several months had zero demand, and the others had very large demand. Normal
distribution may not appropriate in this case. Looking at the individual demands on the left
below, we can combine the quantities which occurred within 3 (RP) + 5 (LT) = 8 days in order to
get a worst-case total during an 8-day period (table of the right). Given SL = 97.5%, we can omit
the largest sum in the right table as a rare case where stock-out will be acceptable. Therefore, we
use the second largest sum, 50+25 = 75, as the Min. (Note: 75 is not far from 66)
Posting Date
Feb 8
Feb 18
Feb 23
Mar 14
Mar 14
Mar 26
Apr 4
Apr 5
Apr 13
Apr 25
Jul 4
Jul 30
Aug 10
Aug 10
Cable Hanger
Quantity
25
25
100
25
25
25
25
25
25
25
75
25
50
25
Posting Date
Feb 8
Feb 18 + 23
Mar 14 + 14
Mar 26
Apr 4 + 5
Apr 13
Apr 25
Jul 4
Jul 30
Aug 10 + 10
Cable Hanger
Quantity
25
25+100
25+25
25
25 + 25
25
25
75
25
50 + 25
Max = Min + EOQ = 75 + 129 = 204 units
Current Min (151) is too high.
Pyrolon Coveralls XL Blue w/ HD Elastic
Q
2 DS

iR
2(158)(10) 44.3 round to 44 units

.2(8.05)
Instructor’s Manual, Chapter 12
12-51
Min
 d ( RP  LT )  z
RP  LT
 158(3  8) / 360  1.96(19.88) (3  8) / 30
= 4.83 + 23.59 = 28.42 round to 28 units
However, this item also has several months of zero demand. Normal distribution may not be
appropriate in this case. Looking at the individual demands on the left below, we can combine
the quantities which occurred within 3 (RP) + 8 (LT) = 11 days in order to get a worst-case total
during an11-day period (table of the right). Given SL = 97.5%, we can omit the largest sum in
the right table as a rare case where stock-out will be acceptable. Therefore, we use the second
largest sum, 26, as the Min. (Note: 26 is not far from 28)
Posting Date
Jan 6
Jan 21
Feb 14
May 19
May 23
Jul 6
Jul 7
Jul 10
Aug 17
Sep 1
Sep 2
Sep2
Sep 3
Sep 8
Sep 18
Sep 23
Oct 1
Dec 7
12-52
Coveralls
Quantity
2
1
1
1
25
2
10
14
25
2
2
2
6
50
2
2
9
2
Posting Date
Jan 6
Jan 21
Feb 14
May 19 +23
Jul 6 + 7 + 10
Aug 17
Sep 1+2+2+3+8
Sep 18+23
Oct 1
Dec 7
Coveralls
Quantity
2
1
1
1+25
2+10+14
25
2+2+2+6+50
2+2
9
2
Max = Min + EOQ = 26 + 44 = 70 units
Current Min (4) is too low.
Operations Management, 5/C/E