imchap013.docxChapter 12 - Inventory Management
Chapter 13 - Inventory Management
Inventory Management
Teaching Notes
This is a fairly long and important chapter. Important points are:
1. Good inventory management is important for successful organizations.
2. The key inventory management issues are when to order and how much to order.
3. Because all items are not of equal importance, it is necessary to establish a classification system for allocating resources for inventory control.
4. EOQ models answer the question of how much to order. Variations of the basic EOQ model include the quantity discount model and the economic production quantity (EPQ) model.
5. EOQ models tend to be rather robust: even though one or more of the parameters may be only roughly correct, the model can yield a total cost that is close to the actual minimum.
6. ROP models are used to answer the question of when to order. Different models are used, depending on whether demand, lead time, or both are variable.
7. Other models described are the fixed interval model and the single-period model.
8. All of the models in this chapter pertain to independent demand.
The single-period model is used to handle ordering of perishables (e.g., fresh fruits and vegetables, seafood, and cut flowers) as well as items that have a limited useful life (e.g., newspapers and magazines). Analysis of single-period situations generally focuses on two costs: shortage and excess. Shortage costs may include a charge for loss of customer goodwill as well as the opportunity cost of lost sales or unrealized profit per unit. Excess cost pertains to items left over at the end of the period and is the difference between purchase cost and salvage value. There may be costs associated with disposing of excess items, which would make the salvage value negative and hence increase the excess cost per unit.
Answers to Discussion and Review Questions
1. Inventories are held: (1) to meet anticipated customer demand, (2) to smooth production requirements, (3) to decouple operations, (4) to reduce the risk of stockouts, (5) to take advantage of order cycles, (6) to hedge against price increases, (7) to permit operations, and (8) to take advantage of quantity discounts.
2. Effective inventory management requires: (1) a system to keep track of inventory on hand and on order, (2) a reliable forecast of demand that includes an indication of possible forecast error, (3) knowledge of lead times and lead time variability, (4) reasonable estimates of inventory holding costs, ordering costs, and shortage costs, and (5) a classification system for inventory items.
3. The four costs associated with inventories include the following:
(1) Purchase cost – Amount paid to a vendor or supplier to buy the inventory.
(2) Carrying or holding costs – Cost of physically having items in storage. Costs include interest, insurance, taxes, depreciation, picking, and warehousing costs (heat, light, rent, and security).
(3) Ordering costs – Cost of ordering and receiving purchased items, or the setup costs for items produced in-house.
(4) Shortage costs – Result when demand exceeds supply and can include the opportunity cost of not making a sale, loss of customer goodwill, late charges, backorder costs, and the cost of lost production or downtime.
4. RFID (Radio Frequency Identification) tags offer the following benefits:
(1) They provide real-time information to network-connected RFID readers, which increase a company’s ability to track inventory and sales, and to process shipping containers, parts in warehouses, and items on supermarket shelves.
(2) The real-time information that they provide should enable a company and its suppliers to react more quickly to market changes.
(3) They do not require line-of sight to be scanned (bar codes do).
(4) They can be used to monitor temperature of agricultural products during shipping and to store information on cultivation history, whether the product is organic, and whether fertilizers or chemicals have been used to grow the product.
Risks associated with RFID tags stem from privacy concerns. It is feared that computer pirates will figure out security controls and be able to scan shoppers’ merchandise and determine what they have bought.
5. It may be inappropriate to compare the inventory turnover ratios of companies in different industries because the profit margins, production times, and the time to sell products may differ across industries. The higher the profit margins, the lower the acceptable number of inventory turns. A product that takes a long time to manufacture or sell will have a low turnover rate. In addition, the material delivery lead times may vary between different industries. The higher the variability of lead time and the longer the lead time, the greater the need for inventory and the lower turnover will be. Industries with higher forecast accuracies have less of a need for inventories, which leads to higher turnover also.
6. Managers can use the results of A-B-C analysis to determine how to allocate control efforts to inventory items: A items should receive close attention through frequent reviews of amounts on hand and tight control over withdrawals to ensure that customer service levels are attained, C items would receive only loose control, and B items should receive control that lies between that of A and C items.
7. a. The major assumptions of the EOQ model include:
(1) Only one product is involved.
(2) Annual demand requirements are known.
(3) Demand is spread evenly throughout the year so that the demand rate is reasonably constant.
(4) Lead time is known and constant.
(5) Each order is received in a single delivery.
(6) There are no quantity discounts.
b. Despite issues with values in the EOQ formula being at best educated guesses, the total cost curve is relatively flat near the EOQ, so that there is a “zone” 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 estimation errors. Also, errors may cancel each other out.
8. As the carrying cost increases, holding inventory becomes more expensive. Therefore, to avoid higher inventory carrying costs, the company will order more frequently in smaller quantities because ordering smaller quantities will lead to carrying less inventory.
9. Safety stock is inventory held in excess of expected demand to reduce the risk of stockout presented by variability in either lead time or demand rates.
10. a. Safety stock would be large when large variations in lead time and/or usage are present.
b. Safety stock would be small when small variations in lead time and/or usage are present.
c. Safety stock would be zero when lead time and usage are constant, or when the service level is 50 percent (and hence, z = 0).
11. Service level can be defined in a number of ways. The text focuses mainly on “the probability that demand will not exceed supply during lead time, i.e., that the amount of stock on hand will be sufficient to meet demand.” Other definitions relate to the percentage of cycles per year without a stockout, or the percentage of annual demand satisfied from inventory.
Increasing the service level requires increasing the amount of safety stock.
12. The A-B-C approach refers to the classification of inventory items according to some measure of importance, usually annual dollar value, and allocating control efforts on that basis. Although annual dollar value may be the primary factor, a manager could consider other factors when making exceptions, e.g., the risk of obsolescence, the risk of a stockout, the distance of a supplier, etc.
13. In effect, this situation is a “quantity discount” case with a time dimension. Hence, buying larger quantities will result in lower annual purchase costs, lower ordering costs (due to fewer orders), but increased carrying costs. Because it is unlikely that the compressor supplier announces price increases far in advance, the purchasing agent will have to develop a forecast of future price increases to use in determining order size. Unlike the standard discount approach, the agent may opt to use trial and error to determine the best order size, taking into account the three costs (carrying, ordering and purchasing). In any event, it is reasonable to expect that a larger order size would be more appropriate; although obsolescence also may be a factor.
14. A decrease in setup time will cause a decrease in the numerator of the formula for the economic run quantity (Economic Production Quantity). This will lead to a decrease in the economic run quantity. When the economic run quantity (Qp) decreases, the maximum inventory (Imax) and the average inventory (Imax / 2) will decrease. This decrease in average inventory would be beneficial because Annual Carrying (Holding) Cost would decrease.
15. The single-period model is used to handle ordering of perishables (e.g., fresh fruits, vegetables, seafood, & cut flowers) and items that have a limited useful life (e.g., newspapers, magazines, & spare parts for specialized equipment) and when items are not carried over from one period to the next. It is appropriate to use this model when the previous conditions are met and when shortage and excess costs can be estimated accurately.
16. The optimal stocking level can be less than the expected demand when excess costs are high and shortage costs are low.
17. A company can reduce the need for inventories by:
a. using standardized parts for multiple products
b. improving demand forecasting
c. using preventive maintenance on equipment and machines
d. reducing supplier delivery lead times and increasing delivery reliability
e. improving the relationships and information sharing in the supply chain
f. restructuring the supply chain so that the supplier holds the inventory
g. reducing production lead time by using more efficient manufacturing methods
h. developing simpler product designs with fewer parts.
Taking Stock
1. a. If we buy additional amounts of a particular good to take advantage of quantity discounts, then we will save money on purchasing cost per unit and annual purchasing costs of the item. We also will save on ordering cost because we will place fewer orders per year. However, as a result of ordering larger quantities with fewer orders, we will have to carry larger inventory in stock, which in turn will result in an increase in inventory holding cost.
b. If we treat holding cost as a percentage of the unit price, then as the unit price increases, so will the holding cost. As a result, if we are using the EOQ approach, we will place smaller orders, resulting in reduced average inventory. On the other hand, if we use a constant amount of a holding cost, the inventory decisions will not be affected by the changes in unit cost (price) of the item.
c. Conducting a cycle count once a quarter instead of once a year will result in more frequent counting, which will result in an increase in labor and overhead costs. However, the more frequent counting also would lead to fewer errors in inventory accuracy and more timely detection of errors, which in turn would lead to timely deliveries to customers, less work in process inventory, more efficient operations, improved customer service, and greater assurance of material availability.
2. When making inventory decisions involving holding costs, setting inventory levels, and deciding on quantity discount purchases, the materials manager, the plant manager, the production planning and control manager, the purchasing manager, and in some cases, the planners who work in production planning and control and the buyers should be involved. Accounting and/or finance might be needed for limited assistance in estimating costs. The level and the nature of involvement of all parties will depend on the organizational structure of the company and the type of product being manufactured or purchased.
3. Technology has had a tremendous impact on inventory management. The utilization of bar coding and RFID tags has reduced the cost of taking physical inventory and has enabled real time updating of inventory records. The satellite control systems that are available in trucks and automobiles have enabled companies to determine and track the location of in-transit inventory.
Critical Thinking Exercises
1. The expansion of menu offerings provides fast food companies with a competitive edge in terms of improving customer satisfaction and service. However, it has also complicated inventory management at a company. There are more ingredients and inventory items to order and to control. This means higher labor costs in terms of placing orders, increased storage facility needs, and increased need for coordinating shipments from suppliers so that transportation cost and performance are optimized. Increasing the variety of items on the menu will also cause problems with forecasting. Because there will be more items on the menu, it is likely that the demand for current menu items will decrease. Forecasts for all items will need to be revised. If we are not able to estimate this possible decrease, then the forecasting problem will result in excess or insufficient inventory levels.
2. A supermarket manager could evaluate criticalness of an inventory shortage by answering the following questions:
a. How important is the item? For example, does it relate to a holiday or other important events, e.g., a graduation?
b. Are comparable substitutes readily available within the manager’s supermarket?
c. What alternatives are available at other supermarkets?
d. Is this an occasional occurrence, or indicative of a larger, perhaps ongoing, problem?
3. The relevant considerations related to the purchase of stamps include:
a. How many stamps does he have now? Does he know how many he has? If so, how many?
b. What is his usage rate or current need for stamps?
c. What else does he need the cash for today?
d. Can he get more money at a bank or an ATM?
e. How long will it be before he will return to the post office?
f. Will the post office be closed for a holiday or a Sunday?
g. Can he buy stamps elsewhere in case he runs low?
h. How convenient is it for him to visit the post office?
i. Can he purchase “forever stamps” and temporarily avoid a price increase?
4. Student answers will vary. Some possible answers are provided below:
a. Intentionally over-estimating or under-estimating any inventory costs would violate the Virtue Principle.
b. If a buyer purchased two years’ worth of an item to decrease purchasing costs, this action would violate the Utilitarian Principle due to the increase in carrying costs.
Solutions
1. a. Given:
Determine an A-B-C classification for the following items:
|
Item |
Unit Cost |
Annual Volume (00) |
|
1 |
$100 |
25 |
|
2 |
$80 |
30 |
|
3 |
$15 |
60 |
|
4 |
$50 |
10 |
|
5 |
$11 |
70 |
|
6 |
$60 |
85 |
Step 1:
Determine the Annual Dollar Value (Unit Cost * Annual Volume) for each item and the sum of the individual Annual Dollar Values.
|
Item |
Unit Cost |
Annual Volume (00) |
Annual Dollar Value |
|
1 |
$100 |
25 |
$2,500 |
|
2 |
80 |
30 |
2,400 |
|
3 |
15 |
60 |
900 |
|
4 |
50 |
10 |
500 |
|
5 |
11 |
70 |
770 |
|
6 |
60 |
85 |
5,100 |
|
|
|
|
12,170 |
Step 2:
Arrange the items in descending order based on Annual Dollar Values. Determine the A, B, and C items. Then, determine the percentage of items and the percentage of Annual Dollar Value for each category (round to two decimals).
|
Item |
Annual Dollar Value |
Category |
Percentage of Items |
Percentage of Annual Dollar Value |
|
6 |
$5,100 |
A |
16.67% [(1/6)*100] |
41.91% [($5,100/$12,170)*100] |
|
1 |
2,500 |
B |
33.33% [(2/6)*100] |
40.26% [($4,900/$12,170)*100] |
|
2 |
2,400 |
|
|
|
|
3 |
900 |
C
|
50.00% [(3/6)*100] |
17.83% [($2,170/$12,170)*100] |
|
5 |
770 |
|
|
|
|
4 |
500 |
|
|
|
|
|
12,170 |
|
100.00% |
100.00% |
b. Given:
D = 4,500, S = $36, and H = $10.
Find the EOQ (round to an integer value):
c. Given:
D = 18,000/year, S = $100, H = $40 per unit per year, p = 120 units per day, and u = 90 units/day.
Find the economic production quantity (EPQ) (round to an integer value):
2. a. Given:
The following table contains figures on the monthly volume and unit costs for a random sample of 16 items. Develop an A-B-C classification for these items:
|
Item |
Unit Cost |
Usage |
|
K34 |
$10 |
200 |
|
K35 |
25 |
600 |
|
K36 |
36 |
150 |
|
M10 |
16 |
25 |
|
M20 |
20 |
80 |
|
Z45 |
80 |
200 |
|
F14 |
20 |
300 |
|
F95 |
30 |
800 |
|
F99 |
20 |
60 |
|
D45 |
10 |
550 |
|
D48 |
12 |
90 |
|
D52 |
15 |
110 |
|
D57 |
40 |
120 |
|
N08 |
30 |
40 |
|
P05 |
16 |
500 |
|
P09 |
10 |
30 |
Step 1:
Determine the Annual Dollar Value (Unit Cost * Usage) for each item and the sum of the individual Annual Dollar Values.
|
Item |
Unit Cost |
Usage |
Annual Dollar Value |
|
K34 |
$10 |
200 |
$2,000 |
|
K35 |
25 |
600 |
15,000 |
|
K36 |
36 |
150 |
5,400 |
|
M10 |
16 |
25 |
400 |
|
M20 |
20 |
80 |
1,600 |
|
Z45 |
80 |
200 |
16,000 |
|
F14 |
20 |
300 |
6,000 |
|
F95 |
30 |
800 |
24,000 |
|
F99 |
20 |
60 |
1,200 |
|
D45 |
10 |
550 |
5,500 |
|
D48 |
12 |
90 |
1,080 |
|
D52 |
15 |
110 |
1,650 |
|
D57 |
40 |
120 |
4,800 |
|
N08 |
30 |
40 |
1,200 |
|
P05 |
16 |
500 |
8,000 |
|
P09 |
10 |
30 |
300 |
|
|
|
|
94,130 |
Step 2:
Arrange the items in descending order based on Annual Dollar Values. Determine the A, B, and C items. Then, determine the percentage of items and the percentage of Annual Dollar Value for each category (round to two decimals).
|
Item |
Annual Dollar Value |
Category |
Percentage of Items |
Percentage of Annual Dollar Value |
|
F95 |
$24,000 |
A |
18.75% [(3/16)*100] |
54.83% [($55,000/$94,130)*100] |
|
Z45 |
16,000 |
|
|
|
|
K35 |
15,000 |
|
|
|
|
P05 |
8,000 |
B |
31.25% [(5/16)*100] |
31.55% [($29,700/$94,130)*100] |
|
F14 |
6,000 |
|
|
|
|
D45 |
5,500 |
|
|
|
|
K36 |
5,400 |
|
|
|
|
D57 |
4,800 |
|
|
|
|
K34 |
2,000 |
C |
50.00% [(8/16)*100] |
10.02% [($9,430/$94,130)*100] |
|
D52 |
1,650 |
|
|
|
|
M20 |
1,600 |
|
|
|
|
F99 |
1,200 |
|
|
|
|
N08 |
1,200 |
|
|
|
|
D48 |
1,080 |
|
|
|
|
M10 |
400 |
|
|
|
|
P09 |
300 |
|
|
|
|
|
94,130 |
|
100.00% |
100.00% |
b. Given:
Determine an A-B-C classification for the following items:
|
Item |
Usage |
Unit Cost |
|
4021 |
90 |
$1,400 |
|
9402 |
300 |
12 |
|
4066 |
30 |
700 |
|
6500 |
150 |
20 |
|
9280 |
10 |
1,020 |
|
4050 |
80 |
140 |
|
6850 |
2,000 |
10 |
|
3010 |
400 |
20 |
|
4400 |
5,000 |
5 |
Step 1:
Determine the Annual Dollar Value (Usage * Unit Cost) for each item and the sum of the individual Annual Dollar Values.
|
Item |
Usage |
Unit Cost |
Annual Dollar Value |
|
4021 |
90 |
$1,400 |
$126,000 |
|
9402 |
300 |
12 |
3,600 |
|
4066 |
30 |
700 |
21,000 |
|
6500 |
150 |
20 |
3,000 |
|
9280 |
10 |
1,020 |
10,200 |
|
4050 |
80 |
140 |
11,200 |
|
6850 |
2,000 |
10 |
20,000 |
|
3010 |
400 |
20 |
8,000 |
|
4400 |
5,000 |
5 |
25,000 |
|
|
|
|
228,000 |
Step 2:
Arrange the items in descending order based on Annual Dollar Values. Determine the A, B, and C items. Then, determine the percentage of items and the percentage of Annual Dollar Value for each category (round to two decimals).
|
Item |
Annual Dollar Value |
Category |
Percentage of Items
|
Percentage of Annual Dollar Value |
|
4021 |
$126,000 |
A |
11.11% [(1/9)*100] |
55.26% [$126,000/$228,000)*100] |
|
4400 |
25,000 |
B
|
33.33% [(3/9)*100] |
28.95% [$66,000/$228,000)*100] |
|
4066 |
21,000 |
|
|
|
|
6850 |
20,000 |
|
|
|
|
4050 |
11,200 |
C
|
55.56% [(5/9)*100] |
15.79% [$36,000/$228,000)*100] |
|
9280 |
10,200 |
|
|
|
|
3010 |
8,000 |
|
|
|
|
9402 |
3,600 |
|
|
|
|
6500 |
3,000 |
|
|
|
|
|
228,000 |
|
100.00% |
100.00% |
c. Determine the percentage of items in each category and the annual dollar value for each category.
Reference table above.
3. Given:
D = 1,215 bags per year
S = $10
H = $75
Note: Round the EOQ to an integer value, but round any other values to a maximum of two decimals.
a. Determine the EOQ:
b. Determine the average inventory:
Q/2 = 18/2 = 9 bags
c. Determine the number of orders per year:
d. Determine the total cost of ordering and carrying flour:
TC = Carrying cost + Ordering cost
e. Assuming that holding cost per bag increases by $9/bag/year, what would happen to total cost?
New H = $75 + $ 9 = $84.
Increase in cost = $1,428.71 – $1,350 = $78.71 per year
4. Given:
D = 40/day x 260 days/yr. = 10,400 boxes
S = $60. H = $30.
Note: Round the EOQ to an integer value, but round any other values to a maximum of two decimals.
a. Determine the EOQ:
b. Determine total cost:
TC = Carrying cost + Ordering cost
c. Yes, annual ordering and carrying costs always are equal at the EOQ (except when rounding).
d. Determine the total cost for Q = 200 and compare to current total cost:
$6,120 – $6,118.82 = $1.18 higher per year for Q = 200 (this should be acceptable).
5. Given:
D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr.
C = $2. H = (.30)($2) = $.60/unit/year
S = $20
Note: Round the EOQ to an integer value, but round any other values to a maximum of two decimals.
a. Determine the additional annual cost for using Q = 1,500:
Step 1:
Determine total cost for Q = 1,500.
Step 2:
Determine EOQ.
Step 3:
Determine total cost for Q = 775.
Step 4:
Determine annual savings from using the EOQ.
$570 – $464.76 = $105.24.
b. The benefit of using the EOQ is that about one half of the storage space would be needed.
6. Given:
D = 12 * 800 = 9,600
H = .35($10) = $3.50 per crate per year
S = $28
Note: Round the EOQ to an integer value, but round any other values to a maximum of two decimals.
Step 1:
Determine current total cost for Q = 800 (the manager orders once per month).
Step 2:
Determine EOQ, total cost for EOQ, and annual savings from using the EOQ.
Savings per year from using EOQ = $1,736 – $1,371.71 = $364.29
7. Given:
Demand is projected to be 600 units for the first half of the year and 900 units for the second half. The monthly holding cost is $2 per unit, and it costs an estimated $55 to process an order.
a. Assuming that monthly demand will be level during each six-month period, determine an order size that will minimize the sum of ordering and carrying costs for each six-month period:
Note: We will solve this problem using months, rather than a year, as the period.
First six-month period:
d = monthly demand = 600 / 6 = 100 & H = $2.00 per unit per month.
Second six-month period:
D = monthly demand = 900 / 6 = 150 & H = $2.00 per unit per month.
b. We can use the EOQ only if demand is level (stable).
c. If the vendor is willing to offer a discount of $10 per order for ordering in multiples of 50 units (e.g., 50, 100, 150), would you advise the manager to take advantage of the offer in either six-month period? If so, what order size would you recommend?
First six-month period:
d = monthly demand = 600 / 6 = 100, H = $2.00 per unit per month, S = $55, & EOQ = 74.
Monthly TC (Q = 74):
With discount of $10, S = $55 – $10 = $45:
Monthly TC (Q = 50):
Monthly TC (Q = 100):
Monthly TC (Q = 150):
Conclusion: Yes, the manager should take advantage of the offer and order Q = 50 units during this six-month period.
Second six-month period:
d = monthly demand = 900 / 6 = 150, H = $2.00 per unit per month, S = $55, & EOQ = 91.
Monthly TC (Q = 91):
With discount of $10, S = $55 – $10 = $45:
Monthly TC (Q = 50):
Monthly TC (Q = 100):
Monthly TC (Q = 150):
Conclusion: Yes, the manager should take advantage of the offer and order Q = 100 units during this six-month period.
8. Given:
d = 27,000 jars per month
H = $0.18 per jar per month
S = $60
Company operates 20 days a month
Current Q = 4,000
Note: Round the EOQ to an integer value, but round any other values to a maximum of two decimals.
a. What penalty is the company incurring by its present order size?
Step 1:
Determine current monthly cost for Q = 4,000.
Step 2:
Determine EOQ, total cost for EOQ, and monthly savings from using the EOQ.
Savings per month from using EOQ = $765 – $763.68 = $1.32.
Conclusion: Penalty from placing orders of Q = 4,000 = $1.32 per month.
b. The manager would prefer ordering 10 times each month (every other day) but would have to justify any change in order size. One possibility is to simplify order processing to reduce the ordering cost. What ordering cost would enable the manager to justify ordering every other day?
Using the current Q = 4,000, total monthly cost = $765.
If the manager orders 10 times per month, Q = 27,000 / 10 = 2,700.
Set TC (Q = 2,700) = $765 and solve for S:
S = $522 / 10
S = $52.20 (round to two decimals).
This is the order cost that would enable the manager to justify ordering every other day.
9. Given:
p = 5,000 hotdogs/day
u = 250 hotdogs/day
Factory operates 300 days per year
D = 250 * 300 = 75,000 hotdogs per year
S = $66
H = $0.45 per hotdog per year
Note: Round Qp to an integer value, but round any other values to a maximum of two decimals.
a. Find the optimal run size:
b. Number of runs per year:
D / Qp = 75,000 / 4,812 = 15.59 runs per year
c. Days to produce the optimal run quantity:
Qp / p = 4,812 / 5,000 = 0.96 days
10. Given:
A chemical firm produces 100-pound bags. Demand for the product = 20 tons per day. The capacity = 50 tons per day. Setup cost = $100, and storage and handling costs = $5 per ton a year. The firm operates 200 days a year. Note: 1 ton = 2,000 pounds.
p = 50 tons per day * 2,000 pounds per ton = 100,000 pounds per day = 100,000 pounds per day / 100 pounds per bag = 1,000 bags per day
u = 20 tons per day * 2,000 pounds per ton= 40,000 pounds per day = 40,000 pounds per day / 100 pounds per bag = 400 bags per day
D = 400 bags per day * 200 days per year = 80,000 bags per year
S = $100
H = $5 per ton per year = $5 per ton per year / 20 bags per ton = $0.25 per bag per year
Note: Round Qp to an integer value, but round any other values to a maximum of two decimals.
e. S = $25:
Savings when S = $25 = $1,549.19 – $774.60 = $774.59 per year.
11. Given:
Assembly takes place 5 days a week, 50 weeks a year. It will take a full day to get the machine ready for a production run of the component for the new product.
S = $300
H = $10.00
p = 200/day
u = 80/day
D = 20,000 (250 days * 80/day)
Note: Round Qp to an integer value, but round any other values to a maximum of two decimals.
a. Optimal run quantity to minimize total annual costs:
b. Days to produce the optimal run quantity:
c. Average amount of inventory:
units
units
d. The manager would like to run another job between runs of the component for the new product and needs a minimum of 10 days per cycle (including setup) for the other job:
How much time is available to run the other job? The job must be finished during the pure consumption time for the component for the new product. The end of the pure consumption time is when inventory of the component for the new product falls to 0 units. If the other job takes longer than the pure consumption time, we will run out of inventory of the component for the new product.
This is the time between starting production runs of the component for the new product.
Plugging in values and solving for Pure Consumption Time:
Conclusion: There will not be enough time to run the other job because the other job requires 10 days, which is .39 days (10 – 9.61) days too many.
e. Three options that the manager could consider that will allow this other job to be performed:
1) Try to shorten the setup time of the component for the new product.
2) Increase the run quantity of the component for the new product to allow a longer time between runs, i.e., run the component less often.
3) Reduce the run size of the other job.
f. Determine the additional units to produce of the component for the new product and the increase in total annual cost from this new Q:
We know the following:
The Pure Consumption Time for the component for the new product must equal 10 days to allow the other job to be run.
p = 200/day, u = 80/day, and Pure Consumption Time = 10 days.
Plugging in values and solving for Qp:
Using the least common denominator:
The additional units per run = 1,467 – 1,414 = 53 units per run.
Increase in total cost:
Q = 1,467:
units
Q = 1,414:
Increase = $8,490.98 – $8,485.28 = $5.70 per year
12. Given:
p = 800 units per day
u = 300 units per day
Q = 2,000 units per batch
Company operates 250 days a year
a. Number of batches of heating elements per year:
b. Amount of inventory after 2 days of production:
The number of units produced in 2 days = (2 days)(800 units/day) = 1600 units
The number of units used in 2 days = (2 days) (300 units per day) = 600 units
Inventory build up after the first 2 days of production = 1,600 – 600 = 1,000 units
Current inventory of the heating unit = 0 units
Total inventory after the first 2 days of production = Beginning Inventory + Inventory Buildup after 2 Days of Production = 0 + 1,000 = 1,000 units.
c. Average Inventory:
d. The other component requires 4 days (including setup). Setup time for the heating element = 0.5 days. Is there enough time to run the other component between batches of heating elements?
How much time is available to run the other component? The other component must be finished during the pure consumption time for the heating element. The end of the pure consumption time is when inventory of the heating element falls to 0 units. If the other component takes longer than the pure consumption time, we will run out of inventory of the heating element.
This is the time between starting production runs of the heating element.
Plugging in values and solving for Pure Consumption Time:
Conclusion: There will not be enough time to run the other component because the other component requires 4 days, which is .33 (4 – 3.67) days too many.
13. Given:
D = 18,000 boxes/year
S = $96
H = $.60/box/year
Price Schedule:
|
Number of Boxes |
Price per Box (P) |
|
1,000-1,999 |
$1.25 |
|
2,000-4,999 |
$1.20 |
|
5,000-9,999 |
$1.15 |
|
10,000+ |
$1.10 |
a. Determine the optimal order quantity (round to an integer value):
Step 1:
Compute the common minimum point.
This quantity is feasible in the range 2000-4,999.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit costs.
Conclusion: Optimal order quantity = 5,000 boxes.
b. Determine number of orders per year:
14. Given:
D = 25 stones/day * 200 days/year = 5,000 stones/year
S = $48
Price Schedule:
|
Number of Stones |
Price per Stone (P) |
|
1-399 |
$10 |
|
400-599 |
$9 |
|
600+ |
$8 |
a. H = $2. Determine the optimal order quantity:
Step 1:
Compute the common minimum point.
This quantity is feasible in the range 400-599.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit costs.
Conclusion: Optimal order quantity = 600 stones.
b. H = 30% of unit cost
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find a feasible minimum point.
Minimum point P = $8:
Minimum point P = $9:
Step 2:
Compare the total cost at Q = 422 to Q = 600.
Conclusion: Optimal order quantity = 600 stones.
c. Lead time = 6 working days. Determine ROP:
ROP = 25 stones/day * 6 days = 150 stones.
15. Given:
D = 4,900
S = $50
H = 40% of purchase cost
Price Schedule:
|
Range |
Price per Unit (P) |
|
1-999 |
$5.00 |
|
1,000-3,999 |
$4.95 |
|
4,000-5,999 |
$4.90 |
|
6,000+ |
$4.85 |
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find a feasible minimum point.
Minimum point P = $4.85:
Minimum point P = $4.90:
Minimum point P = $4.95:
Minimum point P = $5.00:
Step 2:
Compare the total cost at Q = 495 to Q = 1,000, 4,000, & 6,000.
Conclusion: Optimal order quantity = 495 units. Note: The total cost for 1,000 units is only $.05 different.
16. Given:
D = 800 * 12 = 9,600
S = $40
H = 25% of purchase cost
Price Schedule Supplier A:
|
Range |
Price per Unit (P) |
|
1-199 |
$14.00 |
|
200-499 |
$13.80 |
|
500+ |
$13.60 |
Price Schedule Supplier B:
|
Range |
Price per Unit (P) |
|
1-149 |
$14.10 |
|
150-349 |
$13.90 |
|
350+ |
$13.70 |
We need to find the optimal quantity for each supplier and select the supplier with the minimum cost.
Supplier A:
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find a feasible minimum point.
Minimum point P = $13.60:
Minimum point P = $13.80:
Step 2:
Compare the total cost at Q = 472 to Q = 500.
Conclusion: Optimal order quantity from Supplier A: 500 units with TC = $132,718.
Supplier B:
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find a feasible minimum point.
Minimum point P = $13.70:
Step 2:
Compute total cost for Q = 474.
Conclusion: Optimal order quantity from Supplier B: 474 units with TC = $133,141.86.
Compare total cost for Q = 500 from Supplier A to total cost for Q = 474 from Supplier B:
Conclusion: Optimal order quantity = 500 units from Supplier A.
17. Given:
D = 3,600 boxes per year
Q = 800 boxes (recommended)
S = $80/order
H = $10/box/year
Price Schedule:
|
Range |
Price per Unit (P) |
|
1-199 |
$1.20 |
|
200-800 |
$1.10 |
|
801+ |
$1.00 |
If the firm decides to order 800 boxes, the total cost is computed as follows:
If the firm decides to order 801 boxes, the total cost is computed as follows:
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. Therefore, when the quantity changes from 800 to 801, we shift to a different total cost curve.
Conclusion: The order quantity of 801 is preferred to the order quantity of 800 because the total cost for Q = 801 is lower.
Determine the optimal Q:
Step 1:
Compute the common minimum point.
This quantity is feasible in the range 200-800.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit costs.
Conclusion: Optimal order quantity = 240 boxes.
18. Given:
Daily usage = 800 feet/day & lead time = 6 days.
Service level desired: 95%.
Stockout risk for various levels of safety stock:
.10 for 1,500 feet; .05 for 1,800 feet; .02 for 2,100 feet; and .01 for 2,400 feet.
Stockout risk should = 1.00 – .95 = .05. This requires a safety stock of 1,800 feet.
ROP = Expected demand during lead time + Safety stock = EDDLT + SS =
(800 feet/day x 6 days) + 1,800 feet = 6,600 feet.
19. Given:
EDDLT = 300 units
dLT = 30 units
a. Determine ROP for 1% risk of stockout:
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .01 = 0.99.
The closest probability is .9901, which corresponds to z = 2.33.
ROP = (round up)
b. SS = 69.9 = 70 units (round up)
c. Stockout risk of 2% is > 1%. Greater stockout risk = smaller z = less safety stock &
smaller ROP.
20. Given:
EDDLT = 600 lb.
dLT = 52 lb.
Stockout risk = 4%
a. Determine SS for 4% risk of stockout.
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .04 = 0.96.
The closest probability is .9599, which corresponds to z = 1.75.
SS = units
b. ROP = EDDLT + SS = 600 + 91 = 691 units
c. With no safety stock, stockout risk is 50% (z = 0.00).
21. Given:
= 21 gallons/week
d = 3.5 gallons/week
LT = 2 days & the dairy is open 7 days a week
Service level= 90%
Hint: Work in terms of weeks
a. Determine ROP and days of supply on hand:
Using Appendix B, Table B, we look for the z value corresponding to .90.
The closest probability is .8997, which corresponds to z = 1.28.
Days of Supply = 9 / (21/7) = 9 / 3 = 3 days of supply on hand at the ROP
b. OI = 10 days & 8 gallons are on hand at the order time:
Determine the probability of experiencing a stockout before this order arrives:
Risk of a stockout at the end of the initial lead time:
Using Formula 13-13, set the ROP equal to the quantity on hand when the order is placed and solve for z:
2 = 1.871z
z = 2 / 1.871
z = 1.07 (round to two decimals)
From Appendix B, Table B, the lead time service level is .8577.
Risk of stockout before this order arrives = 1 - .8577 = .1423 = 14.23%.
c. The manager is using the ROP model described in part a. One day after placing an order with the supplier, the manager receives a call that the order will be delayed and will arrive 3 days from the initial order date. Two gallons have been sold since the order was placed (one day ago).
Determine the probability of a stockout:
ROP = 9 gallons.
After one day, quantity on hand = 9 – 2 = 7 gallons.
Determine the probability of experiencing a stockout before this order arrives (in 2 days):
Risk of a stockout at the end of the initial lead time:
Using Formula 13-13, set the ROP equal to the quantity on hand 1 day after the order was placed and solve for z:
1 = 1.871z
z = 1 / 1.871
z = 0.53 (round to two decimals)
From Appendix B, Table B, the lead time service level is .7019.
Risk of stockout before this order arrives = 1 - .7019 = .2981 = 29.81%.
22. Given:
= 30 gallons/day
ROP = 170 gallons
SS = 50 gallons and provides a stockout risk of 9%
Step 1:
Solve for the standard deviation of demand during the lead time.
We know that SS = 50.
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .09 = .91.
The closest probability is .9099, which corresponds to z = 1.34.
Plug in values and solve for :
gallons (round to three decimals)
Step 2:
Determine the SS.
Stockout risk = 3%.
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .03 = .97.
The closest probability is .9699, which corresponds to z = 1.88.
(round up)
23. Given:
d = 85 boards/day
ROP = 625 boards
LT = 1.1 days
Determine the probability of a stockout:
625 = (85 x 6) + z (85) (1.1)
625 = 510 + 93.5z
115 = 93.5z
z = 115 / 93.5
z = 1.23 (round to two decimals)
Using Appendix B, Table B, we find a probability of .8907.
The risk of a stockout = 1 - .8907 = .1093 = 10.93%.
24. Given:
Service level = 96%
d = 2 units/day
LT = 1 day
a. Determine the ROP:
Using Appendix B, Table B, we look for the z value corresponding to .96.
The closest probability is .9599, which corresponds to z = 1.75.
ROP = 48 + 22.14
ROP = 70.14 = 71 units (round up)
b. The model might not be appropriate if seasonality were present because during the busy times of the year, the ROP would be set too low (causing stockouts) and during the slow times of the year, the ROP would be set too high (causing excess inventory).
25. Given:
LT = 4 x (1 – 0.25) = 4 x 0.75 = 3 days
S = $30
D = 4,500 gallons
H = $3
360 days/year
d = 2 gallons/day
Price List:
|
Quantity |
Unit Price |
|
1-399 |
$2.00 |
|
400-799 |
$1.70 |
|
800+ |
$1.62 |
a. Determine the optimal order quantity:
Step 1:
Compute the common minimum point.
This quantity is feasible in the range 1-399.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit costs.
Conclusion: Optimal order quantity = 400 gallons.
b. Acceptable stockout risk = 1.5%. Determine ROP:
Using Appendix B, Table B, we look for the z value corresponding to 1 - .015 = .985:
z = 2.17.
26. Given:
= 5 boxes/week
d = .5 boxes/week
LT = 2 weeks
S = $2
H= $.20/box/ year
a. Assuming a 52-week year, determine the EOQ:
D = 52 x 5 = 260
b. If ROP = 12, determine risk of a stockout:
Plugging in values and solving for z:
12 = 10 + .707z
2 = .707z
z = 2 / .707 = 2.83 (round to two decimals)
From Appendix B, Table B, the lead time service level is .9977.
Risk of stockout = 1 - .9977 = .0023 = .23%.
c. OI = 7 weeks. Determine the risk of running out before this order arrives (Q = 36) if the copy center orders when amount on hand = 12:
Use Formula 13-13 and solve for z:
Plugging in values and solving for z:
z = 2 / .707 = 2.83 (round to two decimals)
From Appendix B, Table B, the lead time service level is .9977.
Risk of stockout = 1 - .9977 = .0023 = .23%.
27. Given:
d = 10 lb./day
LT = 1 day
Determine the ROP that would provide a stockout risk of 10%:
Service Level = 1 - .10 = .90.
Using Appendix B, Table B, we look for the z value corresponding to .90.
The closest probability is .8997, which corresponds to z = 1.28.
(round up)
28. Given:
d = 2 rolls/day
LT = 3 days
Supermarket is open 360 day a year
S = $1
H = $.40
a. Determine the EOQ:
D = 10 x 360 = 3,600
b. Determine the ROP that will provide a service level of 96%:
Using Appendix B, Table B, we look for the z value corresponding to .96.
The closest probability is .9599, which corresponds to z = 1.75.
29. Given:
D = 1,200 cases
S = $40 per order
H = $3 per case per year
Service level = 99%
a. Determine the optimal order quantity:
(round to an integer value)
b. Determine the level of safety stock if lead time demand is normally distributed with a mean of 80 cases and a standard deviation of 6 cases:
EDDLT = 80
dLT = 6
Using Appendix B, Table B, we look for the z value corresponding to 0.99.
The closest probability is .9901, which corresponds to z = 2.33.
SS = (round up)
30. Given:
ROP = 18 units
Lead time for resupply = 3 days
Usage over the last 10 days:
|
Day |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Units |
3 |
4 |
7 |
5 |
5 |
6 |
4 |
3 |
4 |
5 |
Determine the service level achieved by the current ROP. Hint: Use Formula 13-13.
Step 1:
Calculate the mean and standard deviation of daily demand.
(round to three decimals)
Step 2:
Plug values into Formula 13-13 and solve for z.
18 = 13.8 + 2.191z
4.2 = 2.191z
z = 4.2 / 2.191 = 1.92 (round to two decimals)
From Appendix B, Table B, the lead time service level is .9726 = 97.26%.
31. Given:
A drugstore uses the fixed-order-interval (FOI) model
Service Level = 98%
OI = 14 days
LT = 2 days
d = 3 units/day
On-hand inventory in each cycle:
|
Cycle |
On Hand |
|
1 |
42 |
|
2 |
8 |
|
3 |
103 |
Using Appendix B, Table B, we look for the z value corresponding to .98.
The closest probability is .9798, which corresponds to z = 2.05.
Cycle 1:
Cycle 2:
Cycle 3:
32. Given:
Company operates 50 weeks per year
We have the following information on the two items:
|
P34 |
P35 |
|
= 60 units/week
|
= 70 units/week
|
|
d = 4 units/week |
d = 5 units/week. |
|
LT = 2 weeks |
LT = 2 weeks |
|
Unit cost = $15 |
Unit cost = $20 |
|
H = (.30)($15) = $4.50 |
H = (.30)($20) = 6.00 |
|
S = $70 |
S = $30 |
|
Risk = 2.5% |
Risk = 2.5% |
|
Can be ordered any time |
OI = 4 weeks |
a. Determine when to reorder each item:
P34:
Using Appendix B, Table B, we look for the z value corresponding to 1 - .025 = .975:
z = 1.96.
P35: Order every 4 weeks.
b. Compute the EOQ for P34:
D = 60 x 50 = 3,000 units/year
c. Compute the order quantity for P35 if 110 units are on hand at the time the order is placed:
Using Appendix B, Table B, we look for the z value corresponding to 1 - .025 = .975:
z = 1.96.
33. Given:
We have the following list of items:
|
Item |
Estimated Annual Demand |
Ordering Cost |
Holding Cost (%) |
Unit Price |
|
H4-010 |
20,000 |
50 |
20 |
2.50 |
|
H5-201 |
60,200 |
60 |
20 |
4.00 |
|
P6-400 |
9,800 |
80 |
30 |
28.50 |
|
P6-401 |
14,500 |
50 |
30 |
12.00 |
|
P7-100 |
6,250 |
50 |
30 |
9.00 |
|
P9-103 |
7,500 |
50 |
40 |
22.00 |
|
TS-300 |
21,000 |
40 |
25 |
45.00 |
|
TS-400 |
45,000 |
40 |
25 |
40.00 |
|
TS-041 |
800 |
40 |
25 |
20.00 |
|
V1-001 |
33,100 |
25 |
35 |
4.00 |
a. Classify the items as A, B, or C:
Step 1:
Determine the Annual Dollar Value (Unit Price x Estimated Annual Demand) for each item and the sum of the individual Annual Dollar Values:
|
Item |
Unit Price |
Estimated Annual Demand |
Annual Dollar Value |
|
H4-010 |
2.50 |
20,000 |
50,000 |
|
H5-201 |
4.00 |
60,200 |
240,800 |
|
P6-400 |
28.50 |
9,800 |
279,300 |
|
P6-401 |
12.00 |
14,500 |
174,000 |
|
P7-100 |
9.00 |
6,250 |
56,250 |
|
P9-103 |
22.00 |
7,500 |
165,000 |
|
TS-300 |
45.00 |
21,000 |
945,000 |
|
TS-400 |
40.00 |
45,000 |
1,800,000 |
|
TS-041 |
20.00 |
800 |
16,000 |
|
V1-001 |
4.00 |
33,100 |
132,400 |
|
|
|
|
3,858,750 |
Step 2:
Arrange the items in descending order based on Annual Dollar Values. Determine the A, B, and C items. Then, determine the percentage of items and the percentage of Annual Dollar Value for each category (round to two decimals).
|
Item |
Annual Dollar Value |
Category |
Percentage of Items |
Percentage of Annual Dollar Value |
|
TS-400 |
1,800,000 |
A |
20% |
71.14% |
|
TS-300 |
945,000 |
|
|
|
|
P6-400 |
279,300 |
B |
20% |
13.48% |
|
H5-201 |
240,800 |
|
|
|
|
P6-401 |
174,000 |
C |
60% |
15.38% |
|
P9-103 |
165,000 |
|
|
|
|
V1-001 |
132,400 |
|
|
|
|
P7-100 |
56,250 |
|
|
|
|
H4-010 |
50,000 |
|
|
|
|
TS-041 |
16,000 |
|
|
|
|
|
3,858,750 |
|
100.00% |
100.00% |
Note: An alternate solution could be to include P6-400 through V1-001 in the B category.
b. Determine the EOQ for each item (round to nearest integer):
|
Item |
Estimated Annual Demand |
Ordering Cost |
Unit Holding Cost ($) |
EOQ |
|
H4-010 |
20,000 |
50 |
.50 |
2,000 |
|
H5-201 |
60,200 |
60 |
.80 |
3,005 |
|
P6-400 |
9,800 |
80 |
8.55 |
428 |
|
P6-401 |
14,500 |
50 |
3.60 |
635 |
|
P7-100 |
6,250 |
50 |
2.70 |
481 |
|
P9-103 |
7,500 |
50 |
8.80 |
292 |
|
TS-300 |
21,000 |
40 |
11.25 |
386 |
|
TS-400 |
45,000 |
40 |
10.00 |
600 |
|
TS-041 |
800 |
40 |
5.00 |
113 |
|
V1-001 |
33,100 |
25 |
1.40 |
1,087 |
34. Given:
Demand for jelly doughnuts is shown in the table below. Labor, materials, and overhead are estimated to be $3.30 per dozen, doughnuts are sold for $4.80 per dozen, and leftover doughnuts are sold at half price.
|
Demand (dozens) |
Relative Frequency |
|
19 |
.01 |
|
20 |
.05 |
|
21 |
.12 |
|
22 |
.18 |
|
23 |
.13 |
|
24 |
.14 |
|
25 |
.10 |
|
26 |
.11 |
|
27 |
.10 |
|
28 |
.04 |
|
29 |
.02 |
Cs = Rev – Cost = $4.80 – $3.20 = $1.60
Ce = Cost – Salvage = $3.20 – $2.40 = $.80
|
Demand (dozens) |
Relative Frequency |
Cumulative Frequency |
|
19 |
.01 |
.01 |
|
20 |
.05 |
.06 |
|
21 |
.12 |
.18 |
|
22 |
.18 |
.36 |
|
23 |
.13 |
.49 |
|
24 |
.14 |
.63 |
|
25 |
.10 |
.73 |
|
26 |
.11 |
.84 |
|
27 |
.10 |
.94 |
|
28 |
.04 |
.98 |
|
29 |
.02 |
1.00 |
Because .67 falls between the cumulative frequencies of .63 and .73, Don should stock 25 dozen to attain a service level of at least .67. The resulting service level will be .73 = 73.00%.
35. Given:
Purchase price for spare part X135 = $100 each. Carrying and disposal costs = 145% of the purchase price. Stockout cost = $88,000. Demand for parts will approximate a Poisson distribution with a mean of 3.2 parts.
a. Determine the optimal number of spare parts to order:
Cs = $88,000
Ce = $100 + 1.45($100) = $245
|
[From Poisson Table with = 3.2] |
|
|
x |
Cumulative Probability |
|
0 |
.041 |
|
1 |
.171 |
|
2 |
.380 |
|
3 |
.603 |
|
4 |
.781 |
|
5 |
.895 |
|
6 |
.955 |
|
7 |
.983 |
|
8 |
.994 |
|
9 |
.998 |
|
10 |
1.000 |
Because .997 falls between the cumulative probabilities of .994 and .998, the optimal number of spare parts to order = 9. The resulting service level will be .998 = 99.8%.
b. Determine the range of shortage cost for which carrying 0 spare parts would be the best strategy:
Determine the value of Cs for which the service level = the service level of stocking 0 spare part and solve for Cs:
Service Level for 0 Spare Parts = .041
(round to two decimals)
Conclusion: Carrying 0 spare parts is the best strategy if the shortage cost is less than or equal to $10.47.
36. Given:
Purchase price = $4.20 per pound. Selling price = $5.70 per pound. Salvage price = $2.40 per pound. Daily demand can be approximated by a normal distribution with a mean of 80 pounds and a standard deviation of 10 pounds.
d = 10 pounds/day
Cs = Rev – Cost = $5.70 – $4.20 = $1.50 per pound
Ce = Cost – Salvage = $4.20 – $2.40 = $1.80 per pound
Using Appendix B, Table B, we find that .4545 falls closest to .4562:
z = -0.11.
pounds (assuming that fractional values are possible)
37. Given:
Daily demand can be approximated by a normal distribution with a mean of 40 quarts per day and a standard deviation of 6 quarts per day. Excess cost = $.35 per quart. The grocer orders 49 quarts per day.
d = 6 quarts/day
a. Determine the implied shortage cost per quart:
Cs = Rev – Cost = unknown
Ce = $.35
Step 1:
Determine z value.
Using Appendix B, Table B, we find that z = 1.50 corresponds to a service level = .9332 = 93.32%.
Step 2:
Plug in .9332 and solve for Cs.
per quart (round to two decimals)
b. This might be a reasonable figure because it probably is close to the lost profit per quart during strawberry season.
38. Given:
Demand can be approximated with a Poisson distribution with a mean of 6 per day. It costs $9 to prepare each cake. Fresh cakes sell for $12 each. Day-old cakes sell for $9 each. One half of the day-old cakes are sold and the rest thrown out.
Cs = Rev – Cost = $12 – $9 = $3.00 per cake
Ce = Cost – Salvage = $9 – (1/2)($9.00) = $4.50/cake
|
[From Poisson Table with = 6] |
|
|
x |
Cumulative Probability |
|
0 |
.003 |
|
1 |
.017 |
|
2 |
.062 |
|
3 |
.151 |
|
4 |
.285 |
|
5 |
.446 |
|
6 |
.606 |
Because .4 falls between the cumulative probabilities of .285 and .446, the optimal number of cakes to prepare = 5. The resulting service level will be .446 = 44.6%.
39. Given:
Purchase price = $1.00 per pound. Salvage value = $.80 per pound. Burgers sell for $.60 each. Four hamburgers can be prepared from each pound of beef. Labor, overhead, meat, buns, and condiments cost $.50 per burger. Demand is normally distributed with a mean of 400 pounds per day and a standard deviation of 50 pounds per day. Hint: Shortage costs must be in dollars per pound.
d = 50 pounds/day
Determine the optimal daily order quantity:
Cs = $.10/burger x 4 burgers/pound = $.40/pound
Ce = Cost – Salvage = $1.00 – $.80 = $.20/pound
Using Appendix B, Table B, we find that .6667 falls closest to .6664:
z = 0.43.
pounds (assuming that fractional values are possible)
40. Given:
Demand for rug cleaning machines is shown in the table below. Machines are rented by the day only. Profit on rug cleaners = $10/day. Clyde has 4 rug-cleaning machines.
|
Demand |
Frequency |
|
0 |
.30 |
|
1 |
.20 |
|
2 |
.20 |
|
3 |
.15 |
|
4 |
.10 |
|
5 |
.05 |
|
|
1.00 |
|
Demand |
Frequency |
Cumulative Frequency |
|
0 |
.30 |
.30 |
|
1 |
.20 |
.50 |
|
2 |
.20 |
.70 |
|
3 |
.15 |
.85 |
|
4 |
.10 |
.95 |
|
5 |
.05 |
1.00 |
|
|
1.00 |
|
a. Determine the implied range of excess cost per machine:
Cs = $10
Ce = unknown
For 4 machines to be optimal, the SL ratio must be ≥ .85 and ≤ .95.
Step 1:
Set SL = .85 and solve for Ce:
(round to two decimals)
Step 2:
Set SL = .95 and solve for Ce:
(round to two decimals)
Conclusion: Implied range of excess cost: $.53 ≤ ≤ $1.76.
b. If Clyde protests that the answer from part a is too low, does this suggest an increase or a decrease in the number of machines he stocks?
If the excess cost is supposed to be higher, then the number of machines should be decreased. When excess cost increases, SL decreases along with the optimum stocking level.
c. Suppose now that excess cost per day = $10 and the shortage cost per day is unknown. Assuming that the optimal number of machines is 4, what is the implied range of shortage cost?
Cs = unknown
Ce = $10
For 4 machines to be optimal, the SL ratio must be ≥ .85 and ≤ .95.
Step 1:
Set SL = .85 and solve for Cs:
(round to two decimals)
Step 2:
Set SL = .95 and solve for Cs:
(round to two decimals)
Conclusion: Implied range of shortage cost: $56.67 ≤ ≤ $190.00.
41. Given:
Spares cost $200 each. Unused spares have a salvage value of $50 each. If a part fails and a spare is not available, 2 days will be needed to obtain a replacement and install it. The cost for idle equipment is $500/day.
Probability of usage:
|
Number |
0 |
1 |
2 |
3 |
|
Probability |
.10 |
.50 |
.25 |
.15 |
a. Use the ratio method to determine the number of spares to order:
|
# of Spares |
Probability of Demand |
Cumulative Probability |
|
0 |
.10 |
.10 |
|
1 |
.50 |
.60 |
|
2 |
.25 |
.85 |
|
3 |
.15 |
1.00 |
Cs = Cost of stockout = ($500 per day) (2 days) = $1000
Ce = Cost of excess inventory = Unit Cost – Salvage Value = $200 – $50 = $150
Because .870 is between the cumulative probabilities of .85 and 1.00, we need to order 3 spares.
b. Use the tabular method to determine the number of spares to order:
|
Stocking |
Demand = 0 |
Demand = 1 |
Demand = 2 |
Demand = 3 |
Expected |
|
Level |
Prob. = 0.10 |
Prob. = 0.50 |
Prob. = 0.25 |
Prob. = 0.15 |
Cost |
|
0 |
$0 |
.50(1)($1000)=$500 |
.25(2)($1000)=$500 |
.15(3)($1000)=$450 |
$1,450 |
|
1 |
.10(1)($150)=$15 |
$0 |
.25(1)($1000)=$250 |
.15(2)($1000)=$300 |
$565 |
|
2 |
.10(2)($150)=$30 |
.50(1)($150)=$75 |
$0 |
.15(1)($1000)=$150 |
$255 |
|
3 |
.10(3)($150)=$45 |
.50(2)($150)=$150 |
.25(1)($150)=$37.50 |
$0 |
$232.50 |
We should order 3 spares.
42. Given:
Cakes cost $33 each, and they sell for $60 each. Unsold cakes are reduced to half price on Monday, and typically one-third of those are sold. Any that remain are donated.
Probability of demand:
|
Number |
0 |
1 |
2 |
3 |
|
Probability |
.15 |
.35 |
.30 |
.20 |
a. Use the ratio method to determine the number of cakes to prepare to maximize expected profit:
|
# of Cakes |
Probability of Demand |
Cumulative Probability |
|
0 |
.15 |
.15 |
|
1 |
.35 |
.50 |
|
2 |
.30 |
.80 |
|
3 |
.20 |
1.00 |
Cs = Selling Price – Unit Cost = $60 – $33 = $27
Ce = Unit Cost – Salvage Value = $33 – [(1/3)(1/2)($60)] = $23
Because the service level of .54 falls between the cumulative probabilities of .50 and .80, the supermarket should stock 2 cases of wedding cakes.
b. Use the ratio method to determine the number of cakes to prepare to maximize expected payoff.
Expected Payoff = Expected Profit – Expected Cost
Expected Profit = Probability of Demand * Expected Profit per Cake Sold at Regular Price ($27) * Number of Cakes Sold at Regular Price.
Expected Cost = Probability of Demand * Expected Cost per Cake Left Over (Ce = $23) * Number of Cakes Left Over.
|
Stocking |
Demand = 0 |
Demand = 1 |
Demand = 2 |
Demand = 3 |
Expected |
|
Level |
Prob. = .15 |
Prob. = .35 |
Prob. = .30 |
Prob. = .20 |
Payoff |
|
0 |
[Sell 0, Over 0] (.15 * 0 * $27) – (.15 * 0 * $23) = $0 |
[Sell 0, Over 0] (.35 * 0 * $27) – (.35 * 0 * $23) = $0 |
[Sell 0, Over 0] (.30 * 0 * $27) – (.30 * 0 * $23) = $0 |
[Sell 0, Over 0] (.20 * 0 * $27) – (.20 * 0 * $23) = $0 |
$0 |
|
1 |
[Sell 0, Over 1] (.15 * 0 * $27) – (.15 * 1 * $23) = -$3.45 |
[Sell 1, Over 0] (.35 * 1 * $27) – (.35 * 0 * $23) = $9.45 |
[Sell 1, Over 0] (.30 * 1 * $27) – (.30 * 0 * $23) = $8.10 |
[Sell 1, Over 0] (.20 * 1 * $27) – (.20 * 0 * $23) = $5.40 |
$19.50 |
|
2 |
[Sell 0, Over 2] (.15 * 0 * $27) – (.15 * 2 * $23) = -$6.90 |
[Sell 1, Over 1] (.35 * 1 * $27) – (.35 * 1 * $23) = $1.40 |
[Sell 2, Over 0] (.30 * 2 * $27) – (.30 * 0 * $23) = $16.20 |
[Sell 2, Over 0] (.20 * 2 * $27) – (.20 * 0 * $23) = $10.80 |
$21.50 |
|
3 |
[Sell 0, Over 3] (.15 * 0 * $27) – (.15 * 3 * $23) = -$10.35 |
[Sell 1, Over 2] (.35 * 1 * $27) – (.35 * 2 * $23) = -$6.65 |
[Sell 2, Over 1] (.30 * 2 * $27) – (.30 * 1 * $23) = $9.30 |
[Sell 3, Over 0] (.20 * 3 * $27) – (.20 * 0 * $23) = $16.20 |
$8.50 |
Conclusion: The supermarket should stock 2 cases of wedding cakes. This number of cakes will maximize the expected payoff.
43. Given:
On average, 18 ticket holders cancel their reservations, so the company intentionally overbooks the flight. Cancellations can be described by a normal distribution with a mean equal to 18 and a standard deviation of 4.55. Profit per passenger = $99. If a passenger is bumped, the company pays that passenger $200.
Determine the number of tickets to overbook:
Cs = $99, Ce = $200
Using Appendix B, Table B, we find that .3311 falls closest to .3300:
z = -0.44.
tickets to overbook
Case: UPD Manufacturing
Given:
OI = 6 weeks
S = $32
H = $.08/unit/week
d = 89 units/week
LT = 5 working days = 1 week
1. Students must recognize that without demand variability, the fixed order interval order quantity equation reduces to:
UPD places an order every 6 weeks and the lead-time is 1 week. Therefore, when the order is placed, there will be 89 units on hand (d x LT = 1 week * 89 units/week).
Because A = 89 = d x LT, the fixed order interval order quantity equation further reduces to the following:
units
Therefore, ordering at six-week intervals requires an order quantity of 534 units.
Optimal order quantity as determined by using the basic EOQ equation:
The weekly total cost based on the EOQ is given below:
The weekly total cost based on six-week fixed order interval (FOI) order quantity is given below:
Weekly savings of using EOQ rather than 6-week FOI = $26.69 – $21.35 = $5.34
The annual savings = (52 weeks) ($5.34/week) = $277.68
2. The total annual savings as a result of switching from the six-week FOI to EOQ are relatively small and switching to the optimal order quantity may not be warranted. However, if the FOI approach is used with other parts or components as well, the total potential loss may be significant.
Case: Harvey Industries
To improve the current inventory control system, the new president may want to consider the following:
1. Computerize the inventory control system. Rationale: There are too many parts for the current manual system.
2. Currently, no paperwork is used when items are withdrawn from the stockroom when they are needed on the shop floor. Harvey Industries may either want to establish a procedure for recording the transactions in the stockroom or invest in a bar coding system. If a bar coding system is purchased, it has to be coordinated with the new computerized inventory control system. Establishing a cycle counting procedure may be very helpful also. Rationale: As a result of these actions, the inventory accuracy should improve substantially.
3. It appears that utilization of A-B-C inventory classification system is needed. Rationale: Harvey Industries rarely should experience stockouts in those “A” items that account for $220,684 of $314,673 in annual purchases or for any “A” items for which a stockout leads to significant downtime costs. ABC analysis will allow Harvey Industries to establish an appropriate degree of control over items in terms of order quantity and ordering frequency.
Case: Grill Rite
Recommendations:
1. The president’s stance on steady output conflicts with seasonal demand. However, it is unlikely that this will change. One alternative might be to identify a complementary product that would offset seasonal demand for electric grills.
2. The main problem is inventory management. Therefore, we recommend the following:
a. Having a single, centralized warehouse: This will lower the need for safety stock due to the canceling effect of random variability in orders from the various regions. Conversely, with separate warehouses, each warehouse needs a relatively larger safety stock to guard against variations in demand.
b. What is needed is overall control of the system that would take into account seasonal variations in demand and achieve a better match between regional demand and supply. This might involve making or improving regional forecasts. In any case, improved system visibility is essential: direct access to regional warehouse data by the main warehouse is needed to be able to coordinate and set priorities on inventory shipments to regional warehouses. That should take care of most of the problem.
c. It also may pay to examine the feasibility of shipping from one warehouse to another when a shortage occurs. Relevant costs would include transaction costs and transportation costs versus the potential increase in profit by avoiding the shortage.
Case: Farmers Restaurant
1. Inventory management is crucial not only to Farmers Restaurant, but to businesses in general. Customer satisfaction and customer return is contingent upon proper inventory management. If customers visit the Farmers Restaurant and are unable to receive the food they desire due to a stockout, the customer may be dissatisfied and may not return to Farmers Restaurant. In addition to customer satisfaction, total food costs are important to businesses also. In the restaurant industry, if too much of a product is ordered and not used; it could result in product waste due to the items expiring. This would result in an increase in total food costs when the goal is to keep food costs low! Overstocking products also can negatively affect Farmers Restaurant and other businesses. By having more products on-hand than needed, Farmers Restaurant is tying up funds that might be more productive elsewhere. Overall, it is imperative that Farmers Restaurant try to manage inventory levels successfully due to these potential/existing problems.
2. A fixed-interval ordering system is appropriate given that the manager reviews inventory and places orders once a week from the supplier.
3. Given:
SL = 95%
units/week
units/week
Gravy mix comes in packs of 2
There are currently 3 packs in inventory = 6 units
LT = 2 days = 2/7 weeks
OI = 1 week
Using Appendix B, Table B, we look for the z value corresponding to .95:
.95 falls midway between .9495 (z = 1.64) and .9505 (z = 1.65).
Using z = 1.64:
Using z = 1.65:
4. Given:
A = 12
Determine the risk of a stockout at the end of the initial lead time and at the end of the second lead time:
Use Formula 13-13 to determine the risk of stockout at the end of the initial lead time:
Using Appendix B, Table B, we look for the corresponding service level: .8577.
The risk of a stockout = 1 - .8577 = .1423 = 14.23%.
Determine the risk of a stockout at the end of the second lead time:
Use Formula 13-16 to determine the risk of stockout at the end of the second lead time:
(round to two decimals)
Using Appendix B, Table B, we look for the corresponding service level: .9995.
The risk of a stockout = 1 - .9995 = .0005 = .05%.
5. Kristin may want to consider dealing with a nearby supplier to be able to order more frequently or to reduce transportation costs. In addition, if she ordered from a nearby supplier, she could have the option of sending an employee to the supplier’s facility to pick up emergency orders. On the other hand, she may want to keep her current supplier due to competitive prices and/or exceptional customer service.
Operations Tour: Bruegger’s Bagel Bakery
1. If too little inventory is maintained, there is a risk of a stockout and potential lost sales. In addition, if there is not sufficient work-in-process inventory, the production process may become too inefficient, raising the cost of production. On the other hand, if too much inventory is maintained, the carrying cost may become excessively high.
2. a. Customers judge the quality of bagels by their appearance (size, shape, and shine), taste, and consistency. Customers are also very interested in receiving high service quality.
b. Bruegger’s checks quality at every stage of operation, from choosing suppliers of ingredients, careful monitoring of ingredients, and keeping equipment in good operating condition to monitoring output at each step of the production process. At the stores, employees watch for deformed bagels and remove them.
c. Steps for Bruegger’s Bagel Bakery Operations:
1) Purchase ingredients from suppliers
2) Receive ingredients from suppliers
3) Mix basic ingredients into the dough
4) Shape the dough into individual bagels
5) Ship bagels to stores in refrigerated trucks
6) Unload and store the bagels
7) Boil bagels in kettle of water and malt
8) Bake bagels for 15 minutes
9) Sell bagels to customers
The company can improve quality at each step by monitoring output more carefully and with training and education of the employees.
3. The basic ingredients can be purchased using either fixed order interval or fixed order quantity models, e.g., EOQ. The EPQ model is most appropriate for deciding the size of the production quantity.
4. If there were a bagel-making machine at each store, the company would have to invest in more machinery, more space for production and storage, and more worker training for the production of bagels. However, the lead time to make the bagels would be shortened. The shorter lead time would provide faster, more flexible response to customer demands and fresher bagels.
Enrichment Module: EPQ Problem
This enrichment module consists of an EPQ problem to solidify the concepts associated with the Economic Production Quantity model.
Problem
A company produces plastic powder in lots of 2,000 pounds, at the rate of 250 pounds per hour. The company uses powder in an injection molding process at the steady rate of 50 pounds per hour for an eight-hour day, five days a week. The manager has indicated that the setup cost is $100 for this product, but “We really have not determined what the holding cost is.”
a. What weekly holding cost per pound does the lot size imply, assuming the lot size is optimal?
b. Suppose the figure you compute for holding cost has been shown to the manager, and the manager says that it is not that high. Would that mean the lot size is too large or too small? Explain.
Solution to Enrichment Module Problem
a.
b. Decreasing the value of carrying cost (H) will result in an increase in the lot size. Because holding inventory is not as expensive, the firm could afford to carry more inventory and therefore produce a larger batch.
Enrichment Module 2: Inventory Model with Planned Shortages
In most cases, shortages are undesirable and should be avoided. However, in certain circumstances, it may be desirable to plan and allow for shortages. Planned shortages are implemented for high dollar volume items where the inventory carrying cost is very high. The model discussed in this section refers to the specific type of shortages called backorders. When a customer attempts to purchase an outofstock item, the firm does not lose the sale. The customer waits until the purchased order arrives from the supplier. If there were no additional cost associated with backordering, there would be no incentive for the firm to maintain any inventory. However, there are costs associated with backordering. The tangible part of the backorder cost involves the cost of expediting the delivery (special delivery) and production of the backordered item. The intangible part of the backorder cost involves the loss of goodwill due to the fact that the customers are forced to wait for their orders. The longer the waiting period, the higher the backorder cost due to loss of goodwill will be.
There is a direct tradeoff between the inventory carrying cost and the cost of a planned shortage in the form of backorders. In many cases, the cost of backorders can be offset easily by the reduction in carrying costs. The model discussed in this section will not be valid if a customer decides not to wait for the backorder.
The fixed order quantity inventory model with planned shortages (backorders) is very similar to the basic EOQ model. When the reorder point is reached, a new economic order quantity (Q) is placed. Figure 1 shows the schematic representation of this model. The size of the backorder is B units and the maximum inventory is Q – B units. The average size of the backorder is B/2 for each order cycle. T is defined as the amount of time between two successive orders (a complete order cycle). t1 is the part of the order cycle where the customer orders are met from stock. In other words, during t1 there is positive inventory level. On the other hand, t2 is the period of time in the order cycle where the inventory is depleted and all the customer orders are placed on backorder (stockout period).
Symbol definitions used to explain various concepts are listed below.
H = carrying cost per unit per year
S = ordering cost per batch (lot)
D = annual demand
Q* = optimal order quantity
B = size of the backorder
CB = backorder cost per unit per year
B* = optimal planned backorder quantity
T = Q/D (length of the complete order cycle in years) or
T = Q/d (length of the complete order cycle in days)
t1 = (Q – B)/D or (Q – B)/d (time period during which inventory is positive)
t2 = B/D or B/d (time period during which there is no inventory)
In this model, the average inventory is not Q/2 or not even (Q – B)/2 because during the shortage period there are no units in inventory. The average inventory calculation for this model can be explained with the following example:
A large local car dealership orders a certain brand of automobiles from a car manufacturer located in Detroit. Order quantity (Q) is 500 units, annual demand (D) is 7,500 and the firm operates 300 working days per year. Due to the high holding costs, the company plans to backorder (B) 200 cars per order cycle. Determine the average inventory.
d = (D/number of operational days) = 7,500/300 = 25 units (daily consumption)
T = Q/d = 500/25 = 20 days (time between orders is 20 days)
t1 = (Q – B)/d = (500 – 200)/25 = 300/25 = 12 days (time period during which there is no shortage)
t2 = B/d = 200/25 = 8 days (time period during which there is no inventory)
The dealership will carry an average of (Q – B)/2 units during t1 and no units during t2. Therefore, total number of unit days during the inventory cycle can be computed by multiplying t1 by (Q – B)/2
In other words, an average of 150 units are carried in inventory for 12 days and zero units are carried for 8 days (shortage period). Therefore, total number of unit days of inventory during the complete order cycle is (150)(12) = 1800.
Because there are a total of 20 days in the complete order cycle, the average inventory can be computed by dividing the total number of unit days of inventory by the number of days in the inventory cycle. In this example, the average inventory is equal to 1,800/20 or 90 units. Therefore, the average inventory can be computed by using the following formula:
Using a similar logic, we can also develop the average backlog formula. The dealership will experience shortage (backorders) for 8 days during the order cycle. The average amount of backorder on a given shortage day is B/2. Based on this information, the total number of backorder unit days can be computed using the following equation: (t2) (B/2) = (B/D)(B/2) = B2 /2D.
In our example, there are 8 days of a planned shortage period. During this period, an average of 200/2 = 100 units of backorders are realized. Therefore, the total number of backorder unit days during the order cycle is (8)(100) = 800 units. Because there are a total of 20 days in the order cycle, the average backorder quantity for the complete order cycle can be determined by dividing the total number of backorder unit days by the number of days in the complete inventory cycle. In this example, using the above equation, we obtain an average backorder quantity of 800/20 = 40 units. The general equation for the average backorder quantity is:
Annual inventory carrying cost still is calculated by multiplying the average inventory by the inventory carrying cost per unit per year. The formula for the annual ordering cost is the same as it was for the basic EOQ model. The annual backorder cost is determined by multiplying the average backorder quantity by the backorder cost per unit per year.
The annual inventory carrying cost is given by:
The annual ordering and backordering costs are given by the following respective formulas:
Therefore, the total annual inventory cost (TC) can be expressed by summing the annual inventory carrying cost, the annual ordering cost, and the annual backordering cost as shown in the following formula:
Taking the first total derivative of the above total cost formula with respect to Q, setting the resulting equation to zero, and solving for Q will result in the following optimal quantity (Q*) and optimal backorders (planned shortages) (B*) formulas:
Figure 1
An inventory situation with planned shortages
( t 2 Inventory Time Q – B Stockout B Maximum Inventory Level Q T = Q / d )
Example:
XYZ Company distributes a major part for the F–15 fighter jets. Due to the very high holding cost, the company wants to implement a model with planned shortages. The annual demand is 81,000 and the company operates 300 days per year. The annual carrying cost rate is 10% of the unit cost and the unit cost of this item is $1,000. The ordering cost per batch is estimated at $500.
a. Determine the optimal order quantity and total annual inventory cost (ordering cost + carrying cost) using the basic EOQ model with no backorders.
b. If each unit backordered costs the company $200 per unit per year, what would be the optimal order quantity and the optimal size of the planned backorder?
c. Determine the annual carrying cost, the annual ordering cost, the annual backordering cost, and the annual total inventory cost for the planned shortage model used in part b.
d. Determine the values of t1, t2 and T in days.
e. Should the company adopt the planned backorder model of part b or the basic EOQ model of part a, which does not allow backorders?
D = 81,000 units
S = $500
d = 81,000/300 days = 27 units per day
H = ($1,000) (.10) = $100
CB = $200
Total annual cost = $45,000 + $45,000 = $90,000
b.
c.
Let TC = Total annual inventory cost
TC = $24,511.12 + $36,751.36 + $12,222.23 = $73,484.71
d.
e. The model with planned backorders is preferred because the total annual inventory cost of the basic EOQ inventory model is substantially higher than the total annual inventory cost of the planned backorder model.
TCbasic EOQ = $90,000
TCbackorder = $73,484.71
$90,000 – $73,484.71 = $16,515.29 difference
Problems
The manager of an inventory system believes that inventory models are important decision-making aids. Although the manager often uses an EOQ policy, he has never considered a backorder model because of his assumption that backorders are “bad” and should be avoided. However, with upper management’s continued pressure for cost reduction, you have been asked to analyze the economics of a backordering policy for some products that possibly can be backordered. For a specific product with D = 800 units per year, S = $150, H = $10, and CB = $20, what is the cost difference in the EOQ and the planned shortage or backorder model? If the manager adds constraints that no more than 35% of the units may be backordered and that no customer will have to wait more than 20 days for an order, should the backorder inventory policy be adopted? Assume 250 working days per year.
Solution to Problem
D = 800 units/year
S = $150
H = $10/unit/year
CB = $20/unit/year
Planned shortage model:
EOQ model:
Total cost planned shortage model:
TC = $424.45 + $631.58 + $208.89 = $1,264.92
Total cost regular EOQ model:
TC = $775.00 + $774.19 = $1,549.19
TCDifference = $1,549.19 - $1,264.92 = $284.27
Using the planned shortage model will result in annual savings of $284.27.
Expected annual number of units short = (63)(4.21) = 265.23
12-2
13-1
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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