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Chapter 4

Inventory Management

Contents

The Concept of Inventory Basics of Managing the Average Inventory Balance Inventory Management and the Cash Flow Timeline Monitoring the Inventory Balance Reducing the Size of the Inventory Investment

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Answers to Questions: 1. Making sure that the company does not run out of inventory to satisfy production

or customer needs but doing so at a reasonable cost. 2. Inventory is a difficult item to manage because it crosses so many different lines

of authority. Marketing is concerned about inventory because sales will be hurt if stock outs occur. Production will be hurt if stockouts of raw materials and work in process occur. The financial manager is concerned about the level of investment in inventory and the costs associated with that investment.

3. It serves the role of shock absorber. If inefficiencies were eliminated in the

production flow then less inventory would be needed. 4. Raw materials: shock absorber between the firm and the supplier. Work-in-

process: shock absorber for inefficiencies in the production system. Finished goods: shock absorber between the firm and its customers.

5. The financial manager is concerned with the amount of and cost of capital tied up

in the inventory investment. 6. EOQ stands for the economic order quantity, that order quantity that minimizes

the inventory management total cost function. 7. By adding a safety stock. 8. Variability in demand, the production process, and delivery time will tend to

increase the optimal size of the safety stock. High inventory carrying costs tend to reduce the optimal level of safety stock holding stock out costs constant.

9. The present value timeline solution allows for the cost of capital directly whereas

the EOQ solution does not. 10. Inventory levels can be reduced by reducing the inefficiencies in the firm's

production systems, by increasing the reliance of supplier deliveries, and by increasing the accuracy of forecasts of sales.

11. It is influenced by sales trends. 12. A balance fraction approach is not influenced by sales trends. Thus shifts in

balance fractions is a direct result of changes in inventory holding patterns.

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Solutions to Problems: Chapter 4 1. Ardmore Farm and Seed - EOQ, average inventory balance, and reorder

point. ASSUMPTIONS Order costs (F) $25.00 Holding costs per gal. (H) $0.25 Total annual quantity (T) 80,000 Order Quantity (Q) 10,000 Planning Period 365 Delivery Time (days) 7 a.) Calculating annual inventory costs. Total cost = (F * T/Q) + (H * Q / 2) = (25 * 80,000 / 10,000) + (0.25 * 10,000 / 2) Total Cost = $1,450 b.) Calculating the EOQ. EOQ = SQRT(2 * F * T / H) = (2 * 25 * 80,000 / 0.25)0.5 EOQ = 4,000 Gallons c.) Calculating the number of orders and the average inventory balance. Optimal Number of Orders = T / EOQ = 80,000 / 4000 Optimal Number of Orders = 20 Average Inventory Balance = EOQ / 2 = 4000 / 2 Average Inventory Balance = 2000 Gallons d.) Calculating the reorder point. Daily Usage Rate = T / # of Days in Planning Period = 80,000 / 365 Daily Usage Rate = 219.18 Gallons per day Reorder Point = Daily Usage Rate * Delivery Time = 219.18 * 7 Reorder Point = 1,534.25 Gallons 2. Lott Manufacturing, Inc. - EOQ, average inventory balance, and reorder

point ASSUMPTIONS Order costs (F) $50.00 Holding costs per unit (H) $3.00 Total period quantity (T) 200,000 Order Quantity (Q) 10,000 Planning Period 250 Delivery Time (days) 2

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a.) Calculating the EOQ. EOQ = SQRT(2 * F * T / H) = (2 * 50 * 200,000 / 3.00)0.5 EOQ = 2,581.99 Units b.) Calculating the EOQ savings. Total cost = (F * T/Q) + (H * Q / 2) = (50 * 200,000 / 10,000) + (3.00 * 10,000/2) Total Cost @10,000 units = $16,000 Total Cost EOQ = (F * T / Q) + (H * Q / 2) where Q = 2,581.99 units = (50 * 200,000 / 2,581.99) + (3.00 * 2,581.99 / 2) = $7,746 Savings with EOQ = $8,254 = $16,000 - $7,746 per planning period c.) Calculating the optimal number of orders and average inventory balance. Optimal Number of Orders = T / EOQ = 200,000 / 2,581.99 = Optimal Number of Orders = 77 Average Inventory Balance = EOQ / 2 = 2,581.99 / 2 Average Inventory Balance = 1,290.99 Units d.) Calculating the reorder point. Daily Usage Rate = T / # of Days in Planning Period = 200,000 / 250 Daily Usage Rate = 800 Units per day Reorder Point = Daily Usage Rate * Delivery Time = 800 * 2 Reorder Point = 1,600 Units 3. Ardmore farm and Seed - considering quantity discounts (see problem 1). ASSUMPTIONS Order costs(F) $25.00 Discount options Cost Per Holding costs per gal. (H) $0.25 Quantity ( Q ) Unit ( C' ) Total annual quantity (T) 80,000 0-4,999 $40.00 Planning Period 365 5,000-9,999 $39.00 Delivery Time (days) 7 10,000-19,999 $37.00 20,000+ $35.00 Total Cost = (F * (T / Q) + (H * (Q / 2) ) ) + (C' * T) (The solution is arrived at by trial and error, partially shown below.)

It might be useful in class to plug in four quantities (Q), and show what happens, as below:

Try EOQ = 4,000 gallons Then total cost = order costs + holding costs + purchase costs = (25)(80,000) / 4,000 + (0.25)(4,000) / 2 + (40)(80000) = $500.00 + $500.00 + $3,200,000 = $3,201,000

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Try Q = 10,000 gallons. Then total cost = (25)(80,000) / 10,000 + (0.25)(10,000) / 2 + (37.00)(80,000) = $200 + $1,250 + $2,960,000 = $2,961,450 Try Q = 20,000 gallons. Then total costs = (25)(80,000) / 20,000 + (0.25)(20,000) / 2 + (35.00)(80,000) = $100 + $2,500 + $2,800,000 = $2,802,600 (This is the lowest cost solution with order costs = $100, holding costs =

$2,500, and purchase costs = $2,800,000 for a total of $2,802,600.

Finally, try Q = 30,000 gallons. Then total costs = (25)(80,000) / 30,000 + (0.25)(30,000) / 2 + (35.00)(80,000) = $66.67 + $3,750 + $ 2,800,000 = $2,803,816.67 Notice how the order costs and the purchase costs fall (and then remain constant), but see how the holding costs rise and eventually offset the decline in the other

two costs. This can also be shown by using the disk program below and "selecting" various order quantities to see the effect upon the holding, ordering, and purchase costs.

Select Order Quantity = 20,000 Total cost = (F * (T/Q) + (H * (Q/2) ) ) + (C' * T) Total Order Holding Purchase

Price Quantity Cost Costs Costs Costs $35.00 20,000 $2,802,600 $100 $2,500 $2,800,000 (This solution was arrived at by trial and error.)

4. Lott Manufacturing, Inc. - considering quantity discounts (see problem 2). ASSUMPTIONS Discount options Order costs(F) $50.00 Quantity Cost Per Holding costs per unit (H) $3.00 (Q) Unit (C ' ) Total period quantity (T) 200,000 0-1,999 $5.00 Planning Period 250 2K - 3,999 $4.99 Delivery Time (days) 2 4K - 5,999 $4.98 6K - 7,999 $4.97 8K - 9,999 $4.96 10,000 + $4.95 Total Cost = (F * (T / Q) + (H * (Q / 2) ) ) + (C' * T) (The solution is arrived at by trial and error, partially shown below.)

It might be useful in class to plug in four quantities (Q), and show what happens, as below:

Try Q (actually, EOQ) = 2,581.99 units.

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Then total cost = order costs + holding costs + purchase costs = (50)(200,000) / 2,581.99 + (3.00)(2,581.99) / 2 + (4.99)(200,000) = $3,872.985 + $3,872.985 + $998,000 = $1,005,745.97 Try Q = 4,000 gallons. Then total cost = (50)(200,000) / 4,000 + (3.00)(4,000) / 2 + (4.98)(200,000) = $2,500 + $6,000 + $996,000 = $1,004,500 (This is the lowest cost solution with order costs = $2,500, holding costs = $6,000, and purchase costs = $996,000 for a total of $1,004,500. Try Q = 6,000 gallons. Then total costs = (50)(200,000) / 6,000 + (3.00)(6,000) / 2 + (4.97)(200,000) = $1,666.67 + $9,000 + $994,000 = $1,004,666.67 Finally, try Q = 8,000 gallons. Then total costs = (50)(200,000) / 8,000 + (3.00)(8,000) / 2 + (4.96)(200,000) = $1,250 + $12,000 + $992,000 = $1,005,250

Notice how the order costs and the purchase costs fall, but see how the holding costs rise to eventually offset the decline in the other two costs.

This can also be shown by using the disk program below and "selecting" various order quantities to see the effect upon the holding, ordering, and purchase costs. Select Order Quantity = 4,000 (This solution was arrived at by trial and error.) Total cost = (F * (T/Q) + (H * (Q/2) ) ) + (C' * T) Total Order Holding Purchase

Price Quantity Cost Costs Costs Costs $4.98 4,000 $1,004,500 $2,500 $6,000 $996,000

5. Ardmore Farm and Seed - considering cost of capital (refer to Problems 1

and 3) ASSUMPTIONS Order costs (F) $25.00 Discount options Holding costs per gal. (H) $0.25 Quantity (Q) Cost Per Unit ( C' ) Total annual quantity (T) 80,000 0-4,999 $40.00 Planning Period 365 5,000-9,999 $39.00 Delivery Time (days) 7 10,000-19,999 $37.00 Opportunity Cost 15% 20,000+ $35.50 Results of random trial solutions: Q PV Cost

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4,000 $2,992,916.84 6,000 $2,923,096.43 8,000 $2,928,578.28

10,000 $2,783,501.27 = Lowest level of present value of inventory 12,000 $2,787,725,.32 cost for optimum order quantity, found by 14,000 $2,792,599.50 trial and error 16,000 $2,798,786.37 (Note: this is one-half of the 20,000 EOQ found in problem 3. ) Inventory purchase = 10,000 * $37.00 = $370,000.00 every 45.625 days, beginning Day 0 (annuity due) Number of orders = 80,000 / 10,000 = 8 orders per year 365 days / 8 orders = 45.63 days between orders, beginning Day 0 Holding cost = (0.25)(10,000) / 2 $1,250.00 occurs at end of planning period Ordering cost = (50)(80,000)/10,000 $400.00 occurs at end of planning period Total inventory cost = $2,783,501.27 occurs at end of planning period

Note: Do not be misled by all of the zeros in the spreadsheet printout below. Because it is an interactive spreadsheet, when the order quantity changes, many of the zero cells change to positive numbers to reflect a different order sequence.

PV of PV Factor Holding & PV of Cost Per Purchase (simple Ordering Inventory t Quantity Unit Day interest) Costs Purchase 0 10,000 37.0 0.000 1.0000 0 370,000 1 10,000 37.0 45.625 0.9816 0 363,190 2 10,000 37.0 91.250 0.9639 0 356,627 3 10,000 37.0 136.875 0.9467 0 350,296 4 10,000 37.0 182.500 0.9302 0 344,186 5 10,000 37.0 228.125 0.9143 0 338,286 6 10,000 37.0 273.750 0.8989 0 332,584 7 10,000 37.0 319.375 0.8840 0 327,072 8 0 0 0.0 0.0000 0 0 9 0 0 0.0 0.0000 0 0 10 0 0 0.0 0.0000 0 0 11 0 0 0.0 0.0000 0 0 12 0 0 0.0 0.0000 0 0 13 0 0 0.0 0.0000 0 0 14 0 0 0.0 0.0000 0 0 15 0 0 0.0 0.0000 0 0

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16 0 0 0.0 0.0000 0 0 17 0 0 0.0 0.0000 0 0 18 0 0 0.0 0.0000 0 0 19 0 0 0.0 0.0000 0 0 20 0 0 0.0 0.0000 0 0 21 0 0 0.0 0.0000 0 0 22 0 0 0.0 0.0000 0 0 23 0 0 0.0 0.0000 0 0 24 0 0 0.0 0.0000 0 0 25 0 0 0.0 0.0000 0 0 26 0 0 0.0 0.0000 0 0 27 0 0 0.0 0.0000 0 0 28 0 0 0.0 0.0000 0 0 29 0 0 0.0 0.0000 0 0 80,000 = total annual quantity ( T ) $1,261 $2,782,240 Total Present Value Cost = PV of holding & order costs + PV of inventory purchase costs = $1,261 + $2,780,979 = $2,782,240 TIMELINE ILLUSTRATION OF CASH FLOWS: Day 0 Day 45.6 Day 91.3 Day 136.9 etc. Day 319.4 Day 365 ---|------------|------------------|--------------|---------------------|---------------|------------> $370K $370K $370K $370K etc. $370K $1650 | | purchasing holding $363,190 | | costs and $356,627 | | order etc. | costs etc. (discounting at 15% / year simple interest) | | $327,072 | | | $2,780,979 = PV of sum of purchases | $1,261 = PV of holding and order costs | $2,782,240 = total present value of inventory cost

Note: This is similar to pricing a bond, only the PMTS are in the form of an annuity due, so the formula (using compound interest) would be:

PV = (PMT)(PVIFA k, n )(1 + k) + FV (PVIF k, n ) PV = ($370,000)(PVIFA 15% / 8 , 8 )[1 + (15% / 8)] + ($1,650)(PVIF 15% / 8, 8 ) PV = $2,776,184.74 + $1,422.14 = $2,777,606.88 which is slightly less than the $2,782,240 found when using simple interest. 6. Lott Manufacturing, Inc. - considering cost of capital (refer to problems 2

and 4)

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ASSUMPTIONS Discount options Order costs(F) $50.00 Quantity Cost Per Holding costs per unit (H) $3.00 (Q) Unit (C') Total period quantity (T) 200,000 0-1,999 $5.00 Planning Period 250 2K - 3,999 $4.99 Delivery Time (days) 2 4K - 5,999 $4.98 Opportunity Cost 20% 6K - 7,999 $4.97 8K - 9,999 $4.96 10M + $4.95

Results of random trial solutions: Q PV Cost 2,000 $492,155.00 4,000 $942,110.72 = Lowest level of present value of inventory cost for 5,000 $943,290.82 optimum order quantity, found by trial and error 6,000 $942,793.62 Inventory purchase = 4,000 * $4.98 = $19,920 every 10 days beginning Day 0 (annuity due) Orders / year = 200,000 / 4,000 = 50 orders / planning period 250 days / 50 orders = 5 days between orders, beginning Day 0 Holding cost = $3.00 * 4,000 / 2 = $6,000 occurs at end of planning period Ordering cost = $50 * 200,000 / 4,000 = $2,500 occurs at end of planning period Total inventory cost at end of planning period = $8,500 Cost Inventory PV Holding PV of Inv. Per Unit Purchase & Ordering Inventory t Quantity of Inventory Day PV Factor Cost Purchase 0 4,000 4.98 0 1.0000 0 19,920 1 4,000 4.98 5 0.9973 0 19,866 2 4,000 4.98 10 0.9946 0 19,811 3 4,000 4.98 15 0.9918 0 19,758 4 4,000 4.98 20 0.9892 0 19,704 5 4,000 4.98 25 0.9865 0 19,651 6 4,000 4.98 30 0.9838 0 19,598 7 4,000 4.98 35 0.9812 0 19,545 8 4,000 4.98 40 0.9786 0 19,493 9 4,000 4.98 45 0.9759 0 19,441 10 4,000 4.98 50 0.9733 0 19,389 11 4,000 4.98 55 0.9707 0 19,337

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12 4,000 4.98 60 0.9682 0 19,286 13 4,000 4.98 65 0.9656 0 19,235 14 4,000 4.98 70 0.9631 0 19,184 15 4,000 4.98 75 0.9605 0 19,134 16 4,000 4.98 80 0.9580 0 19,083 17 4,000 4.98 85 0.9555 0 19,034 18 4,000 4.98 90 0.9530 0 18,984 19 4,000 4.98 95 0.9505 0 18,934 20 4,000 4.98 100 0.9481 0 18,885 21 4,000 4.98 105 0.9456 0 18,836 22 4,000 4.98 110 0.9432 0 18,788 23 4,000 4.98 115 0.9407 0 18,739 24 4,000 4.98 120 0.9383 0 18,691 25 4,000 4.98 125 0.9359 0 18,643 26 4,000 4.98 130 0.9335 0 18,595 27 4,000 4.98 135 0.9311 0 18,548 28 4,000 4.98 140 0.9288 0 18,501 29 4,000 4.98 145 0.9264 0 18,454 30 4,000 4.98 150 0.9241 0 18,407 31 4,000 4.98 155 0.9217 0 18,361 32 4,000 4.98 160 0.9194 0 18,314 33 4,000 4.98 165 0.9171 0 18,268 34 4,000 4.98 170 0.9148 0 18,223 35 4,000 4.98 175 0.9125 0 18,177 36 4,000 4.98 180 0.9102 0 18,132 37 4,000 4.98 185 0.9080 0 18,087 38 4,000 4.98 190 0.9057 0 18,042 39 4,000 4.98 195 0.9035 0 17,997 40 4,000 4.98 200 0.9012 0 17,953 41 4,000 4.98 205 0.8990 0 17,908 42 4,000 4.98 210 0.8968 0 17,864 43 4,000 4.98 215 0.8946 0 17,821 44 4,000 4.98 220 0.8924 0 17,777 45 4,000 4.98 225 0.8902 0 17,734 46 4,000 4.98 230 0.8881 0 17,691 47 4,000 4.98 235 0.8859 0 17,648 48 4,000 4.98 240 0.8838 0 17,605 49 4,000 4.98 245 0.8816 0 17,562 200,000 = total annual quantity ( T ) $7,475.90 $934,634.82 Total Present Value Cost = PV of holding & order costs + PV of inventory purchase costs = $7,475.90 + $934,634.82 = $942,110.72 TIMELINE ILLUSTRATION OF CASH FLOWS: Day 0 Day 5 Day 10 Day 15 etc. Day 245 Day 250

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---|-------------|-----------------|--------------|---------------------|---------------|------------> $19,920 $19,920 $19,920 etc. $19,920 $8,500 | | purchasing holding and $19,865.57 | | costs order costs $19,811.44 | | | etc. | | etc. (discounting at 20% / yr. simple interest) | | $17,562.32 | | | $934,634.82 = PV of sum of purchases | $7,475.90 = PV of holding and order costs | $942,110.72 = total value of inventory cost Note: This is similar to pricing a bond, only the PMTS are in the form of an annuity due, so the formula (using compound interest = 20% / 73 5-day periods in a year) would be: PV = (PMT)(PVIFA k, n )(1 + k) + FV (PVIF k, n ) PV = ($19,920)(PVIFA 20% / 73 , 50 )[1 + (20% / 50)] + ($8,500)(PVIF 20% / 73, 50 )

PV = $932,151.23 + $7,413.24 = $939,564.47 which is slightly less than the $942,110.72 found when using simple interest.

7. ERRATA NOTE: This problem as written in the text contains a flaw that

poses a problem for astute students. The problem puts no limit on discounts, such that if one orders sufficient quantity eventually the price falls to zero. Advise students prior to assigning the problem that the supplier’s quantity discount schedule “max’s out” at 2,500 per order = $9.75/oz.

Beverly Cosmetics - EOQ, optimal order quantity and the cost of capital.

ASSUMPTIONS Discount options Order costs ( F ) $75.00 Quantity Cost Per Holding costs per unit ( H ) $0.15 ( Q ) Unit (C' ) Total annual quantity ( T ) 50,000 1-499 $10.00 Planning Period (in days) 365 500-999 $9.95 Opportunity Cost (per year) 25% 1000-1499 $9.90 1500-1999 $9.85 2,000-2,499 $9.80 2500+ $9.75 a.) EOQ = SQRT(2 * T * F / H) EOQ = 7,071 b.) Results of random trial solutions when k = 0%:

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Q PV Cost 4000 $488,875 5000 $488,625 6000 $488,725 7300 $488,573 = Lowest level of present value of inventory cost for 8000 $488,725 optimum order quantity, found by trial and error, 9000 $488,625 when k = 0% 10000 $488,625 c.) Results of random trial solutions when k = 25%: Q PV Cost 1000 $445,876 2000 $440,947 3,000 $439,317 = Lowest level of present value of inventory cost for 4000 $440,091 optimum order quantity, found by trial and error, 5000 $440,941 when k = 25% 6000 $441,886 Total Cost = (C' * T) + (F * (T / Q) + (H * (Q / 2) ) )

C' Q C' * T F H TC $9.75 3,000 $487,500.00 $1,250.00 $225.00 $488,975.00 Select Order Quantity = 3,000 Total PV of inventory cost = $439,317

Note: Do not be misled by all of the zeros in the spreadsheet printout below. Because it is an interactive spreadsheet, when the order quantity changes, many of the zero cells change to positive numbers to reflect a different order sequence.

. Cost Per Inventory PV Holding PV of Unit of Purchases & Ordering Inventory t Quantity Inventory Per Day PV Factor Costs Purchase 0 3,000 9.75 0 1.0000 0 29,250 1 3,000 9.75 22 0.9852 0 28,818 2 3,000 9.75 44 0.9709 0 28,398 3 3,000 9.75 66 0.9569 0 27,990 4 3,000 9.75 88 0.9434 0 27,594 5 3,000 9.75 110 0.9302 0 27,209 6 3,000 9.75 131 0.9174 0 26,835 7 3,000 9.75 153 0.9050 0 26,471 8 3,000 9.75 175 0.8929 0 26,116 9 3,000 9.75 197 0.8811 0 25,771

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10 3,000 9.75 219 0.8696 0 25,435 11 3,000 9.75 241 0.8584 0 25,107 12 3,000 9.75 263 0.8475 0 24,788 13 3,000 9.75 285 0.8368 0 24,477 14 3,000 9.75 307 0.8264 0 24,174 15 3,000 9.75 329 0.8163 0 23,878 16 2,000 9.8 350 0.8065 0 15,806 17 0 0 0 0.0000 0 0 18 0 0 0 0.0000 0 0 19 0 0 0 0.0000 0 0 20 0 0 0 0.0000 0 0 21 0 0 0 0.0000 0 0 22 0 0 0 0.0000 0 0 23 0 0 0 0.0000 0 0 24 0 0 0 0.0000 0 0 25 0 0 0 0.0000 0 0 26 0 0 0 0.0000 0 0 27 0 0 0 0.0000 0 0 28 0 0 0 0.0000 0 0 29 0 0 0 0.0000 0 0 30 0 0 0 0.0000 0 0 31 0 0 0 0.0000 0 0 32 0 0 0 0.0000 0 0 33 0 0 0 0.0000 0 0 34 0 0 0 0.0000 0 0 35 0 0 0 0.0000 0 0 36 0 0 0 0.0000 0 0 37 0 0 0 0.0000 0 0 38 0 0 0 0.0000 0 0 39 0 0 0 0.0000 0 0 40 0 0 0 0.0000 0 0 41 0 0 0 0.0000 0 0 42 0 0 0 0.0000 0 0 43 0 0 0 0.0000 0 0 44 0 0 0 0.0000 0 0 45 0 0 0 0.0000 0 0 46 0 0 0 0.0000 0 0 47 0 0 0 0.0000 0 0 48 0 0 0 0.0000 0 0 49 0 0 0 0.0000 0 0 50 0 0 0 0.0000 0 0 50,000 = Total annual quantity ( T ) $1,200.00 $438,117

Total Present Value Cost = PV of holding & order costs + PV of inventory purchase costs = $391.30 + $455,820 = $439,317.07

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d.) Economic Order Quantity = 7,071 EOQ solution from part a.) Optimal Order Quantity(w/o Cost of Cap.) = 7,300 User determined from part b.) Optimal Order Quantity(with Cost of Cap.) = 3,000 User determined from part c.)

Compare the three answers and discuss whether the answers make sense to you. It seems logical that considering a quantity discount would generally justify a larger optimal order quantity. However, once the cost of capital iis considered, this would work against having a larger investment in inventory, reducing the optimal order quantity.

8. EBCO, Inc. - COGS, inventory invested, and balance matrices. a.) Calculating average daily COGS. Average Daily COGS (quarterly) = (COGS mo. 1 + COGS mo. 2 + COGS mo. 3) / 90 days Average Daily COGS in Inventory = Ending Inventory / Average Daily COGS ASSUMPTIONS January February March April May June COGS 100 150 225 200 125 90 Ending 40 50 62 62 42 28 Inventory Average Daily COGS (Quarterly) 5.28 6.39 6.11 4.61 Average Days COGS in Inventory 11.75 9.70 6.87 6.07 Purchases = EI - BI + COGS 237 200 105 76

Example: March quarterly COGS = (100 + 150 + 225) / 90 = 5.278 Example: March average daily COGS in inventory = 62 / 5.278 = 11.75 Example: March purchases = 62 - 50 + 225 = 237 b.) Interpretation:

It appears as though inventory is being held for a shorter time period with each successive month from 11.75 days in March to only 6.87 days in May.

c.) Calculating a balance fraction matrix. ASSUMPTIONS Balance Amount Matrix Month of Ending inventory balances for purchases made in previous months Purchase Purchases Feb Mar Apr May June

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February 160 31 15 March 237 47 23 April 200 39 19 May 105 23 11 June 76 17 #N/A 62 62 42 28 Balance Fraction Matrix Month of Ending inventory fractions for purchases made in previous months Purchase Purchases Feb Mar Apr May June February 160 19% 9% March 237 20% 10% April 200 20% 10% May 105 22% 10% June 76 22% Example: February balance fraction for February purchases = 31 / 160 = 19.4% Example: March balance fraction for February purchases = 15 / 160 = 9.4% Example: March balance fraction for March purchases = 47 / 237 = 19.8% A larger portion of each month's purchase remains as an inventory balance with

each successive month through March. Balance fractions for the month of purchase increase from 19% in February to 22% in May. Thus, inventory turnover is actually slowing down slightly.

d.) Explaining the difference in answers b and c.

The balance fraction approach relates the level of inventory at a particular point in time to the level of purchases that originally generated that inventory. This provides a more accurate reflection of inventory usage compared to days COGS held in inventory. Days COGS held in inventory is influenced by trends in the activity level of the firm since it uses the average daily COGS in its calculation. Since these two measures approach the monitoring of inventory differently, there is no reason to think that they would give identical results.

9. Wynn Manufacturing, Inc. - COGS, inventory investment, and balance

matrices. a.) Calculating average daily COGS. Average Daily COGS (Quarterly) = (COGS mo. 1 + COGS mo. 2 +

+ COGS mo. 3) / 90 days Average Daily COGS in Inventory = Ending Inventory / Average Daily COGS ASSUMPTIONS Jan Feb Mar Apr May June COGS 1000 1500 2100 2700 3500 4800

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End. Inv. 300 450 630 810 1050 1440 Avg. Daily COGS (quarterly) 51.11 70.00 92.22 122.22 Days COGS Held in Inv. 12.33 11.57 11.39 11.78 Purchases = EI - BI + COGS 1650 2280 2880 3740 5190 Example: March quarterly COGS = (1000 + 1500 + 2100) / 90 = 51.11 Example: March average daily COGS in inventory = 630 / 51.11 = 12.32 Example: March purchases = 630 - 450 + 2100 = 2280 b.) Inventory is being held for a shorter time period with each succeeding month with

average days COGS dropping from 12.33 days in March to 11.39 days in May.

c.) Calculating a balance fraction matrix. ASSUMPTIONS Balance Amount Matrix Month of Ending inventory balances for purchases made in previous months Purchase Purchases Feb Mar Apr May June February 1650 330 174 March 2280 456 234 April 2880 576 302 May 3740 748 402 June 5190 1038 #N/A 630 810 1050 1440 ` Balance Fraction Matrix Month of Ending inventory fractions for purchases made in previous months Purchase Purchases Feb Mar Apr May June February 1650 20% 11% March 2280 20% 10% April 2880 20% 10% May 3740 20% 11% June 5190 20% Example: February balance fraction for February purchases =330 / 1650=20.00% Example: March balance fraction for February purchases = 174 / 360 = 10.54% Example: March balance fraction for March purchases = 456 / 2280 = 20.00% Discussion: There is a generally a constant balance of inventory after each succeeding month of purchase. This differs from the result using days COGS held in inventory. d.) Explaining the disparity between parts b and c.

The balance fraction approach relates the level of inventory at a particular point in time to the level of purchases that originally generated that inventory. This provides a more accurate reflection of inventory usage compared to days COGS

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held in inventory. Days COGS held in inventory is influenced by trends in the activity level of the firm since it uses the average daily COGS in its calculation. Since, these two measures approach the monitoring of inventory differently, there is no reason to think that they would give identical results.

10. Float-Rite - calculating days COGS held in inventory. Month June July August Sales $50,000 $35,000 $20,000 Cost of goods sold $25,000 $17,500 $10,000 Ending inventory $7,000 $5,000 $3,000

30-day averaging period, days COGS held in inventory 9.00 = $3,000 / ($10,000 / 30)) 60-day averaging period, days COGS held in inventory 6.55 = $3,000 / ($17,500 + $10,000) / 60 90-day averaging period, days COGS held in inventory 5.14 = $3,000 / ($25,000 + $17,500 + $10,000) / 90)