eoq problems

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Chapter 8 Problem Summary Prob. # Concepts Covered Level of Diffic ulty Notes 8.1 EOQ – Determining Q* and R 1 8.2 All Units Quantity Discount Model 2 8.3 Production Lot Size Model 2 8.4 Planned Shortage Model, Determining the Percentage of Customers Placed on Backorder 3 8.5 Periodic Review Inventory – Determining the Order Quantity 3 8.6 Single Period Inventory Model 2 8.7 EOQ – Determining Q* and R 1 8.8 Production Lot Size Model 2 8.9 Planned Shortage Model, Determining the Percentage of Customers Placed on Backorder 3 8.10 All Units Quantity Discount Model 2 8.11 Periodic Review Inventory – Determining the Order Quantity 3 8.12 Single Period Inventory Model, Determining the Expected Profit 3 8.13 EOQ – Determining Q* and R 1 8.14 EOQ – Rounding Off the Solution 1 8.15 EOQ – Rounding Off the Solution 1 8.16 EOQ – Comparing Total Costs If the Demand Forecast Is In Error 2 8.17 Periodic Review Policy – Determining the Order Quantity 2 8.18 Periodic Review Policy – Determining the Order Quantity 2 8.19 Forecasting Demand, Determining Safety Stock Based on Cycle Service Level, EOQ 4 8.20 All Units Quantity Discount 3 8.21 Production Lot Size Model 2 8.22 Production Lot Size Model, 3 Chapter 8 - 1

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Page 1: EOQ Problems

Chapter 8

Problem Summary

Prob. # Concepts Covered Level of Difficulty

Notes

8.1 EOQ – Determining Q* and R 18.2 All Units Quantity Discount Model 28.3 Production Lot Size Model 28.4 Planned Shortage Model, Determining the

Percentage of Customers Placed on Backorder3

8.5 Periodic Review Inventory – Determining the Order Quantity

3

8.6 Single Period Inventory Model 28.7 EOQ – Determining Q* and R 18.8 Production Lot Size Model 28.9 Planned Shortage Model, Determining the

Percentage of Customers Placed on Backorder3

8.10 All Units Quantity Discount Model 28.11 Periodic Review Inventory – Determining the

Order Quantity3

8.12 Single Period Inventory Model, Determining the Expected Profit

3

8.13 EOQ – Determining Q* and R 18.14 EOQ – Rounding Off the Solution 18.15 EOQ – Rounding Off the Solution 18.16 EOQ – Comparing Total Costs If the Demand

Forecast Is In Error2

8.17 Periodic Review Policy – Determining the Order Quantity

2

8.18 Periodic Review Policy – Determining the Order Quantity

2

8.19 Forecasting Demand, Determining Safety Stock Based on Cycle Service Level, EOQ

4

8.20 All Units Quantity Discount 38.21 Production Lot Size Model 28.22 Production Lot Size Model, Examining the

Effect of Changes in Holding Cost3

8.23 EOQ, Determining Safety Stock Based on a Cycle Service Level , Determining the Profit of Carrying an Item

4

8.24 All Units Quantity Discount 38.25 Single Period Inventory Model 18.26 Single Period Inventory Model 3 Note that the goodwill cost is

negative8.27 Planned Shortage Model 28.28 Periodic Review Policy – Determining the Order

Quantity and Safety Stock for a Cycle Service Level

3

8.29 All Units and Incremental Quantity Discount 4

Chapter 8 - 1

Page 2: EOQ Problems

Schedule, Calculating Discount Breaks

Chapter 8 - 2

Page 3: EOQ Problems

8.30 Production Lot Size Model 28.31 Planned Shortage Model – Evaluating Different

Customer Satisfaction Plans, Determining the Profit of Carrying an Item

5

8.32 Planned Shortage Model, Determining the Profit of Carrying an Item

3

8.33 Single Period Inventory Model 18.34 Single Period Inventory Model – Sensitivity

Analysis2

8.35 Determining the Order Quantity for a Periodic Review Model, All Units and Incremental Quantity Discount Schedule, Calculating Discount Breaks, Determining the Reorder Point and Safety Stock for a Cycle Service Level, Determining the Optimal Inventory Plan

8

8.36 Comparing the EOQ and Planned Shortage Models

5

8.37 Make or Buy Problem – Comparing the EOQ and Production Lot Size Models

6

8.38 Incremental Discount Schedule 38.39 Single Period Inventory Model 18.40 Single Period Inventory Model 3 Note that the goodwill cost is

negative8.41 Planned Shortage Model 38.42 Production Lot Size Model -- Evaluating

Different Production Systems5

8.43 Make or Buy Problem – Comparing the EOQ and Production Lot Size Models

6

8.44 Single Period Inventory Model 1 Note that the salvage value is negative

8.45 Single Period Inventory Model 38.46 EOQ, Evaluating the Effect of Changing the

Holding Cost3

8.47 Comparing Annual Profits Using the EOQ Model to Evaluate Two Different Purchase Plans

4

8.48 All Units Quantity Discount Schedul 38.49 All Units Quantity Discount Model 7 This problem requires students to

determine the quantity price breaks from knowing the dollar price breaks. They should use the discounted price per unit in calculating the break points.

8.50 Forecasting Demand, Determining Lead Time Demand and Safety Stock Based on Unit Service Level, EOQ

6

Chapter 8 - 3

Page 4: EOQ Problems

Problem Solutions

8.1 See file Ch8.1.xls

Q* = 340 R = 25

Culton should order 340 bottles when the inventory reaches 25 bottles.

Chapter 8 - 4

Page 5: EOQ Problems

8.2 See file Ch8.2.xls

Calculation of Optimal Inventory Policy Under All-Units Quantity Discounts

    OPTIMAL  INPUTS Values OUTPUTS Values

Annual Demand, D = 104000.00 Order quantity, Q* = 10001Per Unit Cost, C = 16.00 Cycle Time (in years), T = 0.096163462Annual Holding Cost Rate, H = 0.15 # of Cycles Per Year, N = 10.3989601Annual Holding Cost Per Unit, Ch = 2.4000 Reorder Point, R =  

Order Cost, Co = 200.00Total Annual Cost, TC(Q*) = 1426680.81

Lead Time (in years), L =   Safety Stock, SS =  

DISCOUNTS    

Level BreakpointDiscount

Price Q* TC(Q*)Modified

Q*0 1 16.0000 4163 1673992.00 41631 1001 15.2000 4271 1590538.99 42712 5001 14.4000 4389 1507160.25 50013 10001 13.6000 4516 1426680.81 10001

Price-Mart.com should order 10,001 pairs of jeans.

Hand calculations:

Level Range Unit Cost Qi

Modified Qi

Ordering Cost

Holding Cost

Purchasing Cost

Total Cost

0 1– 1000 16 4163 Non Optimal1 1001 – 5000 15.2 4271 4271 200(104000)/4271 2.28(4271)/2 15.2(104000) 15905382 5001 – 10000 14.4 4389 5001 200(104000)/5001 2.16(5001)/2 14.4(104000) 15071603 10001 or more 13.6 4516 10001 200(104000)/10001 2.04(10001)/2 13.6(104000) 1426680

Since 4163 falls above the range its unit cost applies it is non optimal

Since 4271 falls inside the qualifying range it stays unmodified and the corresponding ordering, holding, and purchasing cost need to be calculated.

Since 4389 falls to the left of the qualifying range it needs to be increased until at 5001it qualifies for the discounted cost of $14.40. The same concept applies to Q3 = 4516.

Optimal policy: Order 10001 pairs because the minimum total cost of $1,426,680 is attained at Q = 10001.

Chapter 8 - 5

.15(15.2) = 2.28

.15(14.4) = 2.16

Page 6: EOQ Problems

8.3

ADM should manufacture 2,000 units during each production run. There will be 3 production runs per year.

Hand Calculations:Annual demand = D = 12(500) = 6000Annual production = P = 12(2000) = 24000Set up cost = C0 = 2000Unit cost = C = 40Holding rate = H = .2

Chapter 8 - 6

Production lot size = Q* =

N= D/Q = 6000/2000 = 3

Page 7: EOQ Problems

8.4 See file Ch8.4.xls

Playhouse World should order 14 playhouses. Approximately 14% of customers will have to wait for delivery.

Hand Calculations:This is the EOQ model with planned shortages. Cs = 50*52 = 2600; Cb = 10; C = 3500; D = 2(12) = 24; Co = 1500; L = 1 month. Ch = 1500.

After the substitutions of the parameter values we have Q* = 13.68, and S* = 1.94.The percentage of customers who will have to wait is 1.94/13.68*100 = 14.18 (this is so because the cycle demand is 13.68 of which 1.94 units are not sold on time).Note: The reorder point is R = LD – S = (1month)(monthly demand) – shortage = 1(2) – 1.94 = .06

Chapter 8 - 7

Page 8: EOQ Problems

8.5 See file Ch8.5.xls

SportWorld.com should order 145 dozen golf balls.

Chapter 8 - 8

Page 9: EOQ Problems

8.6 See file Ch8.6.xls

Raul’s Bakery should bake 76 loaves of bread each day.

Chapter 8 - 9

Page 10: EOQ Problems

8.7 See file Ch8.7.xls

a. Food Town should order 3,448 boxes of pasta.

b. The order should be placed when the inventory level reaches 420 boxes.

Hand CalculationsD = 320(52) = 16640; C0 = 30; H = .14; C = .60; L = 1 week; SS = 100a. Order size:

b. Re-order pointR = DL + SS= 16640(1/52) + 100 = 420

Chapter 8 - 10

Page 11: EOQ Problems

See file Ch8.8.xls

Determination of Inventory Policy for Production Lot Size Model

    OPTIMAL  INPUTS Values OUTPUTS Values

Annual Demand, D = 54750.00 Order Quantity, Q* = 5919.46Per Unit Cost, C = 2.25 Cycle Time (in years), T =

Annual Holding Cost Rate, H = 0.20 # of Cycles Per Year, N =

Annual Holding Cost Per Unit, Ch = 0.4500 Reorder Point, R =

Set Up Cost, Co = 90.00 Total Annual Variable Cost, TV(Q*) =

Annual Production Rate, P = 146000.00 Total Annual Cost, TC(Q*) =

Lead Time (in years), L =  Safety Stock, SS =  

Mother Smith’s should bake 5925 pies per production lot.

D = 150(365) = 54750; P = 50(8)(365) = 146000; C = 2.25; H = .2; Co =90

Chapter 8 - 11

Page 12: EOQ Problems

8.9 See file Ch8.9.xls

a. Appliance Alley should order 33 washing machines

b. No one will have to wait for their washing machines.

Chapter 8 - 12

Page 13: EOQ Problems

8.10 See file Ch8.10.xls

Zeigler’s Lumber Supply should order 1,000,000 board feet of 2 by 4’s.

Chapter 8 - 13

Page 14: EOQ Problems

8.11 See file Ch8.11.xls

Bank Drugs should order 1713 tablets.

Chapter 8 - 14

Page 15: EOQ Problems

8.12 See file Ch8.12.xls

a. Furr’s should order 277 calendars.

b. Furr’s will have an expected profit of approximately $1,178.

Chapter 8 - 15

Page 16: EOQ Problems

8.13 See file Ch8.13.xls

Hand Calculations:D = 7(365) = 2555; Ch = .2(.8) = .16; Co = 25. SS = 15

a. Buyright should order Stick razors and use a reorder point of R =

LD+SS = (5/365)2555+15 = 50 razors. b. Days between orders = Q*/D years = Q*/D*365 days = .3497*365 = 128 calendar days c. Total Annual Cost (including holding cost of safety stock) = CoD/Q*+Ch(Q*/2)+CD+ChSS = 25(2555/894)+.16(894/2)+.8(2555)+.16(15) = $2,189.37, and the Projected Net Annual Profit = 2,555*($1.49) - $2,189.37 = $1,617.58.

This solution assumes demand occurs at a constant rate and will continue on at this rate forever.

Chapter 8 - 16

Page 17: EOQ Problems

8.14 See file Ch8.14.xls

a. Buyright should order Q* = 864 Stick razors (6 lots of 144) and use a reorder point of R = 55 razors. b. Days between orders = .3382*365 = 123 calendar days c. Total Annual Cost (including holding cost of safety stock) = $2,190.25 and the Projected Net Annual Profit = 2,555*($1.49) - $2,190.25 = $1,616.70.

This solution can also be obtained by adding a cell for NL, number of lots, in G13 and a constraint in cell G14, =$H$5 - $G$14*144. Solver changes are Changing Cell becomes G13 with constraints G13 = Integer and G14 = 0.

Chapter 8 - 17

Page 18: EOQ Problems

8.15 See file Ch8.15.xls

a. The order quantity is Q* = 2,200 boxes or 22 increments when the inventory level reaches R = 425 boxes. b. The number of days between orders = .1303*365 = 47.6 calendar days c. Total Annual Cost (including holding cost of safety stock) = $16,536.25 and the Projected Net Annual Profit = 16,900*($1.29) - $16,536.25 = $5,264.75.

Chapter 8 - 18

Page 19: EOQ Problems

8.16 See files Ch8.16a.xls and Ch8.16b.xls

a.

a. The optimal order quantity is Q* = 2,400 boxes.

Chapter 8 - 19

Page 20: EOQ Problems

8.16b See file Ch8.16b.xls

b. $19,041.71 - $19,040.45 = $1.26

Chapter 8 - 20

Page 21: EOQ Problems

8.17 See file Ch8.17.xls

The optimal order quantity is Q* = 155

Chapter 8 - 21

Page 22: EOQ Problems

8.18 See file Ch8.18.xls

The optimal order quantity is Q* = 135

Chapter 8 - 22

Page 23: EOQ Problems

8.19 See file Ch8.19.xls

D = 52*21 = 1092. From the sample the weekly mean demand based on the 8 weeks moving average approach = 21. So for a 2-week lead time we have L = 21*2 = 42 and L = 5.45 (given).

a. The optimal order quantity is Q* = 177 (rounded) based on the EOQ model. A 96% cycle service level corresponds to a z value of approximately 1.75. Hence, the reorder point equals + z.96*L = 42 + 1.75*5.45 = 51.54 = 52 (rounded up).

b. Days between orders equals Q*/D = .1618 years = .1618*365 = 59 calendar days

c. Total annual inventory cost (including safety stock holding cost) = CoD/Q* + ChQ*/2 + ChSS + CD = $9,946.93 (note that the safety stock = SS = z.96*L = 1.75(5.45) = 9.5375)

Comment: Suppose management decides to set the re-order point at R = 50 units. What is the new cycle service level? From the reorder point formula we have 50 = L+zSLL = 42+zSL(5.45) which yields zSL = (50 – 42)/5.45 = 1.47. The service level is found from the normal table or from =normdist(1.47) = .929. The safety stock for this case is 50 – 42 = 8.

Chapter 8 - 23

Page 24: EOQ Problems

8.20 See file Ch8.20.xls

a. The optimal order quantity is Q* = 400 and the reorder point is R = 63

b. Number of days between orders = .3205*365 = 117 calendar days

c. Total annual cost (including holding cost of safety stock) = $21,143.69

We assume demand occurs a constant rate and will continue on at this rate forever, safety stock was ordered at the discount price and the only cost associated with the safety stock is the holding cost.

Chapter 8 - 24

Page 25: EOQ Problems

8.21 See file Ch8.21.xls

Hand CalculationsD = 25000(52) = 1,300,000; P = 10000(6)(52) = 3,120,000; Co = 325; Ch = .55

a. The optimal production batch is obtained from using the formula = 51,320 bottles.

Total annual ordering and holding cost = $16,465.24

b. Length of a production run = (Q*/P per hour) +2 = 51,320/1000 + 2 = 53.3 hours

c. Number of days between the start time of successive production runs = T = Q*/D = .0395 years = .0395*365 = 14.4 calendar days.

Chapter 8 - 25

Note that hourly production = 10,000/10hours = 1,000

Page 26: EOQ Problems

We assumed that demand occurs at a constant rate and production will not be interrupted by work stoppages.

Chapter 8 - 26

Page 27: EOQ Problems

8.22 See files Ch8.22a.xls and Ch8.22b.xls

a.

a. The optimal production batch size is Q* = 485,917 The length of a production run = 485,917/(2*60*60) + 50/60 = 68.3 hours

b. Total annual inventory cost = $38,895.54

c. Number of days between the start time of successive production runs = .0231*365 = 8.43 calendar days.

Chapter 8 - 27

Page 28: EOQ Problems

8.22 d. See file Ch8.22d.xls

d. The optimal production batch size would increase to Q* = 532,295 The length of a production run would equal 532,295/(2*60*60) +50/60 = 74.76 hours

Chapter 8 - 28

Page 29: EOQ Problems

8.23 See Ch8.23.xls

D = 52*60 = 3120Lead Time Demand ~ N( = 70, = 15)

a. The optimal order quantity is Q* = 77 with a reorder point of 22. The number of inventory cycles per year = 3120/77 = 40.519. Hence, Click would like the cycle service level to be 1 - 1/40.519 = 97.52%. The z value corresponding to this service level is approximately z = 1.96. Hence, the reorder point = + z* = 70 + 1.96*15 = 99.4 = 99 (See the Cycle Service Level worksheet) and the safety stock is z* = 1.96*15 = 29. However, because this reorder point exceeds the order quantity of 77, orders must be placed one inventory cycle early when the inventory level reaches 99 - 77 = 22.

b. There are 52*6 = 312 working days in a year. Hence, the number of days between orders = .0248*312 = 7.74 working days.

c. Total annual cost (including holding cost of safety stock) = $1,632,364.75

d. Projected annual profit = 3120*($649.99) - $1,632,364.75 = $395,604.05

Chapter 8 - 29

Page 30: EOQ Problems

8.24 See file Ch8.24.xls

a. The optimal order quantity becomes Q* = 1000 with a reorder point of 77. The number of inventory cycles per year = 3120/1000 = 3.12. Hence, Click would like the cycle service level to be 1 - 1/3.12 = 67.95%. The z value corresponding to this service level is approximately z = .465. Therefore the reorder point = + z* = 70 + .465*15 = 76.975 = 77 and the safety stock is z* = .465*15 = 7.

b. There are 52*6 = 312 working days in a year. Hence, the number of days between orders = .3205*312 = 100 working days.

c. Total annual cost (including holding cost of safety stock) = $1,519,849.03.

d. Projected annual profit = 3120*($649.99) - $1,519,849.03 = $508,119.79

Chapter 8 - 30

Page 31: EOQ Problems

8.25 See file Ch8.25.xls

Masks-R-Us should order Q* = 1,047. The expected profit would equal $1,534.83.

Chapter 8 - 31

Page 32: EOQ Problems

8.26 See file Ch8.26.xls

a. Skip Gunther should produce 1817 plates (note goodwill cost is -$40). b. Expected profit = $96,364.24

c. We assumed the company pays a royalty on plates produced even if they are destroyed.

Chapter 8 - 32

Page 33: EOQ Problems

8.27 See file Ch8.27.xls

a. The optimal order quantity is Q* = 81 with a reorder point = 23.27 = 23 .

b. The number of days between orders = .1299*365 = 47.4 calendar days.

c. The percentage of customers who will be placed on backorder = 1/81 = .0012 = 1%.

d. Total annual cost = $11,962.82 Projected annual profit = 624*($32.95) - $11,962.82 = $8,597.98.

Chapter 8 - 33

Page 34: EOQ Problems

8.28 See file Ch8.28.xls

For this problem one needs to consider the probability of a stock out over the entire review period (three weeks). The mean demand is 600 and the standard deviation of demand is 51.96. This results in a reorder point of 707 units.

Chapter 8 - 34

Page 35: EOQ Problems

8.28 continued

a. The optimal order quantity is Q* = 677.

b. The safety stock = 107 units.

Chapter 8 - 35

Page 36: EOQ Problems

8.29 See file Ch8.29.xls

a.

With an all-units discount policy Q* = 9,000

Chapter 8 - 36

Page 37: EOQ Problems

8.29b.

b. With an incremental discount policy Q* = 14,190

c. The reorder point is 5769 pounds.

We assumed that demand is constant and will continue on at the same rate forever.

Chapter 8 - 37

Page 38: EOQ Problems

8.30 See files Ch8.30french.xls and Ch8.30amaretto.xls

For French Roast Q* = 750 (Since they will not roast more than a 10-day supply of the coffee)

Note, finding this solution requires the Solver constraint, H5 <= 750.

Chapter 8 - 38

Page 39: EOQ Problems

8.30 continued

For Amaretto Cream Q* = 200 (since the store will not roast more than a 10-day supply of the coffee)

Note, finding this solution requires the Solver constraint, H5 <= 750.

Chapter 8 - 39

Page 40: EOQ Problems

8.31 See files Ch8.31a.xls and Ch8.31b.xls

Plan I

Chapter 8 - 40

Page 41: EOQ Problems

8.31 continued

Plan II

a. Pete's should adopt Plan I and give one pound of coffee free to backordered customers as the annual inventory costs are slightly lower under this plan.

b. Under Plan I -- Q* = 44 and the reorder point is 17.

c. The expected annual profit = 180*($189) - $18,338.71 = $15,681.29

Chapter 8 - 41

Page 42: EOQ Problems

8.32 See file Ch8.32.xls

a. The optimal order quantity is Q* = 111 and the reorder point is R = 39. .

b. The number of weeks between orders = .0285*52 = 1.48.

c. The percentage of customers who will be placed on backorder = 0%.

d. Annual profit on cash registers = 3900*($80) - $7,065.51 = $304,934.49

Note, profit can also be expressed as net sales less total costs as depicted in cell E16.

Chapter 8 - 42

Page 43: EOQ Problems

8.33 See file Ch8.33.xls

Clothesline should order 9591 scarves. We assume Clothesline will only get one order delivered during the season.

Chapter 8 - 43

Page 44: EOQ Problems

8.34 See files Ch8.34a.xls, Ch8.34b.xls, Ch8.34c.xls, Ch8.34d.xls

a. If g = $.10, Q* = 201.

Chapter 8 - 44

Page 45: EOQ Problems

8.34 b.

b. If g = $.50, Q* = 214.

Chapter 8 - 45

Page 46: EOQ Problems

8.34 c.

c. If g = $1.00, Q* = 222.

Chapter 8 - 46

Page 47: EOQ Problems

8.34 d.

d. If g = $5.00, Q* = 234

Chapter 8 - 47

Page 48: EOQ Problems

8.35 See files Ch8.35c.xls and Ch8.35g.xls

a. Lead time demand ~ N( = 60, = 12). Since Circle 7 wants a 99% cycle service level, z = 2.33 and the safety stock = z* = 2.33*12 = 28. Hence, Q* = (T+L)*D + SS - SH = (1)*60 +29 - 22 = 66.

b. Total annual cost = annual holding cost + cost of the cola = (30+28)*($1.0625) + 60*52*4.25 = $13,322.75.

c.

c. Circle 7 should order Q* = 900 cases.

d. The reorder point is 88.

e. Total annual cost = $11,788.08.

f. Yes, the annual costs decrease by approximately $1,500 if Circle 7 takes advantage of the discount schedule.

Chapter 8 - 48

Page 49: EOQ Problems

8.35 g. See file Ch8.35g.xls

g. In this case, with a safety stock of 28, the optimal order quantity is Q* = 250 - 28 = 222, the reorder point equals 60 + 28 = 88, and the total annual cost = $12,479.94.

Chapter 8 - 49

Page 50: EOQ Problems

8.36 See file Ch8.36.xls

a.

a. The optimal order quantity is Q* = 46 R = 2. Profit = $22,332.07.

Chapter 8 - 50

Page 51: EOQ Problems

8.36 b.

b. The optimal order quantity is Q* = 46 with a reorder point R = 17. Profit = $22,484.28.

c. Yes, allowing for backorders increases profit by approximately $150 per year.

Chapter 8 - 51

Page 52: EOQ Problems

8.37 See file Ch8.37.xls

a. The optimal order quantity is Q* = 11,051 and the reorder point is R = 9,632. Cycle Time = .0789*365 = 29 calendar days. Total annual cost (including holding cost of safety stock) = $1,773,527.99.

Chapter 8 - 52

Page 53: EOQ Problems

8.37 b.

b. The optimal batch size is Q* = 44,143. The length of a production run = 44,143/2,000 = 22 working days or 26 calendar days. Cycle Time = .3153*365 = 115 calendar days. Total annual cost (including machine lease cost) = $1,719,144.58 + $5,000 = $1,724,144.58.

c. Company should begin in-house production. It will save about $50,000 per year.

Chapter 8 - 53

Page 54: EOQ Problems

8.38 See file Ch8.38.xls

a. The optimal order quantity is Q* = 37,427 b. Total annual cost = $1,703,822.51

Chapter 8 - 54

Page 55: EOQ Problems

8.39 See file Ch8.39.xls

Johansen’s should purchase Q* = 83 waffle cones daily.

Chapter 8 - 55

Page 56: EOQ Problems

8.40 See File ch8.40.xls

Frank’s should order 5 mowers at the temporary discount.

Chapter 8 - 56

Page 57: EOQ Problems

8.41 See file Ch8.41.xls

a. United Parcel Delivery should order Q* = 13 engines when the stock on hand reaches R = 5 engines.

b. Total annual cost = $143,615.38

Chapter 8 - 57

Page 58: EOQ Problems

8.42 See file Ch8.42II.xls

If the firm uses machine I since P = 2,750*365 = 1,003,750, this is the maximum amount the company could sell and it would not have any inventory. Hence, the annual profit to the firm using machine I is $.20*1,003,750 - $40,000 - $4,000 = $156,750 (note that this assumes the $4,000 setup cost is charged once during the year and not amortized over more than one year).

If the firm uses Machine II we have the following output

Because the frankfurters cannot be kept for more than three weeks, the production quantity must be reduced from 370,032 down to 98,742. With this reduction in the production quantity, the profit is lower using machine II than with machine I.

Note: To determine the maximum number of frankfurters that can be sold over the three weeks we solve the following equation for x: 69,230.77 = (1 - 1,200,000/4,015,000)x, giving a value of x = 98,742. Three week demand is (3/52)*(1,200,000) = 69,230.77. During production, (1,200,000/4,015,000)% of frankfurters satisfy the demand, the remainder (1-1,200,000/4,015,000) go to inventory. The maximum inventory from a production level, x, cannot exceed the three week demand.

Chapter 8 - 58

Page 59: EOQ Problems

8.43 See file Ch8.43.xls

a. If Mercury purchases the laces the optimal order quantity is Q* = 164,702 and the reorder point R = 20,029. The number of days between orders = .3514*365 = 128 calendar days. The total annual cost (including safety stock holding cost) = $17,064.18.

Chapter 8 - 59

Page 60: EOQ Problems

8.43 b.

b. If Mercury produces the laces in-house the optimal production quantity is Q* = 445,730. Production run time = 445,731/(2,000,000/(260*10)) + 4 = 579.45 + 4 = 583.45 hours and the number of days between the start of successive production runs = .9707*365 = 354.3 calendar days. Total annual cost (including machine lease cost) = $16,342.17 + $1,800 = $18,142.17.

c. Mercury should continue to purchase laces from Tiright as the cost would be over a $1,000 less per year.

Chapter 8 - 60

Page 61: EOQ Problems

8.44

The troop should purchase Q* = 204 trees.

Chapter 8 - 61

Page 62: EOQ Problems

8.45 See file Ch8.45.xls

Business Daily should place Q* = 40 newspapers in the kiosk.

Chapter 8 - 62

Page 63: EOQ Problems

8.46 See files Ch8.46a.xls and Ch8.46c.xls

a. Ibex should order Q* = 2,471 chairs when the inventory level reaches R = 1,187 units. The number of days between orders = .0752*365 = 27 calendar days.

b. Total annual costs (including holding cost of safety stock) = $324,142.53 + 500*($1.56) = $324,922.53

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8.46 c. See file Ch8.46c.xls

c. The order quantity would decrease to Q* = 2,230. An order would be placed when the inventory level reaches R = 1,329 units.

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8.47 See file Ch8.47.xls

The cost of the new supplier is $330,942.92. Hence, it is not worth switching manufacturers.

Chapter 8 - 65

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8.48 See file Ch8.48.xls

a. The optimal order quantity is Q* = 5,000.

b. The reorder point is R = 500 + 50 = 550.

c. The number of days between orders is .3846*365 = 140 calendar days.

d. Total annual costs (including holding cost of safety stock) = $24,789.20.

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8.49 See file Ch8.49.xls

a. The optimal order quantity is Q* = 170,941 pounds at a cost of $70,000 based on a price of $0.4095 per pound.

b. R = L*D + SS - Q* = 180,385 - 170,941 = 9,444 and it will occur in the previous inventory cycle.

c. Total annual cost (including holding cost of safety stock) = $1,507,793.56.

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8.50 See file Ch8.50.xls

a. The optimal order quantity is Q* = 158.

b. The number of days between orders = .0362*365 = 13 calendar days.

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8.50 c.

c. Since the sample mean = 84, we would estimate the mean lead time demand as 3*84 = 252. Since the sample variance = 69.5556, we would estimate the variance of the lead time demand as 3*69.5556 = 208.6668. Hence, the standard deviation of the lead time demand = 208.6668 = 14.45.

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8.50 c. continued

c. The reorder point is R = 286 which implies a safety stock = 34 units.

d. Total Inventory Cost = $197,769.52.

Chapter 8 - 70