• Author / Uploaded
• Khánh Đoan Lê Đình

Citation preview

ESI 6323 Sample Questions for Second Test Problem 1. A manufacturing firm in Calgary produces an item using a 3-month time supply rule (i.e., each production lot covers 3 months of demand). An analyst, attempting to introduce a more logical approach to selecting run quantities, has obtained the following estimates of characteristics of the item: Annual demand rate D = 4000 units/year Fixed order cost S = $5 Variable order cost c = $4 per 100 units Holding cost rate h = 25% (annually)

Assume the ordered amount is delivered instantaneously upon ordering. a) What is the EOQ of the item? EOQ = sqrt(2*5*4000/0.01) = 2000 b) What is the time between consecutive replenishments of the item when the EOQ is used? T* = EOQ/D = 2000/4000 = ½ year c) The production manager insists that the S = $5 figure was pulled out of the air, that is, it is only a guess. Therefore, she insists on using her simple 3-month supply rule. Determine the range of S values for which the EOQ (based on S = $5) would be preferable (in terms of lower total ordering and holding costs) to the 3-month supply. 3-month rule: Q = 1000 Cost of 3-month rule = S * 4000/1000 + 0.01*1000/2 = 4S + 5 EOQ at S = 5: 2000 Cost of Q = 2000: S*4000/2000 + 0.01*2000/2 = 2S + 10. We need 2S + 10 ≤ 4S + 5, i.e., 2S ≥ 5, or S ≥ $2.50

1

Problem 2. The Wagner Company supplies electric motors to Electronic Distributors, Inc. on a delivered-price basis. Wagner has the responsibility for providing transportation. The traffic manager has three transportation service choices for delivery—rail, piggyback, and truck. He has compiled the following information: Transport Mode Rail Piggyback Truck

Transit Time, Days 16 10 4

Rate, $/Unit 25.00 44.00 88.00

Shipment Size, Units 10,000 7,000 5,000

Electronic Distributors purchases 50,000 units per year at a delivered contract price of $500 per unit. Inventory carrying costs for both companies are 25 percent per year. Which mode of transportation should Wagner select? Rail Cycle stock cost = (0.25)*(500)*(10,000/2) = 625,000 Average pipeline inventory = (16/365)*(500)*(0.25)*(50,000) = 273,972.60 Transportation cost = 25*50,000 = 1,250,000 Total cost = $2,148,972.60 Piggyback Cycle stock cost = (0.25)*(500)*(7,000/2) = 437,500 Average pipeline inventory = (10/365)*(500)*(0.25)*(50,000) = 171,232.88 Transportation cost = 44*50,000 = 2,200,000 Total cost = $2,808,732.88 Truck Cycle stock cost = (0.25)*(500)*(5,000/2) = 312,500 Average pipeline inventory = (4/365)*(500)*(0.25)*(50,000) = 68,493.15 Transportation cost = 44*50,000 = 4,400,000 Total cost = $4,780,993.15

2

Problem 3. The weighty trash bag company has the following price schedule for its large trash can liners. For orders of less than 500 bags, the company charges 30 cents per bag; for orders of 500 or more but fewer than 1,000 bags, it charges 29 cents per bag; and for orders of 1,000 or more, it charges 28 cents per bag (an all-units discount structure). The weighty trash bag company’s only customer is Publix supermarkets, who consume the bags at a steady rate of 600 per year. It costs Publix $8 to place an order, and the Publix holding cost rate is 20%. (a) What is the optimal trash bag order size for Publix?

( 2 ) ( 8) ( 600 ) ( 0.2 ) ( 0.3) ( 2 ) ( 8) ( 600 ) Q(1) = ( 0.2 ) ( 0.29 ) ( 2 ) ( 8) ( 600 ) Q(2) = ( 0.2 ) ( 0.28) Q(0) =

= 400 = 406 = 414

The only realizable EOQ value is Q(0) = 400. We need to check cost at Q(0) = 400, and at the breakpoints of 500 and 1000. G0(400) = (600)(0.3) + ( 2 ) ( 8 ) ( 600 ) ( 0.2 ) ( 0.3) = $204 G1(500) = (500)(0.29) + (600)(8)/500 + (0.2)(0.29)(500)/2 = $198.10 G2(1000) = (1000)(0.28) + (600)(8)/1000 + (0.2)(0.28)(1000)/2 = $200.80 So the optimal order quantity is 500.

3

(b) It costs the weighty trash bag company $24 each time Publix places an order, it costs them 20 cents to produce a bag, and they also use a holding cost rate of 20%. The weighty trash bag company produces at a steady rate of 600 per year (the same rate as demand) and each time an order is placed by Publix, it is immediately transported to Publix (with negligible lead time; that is, the bag producer builds up inventory at a steady rate until the exact order/delivery quantity is on hand, at which point it is immediately shipped to Publix, and this process repeats itself infinitely many times). What is the Publix order quantity that would maximize the profit of the weighty trash bag company? Q* =

( 2 ) ( 24 ) ( 600 ) = 848.5 ( 0.2 ) ( 0.2 )

(c) What order quantity would minimize total supply chain costs? (assume that the 20 cent production cost includes the per unit transportation cost to Publix.) The optimal order quantity for the system must be between 500 and 848.53, which implies a selling price of 0.29. Let us consider the system costs at this price. Setup Costs = supplier + retailer setup costs = (24 + 8)(600)/Q Holding Costs = supplier + retailer holding costs = (0.2)(0.2 + 0.29)Q/2 Q* =

( 2 ) ( 24 + 8) ( 600 ) = 626 ( 0.2 ) ( 0.2 + 0.29 )

4

Problem 4 Consider a simple two-stage supply chain where a manufacturer directly supplies a retailer with a single product. Assume that the manufacturer is the only supplier for the retailer and the retailer is the manufacturer’s only customer for the product. The retailer observes customer demand for the product at a steady rate of D units per year and no shortages are allowed. The manufacturer’s production rate is exactly matched to the demand rate. Each time the retailer places an order to the manufacturer it costs $S in administrative and processing costs. The retailer incurs a cost of $H for holding a unit in inventory for one year. When the retailer places an order, the manufacturer makes a shipment, which costs the manufacturer $β S in shipping costs (a fixed cost per shipment; assume a constant or negligible lead time). The manufacturer incurs a cost of $ αH for holding a unit in inventory for one year. Each time the retailer places an order it is for the same quantity, Q. (a) Draw the graph of the inventory level over time for both the retailer and manufacturer below.

Retailer’s Inventory level

time Manufacturer’s Inventory level

time

(b) In terms of the parameters given, what is the retailer’s optimal order quantity? QR* = sqrt(2SD/H)

5

(c) In terms of the parameters given, what is the optimal quantity for the manufacturer to ship? QM* = sqrt(2β SD/α H) (d) Write the sum of average annual retailer plus manufacturer order/shipment plus holding costs. ATC = SD/Q + HQ/2 + β SD/Q + α HQ/2 = (β +1)SD/Q + (α +1)HQ/2 (e) What is the optimal order/shipment quantity for minimizing average annual retailer plus manufacturer order/shipment plus holding costs? Q* = sqrt(2(β + 1)SD/((α + 1)H)) (f) In terms of the parameters α and β , what condition is required for the manufacturer’s optimal shipment quantity to equal the retailer’s optimal order quantity?

α=β (g) If the condition required in your answer to part (f) is violated, should the retailer and manufacturer work together to ensure that the order/shipment quantity in part (e) is used? Ideally yes, provided they can agree on an arrangement to share the savings.

6

Problem 5. (a) Given the following data for demand at the XYZ Company, calculate the monthly forecast for 2003 using a 3-month moving average. Calculate the MAD and the tracking signal. Is this a good forecast? Period Oct 02 Nov 02 Dec 02 Jan 03 Feb 03 Mar 03 Apr 03 May 03 Jun 03 Jul 03 Aug 03 Sep 03 Oct 03 Nov 03 Dec 03

Demand 850 950 900 1000 950 1050 850 1100 900 1150 1100 900 1000 800 1000

Forecast

Error

MAD

Bias

TS

Demand 850 950 900 1000 950 1050 850 1100 900 1150 1100 900 1000 800 1000

Forecast

Error

MAD

Bias

TS

900 950 950 1000 950 1000 950 1050 1050 1050 1000 900

-100 0 -100 150 -150 100 -200 -50 150 50 200 -100

100 50 66.7 87.5 100 100 114.3 106.3 111.1 105 113.6 112.5

-100 -100 -200 -50 -200 -100 -300 -350 -200 -150 50 -50

-1 -2 -3 -.57 -2 -1 -2.62 -3.29 -1.8 -1.43 .44 -.44

Answer: Period Oct 02 Nov 02 Dec 02 Jan 03 Feb 03 Mar 03 Apr 03 May 03 Jun 03 Jul 03 Aug 03 Sep 03 Oct 03 Nov 03 Dec 03

The standard deviation of the random element of demand is approximately 140.63. The tracking signal indicates that this forecast method tends to underforecast demand, but not to the extreme.

7

(b) Using data from part (a), calculate the monthly forecast for 2003 using simple exponential smoothing with an α = .2. Calculate the MAD and the tracking signal. How does this forecast compare with the previous one? Period Oct 02 Nov 02 Dec 02 Jan 03 Feb 03 Mar 03 Apr 03 May 03 Jun 03 Jul 03 Aug 03 Sep 03 Oct 03 Nov 03 Dec 03

Demand 850 950 900 1000 950 1050 850 1100 900 1150 1100 900 1000 800 1000

Level

Demand 850 950 900 1000 950 1050 850 1100 900 1150 1100 900 1000 800 1000

Level

Forecast

Error

Forecast

Error

MAD

Bias

TS

983

Answer: Period Oct 02 Nov 02 Dec 02 Jan 03 Feb 03 Mar 03 Apr 03 May 03 Jun 03 Jul 03 Aug 03 Sep 03 Oct 03 Nov 03 Dec 03

983 986.4 979.1 993.3 964.6 991.7 973.4 1008.7 1027.0 1001.6 1001.3 961.0 968.8

983.0 986.4 979.1 993.3 964.6 991.7 973.4 1008.7 1027.0 1001.6 1001.3 961.0

-17.0 36.4 -70.9 143.3 -135.4 91.7 -176.6 -91.3 127.0 1.6 201.3 -39.0

MAD

17.0 26.7 41.4 66.9 80.6 82.4 95.9 95.3 98.8 89.1 99.3 94.3

Bias

TS

-17.0 19.4 -51.5 91.8 -43.5 48.2 -128.5 -219.8 -92.8 -91.3 110.0 71.0

-1.0 0.7 -1.2 1.4 -0.5 0.6 -1.3 -2.3 -0.9 -1.0 1.1 0.8

The simple exponential smoothing has a smaller MAD and therefore a smaller standard deviation than the moving average. The tracking signal indicates that simple exponential smoothing over-forecasts a little, where the moving average under-forecasts a little.

8

Problem 6. The demand for Krispee Crunchies, a favorite breakfast cereal of people born in the 1960s, is experiencing a decline. The company wants to monitor demand for this product closely as it nears the end of its lifecycle, and decides to use Holt’s method with α = 0.1 and β = 0.2. At the end of December, the January estimate for the average number of cases sold per month was 900,000, and the trend was –50,000 per month. The following table shows the actual sales history for January, February, and March. Generate forecasts for February, March, and April. Month January February March

Sales 890,000 800,000 825,000

LJ = 0.1(890,000) + 0.9(900,000 – 50,000) = 854,000 TJ = 0.2(854,000 – 900,000) + 0.8(–50,000) = –49,200 Forecast for Feb: 854,000 – 49,200 = 804,800 LF = 0.1(800,000) + 0.9(854,000 – 49,200) = 804,320 TF = 0.2(804,320 – 854,000) + 0.8(–49,200) = –49,296 Forecast for Mar: 804,320 – 49.296 = 755,024 LM = 0.1(825,000) + 0.9(804,320 – 49,296) = 762,021.6 ≈ 762,022 TM = 0.2(762,022 – 804,320) + 0.8(–49,296) = –47,896.4 ≈ –47,897 Forecast for Apr: 762,022 – 47,897 = 714,125

9

Problem 7 The Northville Post Office experiences a seasonal pattern of daily mail volume every week. The following data for two representative weeks are expressed in thousands of pieces of mail. Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday Total

Week 1 5 20 30 35 49 70 15 224

Week 2 8 15 32 30 45 70 10 210

(a) Calculate a seasonal factor for each day of the week. Day Sunday

Seasonal factor

Monday Tuesday Wednesday Thursday Friday Saturday

(b) If the postmaster estimates that there will be 230,000 pieces of mail to sort next week, forecast the volume for each day of the week. Day Sunday

Forecast

Monday Tuesday Wednesday Thursday Friday Saturday

10

Solution option 1: Get seasonal factors by dividing by average demand in the week: Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday Total Average

Week 1 5 20 30 35 49 70 15 224 224/7 = 32

SF1 5/32 = 0.15625 20/32 = 0.62500 30/32 = 0.93750 35/32 = 1.09375 49/32 = 1.53125 70/32 = 2.18750 15/32 = 0.46875

Week 2 8 15 32 30 45 70 10 210 210/7 = 30

SF2 8/30 = 0.26667 15/30 = 0.50000 32/30 = 1.06667 30/30 = 1.00000 45/30 = 1.50000 70/30 = 2.33333 10/30 = 0.33333

Avg SF 0.21146 0.56250 1.00209 1.04688 1.51563 2.26042 0.40104

Average daily demand is 230,000/7 = 32,857 Day Sunday

Calculation 0.21146(32,857)

Forecast 6,948

Monday

0.56250(32,857)

18,482

Tuesday

1.00209(32,857)

32,926

Wednesday

1.04688(32,857)

34,397

Thursday

1.51563(32,857)

49,799

Friday

2.26042(32,857)

74,271

Saturday

0.40104(32,857)

13,177

Solution option 2a: Using the moving average from the book, filling in first Wed. value back to Sunday, and last Wed. value forward to Saturday. Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Value 5 20 30 35 49 70 15 8 15 32 30 45 70 10

Deseasonalized SF Averaged SFs Forecast 32.00 0.156250 0.205979 6767.8 32.00 0.625000 0.556686 18291.0 32.00 0.937500 0.989680 32517.9 32.00 1.093750 1.046875 34397.2 32.43 1.511013 1.505507 49466.4 31.71 2.207207 2.270270 74594.3 32.00 0.468750 0.401042 13177.0 31.29 0.255708 30.71 0.488372 30.71 1.041860 30.00 1.000000 30.00 1.500000 30.00 2.333333 30.00 0.333333

11

Solution option 2b: Using the moving average from the book, filling in avg of first Wed. & Thurs. values back to Sunday, and last Tues. & Wed. values forward to Saturday. Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Value 5 20 30 35 49 70 15 8 15 32 30 45 70 10

Deseasonalized SF Averaged SFs Forecast 32.21 0.155211 0.205459 6750.8 32.21 0.620843 0.554607 18222.7 32.21 0.931264 0.986562 32415.5 32.00 1.093750 1.046875 34397.2 32.43 1.511013 1.496683 49176.5 31.71 2.207207 2.256545 74143.3 32.00 0.468750 0.399081 13112.6 31.29 0.255708 30.71 0.488372 30.71 1.041860 30.00 1.000000 30.36 1.482353 30.36 2.305882 30.36 0.329412

12

Problem 8 The XYZ Company has an assembly plant in Cincinnati and its parts plant in Indianapolis. Parts are transported from Indianapolis to Cincinnati using trucks. Each shipment costs $100. The Cincinnati plant assembles and sells 300 finished products each day and operates 50 weeks a year (working 5 days per week). Part #456 costs $50 and XYZ Company incurs a holding cost of 20 percent per year. How many of part #456 should XYZ Company put in each shipment? What is the cycle inventory of part #456 at XYZ Company? Answer:

Q* = √2DS/hC = √(2(300 x 50 x 5)$100)/(.2 x $50) = 1224.745 ≈ 1225 Average cycle inventory = Q*/2 = 1225/2 = 612.5

13

Problem 9 Using the information from the previous problem, suppose XYZ gets price breaks on part #456 such that for shipments between 2500 and 3499, the cost is $49.98 per unit, and for shipments of 3500 and greater, the cost is $49.90 per unit (for all units). How many of part #456 should XYZ Company put in each shipment? Answer:

At C0, Q* = √2DS/hC = √(2(300 x 50 x 5)$100)/(.2 x $50) = 1224.745 ≈ 1225

Avg. annual cost = (300*50*5)*50 + sqrt(2*100*(300*50*5)*(0.2*50)) = $3,762,247.45

At C1, Q* = sqrt(2(300 x 50 x 5)$100)/(.2 x $49.98)) ≈ 1225 Therefore, we need to check the cost at Q = 2500: Avg. annual cost = (300*50*5)*49.98 + 100(300*50*5)/2500 + (0.2*49.98)2500/2 = $3,763,995.00 At C2, Q* = sqrt(2(300 x 50 x 5)$100)/(.2 x $49.95)) ≈ 1226 Avg. annual cost = (300*50*5)*49.90 + 100(300*50*5)/3500 + (0.2*49.90)3500/2 = $3,762,107.86 So the optimal lot size is 3500.

14

Problem 10 Formulate a linear programming model for the following aggregate planning problem: Number of time periods: Demand schedule: Current regular time capacity: Overtime capacity: Unit variable cost: Inventory carrying cost: Cost to add one unit of capacity: Cost to reduce capacity by one unit:

3 D1 = 45, D2 = 52, D3 = 60 45 units per period 25% of regular time capacity $800 on regular time, $1,000 on overtime $12 per unit per period $240 $180

No shortages are allowed No initial inventory exists Xt = production in period t during regular time, Ot = production in period t during overtime, It = Inventory at end of period t, Ht = capacity increase in period t, Ft = capacity decrease in period t. Minimize

800(X1 + X2 + X3) + 1000(O1 + O2 + O3) + 12(I1 + I2 + I3) + 240(H1 + H2 + H3) + 180(F1 + F2 + F3)

Subject to:

X1 ≤ 45 + H1 – F1 O1 ≤ 0.25(45 + H1 – F1) X2 ≤ 45 + H1 – F1 + H2 – F2 O2 ≤ 0.25(45 + H1 – F1 + H2 – F2) X3 ≤ 45 + H1 – F1 + H2 – F2 + H3 – F3 O3 ≤ 0.25(H1 – F1 + H2 – F2 + H3 – F3) X1 + O1 = 45 + I1 I1 + X2 + O2 = 52 + I2 I2 + X3 + O3 = 60 + I3 Xt, It, Ot, Ht, Ft ≥ 0, t = 1, 2, …, 3.

15

Problem 11 L-Print manufacturing makes laser printers. One plant assembles the L-4000 model. Standards indicate that one worker can assemble five printers per day. This model costs about $350 to make (including labor), and the company estimates it costs $7 to hold one printer in inventory for one month. Workers earn $1600 per month and can be hired for $2000 each. Laying off a worker costs $2600. Currently, there are 12 workers in the assembly department. If a printer is backordered, the cost is $35 per unit per month. The inventory on hand at the end of June was 300 units (Assume zero inventory at the end of December, no overtime is permitted). Month July August September October November December

Working days 21 22 21 23 19 20

Forecasted Demand 1060 990 840 1040 1290 690

(a) Develop an aggregate plan for L-Print based on a chase (zero inventory) strategy (with no backorders). We determine the capacity (in terms of number of workers) required to meeting demand in each month, and hire/fire accordingly. Month

Working Net Required Hiring/ days Demand Production workers Layoffs

Workforce Holding Cost Cost

Backorder Production Cost Cost

July

21

760

760

8 Layoff 4

$ 12,800

$

-

$

-

$

266,000

August

22

990

990

9 Hire 1

$ 14,400

$

-

$

-

$

346,500

September

21

840

840

8 Layoff 1

$ 12,800

$

-

$

-

$

294,000

October

23

1040

1040

10 Hire 2

$ 16,000

$

-

$

-

$

364,000

November

19

1290

1290

14 Hire 4

$ 22,400

$

-

$

-

$

451,500

December

20

690

690

$ 11,200

$

-

$

-

$

241,500 Grand Total

Total Cost $ 45,200 $ 89,600

$

-

$

-

$

1,963,500 $ 2,098,300

7 Layoff 7

16

(b) Develop an aggregate plan for L-Print using a level strategy, where the production output is the same in each month, the number of workers is the same in each month, backorders are allowed, and all demand is met during the six months.

Month

Working Net Required Hiring/ days Demand Production workers Layoffs

Workforce Holding Cost Cost

Backorder Production Cost Cost

July

21

760

935

10 Layoff 2

$ 16,000

$ 1,225

$

-

$

327,250

August

22

990

935

10

$ 16,000

$

840

$

-

$

327,250

September

21

840

935

10

$ 16,000

$ 1,505

$

-

$

327,250

October

23

1040

935

10

$ 16,000

$

770

$

-

$

327,250

November

19

1290

935

10

$ 16,000

$

-

$ 8,575

$

327,250

December

20

690

935

10

$ 16,000

$

-

$

$

327,250 Grand Total

$

1,963,500 $ 2,077,615

5610

Total Cost $

5,200 $ 96,000

$ 4,340

-

$ 8,575

(c) Which of the two solutions above would you recommend to L-Print Management? Justify your answer. The level strategy is cheaper, and has far less layoffs, although we do have a shortage of 245 units for one month (November). If the $35 is a true reflection of the shortage cost, then we should go with the level strategy.

17

(d) Formulate the aggregate planning problem for L-Print as a linear programming model assuming no backordering is allowed. Be sure to define all variables and include all of the required constraints. Define: Ht = # hires, beginning of pd. t Lt = # layoffs, beginning of pd. t WFt = # workers in pd. t Xt = production, pd. t It = inventory, end pd. t Dt = net demand pd. t WDt = working days, pd. t Min

2000�t =1 H t + 2600�t =1 Lt + 1600�t =1WFt + 7�t =1 I t 6

It-1 + Xt = Dt + It, WFt = WFt-1 + Ht – Lt, Xt ≤ 5WDt*WFt, Xt Ht, It, Lt, WFt ≥ 0,

6

6

6

t = 1, …, 6, t = 1, …, 6, t = 1, …, 6 t = 1, …, 6.

18