HW2-Solutions

PART A: Minimum Inventory Production Plan (i.e., one that allows arbitrary hiring and firing) A B C D E F G H I J K

Forecast Net Cummulative Minimum # # Units Cummulative Inv. above Actual Demand Demand Net Demand of Workers Produced Production Min required Inventory

C/32 E x 32 I - D

July 1250 1050 1050 33 2 0 1056 1056 6 306

Aug 1100 1100 2150 35 0 2 1120 2176 26 326Sep 950 950 3100 30 5 0 960 3136 36 336

Oct 900 900 4000 29 1 0 928 4064 64 364

Nov 1000 1000 5000 32 0 3 1024 5088 88 388

Dec 1150 1250 6250 40 0 8 1280 6368 118 518

8 13 338 2238A B C D E F G H I J K

Forecast Net Cummulative Minimum # # Units Cummulative Inv. above Actual Demand Demand Net Demand of Workers Produced Production Min required Inventory

C/32 E x 32 I - D

July 1250 1050 1050 33 2 0 1056 1056 6 306

Aug 1100 1100 2150 35 0 2 1120 2176 26 326

Sep 950 950 3100 29 6 0 928 3104 4 304

Oct 900 900 4000 28 1 0 896 4000 0 300

Nov 1000 1000 5000 32 0 4 1024 5024 24 324

Dec 1150 1250 6250 39 0 7 1248 6272 22 422

9 13 82 1982A B C D E G H I J Forecast Net Cummulative Cum. Units Monthly Cummulative Inv. Above Actual Demand Demand Net Demand produced Production Production Min required Inventory per worker 32 x 34 H - D

July 1250 1050 1050 32 1088 1088 38 338Aug 1100 1100 2150 64 1088 2176 26 326Sep 950 950 3100 96 1088 3264 164 464Oct 900 900 4000 128 1088 4352 352 652Nov 1000 1000 5000 160 1088 5440 440 740Dec 1150 1250 6250 192 1088 6528 278 6781298 3198PROBLEM # 1

Month Fire Hire Minimum inventory required per month = 300

Ending Inventory = 400

Employee-hours required per hour = 5 hours

Working days per months = 20

Each employee works 8-hour-day

Workforce at the end of June = 35K CALCULATION K (Units produced by one worker in one day)Initial Inventory = 500

TOTAL

K = 8/5 = 1.6# Units Produced by 1 worker in a month = 1.6 x 20 = 32

In this problem there is sufficient inventory on hand to reduce the workforce by one worker in the months of Sep, Oct and Dec.

That decision will allow to reduce the total inventory above minimum required from 338 units to 82.

Month F

Month Fire Hire TOTAL

D/E

PART B: Production Plan that meets demand but does not hire or fire workers during the six-month period

TOTAL

32

32

33

Because there are 35 workers employed at the end of June, the constant workforce plan require fire 1 worker at the

beginning of July. No further hiring or firing of workers are required. The inventory levels that results from a constant

workforce of 34 workers appear in the following table.

33

Ratio 33

34

PROBLEM # 2

Constant Workforce = 137. Therefore, hire 51 workers at the beginning of April.

The inventory levels that results from a constant workforce of 137 shown in the table below.

In 26 days 100 workers produced 12000 cases of beer Holding Cost = $0.75 per case per month Initial Inventory = 4500

Ending Inventory = 3000

Minimum inventory required per month = 1000

PART A: Minimum constant workforce plan

COST = (107) x (300) + (111) x (125) + (8802) x (0.75) = 52576.50

PART B: Zero Inventory Plan

PART B: Zero Inventory Plan

K CALCULATION K (Units produced by one worker in one day)K = 12000 / (26 x 100) = 4.6153

Hiring Cost = $125 per worker Firing Cost = $300 per worker Workforce at the end of March = 86COST = (51) x (125) + (31976) x (0.75) = 30357

COST = (106) x (300) + (110) x (125) + (8285) x (0.75) = 51763.75

In this problem there is sufficient inventory on hand to reduce the workforce by one worker in the months of June and August.

PART C & D

LINDO INPUT

MINIMIZE

125H1+125H2+125H3+125H4+125H5+125H6+300F1+300F2+300F3+300F4 +300F5+300F6+0.75I1+0.75I2+0.75I3+0.75I4+0.75I5+0.75I6

SUBJECT TO

W1-H1+F1=86

W2-W1-H2+F2=0

W3-W2-H3+F3=0

W4-W3-H4+F4=0

W5-W4-H5+F5=0

W6-W5-H6+F6=0

P1-50.7692W1=0

P2-101.5384W2=0

P3-92.3077W3=0

P4-106.1538W4=0

P5-73.8462W5=0

P6-92.3077W6=0

P1-I1=4000

P2+I1-I2=9300

P3+I2-I3=12200

P4+I3-I4=17600

P5+I4-I5=14000

P6+I5-I6=6300

I1>1000

I2>1000

I3>1000

I4>1000

I5>1000

I6>3000

END

GIN W1

GIN W2

GIN W3

GIN W4

GIN W5

GIN W6

LINDO OUTPUT

OBJECTIVE FUNCTION VALUE

24445.35

VARIABLE VALUE REDUCED COST

W199.000000228.461395

W299.000000380.768982

W3158.000000276.923096

W4158.000000363.846039

W5158.000000410.769287

W6158.000000-230.769226

H113.000000 0.000000

H20.0000000.000000

H359.0000000.000000

H40.0000000.000000

H50.000000125.000000

H60.000000425.000000

F10.000000425.000000

F20.000000425.000000

F30.000000425.000000

F40.000000425.000000

F50.000000300.000000

F60.0000000.000000

I11026.1507570.000000

I21778.4522710.000000

I34163.0688480.000000

I43335.3696290.000000

I51003.0689700.000000

I69287.6855470.000000

P15026.1508790.000000

P210052.3017580.000000

P314584.6171880.000000

P416772.3007810.000000

P511667.6992190.000000

P614584.6171880.000000

I NTERPRETATION OF THE SOLUTION

Total Cost = $24445.35

Total # of workers in April and May = 99

Total # of workers in June, July, August and September = 158

Total # of workers hired in April = 13

Total # of workers hired in June = 59

Total # of workers hired in May, July, August and September = 0

Total # of workers fired in April, May, June, July, August and September = 0

Inventory at the end of April = 1026 cases of beer.

Inventory at the end of May = 1778 cases of beer.

Inventory at the end of June = 4163 cases of beer.

Inventory at the end of July = 3335 cases of beer.

Inventory at the end of August = 1003 cases of beer.

Inventory at the end of September = 9287 cases of beer.

Total # of cases of beer produced in April = 5026

Total # of cases of beer produced in May = 10052

Total # of cases of beer produced in June = 14584

Total # of cases of beer produced in July = 16772

Total # of cases of beer produced in August = 11667

Total # of cases of beer produced in September = 14584

P LAN FOR P ART B WOULD BE THE SAME AS FOR P ART D. (N O FIRING)