Question

In order to effectively address the seasonal variation appearing in the annual demand of its products,a bicycle manufacturer starts an aggregate planning.A planning hori- zon of 6 months is used. The (aggregate) demand forecast for the next six months along the number of working days are given below:

Month	Demand Forcast	Number of Working Days
Jan	1800	22
Feb	1500	19
Mar	1100	21
apr	900	21
may	1100	22
june	1600	20

If the manufacturer operates 300 days a year with 50 workers, the annual output is 30000 bicycles.Production costs are fixed over the coming planning horizon and thus can be ig- nored. Inventory holding cost is $5 per unit.At the beginning of each month, workers can be hired or fired. Hiring cost is $350 per worker and firing cost is $500 per worker.Wages of worker is $20 per hour per worker eight hours per day for regular workers.Overtime cost is $30 per worker per hour. The number of units a worker can produce in overtime is same as the number he produces in regular time.At the beginning of January, the number of workers is 40 and the inventory in hand is 560.Formulate a linear programming problem to minimize the total cost for the company for the next 6 months.

Answer 1

Productivity per worker per day = Annual output / (Number of workers * Number of days per year) = 30000/(50*300) = 2 units per worker per day

Considering 8 hours per day, it takes 4 hours to produce per unit. Regular production cost per unit = 20*4 = 80

Overtime cost per unit = 30*4 = 120

______________________________

LP model is following

Let Hi, Fi, and Wi be the number of workers hired, fired and working in month i

Oi be the number of units produced in overtime in month i

Vi be the inventory in month i

Min 350(H1+H2+H3+H4+H5+H6)+500(F1+F2+F3+F4+F5+F6)+5(V1+V2+V3+V4+V5+V6)+20*8*(22W1+19W2+21W3+21W4+22W5+20W6)+30*4*(O1+O2+O3+4+O5+O6)

s.t.

W1-H1+F1 = 40

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

22*2*W1+O1-V1 = 1800-560

19*2*W2+O2-V2+V1 = 1500

21*2*W3+O3-V3+V2 = 1100

21*2*W4+O4-V4+V3 = 900

22*2*W5+O5-V5+V4 = 1100

20*2*W6+O6-V6+V5 = 1600

O1 <= 22*2*W1

O2 <= 19*2*W2

O3 <= 21*2*W3

O4 <= 21*2*W4

O5 <= 22*2*W5

O6 <= 20*2*W6

All variables >= 0