In order to effectively address the seasonal variation appearing in the annual d
ID: 329698 • Letter: I
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:
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.
Month Demand Forcast Number of Working Days Jan 1800 22 Feb 1500 19 Mar 1100 21 apr 900 21 may 1100 22 june 1600 20Explanation / Answer
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
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