Question

Problem 4-23 (Algorithmic)

EZ-Windows, Inc., manufactures replacement windows for the home remodeling business. In January, the company produced 15,000 windows and ended the month with 9,500 windows in inventory. EZ-Windows’ management team would like to develop a production schedule for the next three months. A smooth production schedule is obviously desirable because it maintains the current workforce and provides a similar month-to-month operation. However, given the sales forecasts, the production capacities, and the storage capabilities as shown, the management team does not think a smooth production schedule with the same production quantity each month is possible.

	February	March	April
Sales forecast	15,000	17,000	20,000
Production capacity	14,000	15,500	17,000
Storage capacity	6,000	6,000	6,000

The company’s cost accounting department estimates that increasing production by one window from one month to the next will increase total costs by $1.00 for each unit increase in the production level. In addition, decreasing production by one unit from one month to the next will increase total costs by $0.65 for each unit decrease in the production level. Ignoring production and inventory carrying costs, formulate and solve a linear programming model that will minimize the cost of changing production levels while still satisfying the monthly sales forecasts. If required, round your answers to two decimal places. If an amount is zero, enter "0".

Let:

F = number of windows manufactured in February

M = number of windows manufactured in March

A = number of windows manufactured in April

I_m = increase in production level necessary during month m

D_m = decrease in production level necessary during month m

s_m = ending inventory in month m

Min	I1 + I2 + I3 + D1 + D2 + D3
s.t.
(1)	F - s1 =	February Demand
(2)	s1 + M - s2 =	March Demand
(3)	s2 + A - s3 =	April Demand
(4)	F - I1 + D1 =	Change in February Production
(5)	M - F - I2 + D2 =	Change in March Production
(6)	A - M - I3 + D3 =	Change in April Production
(7)	F ≤	February Production Capacity
(8)	M ≤	March Production Capacity
(9)	A ≤	April Production Capacity
(10)	s1 ≤	February Storage Capacity
(11)	s2 ≤	March Storage Capacity
(12)	s3 ≤	April Storage Capacity

If required, round your answers to the nearest dollar.

Cost: $  

Answer 1

SOLUTION:

Given That data

manufactures replacement windows for the home remodeling business. In January, the company produced 15,000 windows and ended the month with 9,500 windows in inventory. EZ-Windows’ management team would like to develop a production schedule for the next three months. A smooth production schedule is obviously desirable because it maintains the current workforce and provides a similar month-to-month operation. However, given the sales forecasts, the production capacities, and the storage capabilities as shown, the management team does not think a smooth production schedule with the same production quantity each month is possible.

	February	March	April
Sales forecast	15,000	17,000	20,000
Production capacity	14,000	15,500	17,000
Storage capacity	6,000	6,000	6,000

So

Linear programming model is following:

Min	I1 + I2 + I3 + 0.83D1 + 0.83D2 + 0.83D3
s.t.
(1)	F - s1 = 15000	February Demand
(2)	s1 + M - s2 = 17000	March Demand
(3)	s2 + A - s3 = 20000	April Demand
(4)	F - I1 + D1 = 15000	Change in February Production
(5)	M - F - I2 + D2 = 0	Change in March Production
(6)	A - M - I3 + D3 = 0	Change in April Production
(7)	F ≤ 14000	February Production Capacity
(8)	M ≤ 15500	March Production Capacity
(9)	A ≤ 17000	April Production Capacity
(10)	s1 ≤ 6000	February Storage Capacity
(11)	s2 ≤ 6000	March Storage Capacity
(12)	s3 ≤ 6000	April Storage Capacity

EXCEL FORMULAS:

	January	February	March	April	Total	Unit cost	Total Cost
Sales Forecast		15000	17000	20000	=SUM(C2:E2)
Production Capacity		14000	15500	17000	=SUM(C3:E3)
Production Level	15000	12000	14000	17000	=SUM(C4:E4)
Increase in Production		0	2000	2500	=SUM(C5:E5)	1	=F5*G5
Decrease in Production		3000	0	0	=SUM(C6:E6)	0.65	=F6*G6
Ending Inventory	9000	=B7-C2+C4	=C7-D2+D4	=D7-E2+E4	=SUM(C7:E7)
Balance		=C4-B4-C5+C6	=D4-C4-D5+D6	=E4-D4-E5+E6	=SUM(C8:E8)
Storage capacity		6000	6000	6000
						Total Cost =	=SUM(H5:H6)

Cost = $ 8,350