Question

EZ-Windows, Inc., manufactures replacement windows for the home remodeling business. In January, the company produced 15,000 windows and ended the month with 9000 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.

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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.

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: $  

If required, round your answers to the nearest whole number.

	February	March	April
Production Level
Increase in Production
Decrease in Production
Ending Inventory

Answer 1

March 16500 14000 February April Sales Forecasst Production Capacity Storage Capacity 14000 Let us assume the Actual Production notations for each Increase in Production level of the values in the Decrease in production level If Im Dm Sm Da months Month end Invento Sf Sa The Linear programming equation can be written as below Minimize Z 1lf +1lm +1la 0.65Df 0.65Dm 0.65Da also can be rewritten as F-Sf-6000 SftM-Sm=16500| Sm+A-Sa-20000 F-If+Df=15000 M-Im+Df-F Eq.No With constraints Remarks 9000+F-15000 = Sf Sf+M-16500 Sm Sm+A-20000 = Sa F 15000 If-Df Previous months inventory + Present month's production-Present months Sales = Present months inventory Present months production Previous months prodution = Increase/decrease in production If increase, D will become 0 and If decrease I will 2 4 A M- la-Da Fs= 14000 M<=14000 A<=18000 10 Sf<-6000 11 Sm<-6000 12 S<-6000 Capacity constraints of production Capacity constraints of storage

We use Excel solver to solve this equation for minimization, The formulas are shown as below. The 1's are just initial numbers that will change when we use solver. The formulae used to connect each are also shown below 4.95 Actual Production Increase in Production level Decrease in production level Month end Invento Eq.No.1,2,3 Eq.No.4,5,6 1 O8+13 P8+1 Q8+0.65*O9+0.65*P9+0.65*a9 Actual Production Increase in Production level Decrease in production level Month end Invento Eq.No.1,2,3 Eq.No.4,5,6 07-010 010+P7-P10P10+Q7-010 Q7-Q8+09 07-08+09 P7-P8+P9

We use Excel solver to solve this equation for minimization, Enter the conditions in solver as shown belovw 4.95 Actual Production Increase in Production level Decrease in production level Month end Invento Eq.No.1,2,3 Eq.No.4,5,6 0 Solver Parameters sos5 Set Objective To: OMax By Changing Variable Cells: ⓔMin。Yalue Of: SOS7:SQS10 Subject to the Constraints: SOS10 6000 sos11 6000 SOS12 15000 SOS714000 SPS10 6000 SPS11 16500 Add Change Delete SPS7 14000 SQS 10 <= 6000 SOS 11 = 20000 Reset All SQ57 18000 Load/Save ake Unconstrained Variables Non-Negative Select a Solving Method: GRG Nonlinear Options Solving Method Select the GRG Nonlinear engine for Solver Problems that are smooth nonlinear, Select the LP Simplex engine for linear Solver Problems, and select the Evolutionary engine for Solver problems that are non-smooth

6450 Actual Production Increase in Production level Decrease in production level Month end Invento Eq.No.1,2,3 Eq.No.4,5,6 12000 0 3000 6000 6000 14000 2000 0 3500 16500 16500 2500 0 0 14000 Solver Results Solver found a solution. All Constraints and optimality conditions are satisfied Reports Answer Sensitivity Limits O Keep Solver Solution O Restore Original Values Return to Solver Parameters Dialog outline Reports Cancel Save Scenario. Solver found a solution. All Constraints and optimality conditions are When the GRG engine is used, Solver has found at least a local optimal solution. When Simplex LP is used, this means Solver has found a global optimal solution