Question

The Skimmer Boat Company manufactures the Water Skimmer bass fishing boat. The company purchases the engines it installs in its boats from the Mar-gine Company, which specializes in marine engines. Skimmer has the following production schedule for April, May, June, and July:

Month	Production
April	60
May	85
June	100
July	120

Mar-gine usually manufactures and ships engines to Skimmer during the month the engines are due. However, from April through July, Mar-gine has a large order with another boat customer, and it can manufacture only 40 engines in April, 60 in May, 90 in June, and 50 in July. Mar-gine has several alternative ways to meet Skimmer’s production schedule. It can produce up to 30 engines in January, February, and March and carry them in inventory at a cost of $50 per engine per month until it ships them to Skimmer. For example, Mar-gine could build an engine in January and ship it to Skimmer in April, incurring $150 in inventory charges. Mar-gine can also manufacture up to 20 engines in the month they are due on an overtime basis, with an additional cost of $400 per engine. Mar-gine wants to determine the least costly production schedule that will meet Skimmer’s schedule.

a) Formulate a linear programming model

b) Solve this model by using the computer (Excel Solver)

c) If Mar-gine were able to increase its production capacity in January, February, and March from 30 to 40 engines, what would the effect be on the optimal solution?

Answer 1

a) Linear programming model is following:

Let Ri be the number of engines to be produced on regular time in month i , where i =1 to 7 for January to July

Oi be the number of engines to be produced on overtime in month i, where i = 4 to 7 for April to July

Vi be the ending inventory in month i, where i = 1 to 7 for for January to July

Minimize 50V1+50V2+50V3+50V4+50V5+50V6+50V7+400O4+400O5+400O6+400O7

s.t.

R1-V1 = 0

R2-V2+V1 = 0

R3-V3+V2 = 0

R4+O4-V4+V3 = 60

R5+O5-V5+V4 = 85

R6+O6-V6+V5 = 100

R7+O7-V7+V6 = 120

R1 <= 30

R2 <= 30

R3 <= 30

R4 <= 40

R5 <= 60

R6 <= 90

R7 <= 50

O4 <= 20

O5 <= 20

O6 <= 20

O7 <= 20

Ri, Oi, Vi >= 0

b) Solution using Excel Solver is following:

FORMULAS: T3 =SUMPRODUCT(B3:S3,$B$25:$S$25) copy to T5:T22

Optimal production plan :

Month	Regular production	Overtime production	Inventory
January	30	-	30
February	30	-	60
March	30	-	90
April	40	0	70
May	60	0	45
June	90	15	50
July	50	20	0

Total cost = $ 31,250

c) The revised optimal solution with the increase in production capacity in Jan-Mar is following:

Month	Regular production	Overtime production	Inventory
January	40	-	40
February	40	-	80
March	40	-	120
April	40	0	100
May	60	0	75
June	90	0	65
July	50	5	0

Total cost = $ 26,000