# Linear Programming Assignment Essay

1 a) Linear Programing Model

Decision Variables: Let x = acres of watermelon

Let y = acres of cantaloupe

Objective Function:

Maximize Z = 390x + 1300y – 5(20x + 15y) + 5(2x + 2.5y)

= 270x + 300y – 100x + 75y + 10x + 12.5y

= 256x + 284.5y where Z = total profit 390x = profit from watermelons 1300y = profit from cantaloupe 5(20x + 15y) = cost of fertilizer 5(2x + 2.5y) = cost of labour

Identification of Constants:

Maximize Z = 256x + 284.5y

Subject to x + y ≤ 100 (acres) 50x + 75y ≤ 6,000 (water) x ≤ 0 y ≤ 0

1 b) Sketch of

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Subject to r ≤ 600,000 s ≤ 0.1 (a + f + o + s) f + o ≤ a o + s ≤ r a + f + o + s + r = 2,000,000

4) Linear Programing Model

Decision Variables: Let a1 = # of units of component 1 produced by suppler 1

Let a2 = # of units of component 2 produced by suppler 1

Let b1 = # of units of component 1 produced by suppler 2

Let b2 = # of units of component 2 produced by suppler 2

Let c1 = # of units of component 1 produced by suppler 3

Let c2 = # of units of component 2 produced by suppler 3

Objective Function:

Minimum Z = 12a1 + 10a2 + 13b1 + 11b2 + 14c1 + 10c2 where Z = total cost 12a1 = cost of component 1 produced by suppler 1 10a2 = cost of component 2 produced by suppler 1 13b1 = cost of component 1 produced by suppler 2 11b2 = cost of component 2 produced by suppler 2 14c1 = cost of component 1 produced by suppler 3 10c2 = cost of component 2 produced by suppler 3

Identification of Constants:

Maximize Z = 12a1 + 10a2 + 13b1 + 11b2 + 14c1 + 10c2

Subject to a1 + b1 + c1 = 1,000 a2 + b2 + c2 = 800 a1 + a2 ≤ 600 b1 + b2 ≤ 1,000 c1 + c2 ≤ 800

PART B – Decision Analysis

1 a) Payoff Matrix

PAYOFF MATRIX

State of Nature

Decision

No Strike

Partial Strike

Full Strike

Large Stockpile

($50,000)

($50,000)