Question

In: Advanced Math

4. A local company restores cars and trucks for resale. Each vehicle must be processed in...


4. A local company restores cars and trucks for resale. Each vehicle must be processed in the refinishing/paint shop and the machine/body shop. Each car (on average) contributes $3000 to profit, and each truck contributes (on average) $2000 to profit. The refinishing/paint shop has 2400 work-hours available and the machine/body shop has 2500 work-hours available. A car requires 50 work-hours in the machine/body shop and 40 work-hours in the refinishing/paint shop. A truck requires 50 work-hours in the machine/body shop and 60 work-hours in the refinishing/paint shop. determine the sensativity of the optimal solution to a change in the profit for each truck.

Solutions

Expert Solution

Let x1 be the number of cars restored per day, and x2 be the number of trucks restored per day. The goal is to maximize 3000x+2000y subject to

MAX Z = 3000x1 + 2000x2
subject to
50x1 + 50x2 <= 2500
40x1 + 60x2 <= 2400
and x1,x2 >= 0

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '≤' we should add slack variable S1

2. As the constraint-2 is of type '≤' we should add slack variable S2

After introducing slack variables

MAX Z = 3000x1 + 2000x2 + 0S1 + 0S2
subject to
50x1 + 50x2 + S1 = 2500
40x1 + 60x2 + S2 = 2400
and x1,x2,S1,S2 >= 0


Iteration-1 Cj 3000 2000 0 0
B CB XB x1 x2 S1 S2 MinRatio
XBx1
S1 0 2500 (50) 50 1 0 250050=50
S2 0 2400 40 60 0 1 240040=60
Z=0 Zj 0 0 0 0
Cj-Zj 3000↑ 2000 0 0



Positive maximum Cj-Zj is 3000 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 50 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 50.

Entering =x1, Departing =S1, Key Element =50

R1(new)=R1(old)÷50

R2(new)=R2(old) - 40R1(new)

Iteration-2 Cj 3000 2000 0 0
B CB XB x1 x2 S1 S2 MinRatio
x1 3000 50 1 1 150 0
S2 0 400 0 20 -45 1
Z=150000 Zj 3000 3000 60 0
Cj-Zj 0 -1000 -60 0



Since all Cj-Zj≤0

Hence, integer optimal solution is arrived with value of variables as :
x1=50,x2=0

Max Z=150000


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