ider Problem 13 of Section 2.3. Letting x1 and x2 den ote the number of hours of

Question

ider Problem 13 of Section 2.3. Letting x1 and x2 den ote the number of hours of operation of Systems 1 and 2, the resulting mathematical problem is to Minimize 2x1 11x2 subject to xi+ 4x2 S 100 4xi + 20x2 2 480 2x1 + 40x2 800 (a) Show that the dual problem is to Maximize-1 00yi + 480Y2 + 800% subject to -4y1 + 20y2 + 40% 11 (b) A college student, working for the fruit grower for the summer, believes she can have the fruit picked more efficiently than the grower by using her own system and equipment. Fearing that she has nothing to gain financially by simply revealing her plan to the grower, she suggests to the grower that she will supervise the picking of the crop, paying the grower a set amount for each available hour of labor and then selling back to the grower the fruit, using two prices: one for a bushel of choice produce and the other for regular produce. Considering that the student must convince the grower that it is to his advantage to let her supervise the harvest, how should she set these three costs?

Explanation / Answer

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 subtract surplus variable S2 and add artificial variable A1

3. As the constraint 3 is of type '' we should subtract surplus variable S3 and add artificial variable A2

Now standard form of lpp

subject to

and x1,x2,S1,S2,S3,A1,A20

Now table is

Negative minimum Cj-Zj is -60M+11 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 20 and its row index is 3. So, the leaving basis variable is A2.

The pivot element is 40.

Entering =x2, Departing =A2, Key Element =40

R3(new)=R3(old)÷40

R1(new)=R1(old)-4R3(new)

R2(new)=R2(old)-20R3(new)

Negative minimum Cj-Zj is -3M+1.45 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 25 and its row index is 1. So, the leaving basis variable is S1.

The pivot element is 0.8.

Entering =x1, Departing =S1, Key Element =0.8

R1(new)=R1(old)×1.25

R2(new)=R2(old)-3R1(new)

R3(new)=R3(old)-0.05R1(new)

Negative minimum Cj-Zj is -M8+0.0938 and its column index is 5. So, the entering variable is S3.

Minimum ratio is 40 and its row index is 2. So, the leaving basis variable is A1.

The pivot element is 0.125.

Entering =S3, Departing =A1, Key Element =0.125

R2(new)=R2(old)×8

R1(new)=R1(old)-0.125R2(new)

R3(new)=R3(old)+0.0312R2(new)

Since all Cj-Zj0

Hence, optimal solution is arrived with value of variables as :
x1=20,x2=20

So minimum z=260 answer

Now for dual problem

Our primal problem is

Min(z)=2x1+11x2

S.t

x1+4x2<=100

4x1+20x2>=480

2x1+40x2>=800

x1,x2>=0

Now dual of this lpp given by

all constraints can be converted to type by multipling both sides by -1

Dual is

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