Question

(PLEASE WHEN SOLVING DO IT ON A COMPUTER AND NOT IN A SHEET OF PAPER BECAUSE IT IS HARD TO UNDERSTAND)

2. Consider the following linear:

Min 2A + 2B

s.t.     

       

1A + 3B ≤ 12

3A + 1B ≥ 13

1A - 1B = 3

A, B ≥ 0

a. Show the feasible region.

b. What are the extreme points of the feasible region?

c. Find the optimal solution using the graphical solution procedure.

Answer 1

The LPP is:

Min 2A + 2B

S.T.

1. A + 3B <= 12

2. 3A + B >= 13

3. A – B = 3

4. A, B >= 0

Putting the Constraints in the graph:

Blue line shows constraint 1. The area feasible is under this line

Orange line is the Constraint 2. The area feasible is above (right to) it.

Grey line is constraint 3. Which says, the point should be on it.

Hence,

Question – a:

The feasible region is on the grey line between the orange and the blue line

The feasible region is the black marked line.

Question – b:

The feasible points are:

1. The point where the grey line and orange line meets.

3A + B – 13 = 3A – 3B – 9

4B = 4

B = 1

A = 4

The point is (4,1)

2. The point where the grey and the blue line meets.

A + 3B – 12 = A – B – 3

4B = 9

B = 9/4 = 1.25

A = 5.25

So the extreme points are: (4,1) and (5.25,1.25)

Question – c:

Our objective function is Min 2A + 2B

Putting (4,1) objective value is = 2*4 + 2*1 = 8 + 2 = 10

Putting (5.25,1.25) objective value is = 2*5.25 + 2*1.25 = 10.5 + 2.5 = 13

As the function is to minimize,

The solution is:

A = 4, B = 1

.

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