Question

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Q1 Slove the below LP model graphically and answer:

Min z = x1 +2x2

s.t.      x1+x2 >= 300

           2x1 + x2 >= 400

          2x1 + 5x2 <= 1000

       X1, x2 >= 0

1. Find the point of intersection between the following two constraints:

2x1 + x2 >= 400

2x1 + 5x2 <= 1000

• (200, 300)
• (300, 500)
• (250, 175)
• (150, 125)

2. One of the following point represents the optimal solution of the problem:

• (300, 0)
• (166.7, 133,3)
• (150, 175)
• (500, 0)

3. If the cost of two products (X1 & X2) becomes equal to $2, what will happen to the z :

• It will not change
• No optimal solution can be found
• It will double
• Feasible region will be open (unbounded)

Answer 1

Since, such graphs with big numbers are difficult to make on page, I did it in excel.

Point A (166.6, 133.3), B (500,0), C (300,0) bounding the shaded region in blue is bounded feasible region. Look image above

Table 1:

I just depicted the Objective functions i.e. Z= X1+ 2 X2 and the constraints

C1 : X1 + X2 > = 300

C2: 2X1 + X2 >=400

C3: 2X1 + 5X2 <=1000,

Table 2: Contains end co-ordinates of C1 line i.e (0,300) (300,0)

Table 3: Contains end co-ordinates of C2 line i.e (0,400) (200,0)

Table 4: Contains end co-ordinates of C3 line i.e (0,200) (500,0)

Table 5: To calculate intersection value of C1 and C2. Point D

Table 6: To calculate intersection value of C1 and C3. Point A

Table 7: To calculate intersection value of C2 and C3. Point E

The point of intersection between the following two constraints:

2x1 + x2 >= 400 (C2)

2x1 + 5x2 <= 1000 (C3) are pt. E co-ordinates. From graph = (150,125) (4th option) Answer

Point that represents the optimal solution of the problem:

Point C (300,0) (1st option) Answer (Z=300) the most minimum at this point. (The blue cell value)

If the cost of two products (X1 & X2) becomes equal to $2, what will happen to the z :

It will double (Answer - 3rd option)

Below the orange cell i.e. B2 is converted to 2 from 1 (compare above and below images). The cost value of X2 was already 2 (cell C2). Now, we see the blue cell value becomes = 600. Previous image it was 300. Hence, it doubled.

Point D (100,200) is intersection of C1 and C2