Question

1. JOBCO produces two products on two machines. A unit of product 1 requires 2 hours on machine 1 and 1 hour on machine 2. For product 2, a unit requires 1 hour on machine 1 and 3 hours on machine 2. The revenues per unit of products 1 and 2 are $30 and $20, respectively. The total daily processing time available for each machine is 8 hours. Letting x1 and x2 represent the daily number of units of products 1 and 2, respectively, the LP model is given as

max z=30x1+20x2

s.t.

2x1+x2 ≤8

x1+3x2 ≤8

x1,x2 ≥ 0

(a) Solve the LP graphically.

(b) If JOBCO can increase the capacity of both machines, which machine should receive higher priority?

(c) A suggestion is made to increase the capacities of machines 1 and 2 at the additional cost of $10/hr. Is this advisable?

(d) . If the capacity of machine 1 is increased from the present 8 hours to 13 hours, how will this increase impact the optimum revenue?

(e) Suppose that the capacity of machine 1 is increased to 20 hours, how will this increase impact the optimum revenue?

(f) Suppose that the unit revenues for products 1 and 2 are changed to $35 and $25, respectively. Will the current optimum remain the same?

(g) Suppose that the unit revenue of product 2 is fixed at its current value of c2 = $20.00. What is the associated range for c1, the unit revenue for product 1 that will keep the optimum unchanged?

Answer 1

Let product 1 be X1 units and product 2 be X2 units

The objective function is

max z=30x1+20x2

Subject to

2x1+x2 ≤8

x1+3x2 ≤8

x1,x2 ≥ 0

Ans a)

step 1:draw 2x1+x2<=8,

For 2x1+x2=8

When x1=0, x2= 8

When x2=0 ,x1=4. Draw the graph for this line from these 2 points

Step 2: Draw x1+ 3x2<=8

X1+3x2=8

When x1=0, x2= 8/3

When x2=0 ,x1=8

The lines cuts at p ( 3.2,1.6) Thats our optimal value.

So x1 = 3.2, x2 = 1.6 and z = 128

Ans b)

Lets increase capacity of machine 1 by 1 hour.

So, equation becomes 2x1+x2<=9

So new optima is q (3.8,1.4).

So, x1= 3.8, x2=1.4 and z = 142

Rate of revenue change (p to q)= (142- 128)/(9-8)= $14/hr

So that means if capacity of machine 1 increase by 1 unit , revenue will increase by $14 and vice versa.

Similarly, the dual price of machine 2 can be derived as $2. So machine1 should be priority.

Ans c)

its feasible to increase capacity of machine1 as its is profit ( 14-10) of $4.

But for machine 2 it is loss of ( 2-10) = $8.

Ans d)

To find the feasibility range of machine 1 .

Minimum capacity at (0,2.67) =0*2+1*2.67 = 2.67

Maximum capacity at ( 8,0) = 2*8+1*0 =16

Since proposed hour is within range, the change in revenue will be (13-8) * 14 = $70. Total revenue will be 128+ 70 = $198

And no e)

since the range we have is only upto 16 hours, we donot have sufficient information to make a conclusion on this.

Ans no f)

Let the general format of objective function be

Max Z = c1*x1 + c2*x2

As long as Z = c1*x1 + c2*x2 lies between the constrain line, the solution is at P.

So,range of c1/c2 is

1/3<= c1/c2 <=2/1

New objective function, c1=35, c2= 25

C1/c2= 35/25= 1.4 i.e. within range. So the solution P will remain optimal.

Ans no g)

C2= 20.

(1/3)*20 <= c1<= 2*20

Or 6.67 <=c1<= 40