Question

Q3. (a) Solve the following LP model GRAPHICALLY by drawing a specific objective function iso- cost line) and show the feasible area. Minimize Cost $28X + $24Y s.t. (1) 5X + 4Y S 2600 (2)X + Y 2 300 (3) X 2 80 (4) Y 2 100 X,Y 20 (b) Find the optimal solution and calculate the optimal (minimum) value). SHOW YOUR WORK Q4. (Based on the information from Q3 solved above) (a) WRITE below the STANDARD FORM of Q1 (introduce 4 SLACK and/or SURPLUS variables appropriately) and explain briefly what you have done to write this in the Standard Form: (Write the complete Standard Form here) (b) Given the optimal solution of Q3, find the optimal values of S1, S2, S3, and S4 from the Standard Form (NO NEED TO DRAW SEPARATE GRAPH). SHOW YOUR CALCULATIONS Based on the values of S1, S2, S3, and S4, comment on which constraints are binding and which are not binding?

Answer 1

93. The given proplem is, minimize cost $28x + $24% such that, ™) 5* +4 Y < 2600 2) X+Y > 300 3) x > 80 4) Y>,100 X,YZO Now, wrows from 1) 5x+4 Y< 2600 5X 2600 + 2600 = 2.00 +2.00 +1 = x + Y 31 520 * 650 650 @ from a) x+y >,300 Boo + y ), 1 > 0 3 @ . , 3) & 4) in x-Y plane How graphing we get ® 180, 550) (0,300) (80,0) (800,0) | X=80

Now X=80 & X+Y= 300 intersects when, 80tY = 300 Y - 220 & Y=100 & X+Y=300 intersects when, 100+x=300 => x=200 x=80 & 5x +4 Y = 2600 intersects when, 400 +4y=2600 + 4y = 2200 => Y = 550 Y=100 & 5x+4y= 2600 intersects When 5x + 400 = 2600 => 5x = 2200 => x = 440 The The b) shaded region is the feasible region. corner points are, A = ( 80,320) , B = (80,550) C(440,100) , D = (200,100) Now hit z = 28% + 24 Y at A, I ( 80,320) = 28x80 + 24 x 320 9920 at B, z (80,550) = 28% 80 + 24X550 = 5440

at C, za 440,100) = 28 x 440 + 24x100 = 14.720 at D, Z( 200,100) = 28x200 +24 x 100 = 8000 liz is minimum at X=200 Y =100 so the optimal solution is x= $200 a & optimal value = $8000 Aa