Eight of the entries have been deleted from the computer output that follows. Us

Question

Eight of the entries have been deleted from the computer output that follows. Use what you know about linear programming to find values for the blanks. MIN SUBJECT TO 6 X1 +7.5 X2 +10 X3 2) 25 X1 35 X2 +30 X3>-2400 3) 2X1 +4 X2 +8 X3 400 END LP OPTIMUM FOUND AT STEP OBJECTIVE FUNCTION VALUE 2 1) 612.50000 VARIABLE XI X2 X3 1.312500 27.500000 2) -.125000 781250 NO. ITERATIONS 2 RANGES IN WHICH THE BASIS IS UNCHANGED VARIABLEURRENT X2 X3 6.000000 7.500000 10.000000 1.500000 5.000000 3.571429 RHS 2400.000000 400.000000 1100.000000 240.000000 900.000000 25.714300

Explanation / Answer

1. FOR THE VALUE OF X1

Since in the report reduced cost is given for variable X1, it means the variable X1 is not included in optimal solution. So has to include the variable in solution, the objective coefficient of X1 should be reduced by 1.3125

Value of X1 = 0

2. Value of X2

Since allowable increase and decrease for variable X2 and X3 is provided in report, variable X2 and X3 are included in basis solution (optimal). X3 value is given 27.5 and X1 = 0, the objective function is given by following equation:

Z = 6*0 + 7.5*X2 + 10*27.5

Objective function value = Z = 612.5

612.5 = 6*0 + 7.5*X2 + 10*27.5

7.5X2 = (612.5 – 275)

X2 = 337.5/7.5 = 45

Optimal Value of X2 = 45

3. Reduced Cost of X2 and X3

Variable X2 and X3 are included in basis or optimal solution, thus it is not required to reduce the objective coefficient of X2 and X3. Reduced cost value for X2 and X3 will be zero

Reduced cost value of X2 = 0

Reduced cost value of X3 = 0

4. Slack or Surplus value of row 2 and 3

In report dual prices of the both the constraints are provided, thus both the constraints are utilized till minimum limit. Thus there is no surplus value remaining for both the constraints.

Slack or Surplus of row 2 = 0

Slack or Surplus of row 3 = 0

5. Allowable objective coefficient increase and decrease for variable X1:

Since the variable X1 is not included in optimal solution, the cost of can be increased to infinite with out changing the optimal solution mix.

Allowable decrease will be given as actual value minus reduced cost value, since beyond this value the optimal solution mix will change.

Allowable decrease = 6 – 1.3125 = 4.6875

Allowable increase in X2 value = infinite

Allowable decrease in X2 value = 4.6875

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