Question

Solve the following linear programming problems as directed. Put in a box the values of all the variables you use in your solution, as well as the optimal value of the objective function.

a)  SIMPLEX METHOD

Max Z = 11X1 + 10X2

s.t.

2 X1 + X2 <= 150

4 X1 + 3 X2 <= 200

X1 + 6 X2 <= 175

X1, X2 >= 0

b) GRAPHIC METHOD (do not forget to indicate the feasible region)

Min Z = 30 X1 + 25 X2

s.t.

7 X1 + 4 X2 >= 28

5 X1 + 8 X2 >= 40

X1 + 4 X2 >= 12

X1, X2 >= 0

Answer 1

Answer:

I have solved in excel due to limited time availability

Solution 1:

Simplex method

The box in yellow shows the value for variables X1 and X2 (Cell B4 and C4) = 32.14 and 23.80

The box in green, shows value of optimal value of objective function (cell D12) Z = 591.66 maximum value

Put the above values of X1, X2 and Z in box.

A D E F B С Simplex Method to find X1 and X2 1 2 X1 X2 32.14285714 23.80952381 Limits 150 3 4 Values 5 6 7 Constraint 1 8 Constraint 2 9 Constraint 3 10 11 12 Coefficients Coefficients X1 Coefficients X2 2 1 4 1 6 sum product 88.0952381 <= 200 <= 175 <= 3 200 175 Maximized value 10 591.6666667 11

Steps:

Make a table like this first

A D E F B С Simplex Method to find X1 and X2 1 2 3 X1 X2 4 Values 5 6 7 Constraint 1 8 Constraint 2 9 Constraint 3 10 11 12 Coefficients Coefficients X1 Coefficients X2 2 1 4 3 1 6 sum product 0 <= 0 <= 0 <= Limits 150 200 175 Maximized value 10 0 11

Look at the formula bar below and put the formulas and now copy and drag the formula till the row D9.

D7 Х fx =SUMPRODUCT($B$4:$C$4,37:07) А D E F B С Simplex Method to find X1 and X2 1 2 X1 X2 Limits 150 3 4 Values 5 6 7 Constraint 1 8 Constraint 2 9 Constraint 3 10 11 12 Coefficients Coefficients X1 Coefficients X2 2 1 4 1 6 sum product 0 <= 0 <= 0 <= 3 200 175 Maximized value 10 0 11

And also copy the formula in the green colored cell to get this below image

D12 --- Х fr =SUMPRODUCT($B$4:$C$4,B12:012) A D E F B C Simplex Method to find X1 and X2 1 2 2 3 X1 X2 4 Values 5 6 7 Constraint 1 8 Constraint 2 9 Constraint 3 10 11 12 Coefficients Coefficients X1 Coefficients X2 2 1 4 3 1 6 sum product 0 <= 0 <= 0 <= Limits 150 200 175 3 Maximized value 10 이 11 12

Now, go to Data > Solver and fill the dialogue box as shown and click solve and tick unconstrained variable non-negative to make X1 and X2 > = 0.

File Home Insert Page Layout Formulas Data Review View Help Tell me what you want to do From Text/CSV Queries & Solver Parameters Х Data Ana Lo Recent Sources Existing Connections Le From Web ?, Solver Get Data From Table/Range Properties Refresh All - Edit Links Queries & Connect Set Objective: $D$12 Get & Transform Data Analyze To: Max Min Value Of: 0 D12 fr =SUMPRODUCT($B$4:$C$4,B12: A D E By Changing Variable Cells: $B$4:$C$4 o P B С Simplex Method to find X1 and X2 1 1 2 الا X1 X2 Subject to the Constraints: $D$7 <= $F$7 $D$8 <= $F$8 $D$9 <= $F$9 Add 4 Values 5 Change Coefficients X1 Coefficients X2 2 1 sum product 0 <= Delete 4 3 0 <= 6 7 Constraint 1 8 Constraint 2 9 Constraint 3 0 11 12 Coefficients 1 6 0 <= Reset All Load/Save Maximized value 10 0: Make Unconstrained Variables Non-Negative 11 w Select a Solving Method: Simplex LP Options 4 15 16 17 Solving Method Select the GRG Nonlinear engine for Solver Problems that are smooth nonlinear. Select the LP Simplex engine for linear Solver Problems, and select the Evolutionary engine for Solver problems that are non-smooth. 18 19 20 Help Solve Close 21 2

Solution 2:

The A co-ordinate which is intersection of constraint 1 and constraint 2 are the ones giving the minimum value for objective solution, A = (X1,X2) = (1.78,3.89)

Z = 150.56 (Answer)

Write A and Z in box

The feasible region is indicated by green lines

с F H - J K L M N o P Q R S T V w х u Table 4 DE Table o 25 4]>= 28 8 >= 40 4>= 12 A co-ordinates Constraint 1 Constraint 2 X1 X2 7 4>= 5 8]> 28 40 X1 X2 Z 1.78 3.89 150.56 Minimum Answer 1 Table 5 8 X1 Not in feasible region B co-ordinates Constraint 1 Constraint 3 X1 X2 7 4>= 1 4>= 28 12) 7 2.67 2.33 138.33 X2 Z 6 Table 6 А B 1 X1 X2 2 z 30 3 Constraint 1 7 4 Constraint 2 5 5 Constraint 3 6 7 Constraint 1 Table 1 8 X1 X2 9 4 0 10 0 7 11 12 Constraint 2 Table 2 13 X1 X2 14 8 0 15 0 5 16 17 Constraint 3 Table 3 18 X1 X2 19 12 0 20 0 3 21 22 23 X1 X2 5 C co-ordinates Constraint 3 4 >= 1 5 12 40 X1 X2 Z 5.33 1.67 201.67 4 A Constraint 2 8 >= -Constraint 3 -Constraint 2 - Constraint 2 B 3 с 1 0 0 2 4 6 8 10 12 14 X1 24

Table 0 is the representation of question 2

Table 1 is to make the constraint 1 line in graph

Table 2 is to make the constraint 2 line in graph

Table 3 is to make the constraint 3 line in graph

Table 4 is to show the resulting value of intersection of constraint 1 and constraint 2, it gives the co-ordinates of A, which is inside the feasible region and is the answer, as the value of objective function is minimum Z = 150.56

Table 5 is to show the resulting value of intersection of constraint 1 and constraint 3, it gives the co-ordinates of B, which is outside the feasoble region and is not the answer

Table 6 is to show the resulting value of intersection of constraint 3 and constraint 2, it gives the co-ordinates of C, which is inside the feasible region and is not the answer because the value of objective function at this point is not minimum i.e. Z = 201.67