Solve the following linear programming problem graphically. Then set up a simplex tableau and solve the problem, using the simplex method. Indicate the corner points generated at each iteration by the simplex on your graph. Maximize profit = $3X1 + $5X2 Subject to: X2 ≤ 6 3X1 + 2X2 ≤ 18 X1, X2 ≥ 0

Answer:

Following steps are followed to draw LPP on graph:

Step 1. Formulate the LP problem.

Step 2. Construct a graph and plot the constraint lines

Step 3. Determine the valid side of each constraint line.

Step 4. Identify the feasible solution region.

Step 5. Plot two objective function lines to determine the direction of improvement.

Step 6. Find the most attractive corner.

Step 7. Determine the optimal solution by algebraically calculating coordinates of the most attractive corner.

Step 8. Determine the value of the objective function for the optimal solution.

In this case :

X2 ? 6 or X2 = 6

3X1 + 2X2 ? 18 or 3X1 + 2X2 = 18

Taking value of X1 and X2 as 0 one at a time we will find the cordinates:

3X1 + 2X2 = 18

(0,6) and (9,0)

Similarly :

X2=6

Refer to thr graphical solution

Substituting values of corner points in the maximise equation:

(0,0) ,(0,6),(3,6),(6,0)

3X1 + 5X2

Z=0

Z=30

Z=39

Z=18

Since our object is to maximize Z and Z has maximum at 39 (3,6)

The simplex method is carried out by performing elementary row operations on a matrix that we call the simplex tableau. This tableau consists of the augmented matrix corresponding to the constraint equations together with the coefficients of the objective function written in the form.

X2 + S1= 6

3X1 + 2X2 +S2 = 18

The numbers S1 S2 and are called slack variables because they take up the “slack” in each inequality.

For this initial simplex tableau, the basic variables are S1 and S2 and the nonbasic variables (which have a value of zero) are X1 and X2 and . Hence, from the two columns that are farthest to the right, we see that the current solution is

X1, X2,S1,S2 = 0,0,6,18

The entry in the lower–right corner of the simplex tableau is the current value of z. Note that the bottom–row entries under and are the negatives of the coefficients of and in the objective function Z=3X1+5X2

To perform an optimality check for a solution represented by a simplex tableau, we look at the entries in the bottom row of the tableau. If any of these entries are negative (as above), then the current solution is not optimal.

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