Simplex method : Tableau Form Maximise Z= 10X1+9X2+0S1+0S2+0S3+0S4 subject to: 7/10X1+1X2+1S1 =630 1/2X1+5/6X2+ +1S2 =600 1X1+2/3X2 +1S3 =708 1/10X1+1/4X2 +1S4 =135 X1,X2,X3,X4,S1,S2,S3,S4 > 0 Constraints equations make more variables (six) than equations (four), the simplex method find solutions for these equations by assigning zero values to variables A basic solution can be either feasible or infeasible. A basic feasible solution is a solution which is both basic and satisfy non negativity condition.

cj = objective function coefficient for variables j Simplex Form cj = objective function coefficient for variables j bj = right hand side coefficient for constraints i aij = Coefficient associated with variable j in constraint i c1 c2 …………………………………………….cn -------------------------------------------------------------------------------------- a11 a12 ……………………………………………… a1n b1 a21 a22 ………………………………………………..a2n b2 …………………………………………………………………………. am1 am2 amn bm C(row) = row of objective function coefficient b(column)= column of right hand side value of the constraint equation A(matrix) = m-row and n-column of coefficient of the variables in the constraint equations

Simplex Tableau X1 X2 S1 S2 S3 S4 bj/aij Basis Cj 10 9 7/10 1 630 ½ 7/10 1 630 ½ 5/6 600 2/3 708 1/10 ¼ 135 Profit Zj Cj-Zj

Improving the Solution Zj represents the decrease in profit that will result if one unit of X1 is brought into the solution Cj-Zj - net evaluation row Values in Zj row is calculated by multiplying the element in the cj column by the corresponding elements in the column of the A matrix and summing them. The process of moving from one basic feasible solution to another is called iteration. We select the entering variable that has the highest coefficient in the net evaluation row. Criteria of moving a variable from the current basic: For each row i compute the ratio bj/aij> 0. This ratio tells us the maximum amount of the variable Xj that can be brought into the solution and still satisfy the constraint equation represented by that row. Select the basic variable corresponding to the minimum of these ratios as the variable to leave.

Optimality Optimality Condition: The optimal solution to a LP problem has been reached when there are no positive values in the net evaluation row of the simplex tableau. Pivot Row : Replace the leaving variable In the basic column with the entering variable. New pivot row = current row divided by pivot element All other rows including Z : New row : Current row – (pivot column element) X ( New pivot row)

X1 X2 S1 S2 S3 S4 bj/aij Basis Cj 10 9 16/30 1 -7/10 134.4 ½ -1/2 246 2/3 708 22/120 -1/10 64.2 Profit Zj 20/3 7080 Cj-Zj 7/3 -10

X1 X2 S1 S2 S3 S4 bj/aij Basis Cj 10 9 1 30/16 -21/16 252 -15/16 5/32 120 -20/16 50/16 540 -11/32 9/64 18 Profit Zj 70/16 111/16 7668 Cj-Zj -70/16 -111/16

Condition for Optimal Solution Cj-Zj< 0 for both of the non-basic variables S1 & S2, an attempt to bring non- basic variables into basic will lower the current value of the objective function. The optimal solution of a LP problem has been reached when there are no positive values at the net evaluation row. X1 540 X2 252 S1 0 S2 120 S3 0 S4 18

Sample Problem Max. 4X1+6X2+3X3+1X4 s.t 3/2X1+2X2+4X3+3X4<550 Problem in standard form: Max> 4X1+6X2+3X3+1X4+0S1+0S2+0S3 s t 3/2X1+2X2+4X3+3X4+1S1 = 550 4X1+1X2+2X3+1X4 +1S2 = 700 2X1+3X2+1X3+2X4 +1S3 = 200 X1, X2, X3, X4, S1, S2, S3 > 0

Initial Simplex Tableau Basis Cj 4 6 3 1 3/2 2 550 700 200 Zj Cj-Zj

X1 X2 X3 X4 S1 S2 S3 Cj 4 6 3 1 1/6 5/3 -2/3 1/3 -1/3 x2 2/3 Zj 6/3 Iteration :1… X1 X2 X3 X4 S1 S2 S3 Basis Cj 4 6 3 1 1/6 10/3 5/3 -2/3 416(2/3) 1/3 -1/3 633(1/3) x2 2/3 66(2/3) Zj 12/3 6/3 400 Cj-Zj 3/3 -9/3 -6/3

x1 x2 x3 x4 S1 S2 S3 Cj 4 6 3 1 Zj 525 3/60 3/10 Basis -5/10 425 -1/10 Iteration 2…. x1 x2 x3 x4 S1 S2 S3 Basis Cj 4 6 3 1 3/60 5/10 3/10 -2/10 125 39/12 -15/30 -5/10 425 39/60 15/30 -1/10 12/30 25 Zj 81/20 9/2 54/30 525 Cj-Zj -1/20 -7/2 -3/10 -54/30

Eliminating negative right hand side: Treating Constraints Eliminating negative right hand side: The number of std bag had to be less or X1 < X2-25 equal to the number of deluxe bag after 25 1X1 - 1X2 < -25 bags had been set aside for displaying -X1 + X2 > 25 purpose Greater than equal to constraint : 6X1 + 3 X2 -4X3 > -20 -6X1 – 3X2 + 4 X3 < 20

Simplex Problem Max. 10X1 + 9 X2 s t 7/10X1+1X2 < 630 1X1 >100 a5, a6 (artificial variables) 1X2 >100 very large cost -M Max. 10X1+9X2+ 0S1+S2+0S3+0S4+0S5+0S6-Ma5-Ma6 s t. 7/10X1+1X2+1S1 = 630 1/2X1+5/6X2 +1S2 =600 1X1+2/3X2 +1S3 =708 1/10X1+1/4X2 +1S4 =135 1X1 -1S5+1a5 =100 1X2 -S6+1a6 =100

X1 X2 S1 S2 S3 S4 S5 S6 a5 a6 630 600 708 135 Basis Cj 10 9 -M 7/10 1 -M 7/10 1 630 1/2 5/6 600 2/3 708 1/10 ¼ 135 -1 100 Zj Cj-Zj 10+M 9+M M -200M

X1 X2 S1 S2 S3 S4 S5 S6 a5 a6 Cj 10 9 -M 1 -1 ¼ Zj M 560 5/6 550 2/3 Basis Cj 10 9 -M 1 7/10 -7/10 560 5/6 1/2 -1/2 550 2/3 -1 608 ¼ 1/10 -1/10 125 100 Zj Cj-Zj 9+M -10 M -M-10 1000- 100M

X1 X2 S1 S2 S3 S4 S5 S6 a5 a6 Basis Cj 10 9 -M 30/16 -210/160 1 -1 152 -15/16 25/160 120 -20/16 300/160 440 -11/32 45/320 18 540 252 Zj Cj-Zj 9+M 70/16 -70/16 111/16 -111/16 7668

Max. 6X1+3X2+4X3+1X4 s t -2X1-1/2X2+1X3+-6X4 = -60 Equality Constraints Max. 6X1+3X2+4X3+1X4 s t -2X1-1/2X2+1X3+-6X4 = -60 1X1 +1X3+2/3X4<20 -1X2-5X3 <-50 X1, X2, X3, X4> 0 Max. 6X1+3X2+4X3+1X4+0S2+0S3-Ma1-Ma2 st 2X1+1/2X2-1X3+6X4 +1a1 = 60 1X1 +1X3-2/3X4 +1S1 = 20 1X2+5X3 -1S3 +1a5= 50 X1, X2, X3, X4, S2, S3, a1, a3 > 0

Sensitivity Analysis Study of how optimal solution and value of optimal solution to a LP given a changes in the various coefficients of the problem: 1. What effect on the optimal solution will a change in a coefficient of the objective function (Cj) have? 2. What effect on the optimal solution will a change in constraint in right hand side value (bj) have? 3. What effect on the optimal solution will a change in a coefficient of a constraint equation (aij) have?