LINEAR PROGRAMMING Example 1 Maximise I = x + 0.8y subject to x + y  1000 2x + y  1500 3x + 2y  2400 Initial solution: I = 0 at (0, 0)

LINEAR PROGRAMMING Example 1 Maximise I = x + 0.8y subject to x + y  1000 2x + y  1500 3x + 2y  2400 Maximise I where I - x - 0.8y = 0 subject to x + y + s1 = 1000 2x + y + s2 = 1500 3x + 2y + s3 = 2400

SIMPLEX TABLEAU Initial solution I x y s1 s2 s3 RHS -1 -0.8 1000 2 1500 3 2400 I = 0, x = 0, y = 0, s1 = 1000, s2 = 1500, s3 = 2400

Most negative number in objective row PIVOT 1 Choosing the pivot column I x y s1 s2 s3 RHS 1 -1 -0.8 1000 2 1500 3 2400 Most negative number in objective row

Ratio test: Min. of 3 ratios gives 2 as pivot element Choosing the pivot element I x y s1 s2 s3 RHS 1 -1 -0.8 1000 1000/1 2 1500 1500/2 3 2400 2400/3 Ratio test: Min. of 3 ratios gives 2 as pivot element

Divide through the pivot row by the pivot element Making the pivot I x y s1 s2 s3 RHS 1 -1 -0.8 1000 0.5 750 3 2 2400 Divide through the pivot row by the pivot element

Objective row + pivot row Making the pivot I x y s1 s2 s3 RHS 1 -0.3 0.5 750 1000 3 2 2400 Objective row + pivot row

First constraint row - pivot row Making the pivot I x y s1 s2 s3 RHS 1 -0.3 0.5 750 -0.5 250 3 2 2400 First constraint row - pivot row

Third constraint row – 3 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 -0.3 0.5 750 -0.5 250 -1.5 150 Third constraint row – 3 x pivot row

PIVOT 1 New solution I x y s1 s2 s3 RHS -0.3 0.5 750 -0.5 250 -1.5 150 I = 750, x = 750, y = 0, s1 = 250, s2 = 0, s3 = 150

LINEAR PROGRAMMING Example Maximise I = x + 0.8y subject to x + y  1000 2x + y  1500 3x + 2y  2400 Solution after pivot 1: I = 750 at (750, 0)

Most negative number in objective row PIVOT 2 Choosing the pivot column I x y s1 s2 s3 RHS 1 -0.3 0.5 750 -0.5 250 -1.5 150 Most negative number in objective row

Ratio test: Min. of 3 ratios gives 0.5 as pivot element Choosing the pivot element I x y s1 s2 s3 RHS 1 -0.3 0.5 750 -0.5 250 250/0.5 750/0.5 -1.5 150 150/0.5 Ratio test: Min. of 3 ratios gives 0.5 as pivot element

Divide through the pivot row by the pivot element Making the pivot I x y s1 s2 s3 RHS 1 -0.3 0.5 750 -0.5 250 -3 2 300 Divide through the pivot row by the pivot element

Objective row + 0.3 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 -0.4 0.6 840 0.5 -0.5 250 750 -3 2 300 Objective row + 0.3 x pivot row

First constraint row – 0.5 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 -0.4 0.6 840 -1 100 0.5 750 -3 2 300 First constraint row – 0.5 x pivot row

Second constraint row – 0.5 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 -0.4 0.6 840 -1 100 2 600 -3 300 Second constraint row – 0.5 x pivot row

PIVOT 2 New solution I x y s1 s2 s3 RHS -0.4 0.6 840 -1 100 2 600 -3 300 I = 840, x = 600, y = 300, s1 = 100, s2 = 0, s3 = 0

LINEAR PROGRAMMING Example Maximise I = x + 0.8y subject to x + y  1000 2x + y  1500 3x + 2y  2400 Solution after pivot 2: I = 840 at (600, 300)

Most negative number in objective row PIVOT 3 Choosing the pivot column I x y s1 s2 s3 RHS 1 -0.4 0.6 840 -1 100 2 600 -3 300 Most negative number in objective row

Ratio test: Min. of 2 ratios gives 1 as pivot element Choosing the pivot element I x y s1 s2 s3 RHS 1 -0.4 0.6 840 -1 100 100/1 2 600 600/2 -3 300 Ratio test: Min. of 2 ratios gives 1 as pivot element

Divide through the pivot row by the pivot element Making the pivot I x y s1 s2 s3 RHS 1 -0.4 0.6 840 -1 100 2 600 -3 300 Divide through the pivot row by the pivot element

Objective row + 0.4 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 0.4 0.2 880 -1 100 2 600 -3 300 Objective row + 0.4 x pivot row

Second constraint row – 2 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 0.4 0.2 880 -1 100 -2 400 -3 2 300 Second constraint row – 2 x pivot row

Third constraint row + 3 x pivot row Making the pivot I x y s1 s2 s3 RHS 1 0.4 0.2 880 -1 100 -2 400 3 600 Third constraint row + 3 x pivot row

PIVOT 3 Optimal solution I x y s1 s2 s3 RHS 0.4 0.2 880 -1 100 -2 400 3 600 I = 880, x = 400, y = 600, s1 = 0, s2 = 100, s3 = 0

LINEAR PROGRAMMING Example Maximise I = x + 0.8y subject to x + y  1000 2x + y  1500 3x + 2y  2400 Optimal solution after pivot 3: I = 880 at (400, 600)