This presentation shows how the tableau method is used to solve a simple linear programming problem in two variables: Maximising subject to three  constraints.

LINEAR PROGRAMMINGExample 1 MaximiseI = x + 0.8y subject tox + y  x + y  x + 2y  2400 Initial solution: I = 0 at (0, 0)

LINEAR PROGRAMMINGExample 1 MaximiseI = x + 0.8y subject to x + y  x + y  x + 2y  2400 MaximiseI whereI - x - 0.8y = 0 subject to x + y + s 1 = x + y + s 2 = x + 2y + s 3 = 2400

Basic variable xys1s1 s2s2 s3s3 RHS s1s s2s s3s I SIMPLEX TABLEAU I = 0, x = 0, y = 0, s 1 = 1000, s 2 = 1500, s 3 = 2400 Initial solution

Basic variable xys1s1 s2s2 s3s3 RHS s1s s2s s3s I PIVOT 1Choosing the pivot column Most negative number in objective row

Basic variable xys1s1 s2s2 s3s3 RHS  -values s1s /1 s2s /2 s3s /3 I PIVOT 1Choosing the pivot element Minimum of  -values gives 2 as pivot element

Basic variable xys1s1 s2s2 s3s3 RHS s1s s2s s3s I PIVOT 1Making the pivot Divide through the pivot row by the pivot element

Basic variable xys1s1 s2s2 s3s3 RHS s1s s2s s3s I PIVOT 1Making the pivot Objective row + pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s s2s s3s I PIVOT 1Making the pivot First constraint row - pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s s2s s3s I PIVOT 1Making the pivot Third constraint row – 3 x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s x s3s I PIVOT 1New solution I = 750, x = 750, y = 0, s 1 = 250, s 2 = 0, s 3 = 150

LINEAR PROGRAMMINGExample MaximiseI = x + 0.8y subject tox + y  x + y  x + 2y  2400 Solution after pivot 1: I = 750 at (750, 0)

Basic variable xys1s1 s2s2 s3s3 RHS s1s x s3s I PIVOT 2 Most negative number in objective row Choosing the pivot column

Basic variable xys1s1 s2s2 s3s3 RHS  -values s1s /0.5 x /0.5 s3s /0.5 I PIVOT 2Choosing the pivot element Minimum of  -values gives 0.5 as pivot element

Basic variable xys1s1 s2s2 s3s3 RHS s1s x s3s I PIVOT 2Making the pivot Divide through the pivot row by the pivot element

Basic variable xys1s1 s2s2 s3s3 RHS s1s x s3s I PIVOT 2Making the pivot Objective row x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s x s3s I PIVOT 2Making the pivot First constraint row – 0.5 x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s x s3s I PIVOT 2Making the pivot Second constraint row – 0.5 x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s x y I PIVOT 2New solution I = 840, x = 600, y = 300, s 1 = 100, s 2 = 0, s 3 = 0

LINEAR PROGRAMMINGExample MaximiseI = x + 0.8y subject tox + y  x + y  x + 2y  2400 Solution after pivot 2: I = 840 at (600, 300)

Basic variable xys1s1 s2s2 s3s3 RHS s1s x y I PIVOT 3Choosing the pivot column Most negative number in objective row

Basic variable xys1s1 s2s2 s3s3 RHS  -values s1s /1 x /2 y I PIVOT 3Choosing the pivot element Minimum of  -values gives 1 as pivot element

Basic variable xys1s1 s2s2 s3s3 RHS s1s x y I PIVOT 3Making the pivot Divide through the pivot row by the pivot element

Basic variable xys1s1 s2s2 s3s3 RHS s1s x y I PIVOT 3Making the pivot Objective row x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s x y I PIVOT 3Making the pivot Second constraint row – 2 x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s1s x y I PIVOT 3Making the pivot Third constraint row + 3 x pivot row

Basic variable xys1s1 s2s2 s3s3 RHS s2s x y I PIVOT 3Optimal solution I = 880, x = 400, y = 600, s 1 = 0, s 2 = 100, s 3 = 0

LINEAR PROGRAMMINGExample MaximiseI = x + 0.8y subject tox + y  x + y  x + 2y  2400 Optimal solution after pivot 3: I = 880 at (400, 600)