Farming Example. Lecture 22. Solving a Linear Program. withthe Simplex Algorithm and with Excel s Solver

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1 Lecture 22 Solving a Linear Program withthe Simplex Algorithm and with Excel s Solver m j winter, 2 Farming Example Constraints: acreage: x + y < money: x + 7y < 6 time: x + y < 3 y x + y = B (, 8.7) x + 7y = 6 x + y = 3 C (, ) D (62., 37.) Objective function represents net profits: $ per acre of corn $2 per acre of wheat = x + 2 y A (, ) E(7, ) 2 x

2 What if we don t have a graph? We could take all possible intersections of two equations. We d have to test each one for feasibility (is it a corner of the region or away from it?), then evaluate the objective function at each feasible intersection point. For the farm example: constraints, means lines. How many combinations of two from are there? C 2 = possible intersection points. Too many points to try. Situation is even worse if there are 3 (or more) variables. 3 So, what do we do? There is a systematic method of starting at one vertex, and moving to an adjacent vertex, each time finding a better value of the objective function. The method is call the Simplex Method. Illustrate with the corn-wheat problem. Ideas are more concrete. Begin by introducing some new variables. Every constraint (inequality) will be made into an equality. 4 2

3 Slack Variables Introduce new variables Use x and x 2 instead of x and y. Define s = x x 2 s 2 = 6 x 7 x 2 s 3 = 3 x x 2 = x + 2x 2 x and x 2 are the independent x = variables. When x = x 2 = A (,) we are at corner A. We want to go around the boundary of the region. B (,6/7)) s = s = 2 x 2 = s = 3 C (, ) D (62., 37.) E (7, ) Each vertex is characteried by two variables set = A: x =, x 2 = B: x =, s 2 = C: s 2 =, s = etc. We are at A, want to move to B. Must make x 2 dependent and s 2 independent. Number the equations: Solve eq n 2 for x 2. Substitute into the other equations : s = x x 2 2: s 2 = 6 x 7 x 2 3: s 3 = 3 x x 2 4: = x + 2x 2 x = 6 x s

4 First Exchange Stage : s = x x 2 2: s 2 = 6 x 7 x 2 3: s 3 = 3 x x 2 4: = x + 2x 2 Stage 2 : s = x : x 2= x : s = x + s : = x s For this example, one stage was enough. Why? Stage 2 : s = x : x 2= x : s = x + s : = x s x =, s 2 = is feasible: all variables are non-negative. If either x or s 2 were positive, would be diminished. So we are done! 8 4

5 Questions: how do we know which variable to swap? how do we know which equation to use? how do we know when we are done? Answer: there are rules First make a table (tableau) of the linear program. 9 Stage : x + x 2 + s = 2: x + 7 x 2 + s 2 = 6 Matrix notation - rewrite with all variables on the left hand side. 3: x + x 2 + s 3 = 3 4: x 2x 2 + = x s s

6 Reading the Matrix x s s Set x and =. Read off value of other variables: s =, etc. = Choosing the Pivot Element (deciding how to make an exchange) x s s Rule : pick most negative element in bottom row. This determines the column. will go. = is the y-axis 2 6

7 Going to the next stage x s s3 / 7 6 6/7 3 3/ 2 Rule 2:for each non-ero element in the pivot column, calculate the quotient: last column/pivot column. Pick the smallest, non-negative quotient. This is the pivot row. The pivot element is at the intersection. Divide every element of the pivot row by the pivot. 3 Make the pivot element itself a x s s3 /7 /7 6/7 3 2 Divide every element of the pivot row by the pivot. Next: use matrix methods to make rest of pivot column all s. 4 7

8 Use the pivot row to make the rest of the pivot column all s x s s3 /7 /7 6/7 3 2 new R: R R2 new R3: R3 R2 new R4: R4 + 2R2 Next stage x s s3 2/7 /7 /7 /7 /7 6/7 3/7 /7 8/7 3/7 2/7 2/7 There will be two non-ero entries in the bottom row - they correspond to the independent variables. Setting these variables to tells you what happens at the vertex. 6 8

9 Reading the Vertex x s s3 2/7 /7 /7 /7 /7 6/7 3/7 /7 8/7 3/7 2/7 2/7 set x = and = = 6/7; s = /7; s3 = 8/7 = 2/7 7 Cannot continue - no negative numbers in bottom line (3/7)x + (2/7)() + = 2/7 or = 2/7 (3/7)x (2/7) The maximum possible value of is 2/7 = We are done. 8 9

10 Choose the next pivot Look at the bottom row. choose the most negative element. etc. If there are no negative elements, you are done! 9 You should be able to: determine initial tableau, find pivot element calculate first stage. Example (3 variables) Find the maximum of = 6x + 4x 2 + x 3 subject to the constraints: x + x 2 + x 3 < 2 2x + 2x 2 + 3x 3 < x, x 2, x 3 > 2

11 x Problem - 2 constraints means 2 slack variables x3 s Find the maximum of = 6x + 4x 2 + x 3 subject to the constraints: x + x 2 + x 3 < 2 2x + 2x 2 + 3x 3 < x, x 2, x 3 > Problem - 2 x x3 s

12 Problem - 3 x x3 s Interpret the results: 23 Problem - 4 Interpret the results: x 7 26 x3 2 s At solution, which constraint(s) are equalities? Maximum value of is: Occurs when are ero. Optimal values of x,, x3 are: 24 2

13 Problem - 4 Interpret the results: x 7 26 x3 2 s At solution, which constraint(s) are equalities? Maximum value of is: Occurs when, x3,s are ero. 2 Optimal values of x,, x3 are: 2,, 2 Notice: we are assuming we know the coordinates of a vertex to start with. Using the simplex algorithm with a minimiation problem usually requires two phases: first find a vertex, then proceed as we have done. 26 3

14 Set up the Problem Use The Simplex Method What you need to be able to do: Convert a description to a system of inequalities with an objective function. Decide on the pivot element. Carry out one step of the simplex method by hand - get new bottom row Be able to use Excel to do many steps. Watch the avi demonstration, but print out the next slides for reference. 27 Using Solver in Excel - Set Up Constraint corn wheat limit actual used acres... <=. money for seed <=. man-hrs.. 3. <=. variables.. current values of the varia objective function. 2. coefficients in objective f value of objective function. Enter the formula objective function T You'll need labels in row and in column Numbers entered in the yellow region here are constants - from the problem. For the blue cells you'll enter formulas Put eros in the green cells. The orange cells are just to remind you of the direction of the inequality. 28 4

15 A A B C D E F Constraint corn wheat limit actual used acres... <=. money for seed <=. man-hrs.. 3. <=. variables.. current values of the varia objective function. 2. coefficients in objective f value of objective function. Enter the formula objective function T Objective value is in B Constraints: F3 < D3 F4 < D4 F < D non-negativity: B8 > C8 > 29 Using Solver - Tools Menu (Add-In) 3

16 Setting Constraints in Solver Don t forget to make variables >= Click Add to add another constraint Click OK when done 3 Go to Options Select Assume Linear Model Click OK 32 6

17 Solve and get Answer Report Click OK 33 p Answer Report on Separate Sheet Target Cell (Max) Cell Name Original Value Final Value $B$ value of objective function corn Adjustable Cells Cell Name Original Value Final Value $B$8 variables corn.. $C$8 variables wheat Constraints Cell Name Cell Value Formula Status Slack $F$3 <= actual used 8.7 $F$3<=$D$3 Not Binding 4.29 $F$4 <= actual used 6. $F$4<=$D$4 Binding. $F$ <= actual used 8.7 $F$<=$D$ Not Binding $C$8 variables wheat 8.7 $C$8>= Not Binding 8.7 $B$8 variables corn. $B$8>= Binding. 34 7