1. Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex

2. Linear Programming Fundamentals.. Nice Property of Convex Shapes: Intersection of convex shapes is convex

3. Linear Programming Fundamentals… Every Half-space is convex  Set of feasible solutions is a convex polyhedron Each constraint of an LP  half space

4. Some definitions y ≥ 0 x ≥ 0 x ≤ 500 2x+y ≤ 1500 x + y ≤ 1200 (0,0) 1000 1500 500 1000 1500 500 y x Feasible region Boundary Feasible Solution Corner points

5. Important Properties (0,0) 1000 1500 500 1000 1500 500 y x Feasible region Property 1. IF: only one optimum solution => must be a corner point (0,0) 1000 1500 500 1000 1500 500 y x Feasible region

6. Important Properties.. (0,0) 1000 1500 500 1000 1500 500 y x Feasible region Property 2. IF: multiple optimum solutions => must include 2 adjacent corner pts

7. Important Properties… (0,0) 1000 1500 500 1000 1500 500 y x Feasible region Property 4. Total number of corner pts is finite Property 3. If a corner point is better than all its adjacent corners, it is optimal !

8. Important Properties.... (0,0) 1000 1500 500 1000 1500 500 y x Feasible region Note: Corner point  intersection of some constraint boundaries (lines) Property 5. Moving from a corner point to any adjacent corner point  Exactly one constraint boundary is exchanged.

9. The algebra of Simplex Good news We only need to search for the best feasible corner point Good news We can use property 5 to guide our searching Bad news A problem with 50 variables, 100 constraints => 100 ! 50 ! (100-50) ! => ~ 10 29 corner points

10. The outline of Simplex 1.Start at a corner point feasible solution 2. If (there is a better adjacent corner feasible point) then go to one such adjacent corner point; repeat Step 2; else (report this point as the optimum point). Why does this method work ? 1. Constant improvement (=> no cycling) 2. Finite number of corners

11. The Algebra of Simplex: Gauss elimination Solving for corner points   solving a set of simultaneous equations Method: Gaussian elimination Example: x + y = 2[1] x – 2y = 1[2] Solve by: 2x[1] + [2]:3x = 5 => x = 5/3 [1] - 1x[2]:3y = 1 => y = 1/3

12. The Algebra of Simplex: need for Slack ! Cannot add (multiples of) INEQUATIONS !! Consider: x >= 0[1] y >= 0[2] [1] + [2]: x + y >= 0[3] x y x + y >= 0

13. The Algebra of Simplex: Slack To allow us to use Gaussian elimination, Convert the inequalities to equations by using SLACK Variables 2x + y ≤ 1500, x, y ≥ 0 2x + y + s = 1500, x, y, s ≥ 0 x + y ≥ 200 x, y ≥ 0 x + y – s = 200, x, y, s ≥ 0. InequalityEquality, with slack variable

14. The Algebra of Simplex.. Conversion of the Product mix problem ORIGINAL Maximize z( x, y) = 15 x + 10y subject to 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0 (almost) STANDARD FORM Maximize Z = 15 x 1 + 10x 2 subject to 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i.

15. ORIGINAL Maximize z( x, y) = 15 x + 10y subject to 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0 (almost) STANDARD FORM Maximize Z = 15 x 1 + 10x 2 subject to 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. Original problem: n variables, m constraints Standard form: (m+n) variables, m constraints The Algebra of Simplex…

16. Original problem: n variables, m constraints Standard form: (m+n) variables, m constraints How to use Gaussian elimination ?? Force some variables = 0 How many to be forced to be 0 ? The Algebra of Simplex…

17. Definitions: augmented solution Solution:Augmented solution (x 1 x 2 ) = (200, 200)(200, 200, 900, 800, 300) Max Z = 15 x 1 + 10x 2 S.T. 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i.

18. Definitions: BS, BFS Basic solution  Augmented, corner-point solution Max Z = 15 x 1 + 10x 2 S.T. 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. Examples basic infeasible solution : ( 500, 700, -200, 0, 0) basic feasible solution: (500, 0, 500, 700, 0)

19. Definition: non-basic variables Max Z = 15 x 1 + 10x 2 S.T. 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. Corner Feasible solution Defining eqnsBasic solutionnon-basic variables (0, 0)x 1 = 0 x 2 = 0 (0, 0, 1500, 1200, 500) x 1, x 2 (500, 0)x 1 = 500 x 2 = 0 (500, 0, 500, 700, 0) x 2, x 5 (500, 500)x 1 = 500 2x 1 +x 2 = 1500 (500, 500, 0, 200, 0) x 5, x 3 (300, 900)x 1 + x 2 = 1200 2x 1 +x 2 = 1500 (300, 900, 0, 0, 200) x 3, x 4 (0, 1200)x 1 = 0 x 1 + x 2 = 1200 (0, 1200, 300, 0, 500) x 4, x 1

20. The Algebra of Simplex Max Z = 15 x 1 + 10x 2 S.T. 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. Corner infeasible solution Defining eqnsBasic infeasible solution non-basic variables (750, 0)x 2 = 0 2x 1 + x 2 = 1500 (750, 0, 0, 450, -250) x 2, x 3 (1200, 0)x 2 = 0 x 1 + x 2 = 1200 (1200, 0, -900, 0, -700) x 2, x 4 (500, 700)x 1 + x 2 = 1200 x 1 = 500 (500, 700, -200, 0, 0) x 4, x 5 (0, 1500)x 1 = 0 2x 1 + x 2 = 1500 (0, 1500, 0, -300, 500) x 1, x 3

21. The Algebra of Simplex: Standard form STANDARD FORM Max Z S.T. Z-15 x 1 -10 x 2 = 0 2 x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i.

22. The Algebra of Simplex: Step 1. Initial solution Max Z S.T. Z-15 x 1 -10 x 2 = 0 2 x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. ( 0, 0) Initial non-basic variables: (x 1, x 2 ) Initial basic variables: (x 3, x 4, x 5 ) Initial BFS: (0, 0, 1500, 1200, 500)

23. The Algebra of Simplex: Step 2. Iteration Step Entering variable: Z = 15 x 1 + 10x 2 Fastest rate of ascent

24. The Algebra of Simplex: Step 2. Iteration Step Leaving variable MaxZ S.T. Z-15 x 1 - 10x 2 = 0 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. Consider first constraint: Entering 2x 1 + x 2 + x 3 = 1500 x 3 = 1500 - 2x 1 - x 2 Bound on x 1 : 1500/2 = 750

25. The Algebra of Simplex: Step 2. Iteration Step MaxZ S.T. Z-15 x 1 - 10x 2 = 0 2x 1 + x 2 + x 3 = 1500 x 1 + x 2 + x 4 = 1200 x 1 + x 5 = 500 x i ≥ 0 for all i. Leaving variable

26. The Algebra of Simplex: Step 2. Iteration Step Enter: x 1 Leave: x 5 Z-15 x 1 -10x 2 = 0 [0-0] 2 x 1 + x 2 + x 3 = 1500 [0-1] x 1 + x 2 + x 4 = 1200 [0-2] x 1 + x 5 = 500 [0-3] ROW OPERATIONS so that Each row has exactly one basic variable The coefficient of each basic variable = +1 - [0-3] - 2x [0-3] + 15x [0-3]

27. The Algebra of Simplex: Step 2. Iteration Step Z-15 x 1 -10x 2 = 0 [0-0] 2 x 1 + x 2 + x 3 = 1500 [0-1] x 1 + x 2 + x 4 = 1200 [0-2] x 1 + x 5 = 500 [0-3] Z-10x 2 +15 x 5 = 7500[1-0] x 2 + x 3 - 2 x 5 = 500[1-1] x 2 + x 4 - x 5 = 700[1-2] x 1 + x 5 = 500[1-3] Row operations Basic Feasible Solution: ( 500, 0, 500, 700, 0)

28. The Algebra of Simplex: Step 2. Iteration Step Basic Feasible Solution: ( 500, 0, 500, 700, 0) Objective value: 7500

29. The Algebra of Simplex: Step 3. Stopping condition Z-10x 2 +15 x 5 = 7500[1-0] x 2 + x 3 - 2 x 5 = 500[1-1] x 2 + x 4 - x 5 = 700[1-2] x 1 + x 5 = 500[1-3] Stopping Rule: Stop when objective cannot be improved. New entering variable: x 2

30. The Algebra of Simplex: Second Iteration.. Z-10x 2 +15 x 5 = 7500[1-0] x 2 + x 3 - 2x 5 = 500[1-1] x 2 + x 4 - x 5 = 700[1-2] x 1 + x 5 = 500[1-3] New entering variable: x 2 Analysis for leaving variable:

31. The Algebra of Simplex: Step 2. Iteration Step Enter: x 2 Leave: x 3 ROW OPERATIONS so that Each row has exactly one basic variable The coefficient of each basic variable = +1 - [1-1] + 10x [1-1] Z-10x 2 +15 x 5 = 7500 [1-0] x 2 + x 3 - 2x 5 = 500 [1-1] x 2 + x 4 - x 5 = 700 [1-2] x 1 + x 5 = 500 [1-3]

32. The Algebra of Simplex: Step 2. Iteration Step Basic Feasible Solution: After row operations: Z+10 x 3 -5 x 5 = 12,500[2-0] x 2 + x 3 - 2x 5 = 500[2-1] - x 3 + x 4 + x 5 = 200[2-2] x 1 + x 5 = 500[2-3] ( 500, 500, 0, 200, 0) Have we found the OPTIMUM ?

33. The Algebra of Simplex: Third Iteration New entering variable: x 5 Analysis for leaving variable: Z+10 x 3 -5 x 5 = 12,500[2-0] x 2 + x 3 - 2x 5 = 500[2-1] - x 3 + x 4 + x 5 = 200[2-2] x 1 + x 5 = 500[2-3] Basic variable EquationUpper bound for x 5 x2x2 x 2 = 500 + 2x 5 - x 3 no limit on x 5 x4x4 x 4 = 200 + x 3 - x 5 x 5 ≤ 200  minimum x1x1 x 1 = 500 – x 5 x 5 ≤ 500

34. The Algebra of Simplex: Step 2. Third iteration.. ROW OPERATIONS so that Each row has exactly one basic variable The coefficient of each basic variable = +1 - [2-2] + 5x [2-2] Enter: x 5 Leave: x 4 Z+10 x 3 -5 x 5 = 12,500 [2-0] x 2 + x 3 - 2x 5 = 500 [2-1] - x 3 + x 4 + x 5 = 200 [2-2] x 1 + x 5 = 500 [2-3] + 2x [2-2]

35. The Algebra of Simplex: Step 2. Iteration Step Basic Feasible Solution: ( 300, 900, 0, 0, 200) Have we found the OPTIMUM ? After row operations: Z+5 x 3 + 5x 4 = 13,500[3-0] x 2 - x 3 + 2x 4 = 900[3-1] - x 3 + x 4 + x 5 = 200[3-2] x 1 + x 3 - x 4 = 300[3-3]

36. The Algebra of Simplex: Some important points 1. Higher dimensions 2. Many candidates for entering variable Each step: ONE entering variable ONE leaving variable  Each constraint equation has has same format as our example e.g. Z = 200 + 10x 1 + x 2 + 10x 3 Just pick any one

37. The Algebra of Simplex: Some important points.. 3. Many candidates for leaving variable 4. No candidate for leaving variable 5. Minimization problems Degenerate case, rare. Unbounded objective !? Minimize Z == Maximize -Z next topic: inventory control