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LINEAR PROGRAMMING

LINEAR PROGRAMMING.

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LINEAR PROGRAMMING

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  1. LINEAR PROGRAMMING

  2. “A certain wide class of practical problems appears to be just beyond the range of modern computing machinery. These problems occur in everyday life; they run the gamut from some very simple situations that confront an individual to those connected with the national economy as a whole. Typically, these problems involve a complex of different activities in which one wishes to know which activities to emphasize in order to carry out desired objectives under known limitations.” George B. Dantzig, 1948

  3. LP History • George Dantzig, 1947 • Introduced LP and recognized it as more than a conceptual tool: Computing answers important. • Invented “primal” simplex algorithm. • First LP solved: Laderman, 9 cons., 77 vars., 120 MAN-DAYS. • First computer code – 1951 • LP used commercially – Early 60s • Powerful mainframe codes introduced – Early 70s • Computational progress stagnated – Mid 80s • Remarkable progress last 15 years (PCs, new computer science and mathematics)

  4. Diet Problem* Bob wants to plan a nutritious diet, but he is on a limited budget, so he wants to spend as little money as possible. His nutritional requirements are as follows: • 2000 kcal • 55 g protein • 800 mg calcium * From Linear Programming, by Vaŝek Chvátal

  5. Diet Problem Nutritional values Bob is considering the following foods:

  6. Diet Problem Variables We can represent the number of servings of each type of food in the diet by the variables: x1 servings of oatmeal x2servings of chicken x3servings of eggs x4servings of milk x5 servings of cherry pie x6 servings of pork and beans

  7. Diet Problem Nutritional values Bob is considering the following foods: x1 x2 x3 x4 x5 x6 KCAL constraint: 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6  2000 (110x1 = kcals in oatmeal)

  8. Diet Problem LP formulation Minimize Cost 0.3x1 + 2.40x2 + 1.30x3 + 0.90x4 + 2.0x5 + 1.9x6 subject to: Nutritional requirements 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6  2000 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 Bounds

  9. Diet Problem Solution When we solve the preceding LP (using CPLEX, of course) we get a solution value of $6.71, which is achieved with the following menu: 14.24 servings of oatmeal 0servings of chicken 0servings of eggs 2.71servings of milk 0 servings of cherry pie 0 servings of pork and beans

  10. A linear program (LP) is an optimization problem of the form

  11. Solving Linear Programs

  12. (0,6) Objective = 0.9 * 0.882 + 0.73 * 0.706 = 13.1 million Constraint 1 Objective = 0.9 x + 0.73 y = 40 million Objective = 30 million Constraint 3 (0,1.5) (0.882,0.706) (0,1) Feasible Solutions Constraint 2 (1,0) (2,0) (3,0) y A Graphical Solution Maximize 0.90 x + 0.73 y (OBJECTIVE) Subject To Constraint 1: 0.42 x + 0.07 y <= 4200000 Constraint 2: 0.13 x + 0.39 y <= 3900000 Constraint 3: 0.35 x + 0.44 y <= 7000000 x >= 0 y >= 0 x (0,0)

  13. The Simplex Algorithm A graphical representation We now look at a graphical representation of the simplex method as it solves the following problem: Maximize 3x1 + 2x2 + 2x3 Subject to x1 + x3 8 x1 + x2  7 x1 + 2x2  12 x1, x2, x3  0

  14. z = 28 z = 0 z = 23 z = 21 The Simplex Algorithm Maximize z = 3x1 + 2x2 + 2x3 x3 Optimal! (0,0,8) (0,6,8) (2,5,6) (0,6,0) x2 (2,5,0) (7,0,1) (7,0,0) x1

  15. A Real LP: Sample Run Log started (V8.2.0a1) Tue Apr 8 21:59:46 2003 Problem name: supply_chain.lp.gz Constraints : 10810259 Variables : 19916187 Reduced LP has 3140439 rows, and 7314332 variables. Presolve time = 339.43 sec. Iteration log . . . Iteration: 0 Infeasibility = 1534390803.049594 Iteration: 1247301 Infeasibility = 437397607.938387 ... Iteration: 2348954 Infeasibility = 10965.599999 Elapsed time = 3776.48 sec. (2349000 iterations) Iteration: 2349222 Objective = 1775629743606097400.000000 ... Elapsed time = 5396.65 sec. (2687000 iterations) Iteration: 2687616 Objective = 1403833763253068800.000000 Removing shift (10). Primal simplex - Optimal: Objective = 1.4038337633e+18 Solution time = 5410.00 sec. Iterations = 2687616 (2348954)

  16. MIXED INTEGERPROGRAMMING

  17. Mixed-Integer Programming (MIP)(= linear programming + restriction that some [MIP] or all variables must take on integer values.)

  18. Manufacturing Cereal

  19. Cereal Problem:With setup cost Minimize Total_Cost: 283 x1 + 266 x2 + 187 x3 + 188 x4 + 182 x5 + 181 x6 + 200 y1 + 250 y2 + 400 y3 + 350 y4 + 800 y5 + 1000 y6 Subject To Boxes: x1 + x2 + x3 + x4 + x5 + x6 = 200 S1: - 50 y1 + x1 <= 0 S2: - 70 y2 + x2 <= 0 S3: - 40 y3 + x3 <= 0 S4: - 70 y4 + x4 <= 0 S5: - 100 y5 + x5 <= 0 S6: - 90 y6 + x6 <= 0 Bounds x1, x2, x3, x4, x5, x6 >= 0 y1, y2, y3, y4, y5, y5 = 0 or 1 (binary) End

  20. Cereal Problem:With setup cost – Solution log Problem 'cereal.lp' read. Nodes Cuts/ Node Left Objective IInf Best Integer Best Node ItCnt 0 0 38220.0000 1 38220.0000 6 * 0+ 0 0 38520.0000 38220.0000 6 38410.0000 2 38520.0000 Cuts: 4 8 38495.7143 2 38520.0000 Cuts: 4 12 * 0+ 0 0 38510.0000 38495.7143 12 cutoff 38510.0000 Flowcuts: 2 13 Cover cuts applied: 1 Flow cuts applied: 5 Gomory fractional cuts applied: 2 Integer optimal solution: Objective = 3.8510000000e+004 Solution time = 0.01 sec. Iterations = 13 Nodes = 0 Variable Name Solution Value x3 40.000000 x4 60.000000 x5 100.000000 y3 1.000000 y4 1.000000 y5 1.000000 All other variables in the range 1-12 are zero.

  21. LP optimum y objective feasible solutions x Linear Program

  22. optimum of LP relaxation IP optimum rounding down optimum of LP relaxation y objective feasible solutions = x Integer Program

  23. Solving MIPs: Branch and Bound Consider the following integer program: Maximize x + y + 2z Subject to 7x + 2y + 3z  36 5x + 4y + 7z  42 2x + 3y + 5z  28 x, y, z  0, integer (IP0)

  24. IP0 / / 10 11 IP2 IP1 Integral IP4 IP3 Infeasible IP6 IP5 Integral IP8 IP7 Obj.  incumbent Obj.  incumbent Branch & Bound: Example Best IP value Best IP solution

  25. Diet Problem The Pill Salesman A pill salesman offers Bob energy, protein, and calcium pills to fulfill his nutritional needs. We will represent the costs of each of the pills as follows: y1cost (in dollars) of 1 kcal energy pill y2 cost (in dollars) of 1 g protein pill y3 cost (in dollars) of 1mg calcium pill

  26. Diet Problem LP constraints Nutritional requirements 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6  2000 y1 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 y2 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 y3

  27. Diet Problem LP constraints Minimize Cost 0.3x1 + 2.40x2 + 1.30x3 + 0.90x4 + 2.0x5 + 1.9x6 Nutritional requirements 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6  2000 y1 kcal 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 y2 protein 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 y3 calcium x1 = servings of oatmeal: The cost of the nutrients in one serving of oatmeal shouldn’t exceed the cost of just buying one serving of oatmeal: 110y1 + 4y2 + 2y3 0.3 (4 y2 = cost of protein in oatmeal)

  28. Diet Problem The salesman’s requirements The pill salesman wants to make as much money as possible, given Bob’s constraints. He knows Bob wants 2000 kcal, 55g protein, and 800 mg calcium, so his problem is as follows: Maximize 2000y1 + 55y2 + 800y3 Subject to 110y1 + 4y2 + 2y3 0.3 205y1 + 32y2 + 12y3 2.4 160y1 + 13y2 + 54y3 1.3 160y1 + 8y2 + 285y3 0.9 420y1 + 4y2 + 22y3 2.0 260y1 + 14y2 + 80y3 1.9 y1, y2, y3  0

  29. Diet Problem Solution Solving this LP gives us the following pill prices: $0.27 for 1 kcal energy pill $0.00 for 1 g protein pill $0.16 for 1mg calcium pill $6.71 Total cost = 0.27 (2000) + 0.16 (800) = THE SAME AS THE LOWEST COST DIET!

  30. Diet Problem Weak Duality: Cost of any feas. diet  Price of any feas. pills Strong Duality: (Von Neumann, 1947) Optimal cost of diet = Optimal cost of pills

  31. Diet Problem MORAL: The values of the dual variables, which are always available as part of LP solution, equal the marginal value of corresponding resources.

  32. The Simplex Algorithm Consider the following linear program: Maximize 5x1 + 4x2 + 3x3 Subject to 2x1 + 3x2 + x3  5 4x1 + x2 + 2x3  11 3x1 + 4x2 + 2x3  8 x1, x2, x3  0.

  33. The Simplex Algorithm Slack Variables We add slack variables to each constraint to get equations: Maximize 5x1 + 4x2 + 3x3 Subject to 2x1 + 3x2 + x3 + x4 = 5 4x1 + x2 + 2x3 + x5 = 11 3x1 + 4x2 + 2x3 + x6 = 8 x1, x2, x3, x4, x5, x6  0.

  34. The Simplex Algorithm Getting started Rewriting we can express the slack variables and the objective, denoted z, in terms of the original variables. Setting x1=x2=x3=0, we easily read off the solution x4=5, x5=11, x6=8, z=0. z= 5x1 + 4x2 + 3x3 x4 = 5 - 2x1 - 3x2- x3 x5 = 11 - 4x1 - x2- 2x3 x6 = 8 - 3x1- 4x2 - 2x3

  35. The Simplex Algorithm 1st Iteration We want to increase the value of z from its current value of 0. Clearly, increasing the value of x1will increase the value of z. By how much can we increase it? z= 5x1 + 4x2 + 3x3 x4 = 5 - 2x1 - 3x2- x3 x5 = 11 - 4x1 - x2- 2x3 x6 = 8 - 3x1- 4x2 - 2x3 x4 0  5 – 2x1  0  5  2x1x1 5/2 x5 0  11 – 4x1  0  11  4x1x1 11/4 x6 0  8 – 3x1  0  8  3x1x18/3

  36. The Simplex Algorithm 2nd Iteration Now we see that increasing the value of x3 will improve the value of z. z= 25/2 - 7/2x2 + 1/2x3 - 1/2x4 x1 = 5/2 - 3/2x2 - 1/2x3- 1/2x4 x5 = 1 + 5x2 + 2x4 x6 = 1/2 + 1/2x2- 1/2x3 + 3/2x4 x1 0  5/2 – 1/2x3  0  5/2  1/2x3x3 5 x5 0 x3 unbounded x6 0  1/2 – 1/2x3  0  1/2  1/2x3x1 1

  37. The Simplex Algorithm Optimal!!! z= 13 - 3x2 - x4 - x6 x1 = 2 - 2x2 - 2x4+ x6 x5 = 1 + 5x2 + 2x4 x3 = 1 + x2+ 3x4 - 2x6

  38. Barrier central path • Predictor • Corrector Barrier Algorithms Simplex solution path Optimum

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