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Linear Programming: The Carpenter's Problem in Standard Form

Explore the application of linear programming in solving the Carpenter's Problem, demonstrating how to transform the problem into standard form and utilize the Simplex algorithm for optimization solutions.

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Linear Programming: The Carpenter's Problem in Standard Form

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  1. CS 575Design and Analysis ofComputer AlgorithmsProfessor Michal CutlerLecture 20April 18, 2005

  2. This class • The problem • Graphical solutions • Standard and slack forms • The simplex algorithm • Finding an initial feasible solution

  3. Linear Programming • Among most important advances of 20th century • Generalizes many classical problems • Shortest path, max flow, multi-commodity flow, MST, traveling salesman • Often used for optimal allocation of scarce resources • Helps to find “good” solutions to NP-hard optimization problems

  4. The carpenter’s problem • A carpenter manufactures 2 types of tables. • Production is limited because of availability of labor, glass and wood.

  5. Profits • Only round tables – 10 with profit of $700 • Only rectangular tables – 10 with profit of $900 • 5 round and 8 rectangular – 5*70 + 8*90= $1070

  6. Standard form for LP • Note that the objective function and the inequalities are linear

  7. The carpenter’s problem in standard form • In the following slides we solve the problem using geometry • Show feasible region and extreme points

  8. Feasible regionfor carpenter problem Rectangular table Glass 5x+3y=50 (0, 10) (5, 8) Labor 3x+2y=31 (7, 5) Feasible region Wood 4x+10y=100 (0,0) (10,0) Round table

  9. Objective function70x+90y Rectangular table Profit (0, 10) (5, 8) 70x+90y=$700 (7, 5) Feasible region 70x+90y=$1070 (0,0) (10,0) Round table

  10. Integer linear programming • The carpenter’s problem requires integer solutions • Typically linear programming algorithms do not produce integer solution! • Integer programming in NP-hard!

  11. (0, 10) (5, 8) (7, 5) LP:Geometry Rectangular table Extreme Points of feasible region (0,0) (10,0) Round table

  12. LP: Geometry in general • The solutions of a linear program form an n-dimensional convex polyhedron • Convex : if y and z are feasible solutions, so is ay + (1 - a)z where 0<a<1 • When the solutions of a problem are a convex set and the object function is convex on the convex set a localoptimal solution is a global one • No need to worry about local optima!

  13. LP: Geometry in general • Extreme point: vertex of the polyhedron (solution cannot be expressed as a linear combination of two different solutions y and z satisfying ay + (1 - a)z where 0<a<1) • Extreme point theorem: If there exists an optimal solution to standard form LP, then there exists one that is an extreme point • Need to consider only a finite number of solutions

  14. LP: Main idea of the simplex algorithm • Main idea of the simplex algorithm: search for optimal solution by moving from one extreme point of the polyhedron to a neighboring extreme point without decreasing the objective function

  15. George Danzig – invented Simplex algorithm • First algorithm for solving linear programming problems. (Danzig 1947) • Developed after WWII in response to logistic problems • Used for 1948 Berlin airlift • Finite (exponential) complexity

  16. A “large scale” problem • The first "large-scale" problem was a 9 by 77 instance of the "diet problem" (find a min-cost diet that meets or exceeds nutritional requirements), solved by 9 clerks using hand- operated desk calculators, in about 120 man-days. • Current problems solved with thousands of unknowns and inequalities

  17. Polynomial algorithms • Ellipsoid (Khachian 1979, 1980) • Resolved question of whether linear programming is NP-hard (it is not) • Not practical algorithm • Karmarkar’s algorithm (1984) and other interior point algorithms • Competitive with simplex • Likely to dominate on large problems

  18. Transforming to standard form (1) • Often a linear programming problem is not in standard form • Inequality with  (multiply by –1) x+5y-6z22  -x-5y+6z-22 • Equality = (transform to one inequality with , and one with ) x+5y-6z=22  x+5y-6z  22 and (from x+5y-6z  22 to) -x-5y+6z  -22

  19. Transforming to standard form (2) • Objective function is minimum Multiply by –1 and replace by maximum min x+5y-6z  max -x-5y+6z • x is unrestricted Replace x in problem by y-z, and add y0, z0 After finding the optimal solution x can be calculated as y-z. If y  z in the optimal solution then x  0, otherwise if y < z in the optimal solution then x < 0.

  20. Transforming to standard form (3) • x  c for c < 0 for example x  -3 Replace x in problem by -y + c, add y0 After finding the optimal solution, x can be calculated as –y+c.Since y0, x  c is satisfied in optimal solution

  21. Example – converting to  

  22. Example – converting to

  23. Slack form • Given • We introduce a slack variables and rewrite the inequality as: and add • Apply this conversion to each constraint

  24. Example – converting to slack form  • B = {x4, x5, x6} set of basic variables • N = {x1 , x2, x3} set of nonbasic variables

  25. The simplex algorithm  • For this example slack form provides a basic feasible solution: (0, 0, 0, 30, 24, 36) and z = 0 • To increase z select a nonbasic variable xe whose coefficient in z is positive and increase xe as much as possible without violating the constraints

  26. Transforming to standard form (1) • Often a linear programming problem is not in standard form • Inequality with  (multiply by –1) x+5y-6z22  -x-5y+6z-22 • Equality = (transform to one inequality with , and one with ) x+5y-6z=22  x+5y-6z  22 and (from x+5y-6z  22 to) -x-5y+6z  -22

  27. Transforming to standard form (2) • Objective function is minimum Multiply by –1 and replace by maximum min x+5y-6z  max -x-5y+6z • x is unrestricted Replace x in problem by y-z, and add y0, z0 After finding the optimal solution x can be calculated as y-z. If y  z in the optimal solution then x  0, otherwise if y < z in the optimal solution then x < 0.

  28. Transforming to standard form (3) • x  c for c < 0 for example x  -3 Replace x in problem by -y + c, add y0 After finding the optimal solution, x can be calculated as –y+c.Since y0, x  c is satisfied in optimal solution

  29. Example – converting to  

  30. Example – converting to

  31. Slack form • Given • We introduce a slack variables and rewrite the inequality as: and add • Apply this conversion to each constraint

  32. Example – converting to slack form  • B = {x4, x5, x6} set of basic variables • N = {x1 , x2, x3} set of nonbasic variables

  33. The simplex algorithm  • For this example slack form provides a basic feasible solution: (0, 0, 0, 30, 24, 36) and z = 0 • To increase z select a nonbasic variable xe whose coefficient in z is positive and increase xe as much as possible without violating the constraints

  34. The simplex algorithm (1)  • For this example slack form provides a basic feasible solution: (0, 0, 0, 30, 24, 36) and z = 0 • To increase z select a nonbasic variable xe whose coefficient in z is positive and increase xe as much as possible without violating the constraints

  35. The simplex algorithm (2) • This non basic entering variable xe will now become basic and another variable xl non basic. • Let x1 be the selected entering variable • From (1) 0  x4= 30 - x1 x1 30, • From (2) 0 x5= 24 - 2x1x1 12, • From (3) 0 x6= 36 - 4x1 x1 9. • Equation (3) causes the tightest limitation on x1. So x6 is the leaving basic variable

  36. The simplex algorithm (3) • From equation (3) we get: x1 =9-x2 /4-x3/2- x6 /4 • Substituting x1 in (0), (1) and (2) we get:

  37. The simplex algorithm (4) • Select x3 as new basic variable • From 0  x1= 9 - x3/2x3 18, • From 0 x4= 21 – 5x3/2 x3 42/5 • From 0 x5= 6 - 4x3 x3 3/2. • So x5 is the new non basic variable

  38. The simplex algorithm (5) • From the equation for x5 (3’) we get: x3 = 3/2-3x2/8-x5/4+x6/8 • Substituting x3 in the equations (0’,1’ and 2’) for x1 and x4, and z we get:

  39. The simplex algorithm (6) • Select x2 as new basic variable • 0  x1= 33/4 - x2/16x2 132, • 0 x3=3/2 – 3x2/8  x2 4 • 0 x4= 69/4 + 3x2/16x2. • So x3 is the new non basic variable

  40. The simplex algorithm (7) • The optimal solution is (8, 4, 0,18,0,0), z=28 • The value of x4 =18 is the slack between the left side x1+ x2 = 12 of the original inequality x1+ x2 + x330, and the right hand side 30

  41. In Summary: • The entering variablexe is one for which the coefficient ce in z is positive. • The leaving variable is the one that causes the tightest constraint on the entering variable. • If the tightest constraint on the entering variable is infinity, the linear program is unbounded

  42. Unbounded solution example • Let x1 be the entering variable • From (1) and (2), x1 infinity • So x1 can be increased to infinity and solution is unbounded

  43. Graphical solution -x1+x2 =2 -2x1+x2 =1 x2 (1, 3) Feasible region (0, 1) (0,0) objective function x1+2 x2 =5x1

  44. Termination • It is easy to see that no iteration of the algorithm decreases the objective function • Unfortunately, the objective function may remain unchanged. This phenomenon is called degeneracy

  45. Example Next iteration when x2 becomes basic the objective function will grow to 16

  46. Cycling • Simplex cycles if the slack forms at two different iterations are identical (since it is a deterministic algorithm it will cycle through the same series of slack forms forever.) • There are various rules to avoid cycling • Bland’s rule: always choose the variable with the smallest index (for entering the one with the first positive c, for leaving the first most restricting variable)

  47. Finding an initial basic feasible solution • When there are negative b values in standard form the corresponding slack form is not feasible • Let L denote the original problem • To find an initial feasible solution to L if it exists we formulate an auxiliary problem Lmax

  48. Simplex when initial solution not feasible • Formulate an auxiliary problem Lmax • Transform basic solution to make it feasible • Use Simplex to find an optimal solution to Lmax • If solution <0 there is no feasible solution to original problem L • Otherwise in the optimal solution of Lmax: • Substitute the objective function by the objective function of the original problem L after substituting in it all variables that are basic in the optimal solution of Lmax. • Eliminate x0 from the constraints • Apply Simplex to find an optimal solution to original problem L

  49. 1.The auxiliary problem Lmax • L is feasible iff the optimal solution for Lmax is 0 • If L is feasible, combine the feasible solution with x0 = 0 resulting in a feasible solution to Lmax. Since z=0 solution is also optimal • If the optimal solution of Lmax is 0, the solution without x0 = 0 is feasible for L • Conclusions: If the solution for Lmax is not 0 L is not feasible. If the solution for Lmax is 0 we now have a feasible solution for L

  50. 2.Transform basic solution to make it feasible • Since there are negative b values, the current solution is not feasible • Choose the variable xs with the smallest value in the basic solution as the leaving variable, and x0 as the entering one • It is easy to prove the new solution is feasible

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