Download presentation

Presentation is loading. Please wait.

Published byCandace Ashley Modified over 2 years ago

1
1 Introduction to Linear Programming

2
2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. X1X2X3X4X1X2X3X4 Unit: thousand To win at least 50000 urban votes: -2x 1 +8x 2 +0x 3 +10x 4 50. To win at least 100000 suburban votes: 5x 1 +2x 2 +0x 3 +0x 4 100. To win at least 25000 rural votes: 3x 1 -5x 2 +10x 3 -2x 4 25. Want to minimize x 1 + x 2 + x 3 + x 4

3
3 Minimization linear program : minimize x 1 + x 2 + x 3 + x 4 subject to -2x 1 + 8x 2 + 0x 3 + 10x 4 50 5x 1 + 2x 2 + 0x 3 + 0x 4 100 3x 1 - 5x 2 +10x 3 - 2x 4 25 x 1, x 2, x 3, x 4 0. Maximization linear program : maximize x 1 + x 2 subject to 4x 1 - x 2 8 2x 1 + x 2 10 5x 1 - 2x 2 -2 x 1, x 2 0. Standard form, objective function, feasible solution and region, simplex algorithm, ellipsoid algorithm, interior-point method.

4
4 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

5
5 Standard and Slack forms: Standard form: maximize c j x j subject to a ij x j b i for i=1, …, m. x j 0 for j=1, …, n. Objective function Constraints Nonnegative constraints maximize c T x subject to A x b x 0

6
6 Minimization linear program: minimize -2x 1 + 3 x 2 subject to x 1 + x 2 = 7 x 1 - 2x 2 4 x 1 0. Maximization linear program: maximize 2x 1 -3 x 2 subject to x 1 + x 2 = 7 x 1 - 2x 2 4 x 1 0. Converting linear programs into standard form:

7
7 If x j has no nonnegative constraint, we replace x j by x’ j - x’’ j, where x’ j and x’’ j 0. maximize 2x 1 - 3 x’ 2 + 3 x’’ 2 subject to x 1 + x’ 2 - x’’ 2 = 7 x 1 - 2 x’ 2 + 2 x’’ 2 4 x 1, x’ 2, x’’ 2 0.

8
8 Since x = y iff x y and y x, we can replace equality constraint by the pair of inequality constraints. maximize 2x 1 - 3 x 2 + 3 x 3 subject to x 1 + x 2 - x 3 7 x 1 + x 2 - x 3 7 x 1 - 2 x 2 + 2 x 3 4 x 1, x 2, x 3 0. - x 1 - x 2 + x 3 -7

9
9 Converting linear programs into slack form: maximize 2x 1 - 3 x 2 + 3 x 3 subject to x 4 = 7 - x 1 - x 2 + x 3 x 5 = -7 + x 1 + x 2 - x 3 x 6 = 4 - x 1 + 2 x 2 - 2 x 3 x 1, x 2, x 3, x 4, x 5, x 6 0. x 4, x 5, x 6 : basic variables x 1, x 2, x 3 : nonbasic variables z = 2x 1 - 3 x 2 + 3 x 3 x 4 = 7 - x 1 - x 2 + x 3 x 5 =-7 + x 1 + x 2 - x 3 x 6 = 4 - x 1 + 2 x 2 - 2 x 3 slack form:

10
10 N: the set of indices of nonbasic variables, |N|=n. B: the set of indices of basic variables, |B|=m. N B ={1, 2, …, m+n}. A slack form: z = 28 - x 3 /6 - x 5 /6 - 2 x 6 /3 x 1 = 8 + x 3 /6 + x 5 /6 - x 6 /3 x 2 = 4 - 8x 3 /3 - 2 x 5 /3 + x 6 /3 x 4 = 18 - x 3 /2 + x 5 /2

11
11 Formulating problems as LP: Single-pair shortest-path: given G=(V, E), w, s and t, we want to compute d t, which is the weight of a shortest path from s to t. min d t subj. to d v ≤ d u + w(u, v) for each (u, v) in E, d s = 0. Maximum flow: max v f sv - v f vs subj. to f uv ≤ c(u, v) for each u, v in V, f uv ≥ 0 for each u, v in V, v f uv = v f vu for each u in V-{s, t}.

12
12 Minimum-cost flow:

13
13 Minimum cost flow: min (u,v) a(u,v) f uv subj. to f uv c(u, v) for each u, v in V, f uv ≥ 0 for each u, v in V, v f vu - v f uv = 0 for each u in V-{s, t}, v f sv - v f vs = d,

14
14 Multi-commodity flow: Given a flow network and k different commodities, K 1, K 2,..., K k, where each K i = (s i, t i, d i ) and s i is the source of commodity i, t i is the sink and d i is the demand– the desired flow from s i to t i. Define a flow f i for commodity i and f uv = i=1..k f iuv. Want to determine whether it is possible to find such a flow. LP is the only known polynomial-time algorithm for this problem. min 0 subj. to i=1..k f iuv c(u, v) for each u, v in V, f iuv ≥ 0 for each u, v in V and i= 1..k, v f iuv - v f ivu = 0 for each u in V-{s i, t i } and i= 1..k, v f is i v - v f ivs i = d i for each i = 1..k.

15
15 Simplex algorithm: An example: Max 3 x 1 + x 2 + 2x 3 Subject to x 1 + x 2 + 3 x 3 30 2x 1 + 2x 2 + 5x 3 24 4x 1 + x 2 + 2x 3 36 x 1, x 2, x 3 0 Slack form: z = 3 x 1 + x 2 + 2x 3 x 4 = 30 - x 1 - x 2 - 3 x 3 x 5 = 24 - 2x 1 - 2x 2 - 5x 3 x 6 = 36 - 4x 1 - x 2 - 2x 3 x 1, x 2, x 3, x 4, x 5, x 6 0 (0, 0, 0, 30, 24, 36) is a basic solution. Want to find a basic feasible solution (BFS) from basic solution. Goal: in each iteration, reformulate the LP so that BFS has a greater object value. Attempt: select a non-basic variable x e whose coefficient in the object function is positive and increase its value as much as possible without violating any constraint. Then x e becomes basic and some x l becomes non-basic.

16
16 Slack form: z = 3 x 1 + x 2 + 2x 3 x 4 = 30 - x 1 - x 2 - 3 x 3 x 5 = 24 - 2x 1 - 2x 2 - 5x 3 x 6 = 36 - 4x 1 - x 2 - 2x 3 x 1 : entering variable, x 6 : leaving variable. z = 27+ x 2 /4 + x 3 /2 - 3 x 6 /4 x 1 = 9 - x 2 /4 - x 3 /2 - x 6 /4 x 4 = 21- 3 x 2 /4 - 5 x 3 /2 + x 6 /4 x 5 = 6 - 3 x 2 /2 - 4x 3 + x 6 /2. pivoting z = 111/4+ x 2 /16 - x 5 /8 - 11 x 6 /16 x 1 = 33/4 - x 2 /16 + x 5 /8 - 5x 6 /16 x 3 = 3/2 - 3 x 2 /8 - x 5 /4 + x 6 /8 x 4 = 69/4 + 3 x 2 /16 +5x 5 /8 - x 6 /16. z = 28- x 3 /6 - x 5 /6 - 2 x 6 /3 x 1 = 8 + x 3 /6 + x 5 /6 - x 6 /3 x 2 = 4 - 8 x 3 /3 - 2x 5 /3 + x 6 /3 x 4 = 18 - x 3 /2 +x 5 /2.

17
17

18
18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

19
19

20
20 1.How do we determine if a linear program is feasible? 2.What do we do if the LP is feasible, but the initial basic solution is not feasible? 3. How do we determine if a linear program is unbounded? 4. How do we choose the entering and leaving variables?

21
21 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

22
22 Lemma 2: Given a linear program (A, b, c), suppose the call to INITIALIZE-SIMPLEX in line 1 of SIMPLEX returns a slack form for which the basic solution is feasible. Then if SIMPLEX returns a solution in line 16, that solution is a feasible solution to the LP. If SIMPLEX returns “unbounded” in line 10, the LP is unbounded.

23
23 Lemma 3: Let I be a set of indices. For each i in I, let i and i be real numbers and let x i be a real-valued variable. Let be any real number. Suppose that for any settings of the x i we have:

24
24 Lemma 4: Let (A, b, c) be an LP in standard form. Given a set B of basic variables, the associated slack form is uniquely determined. Pf: By contradiction, suppose there are two slack form with the same B:

25
25 Lemma 5: If SIMPLEX fails to terminate in at most (m+n)!/m!n! iterations, then it cycles. Lemma 6: (Bland’s rule) In line 3 and 8 of SIMPLEX, ties are always broken by choosing the variable with the smallest index, then SIMPLEX must terminate.

26
26 Lemma 7: Assuming that INITIALIZE-SIMPLEX returns a slack form for which the basic solution is feasible, SIMPLEX either reports that an LP is unbounded, or it terminates with a feasible solution in at most (m+n)!/m!n! iterations.

27
27 Duality: primal and dual max c j x j subject to a ij x j b i for i=1, …, m. x j 0 for j=1, …, n. Max 3 x 1 + x 2 + 2x 3 Subject to x 1 + x 2 + 3 x 3 30 2x 1 + 2x 2 + 5x 3 24 4x 1 + x 2 + 2x 3 36 x 1, x 2, x 3 0 Min 30 y 1 + 24y 2 + 36y 3 Subject to y 1 + 2y 2 + 4 y 3 3 y 1 + 2y 2 + y 3 1 3y 1 + 5y 2 + 2 y 3 2 y 1, y 2, y 3 0

28
28

29
29

30
30

31
31 Finding an initial solution: Max 2 x 1 - x 2 Subject to 2x 1 - x 2 2 x 1 - 5x 2 - 4 x 1, x 2 0 x 1 =0, x 2 =0 is NOT a feasible solution. L aux : Max -x 0 Subject to 2x 1 - x 2 - x 0 2 x 1 - 5x 2 - x 0 - 4 x 1, x 2, x 0 0 Slack form: z = - x 0 x 3 = 2 - 2x 1 + x 2 + x 0 x 4 = -4 - x 1 +5x 2 + x 0 z = -4 - x 1 +5 x 2 - x 4 x 0 = 4 + x 1 -5 x 2 + x 4 x 3 = 6 - x 1 -4 x 2 + x 4

32
32 z = -4 - x 1 +5 x 2 - x 4 x 0 = 4+ x 1 -5 x 2 + x 4 x 3 = 6- x 1 -4 x 2 + x 4 (4,0,0,6,0) is a BFS. z = - x 0 x 2 = 4/5- x 0 /5 + x 1 /5 + x 4 /5 x 3 = 14/5+ 4x 0 /5 - 9x 1 /5 + x 4 /5 The following has a BFS. z = -4/5 +9x 1 /5 - x 4 /5 x 2 = 4/5 + x 1 /5 + x 4 /5 x 3 = 14/5- 9x 1 /5 + x 4 /5 Take x 0 = 0. Then x 2 = 4/5 + x 1 /5 + x 4 /5, and 2x 1 - x 2 = -4/5 +9x 1 /5 - x 4 /5.

33
33 Lemma 11: Let L be a LP in standard form. Let L aux be the following LP with n+1 variables: maximize - x 0 subject to a ij x j - x 0 b i for i=1, …, m. x j 0 for j = 0, …, n. Then L is feasible iff the optimal object value of L aux is 0.

34
34 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

35
35 Lemma 12: If a LP L has no feasible solution, INITIALIZE-SIMPLEX returns “infeasible.” Otherwise, it returns a valid slack form for which the basic solution is feasible. Theorem 13: (Fundamental theorem of LP) Any linear program L, given in standard form, either 1.has an optimal solution with a finite objective value, 2.is infeasible, or 3.is unbounded. If L is infeasible, SIMPLEX returns “ infeasible.” If L is unbounded, SIMPLEX returns “unbounded.” Otherwise, SIMPLEX returns an optimal solution with a finite objective value.

Similar presentations

OK

Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc. Linear Programming: An Algebraic Approach 4 The Simplex Method with Standard Maximization.

Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc. Linear Programming: An Algebraic Approach 4 The Simplex Method with Standard Maximization.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Field emission display ppt online Morphological opening ppt on mac Ppt on life achievement of nelson mandela Ppt on solar energy utilization Ppt on nationalism in india Ppt on features of ms excel Ppt on c language history Ppt on print media and electronic media Ppt on tamper resistant Download ppt on eddy current brakes