Download presentation

Presentation is loading. Please wait.

Published byCandace Ashley Modified about 1 year 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

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google