AM 121: Intro to Optimization Models and Methods Fall 2017

Transcription

1 AM 121: Intro to Optimization Models and Methods Fall 2017 Lecture 10: Dual Simplex Yiling Chen SEAS Lesson Plan Interpret primal simplex in terms of pivots on the corresponding dual tableau Dictionaries Define dual simplex and interpret it in terms of pivots on the primal tableau Applications Jensen & Bard: 3.9 1

2 Motivation: Dual Simplex Useful for solving integer programs via branch-and-bound search. Find a solution after a new constraint is added, or an existing constraint is modified. A new Phase 1 Phase 2 method (no need for artificial variables!) Naming convention m+n max j=m+1 c j x j m+n s.t. j=m+1 a ij x j b i x j 0 i = 1, 2,, m j = m+1,, m+n primal vars x m+1,,x m+n ; slack vars x 1,,x m m min i=1 b i y i m s.t. i=1 a ij y i c j y i 0 j = m+1,, m+n i = 1,, m dual vars y 1,,y m ; surplus vars y m+1,,y m+n 2

3 Example max 4x 3 + x 4 + 3x 5 s.t. x 3 + 4x 4 1 3x 3 x 4 + x 5 3 x 3, x 4, x 5 0 Dual: min y 1 + 3y 2 s.t. y 1 + 3y 2 4 4y 1 y 2 1 y 2 3 y 1, y 2 0 equivalent maximization problem max -y 1-3y 2 s.t. -y 1-3y y 1 +y 2-1 -y 2-3 y 1, y 2 0 Tableau (Review) A tableau must have isolated variables but does not need to have non-negative right hand side values. If the RHS is negative, the tableau is not (primal) feasible. 3

4 Primal <> Dual RHS non-basic Initial primal tableau: z 4x 3 x 4 3x 5 = 0 x 1 + x 3 + 4x 4 = 1 x 2 + 3x 3 x 4 + x 5 = 3 B={1,2} B ={3,4,5} feasible Initial dual tableau: z +y 1 +3y 2 = 0 -y 1-3y 2 + y 3 = -4-4y 1 + y 2 + y 4 = -1 -y 2 + y 5 = -3 B={3,4,5} B ={1,2} infeasible basic z x 1 x 2 basic z y 3 y 4 y 5 x 3 x 4 x Primal dictionary y 1 y Dual dictionary RHS (negated columns) non-basic (negated columns) Primal <> Dual Initial primal tableau: z 4x 3 x 4 3x 5 = 0 x 1 + x 3 + 4x 4 = 1 x 2 + 3x 3 x 4 + x 5 = 3 B={1,2} B ={3,4,5} Feasible; Neg shadow costs Initial dual tableau: z +y 1 +3y 2 = 0 -y 1-3y 2 + y 3 = -4-4y 1 + y 2 + y 4 = -1 -y 2 + y 5 = -3 B={3,4,5} B ={1,2} Infeasible; Non-neg shadow costs Primal dictionary B={1,2} B={3,4,5} Dual dictionary -ve transpose of primal; B (dual) = B (primal) 4

5 A Dictionary Definition. The dictionary corresponding to a tableau with basis B is v -c T B (where B is the set of non-basic) b -A B with rows of A B and b ordered by indices of basic vars and columns of c T B ordered by indices of non-basic vars Suppose tableau was reordered as: z x 4 4x 3 3x 5 = 0 x 2 x 4 + 3x 3 + x 5 = 3 x 1 + 4x 4 + x 3 = 1 Corresponding dictionary is still: B={1,2} A dictionary and basis B completely defines a tableau! A simple primal-dual correspondence Primal dictionary Basic B, non-basic B negated transpose çè Dual dictionary Basic B, non-basic B So: (1) primal tableau has corresponding dual tableau (2) can track progress of simplex in the dual! (3) primal obj value = - dual obj value at each step (negated because we converted min f(y) into max f(y)) 5

6 Primal pivot (track in dual) z 4x 3 x 4 3x 5 = 0 x 1 + x 3 + 4x 4 = 1 x 2 + 3x 3 x 4 + x 5 = 3 B={1,2} B ={3,4,5} pivot (2,5); 2 out 5 in pivot z +3x 2 +5x 3 4x 4 = 9 x 1 +x 3 + 4x 4 = 1 x 2 +3x 3 x 4 +x 5 = 3 B={1,5} B ={2,3,4} z +y 1 +3y 2 = 0 -y 1-3y 2 +y 3 = -4-4y 1 + y 2 +y 4 = -1 - y 2 +y 5 = -3 B={3,4,5} B ={1,2} pivot (5,2); 5 out 2 in pivot z +y 1 +3y 5 = -9 -y 1 +y 3-3y 5 = 5-4y 1 +y 4 +y 5 = -4 y 2 -y 5 = 3 B={2,3,4} B ={1,5} still dual infeasible New dictionaries z +3x 2 +5x 3-4x 4 = 9 x 1 +x 3 +4x 4 = 1 x 2 +3x 3 x 4 +x 5 = 3 B={1,5} B ={2,3,4} z x 1 x 5 x 2 x 3 x Primal dictionary z +y 1 +3y 5 = -9 -y 1 +y 3-3y 5 = 5-4y 1 +y 4 +y 5 = -4 y 2 -y 5 = 3 B={2,3,4} B ={1,5} z y 2 y 3 y 4 y 1 y Dual dictionary (negated transpose) 6

7 Primal pivot (track in dual) z +3x 2 +5x 3-4x 4 = 9 x 1 +x 3 +4x 4 = 1 x 2 +3x 3 x 4 +x 5 = 3 B={1,5} B ={2,3,4} Pivot (1,4); 1 out 4 in z +y 1 +3y 5 = -9 -y 1 +y 3-3y 5 = 5-4y 1 +y 4 +y 5 = -4 y 2 -y 5 = 3 B={2,3,4} B ={1,5} Pivot (4,1); 4 out 1 in z +x 1 +3x 2 +6x 3 = 10 ¼x 1 +¼x 3 + x 4 = ¼ ¼x 1 +x 2 +3¼x 3 + x 5 = 3¼ B={4,5} B ={1,2,3} pivot z +¼y 4 + 3¼y 5 = -10 y 1 ¼y 4 ¼y 5 = 1 y 2 y 5 = 3 y 3 ¼y 4 3¼y 5 = 6 B={1,2,3} B ={4,5} pivot Primal feasible, Primal optimal. Dual feasible, Dual optimal Final tableaus; Final dictionaries z +x 1 +3x 2 +6x 3 = 10 ¼x 1 +¼x 3 + x 4 = ¼ ¼x 1 +x 2 +3¼x 3 + x 5 = 3¼ B={4,5} B ={1,2,3} z +¼y 4 + 3¼y 5 = -10 y 1 ¼y 4 ¼y 5 = 1 y 2 y 5 = 3 y 3 ¼y 4 3¼y 5 = 6 B={1,2,3} B ={4,5} z x 4 x 5 z y 1 y 2 y 3 x 1 x 2 x ¼ -¼ 0 -¼ 3¼ -¼ -1-3¼ Primal dictionary y 4 y ¼ -3¼ 1 ¼ ¼ ¼ 3¼ Dual dictionary (negated transpose) 7

8 From Initial to Final tableau Initial primal tableau Final primal tableau z 4x 3 x 4 3x 5 = 0 z +x 1 +3x 2 +6x 3 =10 x 1 + x 3 + 4x 4 = 1 ¼x 1 +¼x 3 + x 4 =¼ x 2 + 3x 3 x 4 + x 5 = 3 ¼x 1 +x 2 +3¼x 3 + x 5 =3¼ B={1,2} B ={3,4,5} B={4,5} B ={1,2,3} Initial dual tableau z +y 1 +3y 2 = 0 -y 1-3y 2 +y 3 =-4-4y 1 + y 2 +y 4 =-1 -y 2 +y 5 =-3 B={3,4,5} B ={1,2} Final dual tableau z +¼y 4 +3¼y 5 =-10 y 1 ¼y 4 ¼y 5 =1 y 2 y 5 =3 y 3 ¼y 4 3¼y 5 =6 B={1,2,3} B ={4,5} Primal Simplex: Effect On Dual Solution Maintains primal feasibility, terminates with primal optimality Non-negative RHS in primal <> non-negative reduced costs in dual Free reduced costs in primal <> free RHS values in dual (and dual infeasible for a while) Terminates with non-negative reduced costs in primal <> non-neg RHS in dual (and dual feasible) Pivot in primal, track in dual. Corresponding dual solution always has nonnegative reduced costs, but is initially infeasible. Note: pair (x,y) satisfy complementary slackness during pivots, but y is infeasible until termination. 8

9 Complementary Slackness z 4x 3 x 4 3x 5 = 0 x 1 + x 3 + 4x 4 = 1 x 2 + 3x 3 x 4 + x 5 = 3 B={1,2} B ={3,4,5} pivot (2,5) pivot z +3x 2 +5x 3 4x 4 = 9 x 1 +x 3 + 4x 4 = 1 x 2 +3x 3 x 4 +x 5 = 3 B={1,5} B ={2,3,4} z +y 1 +3y 2 = 0 -y 1-3y 2 +y 3 = -4-4y 1 + y 2 +y 4 = -1 - y 2 +y 5 = -3 B={3,4,5} B ={1,2} pivot (5,2) pivot z +y 1 +3y 5 = -9 -y 1 +y 3-3y 5 = 5-4y 1 +y 4 +y 5 = -4 y 2 -y 5 = 3 B={2,3,4} B ={1,5} It is because of CS that we get this particular dual <> primal construction Primal dictionary çè Dual dictionary Basic B, non-basic B negated transpose Basic B, non-basic B In particular, it is CS that gives us that: primal obj value = - dual obj value at each step (negated because we converted min f(y) into max f(y)) 9

10 Different kinds of pivots V1: pick non-basic x j with c j <0 to enter, and look at positive column entries to find basic var to leave V2: pick basic x j with b j <0 to exit, and look at negative row entries to find non-basic var to enter V1 V2 Primal simplex Primal tableau Dual tableau Dual simplex Dual tableau Primal tableau Dual Simplex Maintain dual feasibility, terminate with dual optimality Rather than work in dual tableau, can track in primal tableau: perform "dual pivots. 10

11 Dual Simplex Maintain dual feasibility, terminate with dual optimality Rather than work in dual tableau, can track in primal tableau: perform "dual pivots. Reduced costs Primal tableau properties Primal feasible Dual feasible Primal and Dual optimal Free Non-neg Non-neg RHS Non-neg Free Non-neg primal pivots to solve dual pivots to solve Example: Dual Simplex Primal problem max x 4 x 5 s.t. -2x 4 x 5 4-2x 4 + 4x 5-8 -x 4 + 3x 5-7 x 4, x 5 0 Dual problem min 4y 1-8y 2-7y 3 s.t. -2y 1 2y 2 y 3-1 -y 1 + 4y 2 +3y 3-1 y 1, y 2, y 3 0 max -4y 1 + 8y 2 +7y 3 s.t. 2y 1 + 2y 2 + y 3 1 y 1 4y 2 3y 3 1 y 1, y 2, y 3 0 equivalent maximization problem 11

12 Initial Tableau max x 4 x 5 s.t. -2x 4 x 5 4-2x 4 + 4x 5-8 -x 4 + 3x 5-7 x 4, x 5 0 max -4y 1 + 8y 2 +7y 3 s.t. 2y 1 + 2y 2 + y 3 1 y 1 4y 2 3y 3 1 y 1, y 2, y 3 0 z + x 4 + x 5 = 0 x 1-2x 4 x 5 = 4 x 2-2x 4 + 4x 5 = -8 x 3 x 4 + 3x 5 = -7 z +4y 1-8y 2-7y 3 =0 2y 1 +2y 2 + y 3 +y 4 =1 y 1 4y 2 3y 3 +y 5 =1 Dual Pivot (track in primal) z +x 4 + x 5 = 0 x 1-2x 4 x 5 = 4 x 2-2x 4 + 4x 5 = -8 x 3 x 4 + 3x 5 = -7 B={1,2,3} pivot (2,4) (2 out, 4 in) z +4y 1-8y 2-7y 3 =0 2y 1 +2y 2 + y 3 +y 4 =1 y 1 4y 2 3y 3 B={4,5} +y 5 =1 pivot (4,2) (4 out, 2 in) pivot z +½x 2 +3x 5 = -4 x 1 x 2 5x 5 = 12 -½x 2 +x 4 2x 5 = 4 -½ x 2 + x 3 + x 5 = -3 B={1,3,4} z+12y 1-3y 3 +4y 4 =+4 y 1 +y 2 +½y 3 +½y 4 =½ 5y 1 y 3 + 2y 4 +y 5 =3 B={2,5} 12

13 Dual Pivot (track in primal) z +½x 2 +3x 5 =-4 x 1 x 2 5x 5 =12 -½x 2 +x 4 2x 5 =4 -½ x 2 +x 3 + x 5 =-3 B={1,3,4} pivot (3,2) z +12y 1-3y 3 +4y 4 =+4 y 1 +y 2 +½y 3 +½y 4 =½ 5y 1 - y 3 +2y 4 +y 5 =3 B={2,5} pivot (2,3) pivot z + x 3 + 4x 5 =-7 x 1-2x 3-7x 5 =18 - x 3 +x 4-3x 5 =7 x 2-2x 3-2x 5 =6 B={1,2,4} z +18y 1 +6y 2 +7y 4 =+7 2y 1 +2y 2 +y 3 +y 4 =1 7y 1 +2y 2 +3y 4 +y 5 =4 B={3,5} OPTIMAL! Upside Down Pivoting Dual pivots in primal tableau. Maintain dual feasibility, terminate with dual optimality (nonnegative RHS values in primal tableau). Primal infeasible for a while, but non-negative (primal) reduced costs. Terminate with feasible primal solution. Choose a variable with a ve RHS to exit, and do ratio test on strictly negative entries in row with negated reduced cost as numerator. 13

14 Dual simplex Assume have a dual feasible tableau (c B 0) Step 1. Pick a basic variable x r to leave with a strictly negative RHS b r < 0. If no such variable then optimal and STOP. Step 2. Pick a nonbasic variable x k to enter by considering row r and non-basic variables with strictly negative coefficients. Ratio test: k should satisfy -c k /a rk = min {-c j /a rj : j B, a rj <0}. If all a rj on j B nonnegative, then dual unbounded and primal infeasible and STOP. Step 3. Pivot on (r,k) and go to step 1. In corresponding dual tableau: basic variable y r enters and nonbasic variable y k leaves. Example: A dual simplex pivot z +½x 3 + ½x 4 = +1½ x 1 +x 3 = 2 x 2 ½x 3 +½x 4 = - ½ z +x 2 + x 4 = 1 x 1 + 2x 2 + x 4 = 1-2 x 2 + x 3 x 4 = 1 (Pivot on row with ve RHS, picking nonbasic variable to enter with strictly negative a rk and minimal ratio -c k / a rk ) OPTIMAL! 14

15 Example: Dual unbounded Suppose get to a (primal) tableau with row x 1 + 2x 2 + 2x 3 = -5 Dual unbounded, since coefficients on x 2 and x 3 are non-negative. Conclude that primal is infeasible (from weak duality theorem). Making use of the Dual Simplex 1. Find a solution after a new constraint is added (useful for sensitivity analysis, and solving IPs via branch-and-bound search) 2. Find a new solution after a change in a RHS coeff that is outside the allowable range 3. A new Phase 1 Phase 2 method (no need for artificial variables!) 15

16 1. Adding a New Constraint Recall the furniture example max z= 60x x x 3 s.t. 8x 1 + 6x 2 + x 3 + x 4 = 48 (lumber) 4x 1 + 2x x 3 + x 5 = 20 (finishing) 2x x x 3 + x 6 = 8 (carpentry) x 1,, x 6 0 x 1 desks; x 2 tables; x 3 chairs B={4,3,1}. x * =(2,0,8,24,0,0), z=280 Three kinds of new constraints: (i) Current optimal solution still feasible; e.g., add constraint x 1 + x 2 + x Easy: just check. (ii) Current basis becomes infeasible (iii) LP becomes infeasible 1a. New constraint makes optimal basis infeasible. Add: x 2 1. Can just add directly to optimal tableau: z +5x 2 +10x x 6 = 280-2x 2 + x 4 + 2x 5 8x 6 = 24-2x 2 + x 3 + 2x 5 4x 6 = 8 x x 2 0.5x x 6 = 2 -x 2 + x 7 = -1 (primal) B={1,3,4,7}. Dual feasible. Dual pivot in primal (7 out, 2 in). And so pivot (7,2): z +10x 4 +10x 5 + 5x 7 = 275 x 4 +2x 5-8x 6 2x 7 = 26 x 3 +2x 5-4x 6 2x 7 = 10 x 1 ½x 5 +1½x 6 +1¼x 7 = ¾ x 2 x 7 = 1 Find an optimal solution: x * =(¾, 1, 10, 26, 0, 0); z=

17 Remark Could add x 2 1 directly to the initial tableau: x 2 x 7 + x 8 = 1, with artificial variable x 8, and re-solve with primal simplex. By adding the constraint to the final tableau and using dual simplex we just need a single pivot. Large computational gain! 1b. New constraint makes problem infeasible. Consider: x 1 + x 2 12 z +5x x 5 +10x 6 = 280-2x 2 + x 4 + 2x 5 8x 6 = 24-2x 2 + x 3 + 2x 5 4x 6 = 8 x x 2 0.5x x 6 = 2 -x 1 x 2 + x 7 = -12 x 1 no longer isolated. Adopt: (5 ):=(4)+(5) 0.25x 2 0.5x x 6 + x 7 = -10 (5 ) (Primal) B={4,3,1,7}. Dual feasible. Dual pivot in primal: (7 out, 5 in). Get tableau on next slide. (4) (5) 17

18 1b. New constraint makes problem infeasible. z +10x 2 +40x 6 +20x 7 = 80 -x 2 +x 4-2x 6 + 4x 7 = -16 -x 2 + x 3 + 2x 6 + 4x 7 = -32 x 1 + x 2 x 7 = 12 -½x 2 + x 5 3x 6 2x 7 = 12 -½x 2 + x 5 3x 6 2x 7 = 20 (Primal) B={4,3,1,5} Dual pivot in primal: 3 out, 2 in Get tableau on next slide 1b. New constraint makes problem infeasible. z +10x 3 +60x 6 +60x 7 = x 3 +x 4-4x 6 = -16 x 2 x 3-2x 6-4x 7 = 32 x 1 +x 3 +2x 6 +3x 7 = -20 (Primal) B={4,2,1,5} -½x 3 +x 5 4x 6 4x 7 = 36 Dual pivot in primal: 1 out. But ratio test fails, no variable to enter. Dual unbounded, and primal infeasible (by weak duality). 18

19 Making use of the Dual Simplex 1. Find a solution after a new constraint is added (useful for sensitivity analysis, and solving IPs via branch-and-bound search) 2. Find a new solution after a change in a RHS coeff that is outside the allowable range 3. A new Phase 1 Phase 2 method (no need for artificial variables!) 2. Changing a RHS Recall BFS for B={1,3,4} in furniture problem remains feasible for 16 b Suppose b 2 :=30. Let s consider what happens at same basis. z=y T b=( ) 48 = b = A B -1 b = = Construct tableau. Next slide. 19

20 z +5x 2 +10x x 6 = 380-2x 2 + x 4 + 2x 5 8x 6 = 44-2x 2 + x 3 + 2x 5 4x 6 = 28 x x 2 0.5x x 6 = -3 B={1,3,4}. Primal infeasible basis. Do a dual pivot in primal (1 out, 5 in.) z +20x 1 +30x 2 +40x 6 = 320 4x 1 + 3x 2 +x 4-2x 6 = 32 4x 1 + 3x 2 +x 3 +2x 6 = 16-2x 1 2½x 2 +x 5 3x 6 = 6 B={3,4,5} OPTIMAL! Solution x * =(0,0,16,32,6,0). Only make chairs! Making use of the Dual Simplex 1. Find a solution after a new constraint is added (useful for sensitivity analysis, and solving IPs via branch-and-bound search) 2. Find a new solution after a change in a RHS coeff that is outside the allowable range 3. A new Phase 1 Phase 2 method (no need for artificial variables!) 20

21 3. New Phase 1 - Phase 2 method (a) b 0: use primal simplex (no phase 1) (b) c 0, some b i < 0: use dual simplex (no phase 1) (c) Some b i < 0 and some c j < 0 E.g., suppose max x 1-2x 2. First solve max x 1-2x 2. This has z+x 1 +2x 2 first row, and so dual feasible. Use dual simplex to find a primal feasible solution. Second, modify tableau to introduce correct objective. Use primal simplex. ALTERNATIVE: Replace RHS b with b 0, use primal simplex to find a dual feasible solution, then bring back correct RHS and use dual simplex. Summary: Dual Simplex Dual simplex: simplex method applied to the dual problem. Maintains a dual feasible basis, terminates with an optimal dual basis. We track the dual simplex on primal tableau. Upside down pivots : first which basic var exits (-ve RHS), then which non-basic enters? Flexible approach: can work in primal tableau and move seamlessly from primal to dual pivots. Useful when constraints change, for phase1- phase 2, and we ll use for solving IPs! 21