4.5 Simplex method. min z = c T x s.v. Ax = b. LP in standard form

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1 4.5 Simplex method min z = c T x s.v. Ax = b x 0 LP in standard form Examine a sequence of basic feasible solutions with non increasing objective function value until an optimal solution is reached or the LP is found to be unbounded (Dantzig 1947). At each iteration, we go from a basic feasible solution to a nearby one E. Amaldi Fondamenti di R. O. Politecnico di Milano 1

2 Geometrically: adjacent vertices path along the edges of the polyedron of the feasible solutions until an optimal vertex is reached NB: In the worst case the method can examine all the basic feasible solutions (exponential number in n and m) but in practice it turns out to be very efficient. E. Amaldi Fondamenti di R. O. Politecnico di Milano 2

3 We must know how to determine an initial basic feasible solution or establish that the LP is infeasible by applying the Simplex method to an auxilliary LP verify that the current basic feasible solution is optimal go from the current basic feasible solution to a better neighboring basic feasible solution or establish that the LP is unbounded. E. Amaldi Fondamenti di R. O. Politecnico di Milano 3

4 Reduced costs Given a LP min{ c T x : Ax = b, x 0 } and a feasible basis B, Ax = b can be written as B x B + N x N = b x B = B 1 b B 1 N x N with B 1 b 0. Basic feasible solution: x B = B 1 b, x N = 0 E. Amaldi Fondamenti di R. O. Politecnico di Milano 4

5 By substitution in the objective function: x B x N c T x = (c T B ct N ) = (ct B ct N ) c T x = c T B B 1 b c T B B 1 Nx N + c T N x N = c T B B 1 b + ( c T N ct B B 1 N ) x N z 0 = cost of the basic feasible solution x B = B 1 b, x N = 0 B 1 b B 1 Nx N x N c T N reduced costs only expressed in terms of the nonbasic variables E. Amaldi Fondamenti di R. O. Politecnico di Milano 5

6 c j = change in the objective function value if the value of the non basic variable x j was increased by 1 unit while the other non basic variables are kept to 0 Def.: = 0 T c T N c T c T c T B B 1 A = [c T B ct B B 1 B, c T N ct B B 1 N] = I reduced costs vector with respect to the base B E. Amaldi Fondamenti di R. O. Politecnico di Milano 6

7 Optimality test Consider a LP with min (max) and a feasible basis B. If all the reduced costs of the nonbasic variables are nonnegative (non positive), the basic feasible solution x B = B 1 b 0, x N = 0 with a cost c T B B 1 b is optimal. Indeed c T 0 T implies that c T x = c T B B 1 b + c T N x N ct B B 1 b x 0, A x = b NB: This optimality condition is sufficient but not necessary E. Amaldi Fondamenti di R. O. Politecnico di Milano 7

8 Basis change (min) Given a feasible basis B and a nonbasic variable x j (in x N ) with reduced cost c j < 0. Increase x j as much as possible ( it enters in the basis ) while keeping the other nonbasic variables to 0. The basic variable x i (in x B ), whose nonnegativity constraint x i 0 implies the tighter upper bound on the increase in x j, goes to 0 ( it leaves the basis ). The new basis differs from the previous one by a single column (adjacent vertices) E. Amaldi Fondamenti di R. O. Politecnico di Milano 8

9 Canonical form At each iteration we must put the system n j=1 into canonical form x i n j=m+1 a ij x j =b i i=1,...,m x j a ij = b i Bx B + Nx N = b i=1,...,m I x B + N x N = b so as to highlight the basic variables and express them in terms of the nonbasic variables. x B = b N x N E. Amaldi Fondamenti di R. O. Politecnico di Milano 9

10 The canonical form associated to a basis B is obtained by computing a ij and b i in terms of B 1 B 1 B x B + B 1 N x N = B 1 b I N b applying a sequence of pivoting operations I x B + N x N = b highlights the solution basis x B = b e x N = 0! E. Amaldi Fondamenti di R. O. Politecnico di Milano 10

11 Pivoting operation 1. Select a coefficient a rs 0 (the pivot ) 2. Divide the r th row by a rs 3. For each row i r, substract the r th row multiplied by a is NB: The same operations (which do not affect the set of feasible solutions) are used in the Gaussian elemination method to solve systems of linear equations. E. Amaldi Fondamenti di R. O. Politecnico di Milano 11

12 r Example: min z = + + x 3 pivot A x 3 x 4 = 3 4 x 3 + x 5 = x 3 + x 4 = 5 b x i 0 i = 1,,5 ½ 1 ½ ½ /2 4 5 s columns of the basic variables 0 1 ¼ ¾ /2 ½ 0 ¼ 1 5/2 E. Amaldi Fondamenti di R. O. Politecnico di Milano 12

13 The pivoting operations allow to: put the LP in canonical form with respect to a basis B change the canonical form, namely the current basis. Adavantage: B 1 of the new basis B is not computed from scratch but incrementaley by applying to the inverse of the previous basis (which differs from the current basis by a single column) a single pivoting operation! E. Amaldi Fondamenti di R. O. Politecnico di Milano 13

14 Select a neighboring feasible basis (adjacent vertex) so as to: improve the objective function value maintain feasibility 1) Nonbasic variable to enter the basis with reduced cost c j < 0 (NB: decrement z also depends on the upper bound on the increase of x j!) that yields a largest z with respect to z = c BT B 1 b Bland's rule: s = min{ j : c j < 0} For maximization problems: c j > 0 Choice of the pivot column s E. Amaldi Fondamenti di R. O. Politecnico di Milano 14

15 Choice of the pivot row r 1) Basic variable to leave the basis otherwise no upper bound! index i with smallest = θ* among those with a is >0 tighter upper bound on increase of x s Bland's rule: r = min{ i : = θ*, a is >0 } a caso b i a is NB: if c j < 0 with a ij 0 i, no element of the column can play the role of pivot the objective function value is unbounded! b i a is E. Amaldi Fondamenti di R. O. Politecnico di Milano 15

16 Tableau form z = c T x system Ax = b with (implicit) nonnegativity constraints Initial tableau: m righe right hand side of the objective function 0 c T b A x n objective function right hand sides E. Amaldi Fondamenti di R. O. Politecnico di Milano 16

17 Consider a basis B and the partition A = [B N] x m x m+1 x n 0 c T B c T N b B N z = c T x z by pivoting operations (or by pre multiplying by B 1 ) we obtain the tableau in canonical form w.r.t the basis B: z x B[1] x B[m] x m x m+1 x n z c T N b I N basic variables z = c T B B 1 b + c T N x N b = B 1 b E. Amaldi Fondamenti di R. O. Politecnico di Milano 17 z 0

18 Example: min z = x 3 = x 4 = 6 x i 0 i =1,, 4 Tableau w.r.t. the basis with columns 3 and 4: r s x 3 x 4 z x x I 2x2 Pivoting w.r.t. amounts to deriving an expression for from the pivot row and substituting it in all other rows pivot base enters the basis and x 4 leaves the basis E. Amaldi Fondamenti di R. O. Politecnico di Milano 18

19 Tableau w.r.t. new basis: x 3 x 4 z 2 0 5/3 0 1/3 x /3 0 1/3 basis Basic feasible solution: = 2, x 3 = 12, = x 4 = 0 reduced costs with z = 2 E. Amaldi Fondamenti di R. O. Politecnico di Milano 19

20 only nonbasic variable to enter the basis (c 2 = 5/3 < 0) x 3 only basic variable that can leave the basis (a rs = 8 > 0) r s x 3 x 4 z 2 0 5/3 0 1/3 x /3 0 1/3 x 3 x 4 z 6 ½ 0 ¼ 0 6 3/2 1 ¼ 0 x ½ 1 x 3 x 4 z 9/ /24 1/12 3/ /8 1/ /12 1/6 s All reduced costs are 0 r optimal basic (feasible) solution: x * = 0, 1 x* = 6, 2 x* = 0, 3 x* = 18 4 with z * = 6 E. Amaldi Fondamenti di R. O. Politecnico di Milano 20

21 BEGIN Simplex algorithm (LP with min) Let B[1],,B[m] be the column indices of the inital feasible basis B; Costruct the initial tableau A = {a[i,j]: 0 i m, 0 j n} in canonical form w.r.t. B; unbounded:=false; optimum:=false; WHILE (optimum = false) AND (unbounded = false) THEN END IF a[0,j] 0 j=1,,m THEN optimum := true; /* for LP with min */ ELSE Select a nonbasic x s with a[0,s] < 0; IF a[i,s] 0 i=1,,m THEN unbounded := true; ELSE END-IF Determine index r that minimizes with 1 i m e a[i,s] > 0; pivot(r,s) /* update tableau */ B[r] := s; END-IF reduced costs a [ i,0 ] a [ i,s] Procedure pivot(r,s)? E. Amaldi Fondamenti di R. O. Politecnico di Milano 21

22 Degenerate solutions and convergence Def. A basic feasible solution x is degenerate if it contains at least one basic variabile = 0. x with more than n m zeroes corresponds to more than one basis! Same vertex: x More than n constraints ( the m of Ax = b and more than n m among the n of x 0 ) are satisfied with equality ( active ). E. Amaldi Fondamenti di R. O. Politecnico di Milano 22

23 In the presence of degenerate basic (feasible) solutions, the basis change may not decrease the value of the objective function: If the current basic solution is degenerate, θ * can be = 0 and hence the new basic solution is identical (same vertex). Also if θ * > 0, several basic variables may go to 0 when x s is increased to θ *. Hence the new basic solution is degenerate. We could cycle through a sequence of degenerate bases associated to the same vertex. E. Amaldi Fondamenti di R. O. Politecnico di Milano 23

24 Several anticycling rules have been proposed for the choice of the variables that enter and leave the basis (indices r and s) Bland's rule: among all candidate variables to enter/leave the basis (x s and x r ) always select the one with smallest index. Proposition: The Simplex algoreithm with Bland's rule terminate after n m iterations. finite # of pivots E. Amaldi Fondamenti di R. O. Politecnico di Milano 24

25 In some pathological cases (see e.g. Klee & Minty 72) the number of iterations grows exponentially in n and/or m, but the algorithm is overall very efficient. Extensive experimental campaigns: The number of iterations grows linearly with m (m. 3m) and only very slowly ( logarithmically) with n! E. Amaldi Fondamenti di R. O. Politecnico di Milano 25

26 Two phase simplex method Phase 1: Determine an initial basic feasible solution Example: min z = +x 3 a submatrix I 2x2 of A! x 3 = 6,, x x 4 = 5 x 4 0 Given (P) min z = c T x Ax = b x 0 Assumption: b 0 E. Amaldi Fondamenti di R. O. Politecnico di Milano 26

27 Auxilliary problem with artificial variables y i, 1 i m (P A ) min m v= i=1 y i A x + I y = b x 0, y 0 an obvious initial basic feasible solution y = b 0 1) If v * > 0, (P) is infeasible 2) If v * = 0, cleraly y * = 0 and x * is a basic feasible solution of (P) E. Amaldi Fondamenti di R. O. Politecnico di Milano 27

28 For 2) there are two sub cases: If there are nonbasic y i, with i, 1 i m, delete the corresponding columns and obtain a tableau in canonical form w.r.t. A basis; the row of z vmust be determined by substitution. If a basic y i (the b. f. solution is degenerate), we perform a pivot w.r.t. a coefficient 0 of the row of y i and exchange y i with any nonbasic variabile x j. cf. example E. Amaldi Fondamenti di R. O. Politecnico di Milano 28

29 Example: min v = y 1 + y 2 min z = x 3 (P) + 4x 3 = x 3 = 2 (P A ) + 4x 3 + y 1 = x 3 + y 2 = 2,, x 3 0,, x 3, y 1, y 2 0 Put v = y 1 +y 2 in canonical form by substituting the expression of y 1 and y 2 in terms of, and x 3 x 3 y 1 y 2 v y y E. Amaldi Fondamenti di R. O. Politecnico di Milano 29

30 x 3 y 1 y 2 v y y x 3 y 1 y 2 v y optimum v * = 0 E. Amaldi Fondamenti di R. O. Politecnico di Milano 30

31 By selecting as pivot the coefficient 2 of the row of y 2, we obtain: Equivalent optimal basis x 3 y 1 y 2 v ½ ½ 0 the column 1 of I has been trasferred in the area of the original variables optimal basic feasible solution of (P A ) = 0, = 2, x 3 = 0 is a basic feasible solution for (P) E. Amaldi Fondamenti di R. O. Politecnico di Milano 31

32 z = x 3 By substituting canonical form non basic variable = 2 4x 3 = 5x 3 z = 2 + x 3 x 3 z Tableau corresponding to an initial basic feasible solution of (P) Since the basic feasible solution found is (already) optimal, no need for second phase! E. Amaldi Fondamenti di R. O. Politecnico di Milano 32