OR-1 2014 1

Simplex method (algebraic interpretation)

Add slack variables( 여유변수 ) to each constraint to convert them to equations. (We may refer it as an augmented LP)

321 345 maximize xxx

0,, 8243 11 2 4

5 32 subject to

321

321

321

321

xxxxxxxxxxxx

321 345 maximize xxx

0 ,,,,, 8 243 11 2 4

5 32 subject to

654321

6321

5321

4321

xxxxxxxxxx

xxxxxxxx

(1)

(2)

OR-1 2014 2

Every feasible solution of (1) can be extended, in the unique way as given above, into a feasible solution of (2). Every fea-sible solution of (2) can be restricted, by deleting slack vari-ables, into a feasible solution of (1). The correspondence be-tween feasible solutions of (1) and (2) carries optimal solu-tions of (1) onto optimal solutions of (2), and vice versa. So solve (2) instead of (1) and disregard the values of slack variables to obtain an optimal solution to the original prob-lem.

same theare valuesobjective and 2438

2411

325 where

(2) osolution t ),,,,,( (1) osolution t ),,(

*3

*2

*1

*6

*3

*2

*1

*5

*3

*2

*1

*4

*6

*5

*4

*3

*2

*1

*3

*2

*1

xxxx

xxxx

xxxx

xxxxxxxxx

same theare valuesobjective and (1) osolution t ),,( (2) osolution t ),,,,,( *

3*2

*1

*6

*5

*4

*3

*2

*1 xxxxxxxxx

OR-1 2014 3

Surplus variable ( 잉여변수 ) :

Remark: If LP includes equations in the constraints, we need to replace each equation with two inequalities to express the prob-lem in standard form as we have seen earlier. Then we may add slack or surplus variables to convert them to equations. How-ever, this procedure will increase the number of constraints and variables.Equations in an LP can be handled directly without changing them to inequalities. Detailed method will be explained in Chap8. General LP Problems. For the time being, we assume that we follow the standard pro-cedure to replace each equation with two inequalities to obtain a standard form.

OR-1 2014 4

Next let Then find solution to the following system which maximizes z (tableau form)

In the text, dictionary form is used, i.e. each dependent variable (including z) (called basic variable) is expressed as linear com-binations of indep. var. (called nonbasic variable).

0345or 345 321321 xxxzxxxz

0 ,,,,, 8 24311 2 4

5 32

654321

6321

5321

4321

xxxxxxxxxx

xxxxxxxx

0 345 321 xxxz

0 ,,,,, 2438

2 411 325

654321

3216

3215

3214

xxxxxxxxxxxxxxxxxx

321 3450 xxxz (Note that, unlike the text, we place the objective function in the first row. Such presentation style is used more widely and we follow that conven-tion)

OR-1 2014 5

From previous lectures, we know that if the polyhedron P has

at least one extreme point and the LP over P has a finite opti-mal value, the LP has an extreme point optimal solution. Also an extreme point of P for our problem is a basic feasible solu-tion algebraically.We obtain a basic solution by setting x1 = x2 = x3 = 0 and find-ing the values of x4, x5, and x6 , which can be read directly from the dictionary. (also z values can be read.) If all values of x4, x5, and x6 are nonnegative, we obtain a basic feasible so-lution.

OR-1 2014 6

Now, we look for another basic feasible solution (extreme point of the polyhedron) which gives a better objective value than the current solution. Such solution can be examined by setting 7 – 4 = 3 variables at 0 (called nonbasic variables) and solve the equations for the remaining 4 variables (called basic variables). Here z may be regarded as a basic variable and it remains basic at any time during the simplex iterations.

OR-1 2014 7

Initial feasible solution

To find a better solution, find a nonbasic variable having posi-tive coefficient in z row (say x1) and increase the value of the chosen nonbasic variable while other nonbasic variables remain at 0. We need to obtain a solution that satisfies the equations. Since increases and other nonbasic variables remain at 0, the values of basic variables must change so that the new solution satis-fies the equations and nonnegativity. How much can we in-crease x1?

0 ,,,,, 2438

2 411 325

654321

3216

3215

3214

xxxxxxxxxxxxxxxxxx

321 3450 xxxz

0,8,11,5 , var.)(nonbasic 0,, 654321 zxxxxxx

2/51 x4/111 x

3/81 x

OR-1 2014 8

(continued)x1 (5/2) most binding (called ratio test), get new solution x1 = (5/2), x2, x3 = 0, x4 = 0, x5 = 1, x6 = (1/2), z = 25/2

This is a new basic feasible solution since x4 now can be treated as a nonbasic variable (has value 0) and x1 is basic.(We need a little bit of caution here in saying that the new solu-tion is a basic feasible solution since we must be able to obtain it by setting x2, x3, and x4 at 0 and obtain a unique solution after solving the remaining system of equations)

OR-1 2014 9

Change the dictionary so that the new solution can be directly read off x1 : 0 (5/2), x4 : 5 0So change the role of x1 and x4 . x4 becomes independent (nonbasic) variable and x1 becomes dependent (basic) variable.

Why could we find a basic feasible solution easily?1) all independent(nonbasic) variables appear at the right of equality (have value 0)2) each dependent (basic) variable appears in only one equation3) each equation has exactly one basic variable appearing ( z variable may be interpreted as a basic variable, but usually it can be treated separately since it always remains basic and it is irrelevant to the description of the feasible solutions)

So change the dictionary so that it satisfies the above proper-ties.

OR-1 2014 10

3216

3215

3214

2438 2 411

325

xxxxxxxxxxxx

321 3450 xxxz

421

321

223

25

1 xxxx

42

32421

321

223

25

5

251

2)(411

xx

xxxxxx

423

321

221

21

32421

321

223

25

6

24)(38

xxx

xxxxxx

425

321

227

225

22421

321

223

25

34)(5

xxx

xxxxxz

OR-1 2014 11

421

321

223

25

1 xxxx

425 2 51 xxx

423

321

221

21

6 xxxx

425

321

227

225 xxxz

421

321

223

25

1 xxxx

425 2 51 xxx

423

321

221

21

6 xxxx

425

321

227

225 xxxz

3216

3215

3214

2438 2 411

325

xxxxxxxxxxxx

321 3450 xxxz

3261

3251

4321

248 3 2 11 4

3 5 2

xxxxxxxx

xxxx

321 340 5 xxxz

OR-1 2014 12

3261

3251

421

321

223

25

1

2 48 3 2 11 4

xxxxxxxx

xxxx

321 340 5 xxxz

3261

3251

4321

248 3 2 11 4

3 5 2

xxxxxxxx

xxxx

321 340 5 xxxz

421

321

223

25

1 xxxx

425 2 51 xxx

423

321

221

21

6 xxxx

425

321

227

225 xxxz

Equivalent to performing row operations

OR-1 2014 13

Note that the previous solution, , , , and the new solution

satisfies the updated system of equations. Only difference is that the new solution can be read off directly from the new dic-tionary.We update the dictionary to read a new basic solution directly, but the set of solutions is not changed.

421

321

223

25

1 xxxx

425 251 xxx

423

321

221

6 xxxx 21

425

321

227

225 xxxz

OR-1 2014 14

Next iteration: Select as the nonbasic variable to increase the value (called entering nonbasic variable). becomes 0 (changes status from basic variable to nonbasic variable)

Perform substitutions (elementary row operations)

421

321

223

25

1 xxxx

425 2 51 xxx

423

321

221

21

6 xxxx

425

321

227

225 xxxz

53 x

13 x

nrestrictio no gives

OR-1 2014 15

New solution is

It is optimal since any feasible solution must have nonnega-tive values and implies that z 13 for any nonnegative fea-sible solution.

Hence if the coefficients of the nonbasic variables in z- row are all non-positive, current solution is optimal (note that it is a sufficient condition for optimality but not a necessary condi-tion)

13,0,1,0,1,0,2 654321 zxxxxxx

6423

425

6421

642

231251222

313

xxxxxxx

xxxxxxxz

OR-1 2014 16

Moving directions in Rn in the example

3216

3215

3214

2438 2 411

325

xxxxxxxxxxxx

321 3450 xxxz

421

321

223

25

1 xxxx

425 2 51 xxx

423

321

221

21

6 xxxx

425

321

227

225 xxxz

0,8,11,5 , var.)(indep. 0,, 654321 zxxxxxx

x1 = (5/2), x2, x3 = 0, x4 = 0, x5 = 1, x6 = (1/2), z = 25/2

1/2) 1, 0, 0, 0, (5/2, 8), 11, 0,5, ,0 ,0(Let 10 xx

Then we obtained , where (1, 0, 0, -2, -4, -3) and Note that the vector can be found from the dictionary.( the column for )We make as large as possible while .

OR-1 2014 17

Geometric meaning of an iteration Notation

0,, 1

321

321

xxxxxx

x1

x3

x2

x1=0x2=0

x3=0

OR-1 2014 18

Our example : assume x2 does not exist. It makes the polyhe-dron 2 dimensional since we have 5 variables and 3 equations (except nonnegativity and objective row)

3216

3215

3214

2438 2 411

325

xxxxxxxxxxxx

321 3450 xxxz

421

321

223

25

1 xxxx

425 2 51 xxx

423

321

221

21

6 xxxx

425

321

227

225 xxxz

x1=0

x4=0

x3=0 x6=0

d

A

B

We move from A, which isan extreme point defined by 3 eq. and , to B defined by the 3 eq. and .

OR-1 2014 19

Terminology

Assume that we have max , where is and full row rank.

A solution is called a basic solution ( 기저해 ) if it can be ob-tained by setting of the variables equal to 0 and then solving for the remaining m variables, where the columns of the ma-trix corresponding to the variables are linearly independent. (Hence provides a unique solution.)

In the text, basic solution is defined as the solution which can be obtained by setting the right-hand side variables (indepen-dent var.) at zero in the dictionary. This is the same definition as the one given above. But the text does not make clear dis-tinction between basic solution and basic feasible solution.

OR-1 2014 20

For a basic solution , the variables which are set to 0 are called nonbasic variables ( 비기저변수 ) (independent var.) and the remaining variables are called basic variables (기저변수 ) (dependent var.)

(1) The z-row may be considered as part of the system of equations. In that case, z variable is regarded as basic vari-able. It always remains basic during the simplex iterations.(2) On the other hand, z-row may be regarded as a separate equation which is used to read off the objective function val-ues and other equations and nonnegativity describes the fea-sible solution set. Note that the set of feasible solutions does not change although we perform simplex iterations. Both viewpoints are useful.

A solution is called a basic feasible solution ( 기저가능해 ) if it is a basic solution and satisfies . (feasible solution to the augmented LP)

OR-1 2014 21

The set of basic variables is called a basis ( 기저 ) of the basic

solution. (note that the set of columns in matrix correspond-ing to basic variables spans the subspace generated by the columns of matrix (which is ).)

In a simplex iteration, the nonbasic variable which becomes basic in that iteration is called entering (nonbasic) variable ( 도입변수 ) and the basic variable which becomes nonbasic is called leaving (basic) variable ( 탈락변수 )

Minimum ratio test ( 최소비율검사 ) : test to determine the leav-ing basic variable

Pivoting : computational process of constructing the new dic-tionary (elementary row operations)

OR-1 2014 22

Remarks

Maximum number of b.f.s. in augmented form is .

In the simplex method, one nonbasic variable becomes basic and one basic variable becomes nonbasic in each iteration (except the z variable, it always remains basic.)

In real implementations, we do not update entire dictionary (or tableau). We maintain information about the current basis. Then entire tableau can be constructed from that information and the simplex iteration can be performed (called revised simplex method, details later in Chapter 7).

OR-1 2014 23

Obtaining all optimal solutions

425

6421

6423

2512-22

231

xxxxxxxxxxx

642313 xxxz

For any feasible solution , if any of (nonbasic) is greater than 0, then the ob-jective value is less than 13. (from row)Hence we must have for any optimal solution. Then the current optimal solu-tion is the only one solution which satisfies the constraints, which means it is a unique optimal solution.If all coefficients in the z- row are < 0, it gives a sufficient condition for the uniqueness of the current optimal solution.

OR-1 2014 24

Another example

3526

3521

3524

294865172 3

xxxxxxxxxxxx

3 8 xz

Any feasible solution with is an optimal solution. The set of feasible solutions with is given by

0, 0294

065102 3

52

526

521

524

xxxxxxxxxxx

0, 429

16532

52

52

52

52

xxxxxxxx

OR-1 2014 25

Tableau format

824311245320345

6321

5321

4321

321

xxxxxxxx

xxxxxxxz

81002431101021450011320000345

8100243110102142/5002/12/12/310000345

2/1102/32/12/1010120502/5002/12/12/312/25002/52/12/70

OR-1 2014 26

Tableau format only maintains coefficients in the equations.It is convenient to carry out simplex iterations in the tableau.