LP- a review

8/3/2019 LP- a review

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Linear

Programminghttp://endrayanto.staff.ugm.ac.id/courses/MMS2302


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LP: Standard Formc1x1 + c2x2 + + cnxn = z (Min)

a11x1 + a12x2 + + a1nxn = b1

a21x1 + a22x2 +

+ a2nxn = b2...

......

......

am1x1 + am2x2 + + amnxn = bm,

and x1 0, x2 0, . . . , xn 0.


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LP: Matrix NotationMinimize cTxsubject to Ax = b, A : m n,

x 0,

c : column vector (n x 1)

x : column vector (n x 1)

b : column vector (m x 1)


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Definitions

Basic Solutions

Degeneracy Basis

Basic Columns


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Improving a non-optimal BFS

Definition: Reduced Costs (Relatives Cost Factors)

(z) + 2x1 5x3 + x5 = 11+ 2x1 + 3x3 + x4 x5 = 3

x1 + x2 + 2x3 + 2x5 = 4.

z = 11, xB

= (x4, x2) = (3, 4), xN = (x1, x3, x5) = (0, 0, 0).

z = 11 5x3.Ifx3 is increased to any positive value

the value of z would be reduced


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Improving a non-optimal BFS(z) + 2x1 5x3 + x5 = 11

+ 2x1 + 3x3 + x4 x5 = 3 x1 + x2 + 2x3 + 2x5 = 4.

to make x3 as large as possible, z = 11 5x3.

However, x3 cannot be increased indefinitely

x4 = 3

3x3x2 = 4 2x3.

while the other nonbasic variables remain zero


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Improving a non-optimal BFSx4 = 3

3x3

x2 = 4 2x3.if x3 increases beyond 3 3, then x4 becomes negative

x3 increases beyond 4 2 then x2 also becomes negative.

value of x3 is the smaller of these, namely x3 = 1

z = 11 5x3.z = 6, x3 = 1, x2 = 2, x1 = x4 = x5 = 0.

z = 11, xB

= (x4, x2) = (3, 4), xN = (x1, x3, x5) = (0, 0, 0).

new feasible basic solutions

compared to previous solutions


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Improving a non-optimal BFS(z) + 2x1 5x3 + x5 = 11

+ 2x1 + 3x3 + x4 x5 = 3 x1 + x2 + 2x3 + 2x5 = 4.

Choose a pivot term s.t.

basic variables, x2 and x4, could be adjusted without becoming negative,

Pivoting on 3x3,

(z) + 163x1 +

5

3x4

2

3x5 = 6

+2

3x1 + x3 +1

3x4

1

3x5 = 1

7

3x1 + x2

2

3x4 +

8

3x5 = 2.


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Improving a non-optimal BFS(z) + 16

3x1 + 53x4

23x5 = 6

+ 23x1 + x3 +

1

3x4

1

3x5 = 1

7

3x1 + x2

2

3x4 +

8

3x5 = 2.

the basic solution,

z = 6, xB

= (x3, x2) = (1, 2), xN = (x1, x4, x5) = (0, 0, 0),

z = 6 + 23x5

x3 = 1 +1

3x5

x2 = 2 8

3x5. the pivot term,

83x5


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Improving a non-optimal BFS

(z) + 194x1 +

1

4x2 +

3

2x4 =

11

2

+ 38x1 +

1

8x2 + x3 +

1

4x4 =

5

4

7

8x1 +

3

8x2

1

4x4 + x5 =

3

4

he basic feasible solution

z =11

2, x3 =

5

4, x5 =

3

4, x1 = x2 = x4 = 0.

(z) + 16

3x1 + 53x4

23x5 =

6

+ 23x1 + x3 +

1

3x4

1

3x5 = 1

7

3x1 + x2

2

3x4 +

8

3x5 = 2.


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Key component of SM

The optimality test

Introducing Non Basic Variable (NBV) intobasis


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Exercises Exercise 3.2 (Infeasible Problem)

Exercise 3.3 (Unique Minimum)

Exercise 3.4 (Multiple Minima)

Exercise 3.5 (Unbounded Class of Solutions)


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Exercise 3.2It is obvious that the linear program shown below is infeasible.Show algebraically by generating an infeasible inequality thatthis is indeed the case

Minimize x1 + x2subject to x1 + x2 = 2

x1 0

x2 0.


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Exercise 3.3Minimize 7

2x1 = z + 15

subject to x1

+ x2

= 332

x1 + x3 = 4

and x1 0, x2 0, x3 0.

Determine by inspection the basic solution to

Why is it feasible, optimal, and unique?


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Exercise 3.4Prove that the basic solution

is a feasible optimal solution to

Minimize 0x3 = z + 15

subject to x2 2

3x3 =

1

3

x1 +2

3x3 =

8

3

and x1 0, x2 0, x3 0,

z = 15, x1 = 8/3, x2 = 1/3

but that it is not unique.Can you find another optimal basic feasible solution?Are there any nonbasic feasible solutions that are also optimal?


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Exercise 3.5

to feasible canonical form and generate a class of solutions

that in the limit cause the objective function to go to -infinity.

Minimize x1 x2 = zsubject to x1 x2 = 1

x1 0x2 0

Reduce

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