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Simplex Method Meeting 5

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Simplex MethodMeeting 5

Course: D0744 - Deterministic Optimization

Year: 2009

What to learn?

Artificial variables

Big-M method

The Facts

To start, we need a canonical form

If we have a constraint with a nonnegative right-hand side, it will contain an obvious basic variable (which?) after introducing a slack var.

If we have an equality constraint, it contains no obvious basic variable

If we have a constraint with a nonnegative right-hand side, it contains no obvious basic variable even after introducing a surplus var.

2x + 3y 5 2x + 3y + s = 5, s 0 (s basic)

2x + 3y = 5 ??????? Infeasible if x=y=0!

2x + 3y 5 2x + 3y -s = 5, s 0 (??????)

Infeasible if x=y=0!

??????????????????

Compare!

One Equality???

2x + 3y = 5 2x + 3y + a = 5, a = 0 (I)

(s basic, but it should be 0!)

How do we force a = 0? This is of course not feasible if x=y=0, as 0+0+0 5!

2x + 3y = 5 2x + 3y + a = 5, a = 0 (I)

(a basic, but it should be 0!)

How do we force a = 0? This is of course not feasible if x=y=0, as 0+0+0 5

Idea: solve a first problem with

Min {a | constraint (I) + a 0 + other constraints }!

One Equality???

Artificial Variables

Notice: In an equality constraint, the extra variable is called an artificial variable.

For instance, in

2x + 3y + a = 5, a = 0 (I)

a is an artificial variable.

One Inequality ???

2x + 3y 5 2x + 3y - s = 5, s 0 (I)

s could be the basic variable,

but it should be 0

and for x=y=0, it is -5 !

How do we force s 0?

?

2x + 3y 5 2x + 3y - s = 5, s 0 (I)

s could be the basic variable,

but it should be 0

and it is -5 for x=y=0!

How do we force s 0?

By making it 0!

how?

2x + 3y 5 2x + 3y - s = 5, s 0 (I)

s could be basic, but it should be 0

and it is -5 for x=y=0!

How do we force s 0?

By making it 0! But we have to start with a canonical form… so

treat is as an equality constraint!

2x + 3y - s+ a = 5, s 0, a 0 and Min a

Artificial Variables

Notice: In a inequality constraint, the extra variable is called an artificial variable.

For instance, in

2x + 3y – s+ a = 5, s 0, a 0 (I)

a is an artificial variable.

In a sense, we allow temporarily a small amount of cheating, but in the end we cannot allow it!

What if we have many such = and constraints?

7x - 3y – s1+ a1 = 6, s1,a1 0 (I)

2x + 3y + a2 = 5, a2 0 (II)

a1 and a2 are artificial variables, s1 is a surplus variable.

One minimizes their sum:

Min {a1+a2 | a1, a2 0, (I), (II), other constraints}

i.e., one minimizes the total amount of cheating!

Then What?

We have two objectives:

Get a “feasible” canonical form

Maximize our original problem

Two methods:

big M method

phase 1, then phase 2

Big-M Method

Combine both objectives :

(1)Min iai

(2)Max j cjxj

into a single one:

(3)Max – M iai + j cjxj

where M is a large number, larger than anything subtracted from it.

If one minimizes j cjxj

then the combined objective function is

Min M iai + j cjxj

The Big M Method

The simplex method algorithm requires a starting bfs.

Previous problems have found starting bfs by using the slack variables as our basic variables.

If an LP have ≥ or = constraints, however, a starting bfs may not be readily apparent.

In such a case, the Big M method may be used to solve the problem. Consider the following problem.

Example

Bevco manufactures an orange-flavored soft drink called Oranj by combining orange soda and orange juice. Each orange soda contains 0.5 oz of sugar and 1 mg of vitamin C. Each ounce of orange juice contains 0.25 oz of sugar and 3 mg of vitamin C. It costs Bevco 2¢ to produce an ounce of orange soda and 3¢ to produce an ounce of orange juice. Bevco’s marketing department has decided that each 10-oz bottle of Oranj must contain at least 30 mg of vitamin C and at most 4 oz of sugar. Use linear programming to determine how Bevco can meet the marketing department’s requirements at minimum cost.