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Lecture 3

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- Linear Programming:
- Simplex Method
- Computer Solutions
- Based on Excel
- Based on QM software

- Tutorial

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- Last week, we covered on using the graphical approach in deriving solutions for LP problem
- Question:
- Can we solve all LP problems using graphical approach ?

(to p3)

- Why?
Consider the following equation:

x1+x2+x3+x4 = 4

- It is extremely difficult for us to use graph to represent this equation
- Thus, we need another systematic approach to solve an LP problem – known as Simplex Method

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- It is a method which applies Linear Algebra technique to determine values of decision variables of a set of equations
- Steps for simplex method
- Special/irregular cases
- Other cases

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(to p1)

- Step 1
- Convert all LP resource constraints into a standard format

- Step 2
- Form a simplex tableau
- Transfer all values of step 1 into the simple tableau
- Determine the optimal solution for the above tableau by following the simplex method algorithm

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- Convert LP problem into standard format!
- We refer standard format here as.
- All constraints are in a form of “equation”
- i.e. not equation of ≥ or ≤
- Procedural steps

(to p7)

- Consider the following 3 types of possible equation in a LP problem:
- ≤ constraints
- Such as x1 + x2 ≤ 3 ….(e1)

- Such as x1 + x2 ≥ 3 ….(e2)

- Such as x1 + x2 = 3 ….(e3)

(to p8)

- ≤ constraints
- x1 + x2 ≤ 3 x1 + x2 + s1 = 3 ………(e1)

- x1 + x2 ≥ 3 x1 + x2 - s2 + A2= 3 …..(e2)

- x1 + x2 = 3 x1 + x2 + A3 = 3 …….(e3)

We will tell you why we need this format later!

Where, S is slack,

A is artificial variable

(to p9)

Consider the LP problem of ……

- maximize Z=$40x1 + 50x2
subject to

1x1 + 2x2 40 ………(e1)

4x2 + 3x2 120 ……..(e2)

x1 0 ……...(e3)

x2 0 ………(e4)

Convert them into a standard format will be like …

(We can leave e3 and e4 alone as it is an necessary constraints for LP solution!)

More example….

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Max Z=$40x1+50x2+0s1+0s2

subject to

1x1 + 2x2 + s1 = 40 ………(e1)

4x2 + 3x2 +s2 = 120 …..(e2)

x1, x2 0

- The different is …

(to p11)

maximize

Z=$40x1 + 50x2

subject to

1x1 + 2x2 40

4x2 + 3x2 120

x1,x2 0

Max

Z=$40x1+50x2+0s1+0s2

subject to

1x1 + 2x2 + s1 = 40

4x2 + 3x2 + s2 = 120

x1,x2 0

Original LP format

Standard LP format

(to p9)

Extra!

Consider the following:

M refer to very big value, -ve value here means

that we don’t wish to retain it in the final solution

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- A simple tableau is outline as follows:

(to p14)

What are these?

LP Decision variables

- A simple tableau is outline as follows:

Cost in the obj. func.

We will compute this value later

It is known as marginal value

we will discuss onhow to use them very soon!

(to p5)

Max

Z=$40x1+50x2+0s1+0s2

subject to

1x1 + 2x2 + s1 = 40

4x2 + 3x2 + s2 = 120

x1,x2 0

These values read from s1 and s2 here

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(to p16)

Basic variables

- A decision variable is a basic variable in a tableau when
- it is the only variable that has a coefficient of value “1” in that column and that others have values “0”

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S1 has value “1” in this column only!

- Compute zj values
- Compute cj-zj values
- Determine the entering variable
- Determine the leaving variable
- Revise a new tableau
- Introducing cell that crossed by “pivot row” and “pivot column” that has only value “1” and the rest of values on that column has value “0”

- Repeat above steps until all cj-zi are all negative values
- example

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- Compute by multiplying cj column values by the variable column values and summing:

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Z1 is sum of multiple of these two columns

All Cj are listed on this row

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- It is referred to
- The variable (i.e. the column) with the largest positive cj-zj value
- Also known as “Pivot column”

This mean, we will next introduce x2 as a basic variable

In next Tableau

(to p17)

Max value, ie higher marginal cost contribute to the obj fuc

- Min value of ratio of quantity values by the pivot column of entering variable
- Also known as “Pivot row”

This mean, s1 will leave as basic variable in next Tableau

Min value = min (40/2, 120/3) = mins (20,40), thus pick the first value

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Note, this value is copied

This row values

divided by 2

New row x 3 – old row (2), note quantity must > 0

Resume z and c-j computation!

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- The following is the optimal tableau, and the solution is:

And s1 =0 and s2= 0

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All negative values, STOP

- How to realize the following cases from the simplex tableau:
- Multiple/alternative solutions
- Infeasible LP problem
- Unbound LP problem

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Note: S1 is not a basic variable but has value “0” for cj-zj

- Alternative solution is to consider the non-basic variable that has cj-zj = 0 as the next pivot column and repeat the simplex steps

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M value appear in final solution representing infeasible solution

- Infeasible means the LP problem is not properly formulated and that a feasible region cannot be identified.

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(for s1 as pivot column)

- Cannot identifying the Pivot row (i.e. leaving basic variable)

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- Minimizing Z
- When a decision variable is
- either ≤ or ≥

- Degeneracy

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- The tableau of Max Z still applied but we change the last row cj-zj into zj-cj

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Select this as Pivot column

If x1 is either ≤ or ≥, then

We adopt a transformation as such:

let x1 = x’1 – x’’1

And then substitute it into the LP problem, and then follow the normal procedure

Example …

(to p31)

(to p28)

maximize

Z=$40x1 + 50x2

subject to

1x1 + 2x2 40

4x2 + 3x2 120

x1 0

maximize

Z=$40x1 + 50(x’2-x’’2)

subject to

1x1 + 2(x’2-x’’2) 40

4x2 + 3(x’2-x’’2) 120

x1, x’2, x’’2 0

(to p30)

- It refers to the nth tableau and (n+1)th tableau is the same (repeated)
- Two ways
- A tie value when selecting the pivot column
- A tie value when selecting the pivot row
- Example, Degeneracy

- Solution:
- Go back to nth tableau and select the other one tie-value variable as pivot column/row

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Degeneracy

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Xi >= 0

We type them in an Excel file

Then, we select

Tools with “solver”

An solution is obtained

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Then, enter formulate

here

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How to read them?

- Install the QM software
- Loan QM software
- Select option of “Linear Programming” from the “Module”
- Then select “open” from option “file” to type in a new LP problem
- Following instructions of the software accordingly
- See software illustration!

(to p1)

- Chapter 4,
- Edition 8th: #12, #17, #19
- Edition 9th: #7, #11, #12
And from appendix A/B

- P3, p20, p25, p32