Lecture 3

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# Lecture 3 - PowerPoint PPT Presentation

Lecture 3. Linear Programming: Simplex Method Computer Solutions Based on Excel Based on QM software Tutorial. (to p2). (to p34). (to p36). (to p37). Simplex Method. Last week, we covered on using the graphical approach in deriving solutions for LP problem Question:

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Lecture 3
• Linear Programming:
• Simplex Method
• Computer Solutions
• Based on Excel
• Based on QM software
• Tutorial

(to p2)

(to p34)

(to p36)

(to p37)

Simplex Method
• 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

(to p4)

Simplex Method
• 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

(to p5)

(to p24)

(to p28)

(to p1)

Steps for simplex method
• 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

(to p6)

(to p13)

(to p15)

(to p17)

(to p4 )

Step 1
• 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)

Standard format
• Consider the following 3 types of possible equation in a LP problem:
• ≤ constraints
• Such as x1 + x2 ≤ 3 ….(e1)
• ≥ constraints
• Such as x1 + x2 ≥ 3 ….(e2)
• = constraints
• Such as x1 + x2 = 3 ….(e3)
• We need to change all these into a standard format as such:

(to p8)

Standard format
• ≤ constraints
• x1 + x2 ≤ 3 x1 + x2 + s1 = 3 ………(e1)
• ≥ constraints
• x1 + x2 ≥ 3 x1 + x2 - s2 + A2= 3 …..(e2)
• = constraints
• 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 ……

Sample of an LP problem
• 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….

(to p10)

(to p12)

Standard format

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:More example of standard format

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

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

(to p5)

Forming a simplex tableau
• A simple tableau is outline as follows:

(to p14)

What are these?

Forming a simplex tableau

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)

Transfer all values

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

(to p5)

(to p16)

Basic variables

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”

(to p15)

S1 has value “1” in this column only!

simplex method algorithm
• 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

(to p18)

(to p19)

(to p20)

(to p21)

(to p22)

(to p23)

(to p5)

Compute zj values
• Compute by multiplying cj column values by the variable column values and summing:

(to p17)

Z1 is sum of multiple of these two columns

Compute cj-zj values

All Cj are listed on this row

(to p17)

Determine the entering variable
• 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

Determine the leaving variable
• 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

(to p17)

Revise a new tableau

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!

(to p17)

until all cj-zi are all negative values
• The following is the optimal tableau, and the solution is:

And s1 =0 and s2= 0

(to p17)

All negative values, STOP

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

(to p25)

(to p26)

(to p27)

(to p4)

Multiple/alternative solutions

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

(to p24)

Infeasible LP problem

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.

(to p24)

Unbound LP problem

(for s1 as pivot column)

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

(to p24)

Other cases
• Minimizing Z
• When a decision variable is
• either ≤ or ≥
• Degeneracy

(to p29)

(to p30)

(to p32)

(to p4)

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

(to p28)

Select this as Pivot column

either ≤ or ≥

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

Example

(to p30)

Degeneracy
• 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

(to p33)

(to p28)

Example, Degeneracy

Degeneracy

(to p32)

Based on Excel

Xi >= 0

We type them in an Excel file

Then, we select

Tools with “solver”

An solution is obtained

(to p35)

Then, enter formulate

here

Solution using Excel

(to p1)

QM software
• 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)

Tutorial
• Chapter 4,
• Edition 8th: #12, #17, #19
• Edition 9th: #7, #11, #12

And from appendix A/B

• P3, p20, p25, p32