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Lecture 3

• Linear Programming: – Simplex Method – Computer Solutions

• Based on Excel• Based on QM software

• Tutorial

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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)

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Answer is “NO”

• 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)

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Simplex Method

• It is a method which applies Linear Algebra technique to determine values of determine values of decision variablesdecision variables of a set of equations

• Steps for simplex method

• Special/irregular cases

• Other cases

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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 simplex method algorithmalgorithm

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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)

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Standard format

• Consider the following 3 types of possible equation in a LP problem:

1. ≤ constraints• Such as x1 + x2 ≤ 3 ….(e1)

2. ≥ constraints• Such as x1 + x2 ≥ 3 ….(e2)

3. = constraints• Such as x1 + x2 = 3 ….(e3)

• We need to change all these into a standard format as such: (to p8)

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Standard format1. ≤ constraints

• x1 + x2 ≤ 3 x1 + x2 + s1 = 3 ………(e1)

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

3. = constraints• x1 + x2 = 3 x1 + x2 + A3 = 3 …….(e3)

We will tell you why we need this format later!

Consider the LP problem of ……

Where, S is slack, A is artificial variable

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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….

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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)

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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

Extra!(to p9)

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More example of standard format

• Consider the following:

M refer to very big value, -ve value here meansthat we don’t wish to retain it in the final solution

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Forming a simplex tableau

• A simple tableau is outline as follows:

What are these? (to p14)

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Forming a simplex tableau

• A simple tableau is outline as follows:

we will discuss onhow to use them very soon!

Cost in the obj. func.

LP Decision variables

We will compute this value later It is known as marginal value

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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

Basic variables(to p5)

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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”

S1 has value “1” in this column only!

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simplex method algorithmsimplex method algorithm

• Compute zj values• Compute cj-zj values• Determine the enteringentering variable• Determine the leavingleaving 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 zj values

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

Z1 is sum of multiple of these two columns(to p17)

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Compute cj-zj values

All Cj are listed on this row

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Determine the enteringentering variable• It is referred to

– The variable (i.e. the column) with the largest largest positive cpositive cjj-z-zjj value value

– Also known as “Pivot column”

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

This mean, we will next introduce x2 as a basic variableIn next Tableau

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Determine the leavingleaving variable

• Min value of ratio of quantity values by the pivot column of entering variable

• Also known as “Pivot row”

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

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

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Revise a new tableauNote, 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)

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until all cj-zi are all negative values

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

All negative values, STOP

And s1 =0 and s2= 0

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Irregular cases

• How to realize the following cases from the simplex tableau:

1. Multiple/alternative solutions

2. Infeasible LP problem

3. Unbound LP problem

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Multiple/alternative solutions

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

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

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Infeasible LP problem

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

M value appear in final solution representing infeasible solution

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Unbound LP problem

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

(for s1 as pivot column)

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Other cases

1. Minimizing Z

2. When a decision variable is– either ≤ or ≥

3. Degeneracy

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Minimizing Z

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

Select this as Pivot column

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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 …

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Example

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

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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

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Example, Degeneracy

Degeneracy

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Based on Excel

Xi >= 0We type them in an Excel file

Then, enter formulatehere

Then, we select Tools with “solver”An solution is obtained

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Solution using Excel

How to read them?

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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!

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Tutorial

• Chapter 4,

• Edition 8th: #12, #17, #19

• Edition 9th: #7, #11, #12

And from appendix A/B

• P3, p20, p25, p32