Post on 31-Dec-2015
description
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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:
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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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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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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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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