Civil Engineering Systems Analysis
Lecture VI
Instructor: Prof. Naveen Eluru
Department of Civil Engineering and Applied Mechanics
Today’s Learning Objectives
Simplex Method
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Simplex : Example 2
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Max Z = 3x1+5x2
Subject to
x1≤4
2x2 ≤12
3x1+2x2 ≤18
x1≥0,x2 ≥0
Solution
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Augmented form
Max Z = 3x1+5x2
Subject to x1 + x3 =4
2x2 +x4 =12
3x1+2x2 +x5 =18
x1≥0,x2 ≥0
Simplex Table
x1 x2 x3 x4 x5 Solution
Z -3 -5 0 0 0 0
x3 1 0 1 0 0 4
x4 0 2 0 1 0 12
x5 3 2 0 0 1 18
Solution
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Solution (0,0,4,12,18) and Z =0
Do we have an optimal solution?
No
What is the entering variable?
x2
x1 x2 x3 x4 x5 Solution
Z -3 -5 0 0 0 0
x3 1 0 1 0 0 4
x4 0 2 0 1 0 12
x5 3 2 0 0 1 18
Solution
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Now compute the different ratios
We can see that row corresponding to x4 has
the minimum ratio
Hence x4 is the leaving variable
x1 x2 X3 x4 x5 Solution Ratio
Z -3 -5 0 0 0 0 -
x3 1 0 1 0 0 4 ∞
x4 0 2 0 1 0 12 12/2
x5 3 2 0 0 1 18 18/2
Entering variable
Leaving variable Pivot Element
Solution
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Start making the changes to the table
Step 1: for pivot row divide the pivot row elements
by pivot element (in the example 2)
Step 2: for every other row:
New row = current row – Pivot column coefficient * new
pivot row
x1 x2 x3 x4 x5 Solution
Z -3-0 -5+5 0-0 0+5/2 0-0 0+30
x3 1-0 0-0 1-0 0-0 0-0 4-0
x4x2 0/2 2/2 0/2 1/2 0/2 12/2
x5 3-0 2-2 0-0 0-1 1-0 18-12
Solution
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Consolidate
Solution (0,6,4,0,6) and Z = 30;
Are we optimal yet?
No
x1 will enter
Leaving variable
Min(4/1, 6/0, 6/3) => x5 leaves
x1 x2 x3 x4 x5 Solution
Z -3 0 0 5/2 0 30
x3 1 0 1 0 0 4
x2 0 1 0 1/2 0 6
x5 3 0 0 -1 1 6
Solution
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Do the operations
Consolidate
x1 x2 x3 x4 x5 Solution
Z -3+3 0-0 0-0 5/2-1 0+1 30+6
x3 1-1 0-0 1-0 0+1/3 0-1/3 4-2
x2 0 1-0 0-0 ½-0 0-0 6-0
x5x1 3/3 0/3 0/3 -1/3 1/3 6/3
x1 x2 x3 x4 x5 Solution
Z 0 0 0 3/2 1 36
x3 0 0 1 1/3 -1/3 2
x2 0 1 0 ½ 0 6
x1 1 0 0 -1/3 1/3 2
Optimal????? Yes – Solution (2,6,2,0,0)
Summary
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Subject to
x1≤4
2x2 ≤12
3x1+2x2 ≤18
x1≥0,x2 ≥0
x1 = 2, x2 = 6
So Eqn 1 has abundant resources
Eqn 2 and Eqn 3 lead to scarce resources
Minimum Ratio - Notes
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In the simplex table computing minimum ratio
has two components
Coefficients for the entering variable
Have to be >0
RHS
≥ 0
Hence Minimum Ratio ≥ 0 given you meet the above
constraints
So if RHS is 0 and coefficient is –ive that is not a valid
ratio to consider
Insights on simplex
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What if there is a tie for entering variable? At a juncture in the simplex tableau you can have two
variable –ive and of the same magnitude
How do we determine what enters
Choose arbitrarily! Eventually you will reach the solution
What if tie in the Minimum ratio test What does it imply?
Two constraints are such that they yield a same lower limit on entering variable i.e. two current basic variables go to 0 simultaneously
Referred to as degeneracy
How to address it Break tie arbitrarily
You might enter a loop, if so, then the time of the tie use the other variable as leaving variable
Degeneracy example
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Max Z = 3x1+9x2
x1+4x2 ≤ 8
x1+2x2 ≤ 4
x1, x2 ≥ 0
Entering variable?
x2
Leaving variable : Min(8/4, 4/2)
Lets decide as x4
x1 x2 x3 x4 Solution
Z -3 -9 0 0 0
x3 1 4 1 0 8
x4 1 2 0 1 4
Degeneracy example
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x2 enters and x4 leaves
So we reached optimal value
x1 x2 x3 x4 Solution
Z -3+9/2 -9+9 0+0 0+9/2 0+18
x3 1-2 4-4 1-0 0-2 8-8
x4x2 1/2 2/2 0/2 1/2 4/2
x1 x2 x3 x4 Solution
Z 3/2 0 0 9/2 18
x3 -1 0 1 -2 0
x2 ½ 1 0 1/2 2
Degeneracy example
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What if we picked the other variable
Push x3 out in the first iteration
Now x1 enters
What leaves? x4
x1 x2 x3 x4 Solution
Z -3+9/4 -9+9 0+9/4 0+0 0+18
x3x2 ¼ 4/4 1/4 0/4 8/4
x4 1-2/4 2-2 0-2/4 1-0 4-4
x1 x2 x3 x4 Solution
Z -3/4 0 9/4 0 18
x2 ¼ 1 1/4 0 2
x4 1/2 0 -1/2 1 0
Degeneracy example
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Optimal?
x1 x2 x3 x4 Solution
Z -3/4+3/4 0-0 9/4+3/4 0+3/2 18+0
x2 ¼-1/4 1-0 ¼+1/4 0-1/2 2-0
x4x1 ½*2 0*2 -1/2*2 1*2 0*2
x1 x2 x3 x4 Solution
Z 0 0 3 3/2 18
X2 0 1 1/2 -1/2 2
X1 1 0 -1 2 0
Insights on simplex
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Just as you can have two possible leaving
variables – you can have 0 variables for leaving
Happens when Z is unbounded
The basic variable entering can be increased
indefinitely
Unbounded Z
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Max z =2x1 + x2
x1-x2 ≤ 10
2x1 ≤40
x1,x2 ≥ 0
Entering variable x1, leaving variable x3
x1 x2 x3 x4 Solution
Z -2 -1 0 0 0
x3 1 -1 1 0 10
x4 2 0 0 1 40
x1 x2 x3 x4 Soluti
on
Z -2+2 -1-2 0+2 0+0 0+20
x3x1 1 -1 1 0 10
x4 2-2 0+2 0-2 1-0 40-20
x1 x2 x3 x4 Soluti
on
Z 0 -3 2 0 20
x1 1 -1 1 0 10
x4 0 2 -2 1 20
Unbounded Z
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Entering variable x2
Leaving variable x4
Now x3 entering.. But no leaving variable..
x1 x2 x3 x4 Soluti
on
Z 0+0 -3+3 2-3 0+3/2 20+30
x1 1-0 -1+1 1-1 0+1/2 10+10
x4x2 0/2 2/2 -2/2 1/2 20/2
x1 x2 x3 x4 Soluti
on
Z 0 0 -1 1/2 50
x1 1 0 0 ½ 20
x2 0 1 -1 ½ 10
Multiple Optimal Solutions
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Max Z = 2x1+4x2
x1+2x2 ≤5
x1+x2 ≤4
x1,x2≥0
x2 enters … leaving variable x3
Z = 10 and (0,5/2,0,3/2)
Basic Z x1 x2 x3 x4
Solut
ion
Z 1 -2 -4 0 0 0
x3 0 1 2 1 0 5
x4 0 1 1 0 1 4
Basic Z x1 x2 x3 x4
Solut
ion
Z 1-0 -2+2 -4+4 0+4/2 0-0 0+10
x3x2 0/2 ½ 2/2 1/2 0/2 5/2
x4 0-0 1-1/2 1-1 0-1/2 1-0 4-5/2
Z x1 x2 x3 x4 Solu
tion
Z 1 0 0 2 0 10
x2 0 ½ 1 ½ 0 5/2
x4 0 ½ 0 -1/2 1 3/2
Multiple Optimal Solutions
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We reached optimal but see that x1 and x2
have 0 coefficients
Lets try to look at z
We notice that one of the non-basic variables
has a coefficient of 0 .. So without changing z
we can have x1 enter
If x1 enters x4 leaves
Z x1 x2 x3 x4
Solu
tion
Z 1 0 0 2 0 10
x2 0 ½-1/2 1-0 ½+1/
2 0-1
5/2-
3/2
x4
x1
0 ½ *2 0*2 -
1/2*2 1*2 3/2*2
Z x1 x2 x3 x4 Solu
tion
Z 1 0 0 2 0 10
x3 0 0 1 1 -1 1
x1 0 1 0 -1 2 3
Z = 10 and (3,1,0,0)
Summary of special cases
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Entering ties
Degeneracy Results from multiple leaving options
Possibility of a loop
Unbounded Z Simplex is unable to find the corner point
Multiple optimal solutions Results when an edge is the optimal solution
Simplex assumptions
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So far we examined simplex.. But implicitly we
made the following assumptions for the
standard problem
≤ constraints
Slack variables … easily provide basic feasible solution
All the RHS values are non-negative
Ensure the variables are not <0
Maximization
We have –ive values in simplex (these enter)
But we need to know how to adapt the simplex for
other forms also!
References
9/25/2012 CIVE 208: CIVIL ENGINEERING SYSTEMS ANALYSIS 24
Hillier F.S and G. J. Lieberman. Introduction to
Operations Research, Ninth Edition, McGraw-
Hill, 2010
Revelle C.S, E.E. Whitlatch and J. R. Wright.
Civil and Environmental Systems Engineering