MATH2070 - Linear programming · The standard Linear Programming (LP) Problem Graphical method of...
Transcript of MATH2070 - Linear programming · The standard Linear Programming (LP) Problem Graphical method of...
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MATH2070 Optimisation
Linear Programming
Semester 2, 2012Lecturer: I.W. Guo
Lecture slides courtesy of J.R. Wishart
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Review
The standard Linear Programming (LP) Problem
Graphical method of solving LP problem
The simplex procedural algorithm
Extensions to LP Algorithm to include other conditions
Dual Problem
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
The standard Linear Programming (LP) Problem
Graphical method of solving LP problem
The simplex procedural algorithm
Extensions to LP Algorithm to include other conditions
Dual Problem
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Motivation
Where do Linear Programming problems occur?
History and Applications
Became a field of interest during World War II.
Applications
I Allocation Problems
I Scheduling
I Blending of raw materials
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Initial Example
A pharmaceutical company can produce two types of drug usingthree different resources. Each resource is in very limited supply.Production data are listed in the table below:
Resource usage/unit ResourceDrug 1 Drug 2 Availability
Resource 1 1 0 42 3 2 183 0 2 12
Profit/unit 3 5 × $1000
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Formulation
Resource usage/unit ResourceDrug 1 Drug 2 Availability
Resource 1 1 0 42 3 2 183 0 2 12
Profit/unit 3 5 × $1000
Define x1, x2 be the number produced for each type of drug.
Profit : Z = 3x1 + 5x2
Optimisation
We want to find the optimal x∗1 and x∗2 to maximise the totalprofit, Z∗ .
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Standard LP problem
In full generality, the problem is given by,
Maximise:Z = c1x1 + c2x2 + · · ·+ cnxn
such thata11x1 + a12x2 + · · · + a1nxn ≤ b1
a21x1 + a22x2 + · · · + a2nxn ≤ b2
......
...
am1x1 + am2x2 + · · · + amnxn ≤ bm
and x1 ≥ 0, x2 ≥ 0, . . . , xn ≥ 0.
where aij , ci are constants and bi > 0 are constants.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Using matrix notation (slightly abusing notation),
Maximise: Z = cTx
such that: Ax ≤ b and x ≥ 0
where the vectors and matrices are defined,
c =
c1
c2...cn
A =
a11 a12 . . . a1n
a21 a22 . . . a2n...
.... . .
...am1 an2 . . . amn
x =
x1
x2...xn
b =
b1
b2...bm
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Objective Function and Decision Variables
Decision Variables
xT = (x1, x2, . . . , xn)
where xi are the decision variables and it is assumed that xi ≥ 0.
Objective Function
Z = Z(x) = cTx = c1x1 + c2x2 + · · ·+ cnxn,
where ci are constants and are referred to as the cost coefficients.
ci shows the linear increase/decrease in Z for unit increase in xi.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Constraints
Constraint system of equations
Ax ≤ b ←→
a11x1 + a12x2 . . . + a1nxn ≤ b1
a21x1 + a22x2 . . . + a2nxn ≤ b2...
...am1x1 + an2x2 . . . + amnxn ≤ bm
System of m different constraints described by the m equationswith n decision variables.
b is referred to the resource vector and it is assumed that bi > 0are constants.
Positivity condition
Note that the resource vector b must have positive elements.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Feasible Solutions
Feasible Solution
Any point x satisfying Ax ≤ b and x ≥ 0 is called a feasiblesolution.
Infeasible Solution
Conversely, if a point x does not satisfy the above equations it isan infeasible solution.
A feasible solution lies inside a closed region inside then−dimensional decision space.
For the initial example that is a two dimensional region (x1, x2)plane.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Feasible Regions
Empty Region
It is possible for the feasible region to be empty. That is no pointcan be found to satisfy the constraints.
This will be called an ill-posed problem.
Feasible region is empty if constraint equations are inconsistent.
Example
x1
x2
R1
R2
R1 ∩R2 = ∅
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Optimal solution
Optimal solution
A feasible solution that maximises the objective function Z is theoptimal solution.
The optimal solution is usually denoted x∗ = (x∗1, x∗2, . . . , x
∗n).
The maximised objective function is denoted
Zmax = Z∗ = Z(x∗).
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Location of Optimal Solutions
The multivariate objective function Z = cTx is linear.
From introductory lectures, we know that Z∗ must lie on theboundary of the feasible region. Intro Lectures Here
Optimal solution x∗ will lie at either a corner point or an edge ofthe feasible polygon.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Feasible Regions
The feasible region is actually a polygon in n−dimensional space.
Each equation in the constraint describes a plane.
All equations (planes) together describe a region in space. Forexample,
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Number of Solutions
No solution
Feasible region is empty.
One optimal solution
Optimal solution at a corner point.
*
Many optimal solutions
Optimal solutions along an edge.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Unbounded feasible region
Another scenario exists where the feasible region could beunbounded. In an example similar to the empty region.
Example
Unbounded feasible region
x1
x2
x1 − x2 ≤ 1−x1 + x2 ≤ 2
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
The standard Linear Programming (LP) Problem
Graphical method of solving LP problem
The simplex procedural algorithm
Extensions to LP Algorithm to include other conditions
Dual Problem
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Return to example problem
Resource usage/unit ResourceDrug 1 Drug 2 Availability
Resource 1 1 0 42 3 2 183 0 2 12
Profit/unit 3 5 × $1000
Bivariate problem (two dimensional) so can easily be solvedgraphically.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Feasible Region
Feasible region is bounded by the five straight lines:x1 = 0 , x2 = 0 , x1 = 4 , 3x1 + 2x2 = 18 , 2x2 = 12 .
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
Feasible region
3x1+2x2=18
2x2 = 12
x1 = 4
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Find optimal solution(s)
Objective function Z = 3x1 + 5x2 yields a single optimal solution.
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
0
5
10
20
36
Direction of increasing Z(= 3x1 + 5x2)
Optimal solution
(x∗
1, x∗
2) = (2, 6)
Z∗ = 36
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Optimal Solutions
Consider new objective function Z = 6x1 + 4x2
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
C2(4,3)
C1(2,6)
8
16
24
32 36
Direction of increasing Z(= 4x1 + 6x2)
Note: Z = 6x1 + 4x2 isparallel to the boundary
line 3x1 + 2x2 = 18
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Limitations of graphical method
Graphical Limitations
Graphical method is only practical in two dimensions.
Alternative?
The algebraic simplex algorithm.
I Iterative procedure.
I Historically created by George Dantzig in 1947.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
The standard Linear Programming (LP) Problem
Graphical method of solving LP problem
The simplex procedural algorithm
Extensions to LP Algorithm to include other conditions
Dual Problem
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Simplex Algorithm
General overview of method
I Start at an initial feasible corner point.
Usually x = 0.
I Move to an adjacent feasible corner point.
Do this by finding corner with best potential increase in Z.
I Stop when at the optimal solution, Z∗.
Determine optimality when other adjacent corner points resultin decrease in Z.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Start at initial feasible corner point
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
0
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
First corner point iteration
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
12
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Second corner point iteration
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
27
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Third and final corner point iteration
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)36
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Alternative direction: Start at initial point
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
0
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Alternative direction: First corner point
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
30
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Alternative direction: Second and final corner point
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)36
30
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
System of equations
Need to move from inequality constraints to equality constraints.
Introduce new variables to reduce the inequalities to equalities.
Example
2x1 − 3x2 ≤ 5 ⇐⇒ 2x1 − 3x2 + x3 = 5
for some x3 ≥ 0.
Definition
Slack variables: Each introduced variable that reduces a (less than)inequality constraint to an equality constraint is called a slackvariable.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Slack Variables
For each constraint introduce a new variable equal to thedifference between the RHS and LHS of the constraint.
Ax ≤ b −→ Ax = b
That is we have Ax ≤ b,
a11x1 + a12x2 + · · · + a1nxn ≤ b1
a21x1 + a22x2 + · · · + a2nxn ≤ b2...
......
...am1x1 + am2x2 + · · · + amnxn ≤ bm
with: x1 ≥ 0 , x2 ≥ 0 , · · · , xn ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
New constraint matrix
After adding slack variables for each constraint
xs = (xn+1, xn+2, . . . , xn+m) ,
a11x1 + a12x2 + · · · + a1nxn + xn+1 = b1
a21x1 + a22x2 + · · · + a2nxn + xn+2 = b2
......
.... . .
am1x1 + am2x2 + · · · + amnxn + xn+m = bm
with: x1 ≥ 0, x2 ≥ 0, . . . , xn ≥ 0,
and: xn+1 ≥ 0, xn+2 ≥ 0, . . . , xn+m ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Return to example
Initial constraints,
x1 ≤ 4; 3x1 + 2x2 ≤ 18; 2x2 ≤ 12
Introduce a new variable equal to the difference between the RHSand LHS of the constraint.
x3 = 4− x1; x4 = 18− 3x1 − 2x2 ; x5 = 12− 2x2 .
New system of constraints,
x1 ≤ 4; 3x1 + 2x2 ≤ 18; 2x2 ≤ 12.↓ ↓ ↓
x1 + x3 = 4; 3x1 + 2x2 + x4 = 18; 2x2 + x5 = 12.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Algebraic representation
Turn attention to algebraic representation of corner point method.
Full problem with slack variables,
Maximise:Z = c1x1 + c2x2 + · · ·+ cnxn
such thata11x1 + a12x2 + · · · + a1nxn + xn+1 = b1
a21x1 + a22x2 + · · · + a2nxn + xn+2 = b2
......
.... . .
am1x1 + am2x2 + · · · + amnxn + xn+m = bm
andx1 ≥ 0, x2 ≥ 0, . . . , xn ≥ 0, xn+1 ≥ 0, . . . , xn+m ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
In tableau form
Z x1 x2 . . . xn xn+1 xn+2 . . . xn+m RHS
1 −c1 −c2 . . . −cn 0 0 . . . 0 00 a11 a12 . . . a1n 1 0 . . . 0 b1
0 a21 a22 . . . a2n 0 1 . . . 0 b2...
......
. . ....
......
. . ....
...0 am1 am2 . . . amn 0 0 . . . 1 bm
I Choose initial corner point solution at (x1, x2, . . . , xn) = 0
I Slack variables take the values of resource vector
xs = (xn+1, xn+2, . . . , xn+m)T = b
I Initially, slack variables are the basic variables.
I Initially, decision variables are the non-basic variables.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
For the example
Maximise: Z − 3x1 − 5x2 = 0
such that: x1 + x3 = 4
3x1 + 2x2 + x4 = 18
+ 2x2 + x5 = 12
and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, x5 ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Geometric interpretation of slack variables
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
I5
I1 I4
I2
I3 I5
I4
x1=
0
x2 = 0
x3=
0
x4=0
x5 = 0
F5
F1 F2
F3
F4
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
In tableau form
Step 1: Setup problem with initial corner point solution.
Z x1 x2 x3 x4 x5 RHS
1 −3 −5 0 0 0 00 1 0 1 0 0 40 3 2 0 1 0 180 0 2 0 0 1 12
Slack variables are the initial basic variables,
x3 = 4 , x4 = 18 , x5 = 12 .
Decision variables are set to zero,
x = 0
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Step 2: Move to adjacent corner point.
Which direction do we move? There are two choices....
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
(0,6)
(0,0)
(4,0)
(4,3)
(2,6)
0
Direction of increasing Z(= 3x1 + 5x2)
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Choice of movement
I From Objective function, Z − 3x1 − 5x2 = 0I Choose x2 to enter the basis.
I Which basic variable must leave the basis: x3, x4 or x5 ?
I Fix x1 = 0, Increase x2 and see which basic variable reacheszero first.
Constraints equations reduce to
x3 = 42x2 + x4 = 182x2 + x5 = 12
which imply
x3 = 4, x4 = 18− 2x2, x5 = 12− 2x2 .
Thus x3 is unaffected as x2 increases, x4 → 0 as x2 → 9 andx5 → 0 as x2 → 6. Thus x5 must leave the basis, as it goes tozero first.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Algebraic rule
Variable to enter the basis
Choose variable that has the most negative coefficient in the firstrow of the Tableau.
Variable to leave the basis
Given variable xj is to enter the basis.
Find
k = arg mini=1,...,m
biaij
, where aij > 0.
Then variable xk is to leave the basis.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Algebraic Rule in practice
Return to example,
Basis Z x1 x2 x3 x4 x5 RHS Ratio bi/ai2Z 1 −3 −5 0 0 0 0 –x3 0 1 0 1 0 0 4 –x4 0 3 2 0 1 0 18 9x5 0 0 2 0 0 1 12 6 ← min
Pivot on highlighted element.
Basis Z x1 x2 x3 x4 x5 RHSZ 1 −3 0 0 0 5/2 30x3 0 1 0 1 0 0 4x4 0 3 0 0 1 −1 6x2 0 0 1 0 0 1/2 6
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Continue the algorithm
Basis Z x1 x2 x3 x4 x5 RHS Ratio bi/ai1Z 1 −3 0 0 0 5/2 30 –x3 0 1 0 1 0 0 4 4x4 0 3 0 0 1 −1 6 2 ← minx2 0 0 1 0 0 1/2 6 –
x1 needs to enter the basis while x4 leaves the basis.
Basis Z x1 x2 x3 x4 x5 RHS Ratio
Z 1 0 0 0 1 3/2 36x3 0 0 0 1 −1/3 1/3 2x1 0 1 0 0 1/3 −1/3 2x2 0 0 1 0 0 1/2 6
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Step 3: Stop when at Optimal solution.
How do we know when we reach an optimal solution?
Basis Z x1 x2 x3 x4 x5 RHS Ratio
Z 1 0 0 0 1 3/2 36x3 0 0 0 1 −1/3 1/3 2x1 0 1 0 0 1/3 −1/3 2x2 0 0 1 0 0 1/2 6
Stop when all the coefficients in the first row are non-negative.
This is the optimal solution.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Interpret Tableau of Optimal solution
Basis Z x1 x2 x3 x4 x5 RHS Ratio
Z 1 0 0 0 1 3/2 36x3 0 0 0 1 −1/3 1/3 2x1 0 1 0 0 1/3 −1/3 2x2 0 0 1 0 0 1/2 6
Any move from here will cause a decrease in Z.
Objective function is Z = 36− x4 − 32x5.
Optimal solution at x∗ = (2, 6) with Z∗ = 36.
Slack variables are x∗s = (2, 0, 0).
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Tie Breakers: Variable entering basis
It is possible that there is no unique variable to choose for the rules.
Entering Variable
There may be a tied situation when there are two variables withthe most negative coefficient. E.g.
Basis Z x1 x2 x3 x4 x5 RHS
Z 1 0 −a −a 0 0 C
where a > 0 and C > 0.
Which variable should be chosen to enter the basis? x2 or x3?
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Tie Breakers: Variable leaving basis
It is also possible to have a tie for the variable leaving the basis.For example, consider the drug problem, with the secondconstraint modified to 3x1 + 2x2 ≤ 12.
Maximise: Z = 3x1 + 5x2
subject to: x1 ≤ 43x1 + 2x2 ≤ 12
2x2 ≤ 12with: x1, x2 ≥ 0 .
Initial simplex tableau:
Basis Z x1 x2 x3 x4 x5 RHS RatioZ 1 −3 −5 0 0 0 0 –x3 0 1 0 1 0 0 4 –x4 0 3 2 0 1 0 12 6 ← choosex5 0 0 2 0 0 1 12 6 ← either
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Tie breaker (cont.)
Basis Z x1 x2 x3 x4 x5 RHS RatioZ 1 −3 −5 0 0 0 0 –x3 0 1 0 1 0 0 4 –x4 0 3 2 0 1 0 12 6x5 0 0 2 0 0 1 12 6Z 1 -3 0 0 0 5/2 30x3 0 1 0 1 0 0 4x4 0 3 0 0 1 -1 0 ← non-positivex2 0 0 1 0 0 1/2 6
I Notice the resource element is zero.
Assumptions of simplex framework violated
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Tie breaker (cont.)
Basis Z x1 x2 x3 x4 x5 RHS RatioZ 1 −3 −5 0 0 0 0 –x3 0 1 0 1 0 0 4 –x4 0 3 2 0 1 0 12 6x5 0 0 2 0 0 1 12 6Z∗ 1 9/2 0 0 5/2 0 30 optimalx∗3 0 1 0 1 0 0 4x∗2 0 3/2 1 0 1/2 0 6x∗5 0 −3 0 0 −1 1 0 ← degeneracy
I Degenerate point in the basis with x5 = 0I Can cause stalling.I Can cause an endless cycle.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Degenerate points
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
Feasibleregion
Corner points(0, 6) and (4, 0) are
degenerate.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Solution upon an edge
Return to the previous pharmeutical example with Z = 6x1 + 4x2
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
C2(4,3)
C1(2,6)
8
16
24
32 36
Direction of increasing Z(= 4x1 + 6x2)
Note: Z = 6x1 + 4x2 isparallel to the boundary
line 3x1 + 2x2 = 18
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Solution upon an edge (cont)
Final tableau form.
Basis Z x1 x2 x3 x4 x5 RHS RatioZ 1 −6 −4 0 0 0 0 –x3 0 1 0 1 0 0 4 4x4 0 3 2 0 1 0 18 6x5 0 0 2 0 0 1 12 –Z 1 0 -4 0 0 0 24 –x1 0 1 0 1 0 0 4 –x4 0 0 2 -3 1 0 6 2x5 0 0 2 0 0 1 12 6Z∗ 1 0 0 0 2 0 36 optimalx∗1 0 1 0 1 0 0 4x∗2 0 0 1 -3/2 1/2 0 3x∗5 0 0 0 3 −1 1 6
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Solution upon an edge (cont)
Final tableau form.
Basis Z x1 x2 x3 x4 x5 RHS RatioZ∗ 1 0 0 0 2 0 36 optimalx∗1 0 1 0 1 0 0 4x∗2 0 0 1 -3/2 1/2 0 3x∗5 0 0 0 3 −1 1 6
I x1, x2, x5 are in the basis
I x3 and x4 are out of the basis.
I However, x3 has a zero coefficient in the objective row.
I Optimal objective function
Z∗ = 36− 2x∗4
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Solution upon an edge (cont)
Final tableau form.
Basis Z x1 x2 x3 x4 x5 RHS RatioZ∗ 1 0 0 0 2 0 36 optimalx∗1 0 1 0 1 0 0 4x∗2 0 0 1 -3/2 1/2 0 3x∗5 0 0 0 3 −1 1 6
Can parametrise the solution to locate the edge.
I System of equations. Let x∗3 = t ≥ 0.
I
x∗1 = 4− t ≥ 0 ⇒ t ≤ 4.x∗2 = 3 + 3/2t ≥ 0 ⇒ t ≥ −2.x∗5 = 6− 3t ≥ 0 ⇒ t ≤ 2.
I t ∈ [0, 2]
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
The standard Linear Programming (LP) Problem
Graphical method of solving LP problem
The simplex procedural algorithm
Extensions to LP Algorithm to include other conditions
Dual Problem
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Minimising the Objective Function
Consider finding the minimum
To minimizeZ = c1x1 + c2x2 + · · ·+ cnxn.
Define a new objective function Z = −Z. Then
minZ ≡ −max Z .
f(x)
x∗−f(x)
x∗
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Greater than or equal to constraints
If the problem has a constraint of the form
aTx = a1x1 + a2x2 + · · ·+ anxn ≥ b > 0 ,
introduce a surplus variable xn+1 such that
a1x1 + a2x2 + · · ·+ anxn − xn+1 = b , xn+1 ≥ 0 .
As an example, consider
Maximise: Z = 3x1 + 5x2
subject to: x1 ≤ 43x1 + 2x2 ≥ 18
2x2 ≤ 12with: x1, x2 ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Greater than constraint example
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
Feasible
region
Z∗ = 42
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Negative Resource Elements
In the standard problem all resource elements bj (i ≤ j ≤ m) arenon-negative. Suppose bj = −b < 0. Then,
aTx = a1x1 + a2x2 + · · ·+ anxn ≤ −b ,
is equivalent to
−aTx = −a1x1 − a2x2 − · · · − anxn ≥ b .
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Negative Decision Variable
If xk ≤ 0 introduce a new variable xk = −xk. Then xk ≤ 0 isequivalent to xk ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Unrestricted Decision Variable
If xk is unrestricted in sign, introduce two new variables, xk ≥ 0and xk ≥ 0, and let xk = xk − xk.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Two-Phase Method
Return to the problem,
Maximise: Z = 3x1 + 5x2
subject to: x1 ≤ 43x1 + 2x2 ≥ 18
2x2 ≤ 12with: x1, x2 ≥ 0.
Introduce slack and surplus variables,
x1 + x3 = 43x1 + 2x2 − x4 = 18
2x2 + x5 = 12
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Finding an initial feasible corner point
I In regular framework.I Set decision variables to zero.I Slack variables set to resource elements.
x1 + x3 = 43x1 + 2x2 − x4 = 18
2x2 + x5 = 12
If done here, implies that,
x1 = 0 , x2 = 0 , x3 = 4 , x4 = −18 , x5 = 12
Surplus variable violates the positivity constraint!
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Artificial Variables
I Introduce an artificial variable for any constraint that involvesan equality or greater than or equal to condition.
I For example, introduce x6 ≥ 0 in second constraint equation.
The constraint equations become
x1 + x3 = 43x1 + 2x2 − x4 + x6 = 18
2x2 + x5 = 12
I Set decision and surplus variables to be zero.
I Set slack and artificial variables to resource elements (RHS)
Initial corner point
(x1, x2, x3, x4, x5, x6) = (0, 0, 4, 0, 12, 18) .
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Not a corner point of the actual problem.
The initial corner point (x1, x2, x3, x4, x5, x6) = (0, 0, 4, 0, 12, 18)is not in the feasible set.
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6x1
x2
Feasible
region
Z∗ = 42
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Reconcile the initial corner point
Feasible solution
A feasible solution of the problem with artificial variables is afeasible solution of the original problem if and only if the artificialvariables are zero.
In the example, the second constraint gives
3x1 + 2x2 − x4 = 18− x6
RHS is smaller than 18 if x6 > 0.
3x1 + 2x2 − x4 = 18
if and only if x6 = 0. Thus we must force the artificial variables tozero.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Find initial feasible solution
To find a basic feasible solution to the original problem maximize,
W = −∑
artificial
xk ≤ 0 .
If maxW = 0, then all artificial variables are zero.
Then proceed with regular simplex algorithm.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
First Phase: Maximise W
Basis x1 x2 x3 x4 x5 x6 RHS
W 0 0 0 0 0 1 0Z −3 −5 0 0 0 0 0x3 1 0 1 0 0 0 4– 3 2 0 −1 0 1 18x5 0 2 0 0 1 0 12
W −3 −2 0 1 0 0 −18Z −3 −5 0 0 0 0 0x3 1 0 1 0 0 0 4x6 3 2 0 −1 0 1 18x5 0 2 0 0 1 0 12
After Maximisation on W is complete, remove W and all artificialvariables.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
First Phase: Maximise W
Basis x1 x2 x3 x4 x5 x6 RHS Ratio
W −3 −2 0 1 0 0 −18 –Z −3 −5 0 0 0 0 0 –x3 1 0 1 0 0 0 4 4x6 3 2 0 −1 0 1 18 6x5 0 2 0 0 1 0 12 –
W 0 −2 3 1 0 0 −6 –Z 0 −5 3 0 0 0 12 –x1 1 0 1 0 0 0 4 –x6 0 2 −3 −1 0 1 6 3x5 0 2 0 0 1 0 12 6
W 0 0 0 0 0 1 0 –Z 0 0 −9/2 −5/2 0 5/2 27 –x1 1 0 1 0 0 0 4 4x2 0 1 −3/2 −1/2 0 1/2 3 –x5 0 0 3 1 1 −1 6 2
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Second Phase: Maximise Z with new starting point.
Basis x1 x2 x3 x4 x5 RHS Ratio
Z 0 0 −9/2 −5/2 0 27 –x1 1 0 1 0 0 4 4x2 0 1 −3/2 −1/2 0 3 –x5 0 0 3 1 1 6 2
Z 0 0 0 −1 3/2 36 –x1 1 0 0 −1/3 −1/3 2 –x2 0 1 0 0 1/2 6 –x3 0 0 1 1/3 −1/3 2 6
Z 0 0 3 0 5/2 42 Optimalx1 1 0 1 0 0 4x2 0 1 0 0 1/2 6x4 0 0 3 1 1 6
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Big M-method
I An alternative to the Two-Phase
Maximise: Z = −x1 + 2x2
subject to: x1 + x2 ≥ 2−x1 + x2 ≥ 1
x2 ≤ 3with: x1, x2 ≥ 0.
I Introduce slack, surplus and artificial variables as before.
I Modify objective function to penalise the artificial variables.
Z = Z ±M∑
k∈artificial
xk.
I NB: Sign infront of M penalises against the optimatisation.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Big M-method
I Constraint equations
x1 + x2 − x3 + x6 = 2−x1 + x2 − x4 + x7 = 1
x2 + x5 = 3
I Objective function is Maximisation:
Max: Z = −x1 + 2x2−M(x6 + x7)
I NB: If Minimisation was desired consider,
Min: Z = −x1 + 2x2+M(x6 + x7)
I M is considered to be an arbitrarily large number.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Big M - implementation:
Basis x1 x2 x3 x4 x5 x6 x7 RHS Ratio
Z 1 -2 0 0 0 M M 0 –– 1 1 −1 0 0 1 0 2 –– -1 1 0 −1 0 0 1 1 –x5 0 1 0 0 1 0 0 3 –
Z 1 -2M-2 M M 0 0 0 -3M –x6 1 1 −1 0 0 1 0 2 2x7 -1 1 0 −1 0 0 1 1 1x5 0 1 0 0 1 0 0 3 3
I Initialise tableau to ensure artificial variables are basic.
I Continue regular simplex algorithm.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Big M - implementation:
Basis x1 x2 x3 x4 x5 x6 x7 RHS Ratio
Z 1 -2M-2 M M 0 0 0 -3M –x6 1 1 −1 0 0 1 0 2 2x7 −1 1 0 −1 0 0 1 1 1x5 0 1 0 0 1 0 0 3 3
Z -2M-1 0 M -M−2 0 0 2M+2 -M+2 –x6 2 0 −1 1 0 1 -1 1 1
2x2 −1 1 0 −1 0 0 1 1 –x5 1 0 0 1 1 0 -1 2 2
Z 0 0 −0.5 −1.5 0 M+0.5 M+1.5 2.5 –x1 1 0 −0.5 0.5 0 0.5 -0.5 0.5 1x2 0 1 −0.5 −0.5 0 0.5 0.5 1.5 –x5 0 0 0.5 0.5 1 -0.5 -0.5 1.5 3
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Big M - implementation:
Basis x1 x2 x3 x4 x5 x6 x7 RHS Ratio
Z 0 0 −0.5 −1.5 0 M+0.5 M+1.5 2.5 –x1 1 0 −0.5 0.5 0 0.5 -0.5 0.5 1x2 0 1 −0.5 −0.5 0 0.5 0.5 1.5 –x5 0 0 0.5 0.5 1 −0.5 −0.5 1.5 3
Z 3 0 −2 0 0 M+2 M 4 –x4 2 0 −1 1 0 1 -1 1 –x2 1 1 −1 0 0 1 0 2 –x5 −1 0 1 0 1 −1 0 1 1
Z 1 0 0 0 2 M M 6 optx4 1 0 0 1 1 0 −1 2 –x2 0 1 0 0 1 0 0 3 –x5 −1 0 1 0 1 −1 0 1 –
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
The standard Linear Programming (LP) Problem
Graphical method of solving LP problem
The simplex procedural algorithm
Extensions to LP Algorithm to include other conditions
Dual Problem
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Dual Problem
Recall the regular problem in matrix notation,
PrimalMaximise: Z = cTxsubject to: Ax ≤ b
with: x ≥ 0
Refer to it as the Primal problem.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Definition of dual problem
Primal DualMaximise: Z = cTx Minimise: v = yTbsubject to: Ax ≤ b subject to: yTA ≥ cT
with: x ≥ 0 with: y ≥ 0
Return to initial pharmaceutical example:
Maximise : Z = 3x1 + 5x2
such that: x1 ≤ 4
3x1 + 2x2 ≤ 18
2x2 ≤ 12
and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, x5 ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Example: Dual construction
I Find a bound for optimal solution.I From constraints,
Z = 3x1 + 5x2 ≤ 3× 4 + 5× 6 = 42
I Consider a general linear combination of constraints,
(x1)y1 + (3x1 + 2x2) y2 + (2x2)y3 ≤ 4y1 + 18y2 + 12y3
I Define a new objective function v := 4y1 + 18y2 + 12y3
Notice,
Z := 3x1 + 5x2
≤ (y1 + 3y2)x1 + (2y2 + 2y3)x2
≤ 4y1 + 18y2 + 12y3 =: v
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Example: continued
I Last statement true if y ≥ 0 and,
y1 + 3y2 ≥ 3 and 2y2 + 2y3 ≥ 5
I Dual problem is,
Minimise : v = 4y1 + 18y2 + 12y3
such that: y1 + 3y2 ≥ 3
2y2 + 2y3 ≥ 5
and y1 ≥ 0, y2 ≥ 0, y3 ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Useful applications
I Dimension of decision variables and constraint equations swap.
I Optimal values of Max: Z and Min: v coincide under certainconditions.
I Can sometimes simplify the problem to be solved.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Example
Example
Maximise: Z = 8x1 + 30x2 + 7x3
subject to: x1 + 5x2 + 3x3 ≤ 104x1 + 6x2 + x3 ≤ 15
with: x1, x2, x3 ≥ 0.
The corresponding dual problem is:
Minimise: v = 10y1 + 15y2
subject to: y1 + 4y2 ≥ 85y1 + 6y2 ≥ 303y1 + y2 ≥ 7
with: y1, y2 ≥ 0 ,
which can be solved graphically.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Graphical solution
Minimise: v = 10y1 + 15y2
subject to: y1 + 4y2 ≥ 85y1 + 6y2 ≥ 303y1 + y2 ≥ 7
with: y1, y2 ≥ 0 ,
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9 10 11y1
y2
Optimal solution,
v = 435
7, (y1, y2) = (36
7,5
7)
5y1 + 6y2 = 30
y1 + 4y2 = 8
2y1 + y2 = 8
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Second Example
Return to the Pharmaceutical problem. It has the dual problem,
Minimise : v = 4y1 + 18y2 + 12y3
such that: y1 + 3y2 ≥ 3
2y2 + 2y3 ≥ 5
and y1 ≥ 0, y2 ≥ 0, y3 ≥ 0.Standardise the problem for the simplex framework.
Maximise: v = −4y1 − 18y2 − 12y3
subject to:y1 + 3y2 − y4 + y6 = 3
2y2 + 2y3 − y5 + y7 = 5
with: y1 ≥ 0, y2 ≥ 0, y3 ≥ 0.
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Solve using Two-Phase
First Phase: Maximise W
Basis y1 y2 y3 y4 y5 y6 y7 RHS
W 0 0 0 0 0 1 1 0v 4 18 12 0 0 0 0 0– 1 3 0 -1 0 1 0 3– 0 2 2 0 -1 0 1 5
W −1 −5 -2 1 1 0 0 −8v 4 18 12 0 0 0 0 0y6 1 3 0 -1 0 1 0 3y7 0 2 2 0 -1 0 1 5
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Maximise W
Basis y1 y2 y3 y4 y5 y6 y7 RHS Ratio
W −1 −5 -2 1 1 0 0 −8 –v 4 18 12 0 0 0 0 0 –y6 1 3 0 -1 0 1 0 3 1y7 0 2 2 0 -1 0 1 5 5/2
W 2/3 0 -2 -2/3 1 5/3 0 −3 –v −2 0 12 6 0 -6 0 -18 –y2 1/3 1 0 -1/3 0 1/3 0 1 –y7 -2/3 0 2 2/3 -1 -2/3 1 3 3/2
W 0 0 0 0 0 1 1 0 –v 2 0 0 2 6 -2 -6 -36 –y2 1/3 1 0 -1/3 0 1/3 0 1 –y3 -1/3 0 1 1/3 -1/2 -1/3 1/2 3/2 –
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Standard Problem Graphical Method Algebraic Simplex Method Extensions Dual
Phase Two: Discard W
Basis y1 y2 y3 y4 y5 RHS
v 2 0 0 2 6 -36y2 1/3 1 0 -1/3 0 1y3 -1/3 0 1 1/3 -1/2 3/2
I Notice that,
Max Z = 3x1 + 5x2 = 36 = −Max v
I y∗ = (y∗1, y∗2, y∗3) = (0, 1, 3/2) and x∗ = (2, 6)
I The optimal x∗ can be found as the cost coefficients of thesurplus variables y4 and y5 in the objective row.
I Agrees with graphical and simplex methods used before