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8/11/2019 Operations Research- Lecture 3 - Algebraic Simplex Method_2
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Lecture 3 The Simplex Method for LP
Principles of SimplexGeometric Interpretation and Key Concepts
The Algebraic Simplex MethodSlack variablesAugmented Solution, Basic SolutionBasic/Non-basic VariablesCanonical System of EquationsOptimality Test
Learning outcomes:understand the geometric concepts behind theSimplex method; relate the geometric and algebraic interpretationsof the Simplex method; apply the algebraic Simplex method step bystep to solve small LP problems.
Operations Research
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Principles of Simplex
Developed by G. Dantzig in 1947, this is a very efficient
algebraic procedureto solve very large LP problems.
The underlying principles of the Simplex method arebased on the geometric concepts:
constraint boundaries corner-point solutions
corner-point feasible (CPF) solutions
corner-point infeasible solutions
adjacent CPF solutions edges of the feasible region
a CPF solution is optimal if it does not have adjacent CPFsolutions that are better
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0
1
2
3
4
5
6
1 2 3 4 5 6
x1
x2
Geometric Concepts of Simplex
(4)
(5)
(2)
(3)
exampleproblemATLAS
(6)0,0
(5)2
(4)1
(3)62
(2)2446:Subject to
(1)45:Maximise
21
2
21
21
21
21
xx
x
xx
xx
xx
xxZ
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4
0
1
2
3
4
5
6
1 2 3 4 5 6
x1
x2
(4)
(5)
(2)
(3)
The Simplex Procedure
1. Initialisation
2. Optimality Test
3. If Optimal, Stop
4. If not Optimal, performiteration: move to betteradjacent CPF solutionand then go to step 2.
(6)0,0
(5)2
(4)1(3)62
(2)2446
:Subject to
(1)45:Maximise
21
2
21
21
21
21
xx
x
xx
xx
xx
xxZ
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Key Concepts of the Simplex Procedure
It focuses only on CPF solutions (assuming there is at least
one). In each iteration, another CPF solution is explored.
If possible, the origin (xi= 0 i) being a CPF solution, isselected as the initial point.
The procedure moves to adjacent CPF solutions by movingalong the edges of the feasible region.
Better adjacent CPF solutions are identified by positive rateof improvementon the objective value Z.
If the rate of improvement along the edges emanating fromthe current CPF solution are all negative, then the currentCPF solution is optimal.
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Importance of the Simplex Method
Simplex is the standard methodfor solving LP problems
Even modern algorithms to solve large IP problems arebased on successive LP relaxations solved using Simplex
Variants of the Simplex method are tailored for specifictypes of problems
The Primal Simplexmethod is simply known as Simplex
The Dual Simplexmethod is efficient for re-optimizationafter the model has changed (change of coefficients, changeof right-hand side values, addition/deletion of constraints)
The Revised Simplexmethod is useful as part of atechnique called Column Generationfor solving very largeLP problems
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Special Cases in the Simplex Method
The form of the Simplex method explained in this lecture
works for a given format of the LP model: Maximisation objective (starting from solution xi=0 i
is straightforward)
Inequalities in the constraints are of type (adding
slack variables makes sense) The right-hand side values biin the constraints are
positive
Non-negativity constraints in all decision variables.
To deal with variations of the above LP model formal,some algebraic manipulationis required to prepare theaugmented model.
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Unbounded Search Space
No leaving basic variable can be identified in Step 2 of agiven iteration. This will happen if there is no bound onthe increase of the entering basic variable.
Multiple Optimal Solutions
The Simplex method stops immediately after the firstoptimal solution is found. Multiple optimal solutionsexist if at least one of the non-basic variables has acoefficient of zero in the objective function equation.
Additional optimal solutions can be obtained byperforming additional iterations of the Simplex method,each time choosing a non-basic variable with zerocoefficient as the entering basic variable.
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The Algebraic Simplex Method
Convert constraints to equations
Introduce 1 slack variable in each functional constraint
to obtain equalities:0and24462446 332121 xxxxxx
0and6262 442121 xxxxxx
0and11 552121 xxxxxx
0and22 6622 xxxx
Example 5.1Apply the algebraic Simplex method to solve the
ATLAS maximisation problem.
(6)0,0
(5)2
(4)1
(3)62
(2)2446:Subject to
(1)45:Maximise
21
2
21
21
21
21
xx
x
xx
xx
xx
xxZ
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The result is an augmented model
(6)6...1for0
(5)2
(4)1
(3)62
(2)2446:Subject to
(1)045:Maximise
62
521
421
321
21
ix
xx
xxx
xxx
xxx
xxZ
i
0
1
2
3
4
5
6
1 2 3 4 5 6
x1
x2
(4)
(5)
(2)
(3)
For a given solution, thevalue of the slack variableindicates if the solution is: On the constraint
boundary On the feasible region On the infeasible region
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Augmented solutions and basic solutions
0
1
2
3
4
5
6
1 2 3 4 5 6
x1
x2
(4)
(5)
(2)
(3)
For a given solution, theaugmented solutionisobtained by calculating
the value of the slackvariables. A basic solutionis an augmented corner-point solution.
(6)6...1for0
(5)2
(4)1
(3)62
(2)2446:Subject to
(1)045:Maximise
62
521
421
321
21
ix
xx
xxx
xxx
xxx
xxZ
i
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Algebraic procedure from the augmented model
From the set of variables in the augmented model, someare designated as basic variablesand the others are
designated as non-basic variablesduring the solutionprocedure.
The non-basic variables are set to zero and the basic
variables are calculated by solving the equations(augmented constraints).
Initialisation
0
so,variablesbasic-nontheasdesignatedareand
21
21
xx
xx
6543
21
,,,:basic
,:basic-non
xxxx
xx
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Solving for the basic variables from the augmented model:
(6)6...1for0
2(5)2
1(4)1
6(3)62
24(2)2446
:Subject to
(1)045:Maximise
662
5521
4421
3321
21
ix
xxx
xxxx
xxxx
xxxx
xxZ
i
The initial basic feasible (BF) solution: (0,0,24,6,1,2)
Optimality Test
The initial BF solution (0,0,24,6,1,2) is not optimal
6543
21
,,,:basic
,:basic-non
xxxx
xx
(1)45 21 xxZ
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Iteration
1. Determine direction of movement.
Rate of improvement 5 (corresponding to x1) is better.
The non-basic variable x1is now called the entering basicvariable.
2(5)2
boundsnoso1(4)1
6so6(3)62
4so624(2)2446
662
115521
114421
113321
xxx
xxxxxx
xxxxxx
xxxxxx
2. Determine when to stop the movement.
Increase x1as much as possible but maintain the other non-basic variables on zero (x2= 0), maintain the satisfaction ofthe equations, and maintain variables non-negative.
(1)45 21 xxZ
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(cont. Step 2)
Then, x1can be increased up to 4 as indicated by theaugmented constraint (2).
The basic variable x3becomes the leaving basic variable.
(5)2
(4)1
(3)62
(2)2446
(1)045
62
521
421
321
21
xx
xxx
xxx
xxx
xxZ
6541
32
,,,:basic
,:basic-non
xxxx
xx3. Solving for the new BF solution
Initial BF solution: (0,0,24,6,1,2)
New BF solution: (4,0,0,?,?,?)
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(cont. Step 3)
The coefficient pattern of the leaving basic variable x3(0,1,0,0,0) is reproduced for the entering variable x1by
means of algebraic manipulation.
(5)2
(4)1
(3)62
(2)2446
(1)045
62
521
421
321
21
xx
xxx
xxx
xxx
xxZ(1)02
6
5
3
232 xxZ
Given the new non-basic variables
x2=0 and x3=0 then:Z=20, x1=4, x4=2, x5=5, x6=2
The new BF solution: (4,0,0,2,5,2)6541
32
,,,:basic
,:basic-non
xxxx
xx
(2)46
1
3
2321 xxx
(3)261
34 432 xxx
(5)262 xx
(4)56
1
3
5532 xxx
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(1)6
5
3
220 32 xxZ
Optimality Test
The new BF solution (4,0,0,2,5,2) is not optimal.
Iteration 2
1. Determine direction of movement.
Rate of improvement 2/3 (corresponding to x2) is better.
The non-basic variable x2is now the entering basic variable.
2. Determine when to stop the movement.
Increase x2as much as possible but maintain the other non-basic variables on zero (x3= 0), maintain the satisfaction ofthe equations, and maintain variables non-negative.
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(cont. Step 2)
2so2(5)2
3so3
55(4)5
6
1
3
5
4
6so
3
42(3)2
6
1
3
4
6so3
24(2)4
6
1
3
2
22662
225532
224432
221321
xxxxx
xxxxxx
xxxxxx
xxxxxx
Then, x2can be increased up to 6/4=1.5 as indicated by theaugmented constraint (3).
The basic variable x4becomes the leaving basic variable.
6521
43
,,,:basic
,:basic-non
xxxx
xx
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(cont. Step 3)
Since the new non-basic variables x3=0 and x4=0 then:Z=21, x
1
=3, x2
=1.5, x5
=5/2, x6
=1/2
The new BF solution: (3,1.5,0,0,2.5,0.5)
Optimality Test
The new BF solution (3,1.5,0,0,2.5,0.5) is optimal becausein the new objective function both x3and x4have negativecoefficient, so increasing any of them would lead to a worseadjacent BF solution.
(1)21
4321 43 xxZ
6521
43
,,,:basic
,:basic-non
xxxx
xx
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Example 5.2Apply the algebraic Simplex method to solve the WENBUmaximisation problem.
Augmented Model
0,0
(4)60075
(3)150025(2)80001:Subject to
(1)1025:Maximise
21
21
1
2
21
xx
xx
x
x
xxZ
0,,,,0
60075)4(
150025(3)
80001)2(
01025)1(
54321
521
41
32
21
xxxxx
xxx
xx
xx
xxZ
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Example 5.2 (cont.)
Initialisation
543
21
,,:basic,:basic-non
xxx
xx
60060075)4(
1500150025(3)
80080001)2(01025)1(
5521
441
332
21
xxxx
xxx
xxx
xxZ
(1)1024 21 xxZ
Optimality Test
New BF solution (0,0,800,1500,600) with Z = 0 is not optimal.
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Example 5.2 (cont.)
Iteration
120560060075)4(
60251500150025(3)onboundno80080001)2(
01025)1(
115521
11441
1332
21
xxxxxx
xxxxx
xxxx
xxZ
x1can increase up to 60 as given by (3)Leaving basic variable: x
4New BF solution: (60,0,?,0,?)531
42
,,:basic
,:basic-non
xxx
xx
60075)4(150025(3)
80001)2(
01025)1(
521
41
32
21
xxx
xx
xx
xxZ
60251
41 xx
150010 42 xxZ
80010 32 xx
3005
17 542 xxx
New BF solution (60,0,800,0,300) Z = 1500
Entering basic variable: x1
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Example 5.2 (cont.)
(1)101500 42 xxZ
Optimality Test
New BF solution (60,0,800,0,300) with Z = 1500 is not optimal.
531
42
,,:basic
,:basic-non
xxx
xx
Iteration 2
7
30073003005
17)4(
onboundno6060251(3)
801080080001)2(
150010)1(
225542
2141
22332
42
xxxxxx
xxxx
xxxxx
xxZ
Entering basic variable: x2
x2can increase up to 300/7 42.86 as given by (4)
Leaving basic variable: x5New BF solution: (?,300/7,?,0,0)321
54
,,:basic
,:basic-non
xxx
xx
b i
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Example 5.2 (cont.)
3005
17)4(
6025
1(3)
80001)2(
150010)1(
552
41
32
42
xxx
xx
xx
xxZ
6025
141 xx
713500
710
3525
54 xxZ
72600
710
3510
543 xxx
7300
71
351
542 xxx
New BF solution (60,300/7,2600/7,0,0) Z = 13500/7
321
54
,,:basic
,:basic-non
xxx
xx
(1)7
10
35
25
7
1350054 xxZ
Optimality Test
New BF solution (60,42.86,371.42,0,0) with Z = 1928.57 is optimal
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Example 5.4For the following LP model and with reference to theSimplex method in algebraic form, write the augmented model, findthe initial (basic feasible) BF solution and determine if this initialsolution is optimal or not. Then, continue the process and work
through the steps of the Simplex method (in algebraic form) to solvethis model. Please label clearly each of the steps and calculationscarried out.
27
)4(0,0
(3)303(2)303:Subject to
(1):Maximise
21
21
21
21
xx
xx
xx
xxZ