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![Page 1: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/1.jpg)
Linear Programming Fundamentals
Convexity
Definition: Line segment joining any 2 pts lies inside shape
convex NOT convex
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Linear Programming Fundamentals..
Nice Property of Convex Shapes:
Intersection of convex shapes is convex
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Linear Programming Fundamentals…
Every Half-space is convex
Set of feasible solutions is a convex polyhedron
Each constraint of an LP half space
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Some definitions
y ≥ 0
x ≥ 0
x ≤ 500
2x+y ≤ 1500
x + y ≤ 1200(0,0)
1000
1500
500 1000 1500
500
y
x
Feasibleregion
Boundary Feasible Solution
Corner points
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Important Properties
(0,0)
1000
1500
500 1000 1500
500
y
x
Feasibleregion
Property 1. IF: only one optimum solution => must be a corner point
(0,0)
1000
1500
500 1000 1500
500
y
x
Feasibleregion
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Important Properties..
(0,0)
1000
1500
500 1000 1500
500
y
x
Feasibleregion
Property 2. IF: multiple optimum solutions => must include 2 adjacent corner pts
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Important Properties…
(0,0)
1000
1500
500 1000 1500
500
y
x
Feasibleregion
Property 4. Total number of corner pts is finite
Property 3. If a corner point is better than all its adjacent corners, it is optimal !
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Important Properties....
(0,0)
1000
1500
500 1000 1500
500
y
x
Feasibleregion
Note: Corner point intersection of some constraint boundaries (lines)
Property 5. Moving from a corner point to any adjacent corner point Exactly one constraint boundary is exchanged.
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The algebra of Simplex
Good newsWe only need to search for the best feasible corner point
Good newsWe can use property 5 to guide our searching
Bad newsA problem with 50 variables, 100 constraints =>
100 !
50 ! (100-50) !
=> ~ 1029 corner points
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The outline of Simplex
1. Start at a corner point feasible solution
2. If (there is a better adjacent corner feasible point) then
go to one such adjacent corner point;
repeat Step 2;
else (report this point as the optimum point).
Why does this method work ?
1. Constant improvement (=> no cycling)
2. Finite number of corners
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The Algebra of Simplex: Gauss elimination
Solving for corner points solving a set of simultaneous equations
Method: Gaussian elimination
Example:
x + y = 2 [1]x – 2y = 1 [2]
Solve by:
2x[1] + [2]: 3x = 5 => x = 5/3[1] - 1x[2]: 3y = 1 => y = 1/3
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The Algebra of Simplex: need for Slack !
Cannot add (multiples of) INEQUATIONS !!
Consider:
x >= 0 [1]y >= 0 [2]
[1] + [2]: x + y >= 0 [3]
x
y
x + y >= 0
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The Algebra of Simplex: Slack
To allow us to use Gaussian elimination,
Convert the inequalities to equations by using SLACK Variables
2x + y ≤ 1500, x, y ≥ 0
2x + y + s = 1500, x, y, s ≥ 0
x + y ≥ 200 x, y ≥ 0
x + y – s = 200, x, y, s ≥ 0.
Inequality Equality, with slack variable
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The Algebra of Simplex..
Conversion of the Product mix problem
ORIGINAL
Maximize z( x, y) = 15 x + 10y
subject to 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0
(almost) STANDARD FORM
Maximize Z = 15 x1 + 10x2
subject to
2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500
xi ≥ 0 for all i.
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ORIGINAL
Maximize z( x, y) = 15 x + 10y
subject to 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0
(almost) STANDARD FORM
Maximize Z = 15 x1 + 10x2
subject to
2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500
xi ≥ 0 for all i.
Original problem: n variables, m constraintsStandard form: (m+n) variables, m constraints
The Algebra of Simplex…
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Original problem: n variables, m constraints
Standard form: (m+n) variables, m constraints
How to use Gaussian elimination ??
Force some variables = 0
How many to be forced to be 0 ?
The Algebra of Simplex…
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Definitions: augmented solution
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
Solution: Augmented solution(x1 x2) = (200, 200) (200, 200, 900, 800, 300)
Max Z = 15 x1 + 10x2
S.T. 2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500 xi ≥ 0 for all i.
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Definitions: BS, BFS
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
Basic solution Augmented, corner-point solution
Max Z = 15 x1 + 10x2
S.T. 2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500 xi ≥ 0 for all i.
Examplesbasic infeasible solution : ( 500, 700, -200, 0, 0) basic feasible solution: (500, 0, 500, 700, 0)
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Definition: non-basic variables
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
Max Z = 15 x1 + 10x2
S.T. 2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500 xi ≥ 0 for all i.
CornerFeasiblesolution
Defining eqns Basic solution non-basic variables
(0, 0) x1 = 0
x2 = 0
(0, 0, 1500, 1200, 500)
x1, x2
(500, 0) x1 = 500
x2 = 0
(500, 0, 500, 700, 0) x2, x5
(500, 500) x1 = 500
2x1+x2= 1500
(500, 500, 0, 200, 0) x5, x3
(300, 900) x1+ x2 = 1200
2x1+x2= 1500
(300, 900, 0, 0, 200) x3, x4
(0, 1200) x1 = 0
x1+ x2 = 1200
(0, 1200, 300, 0, 500) x4, x1
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The Algebra of Simplex
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
Max Z = 15 x1 + 10x2
S.T. 2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500 xi ≥ 0 for all i.
Corner infeasiblesolution
Defining eqns Basicinfeasiblesolution
non-basic variables
(750, 0) x2 = 0
2x1+ x2 = 1500
(750, 0, 0, 450, -250) x2, x3
(1200, 0) x2 = 0
x1 + x2 = 1200
(1200, 0, -900, 0, -700) x2, x4
(500, 700) x1 + x2 = 1200
x1 = 500
(500, 700, -200, 0, 0) x4, x5
(0, 1500) x1 = 0
2x1+ x2 = 1500
(0, 1500, 0, -300, 500) x1, x3
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The Algebra of Simplex: Standard form
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
STANDARD FORMMax
Z S.T.
Z -15 x1 -10 x2 = 0 2 x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500
xi ≥ 0 for all i.
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The Algebra of Simplex: Step 1. Initial solution
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
MaxZ
S.T.Z -15 x1 -10 x2 = 0
2 x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500
xi ≥ 0 for all i.
( 0, 0)
Initial non-basic variables: (x1, x2)
Initial basic variables: (x3, x4 , x5)
Initial BFS: (0, 0, 1500, 1200, 500)
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The Algebra of Simplex: Step 2. Iteration Step
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
Entering variable:
Z = 15 x1 + 10x2
Fastest rate of ascent
![Page 24: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/24.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Leaving variable
Max Z S.T.
Z -15 x1 - 10x2 = 0 2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500
xi ≥ 0 for all i.
Consider first constraint:
Entering
2x1 + x2 + x3 = 1500
x3 = 1500 - 2x1 - x2
Bound on x1: 1500/2 = 750
![Page 25: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/25.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Basic variable
Equation Upper bound for x1
x3 x3 = 1500 - 2x1 - x2 x2 = 0, so x1 ≤ 1500/2; UB = 750 x4 x4 = 1200 - x1 - x2 x2 = 0, so x1 ≤ 1200; UB = 1200
x5 x5 = 500 - x1 x1 ≤ 500 minimum
Max Z S.T.
Z -15 x1 - 10x2 = 0 2x1 + x2 + x3 = 1500 x1 + x2 + x4 = 1200 x1 + x5 = 500
xi ≥ 0 for all i.
Leaving variable
![Page 26: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/26.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Enter: x1 Leave: x5
Z -15 x1 -10x2 = 0 [0-0]
2 x1 + x2 + x3 = 1500 [0-1]
x1 + x2 + x4 = 1200 [0-2]
x1 + x5 = 500 [0-3]
ROW OPERATIONS so that
Each row has exactly one basic variable
The coefficient of each basic variable = +1
- [0-3]
- 2x [0-3]
+ 15x [0-3]
![Page 27: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/27.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Z -15 x1 -10x2 = 0 [0-0]
2 x1 + x2 + x3 = 1500 [0-1]
x1 + x2 + x4 = 1200 [0-2]
x1 + x5 = 500 [0-3]
Z -10x2 +15 x5 = 7500 [1-0]
x2 + x3 - 2 x5 = 500 [1-1]
x2 + x4 - x5 = 700 [1-2]
x1 + x5 = 500 [1-3]
Row operations
Basic Feasible Solution: ( 500, 0, 500, 700, 0)
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The Algebra of Simplex: Step 2. Iteration Step
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
Basic Feasible Solution: ( 500, 0, 500, 700, 0)
Objective value: 7500
![Page 29: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/29.jpg)
The Algebra of Simplex: Step 3. Stopping condition
Z -10x2 +15 x5 = 7500 [1-0]
x2 + x3 - 2 x5 = 500 [1-1]
x2 + x4 - x5 = 700 [1-2]
x1 + x5 = 500 [1-3]
Stopping Rule: Stop when objective cannot be improved.
New entering variable: x2
![Page 30: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/30.jpg)
The Algebra of Simplex: Second Iteration..
Z -10x2 +15 x5 = 7500 [1-0]
x2 + x3 - 2x5 = 500 [1-1]
x2 + x4 - x5 = 700 [1-2]
x1 + x5 = 500 [1-3]
New entering variable: x2
Basic variable
Equation Upper bound for x2
x3 x3 = 500 – x2 + 2x5 x2 ≤ 500 minimum x4 x4 = 700 – x2 + x5 x2 ≤ 700 x1 x1 = 500 – x5 no limit on x2
Analysis for leaving variable:
![Page 31: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/31.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Enter: x2 Leave: x3
ROW OPERATIONS so that
Each row has exactly one basic variable
The coefficient of each basic variable = +1
- [1-1]
+ 10x [1-1] Z -10x2 +15 x5 = 7500 [1-0]
x2 + x3 - 2x5 = 500 [1-1]
x2 + x4 - x5 = 700 [1-2]
x1 + x5 = 500 [1-3]
![Page 32: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/32.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Basic Feasible Solution:
After row operations:
Z +10 x3 -5 x5 = 12,500 [2-0]
x2 + x3 - 2x5 = 500 [2-1]
- x3 + x4 + x5 = 200 [2-2]
x1 + x5 = 500 [2-3]
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
( 500, 500, 0, 200, 0)
Have we found the OPTIMUM ?
![Page 33: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/33.jpg)
The Algebra of Simplex: Third Iteration
New entering variable: x5
Analysis for leaving variable:
Z +10 x3 -5 x5 = 12,500 [2-0]
x2 + x3 - 2x5 = 500 [2-1]
- x3 + x4 + x5 = 200 [2-2]
x1 + x5 = 500 [2-3]
Basic variable
Equation Upper bound for x5
x2 x2 = 500 + 2x5 - x3 no limit on x5
x4 x4 = 200 + x3 - x5 x5 ≤ 200 minimum
x1 x1 = 500 – x5 x5 ≤ 500
![Page 34: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/34.jpg)
The Algebra of Simplex: Step 2. Third iteration..
ROW OPERATIONS so that
Each row has exactly one basic variable
The coefficient of each basic variable = +1
- [2-2]
+ 5x [2-2]
Enter: x5 Leave: x4
Z +10 x3 -5 x5 = 12,500 [2-0]
x2 + x3 - 2x5 = 500 [2-1]
- x3 + x4 + x5 = 200 [2-2]
x1 + x5 = 500 [2-3]
+ 2x [2-2]
![Page 35: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/35.jpg)
The Algebra of Simplex: Step 2. Iteration Step
Basic Feasible Solution:
x1 = 500
2x1+x2 = 1500
x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
x1 = 500x1 = 500
2x1+x2 = 15002x1+x2 = 1500
x1 + x2 = 1200x1 + x2 = 1200
(0,0)
1000
1500
500 1000 1500
500
x2
x1
( 300, 900, 0, 0, 200)
Have we found the OPTIMUM ?
After row operations:
Z +5 x3 + 5x4 = 13,500 [3-0]
x2 - x3 + 2x4 = 900 [3-1]
- x3 + x4 + x5 = 200 [3-2]
x1 + x3 - x4 = 300 [3-3]
![Page 36: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/36.jpg)
The Algebra of Simplex: Some important points
1. Higher dimensions
2. Many candidates for entering variable
Each step:ONE entering variableONE leaving variable
Each constraint equation hashas same format as our example
e.g. Z = 200 + 10x1 + x2 + 10x3
Just pick any one
![Page 37: Linear Programming Fundamentals Convexity Definition: Line segment joining any 2 pts lies inside shape convex NOT convex.](https://reader036.fdocuments.us/reader036/viewer/2022062407/56649cf95503460f949c9f12/html5/thumbnails/37.jpg)
The Algebra of Simplex: Some important points..
3. Many candidates for leaving variable
4. No candidate for leaving variable
5. Minimization problems
Degenerate case, rare.
Unbounded objective !?
Minimize Z == Maximize -Z
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