Quantitative Analysis
for Management
LP:
Graphical solution Method
Simplex Method
Mohammad T. Isaai
Quantitative Analysis for Management
In symbolic form, the linear programming model is:
Maximize Z c1x1 cnxn Objective Function
subject to
a11x1 a1nxn b1
a21x1 a2nxn b2
am1x1 amnxn bm
Functional Constraints
and
x1 0,, xn 0 Nonnegativity Constraints
for known parameters c1, … , cn ; a11, … , amn ; b1, … , bm.
Mathematical Statement of Linear Programming
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Basic Assumptions of Linear Programming
• Certainty
• Proportionality (1-3, 10-30)
• Additivity (8,3,11)
• Divisibility (10.2)
• Nonnegativity
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Max Z = 300 x1 +500 x2
Subject to
x1 4 Resource 1
2x2 12 Resource 2
3 x1 + 2 x2 18 Resource 3
x1 0, x2 0
Graphical Solution
Solve the following problem. Find the optimal solution.
Product 1 needs 1 unit of resource 1, and 3 units of resource 3.
Product 2 needs 2 units of resource 2 and 2 units of resource 3
There are 4 units of resource 1, 12 units of resource 2, and 18 units of
resource 3
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1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x2
x1
Max
Z = 300 x1 +500 x2
Subject to
x1 4
2x2 12
3 x1 + 2 x2 18
x1 0, x2 0
Solution - Constraints
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1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x2
x1
Max
Z = 300 x1 +500 x2
Subject to
x1 4
2x2 12
3 x1 + 2 x2 18
x1 0, x2 0
Solution – Objective Function
Isoprofit LineOr
Isocost Line
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Terms Used in LP
• Solution
• Feasible Solution (Infeasible Solution)
• Feasible Region
• Optimal Solution
• Boundary Equation (Constraint Line)
• Corner Point Solution
• Adjacent Corner Point Solutions
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Property of an Optimal Solution
• If there is only one optimal solution, then it lies on a
corner point solution.
• If there are more than one optimal solution, then at
least two adjacent corner point solutions are optimal
as well as every point on the line connecting them.
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The Concept of Simplex Method
• Initial Step: Start from a feasible corner point solution
• Iterative Step: Move from the existing corner point
solution to a better, feasible adjacent one
• Stopping Rule: If the existing corner point solution is
better than all of its adjacent feasible corner point
solutions, it is optimal
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Simplex Algorithm Movement
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Standard L.P. Model
• Simplex alg. Must be applied to the standard form in which:
– Objective Function in Maximization Form
– Constraints are less than or equal .
– Decision Variables are nonnegative
In different books standards are defined differently
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Standard L.P. Model
Maximize Z c1 x1 cn xn Objective Function
subject to
a11 x1 a1n xn =b1
a21x1 a2n xn =b2
M
am1 x1 amnxn =bm
Functional Constraints
and
x1 0,, xn 0 Nonnegativity Constraints
How to generate the standard form?
• Slack variables
• Surplus variables
• Artificial variables
• Min Z = 300 x1 +500 x2
Subject to
• x1 + 3x2 4
• - x1 + x2 = 12
• 3 x1 + 2 x2 18
• x1 0, x2 Unrestricted in
sign
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Sensitivity Analysispost optimality analysis
• The impact of changing parameters
• “What-if” questions.
• Sensitive Parameters.
• In the real world, real data is not certain and we use our best estimate. However, they may change.
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Sensitivity Analysis
• Changes in Resources (RHS)
• Changes in the Objective Function Coefficients
• Changes in Technological Coefficients
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Sensitivity Example
• How much are you willing to pay for each
additional Resource Unit?
• How many additional units do you buy?
assume that we can get 1 extra unit of res. 2.
Let’s start with 1 additional unit of resource 2.
Then 2, 3 and more additional units.
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1
2
3
4
5
6
7
8
9
10
x2
x1
2x2 = 12
3 x1 + 2 x2 = 18
x1 = 4
Z = 300 (2) +500(6)
Z = 3600
Main Problem
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1
2
3
4
5
6
7
8
9
10
x2
x1
2x2 = 12 + 1
3 x1 + 2 x2 = 18
x1 =4
x2 = 6.5
x1 = 5/3
Z = 300 (5/3)+500(6.5)
Z = 3750
3750-3600 = 150
One unit increase in b2
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1
2
3
4
5
6
7
8
9
10
x2
x1
2x2 = 12 + 2
3 x1 + 2 x2 = 18
x1 =4
x2 = 7
x1 = 4/3
Z = 300 (4/3) +500(7)
Z = 3900
3900-3750 = 150
Two units increase in b2
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1
2
3
4
5
6
7
8
9
10
x2
x1
2x2 = 12 + a
3 x1 + 2 x2 = 18
x1 =4
x2 = 6+a/2
x1 = 2-a/3
Z = 300 (2-a/3)
+500(6+a/2)
Z = 600-100a
+3000+250a
Z = 3600 +150a
a unit increase in b2
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1
2
3
4
5
6
7
8
9
10
x2
x1
3 x1 + 2 x2 = 18
x1 = 0
x2 = 9
x1 = 0
Z = 300 (0)+500(9)
Z = 4500
we buy at most 6
additional units
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Shadow Price
• Since the resources are limited, the profit is also limited.
• If we increase the limited resources the profit also
increases.
• How much are we ready to pay for one extra unit of each
resource?
• Shadow price for each resource is the maximum amount
one is willing to pay for one additional unit of that
resource.
• Clearly, Shadow price is different from the market price
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Shadow Price-2
• If a resource is not limited, then its Shadow price iszero.
• It means, if the value of the slack variable for a constraint is positive, then its Shadow price is zero.
• In the simplex method for standard case, shadow prices are shown on the objective function row and under the slack variables.
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Given the following model
Min Z = 40 x1 + 50 x2
Subject to
(C1) 2x1 + 3x2 30
(C2) x1 + x2 12
(C3) 2 x1 + x2 20
x1 0, x2 0
Problem
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Min
Z = 40 x1 +50 x2
Subject to
2x1 + 3x2 30
x1+ x2 12
2 x1 + x2 20
x1 0, x2 0
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x2
x1
feasible region
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1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x2
x1
Min
Z = 40 x1 +50 x2
Subject to
2x1 + 3x2 30
x1+ x2 12
2 x1 + x2 20
x1 0, x2 0
2x1 + 3x2 = 30
2 x1 + x2 = 20
x2 = 5
x1 = 7.5
Z = 40 (7.5)+50
(5)
Z = 550
the Optimal Solution
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1
2
3
4
5
6
7
8
9
10
x2
x1
Min
Z = 40 x1 +70 x2
Subject to
2x1 + 3x2 30
x1+ x2 12
2 x1 + x2 20
x1 0, x2 0
2x1 + 3x2 = 30
x2 = 0
x1 = 15
Z = 40 (15)+70 (0)
Z = 600, changed from 550 to 600 (O.F. is Min)
What if the objective function is changed
from 40 x1 +50 x2 to 40 x1 +70 x2
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Min
Z = 40 x1 +50 x2
Subject to
2x1 + 3x2 30
x1+ x2 12
2 x1 + x2 20
x1 0, x2 0
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1
2
3
4
5
6
7
8
9
10
x2
x1
again the original problem
What if the green
constraint is
changed to
2 x1 + x2 15
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Min Z= 40 x1+50 x2
Subject to
2x1 + 3x2 30
x1+ x2 12
2 x1 + x2 15
x1 0, x2 0
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x2
x1
2x1 + 3x2 = 30
x1+ x2 = 12
x2 = 6
x1 = 6
Z = 540 from 550 to 540 (O.F. is Min)
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Special Cases in LP
• Infeasibility
• Redundancy
• More Than One Optimal Solution
• Unbounded Solutions
• Degeneracy
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A Problem with No Feasible Solution
X2
X1
8
6
4
2
02 4 6 8
Region Satisfying
the 3rd Constraint
Region Satisfying First 2 Constraints
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A Problem with a Redundant Constraint
X2
X1
30
25
20
15
10
5
05 10 15 20 25 30
Feasible
Region
2X1 + X2 < 30
X1 < 25
X1 + X2 < 20
Redundant
Constraint
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An Example of Alternate Optimal Solutions
Optimal Solution Consists of All
Combinations of X1 and X2 Along the AB
Segment
Isoprofit Line for $12
Overlays Line Segment
Isoprofit Line for $8A
B
AB
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A Solution Region That is Unbounded to the Right
X2
X1
15
10
5
0
5 10 15
Feasible Region
X1 > 5X2 < 10
X1 + 2X2 > 10
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• Having selected the pivot column, one divides each
quantity column no. (RHS) to the corresponding
pivot column no., if all ratios are negative or
undefined, it indicates that the problem is
unbounded.
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Degeneracy
• Having selected the pivot column, one divides each
quantity column no. (RHS) to the corresponding
pivot column no., if there is a tie for the smallest
ratio, this is a signal that degeneracy exists.
• Cycling may result from degeneracy.
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