4' Sensitivity Analysis: The Graphical Method

1
4. Sensitivity Analysis The Graphical Method
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What is Sensitivity Analysis? Examine the effect
on the optimal solution for changes in
Parameters of the objective function
Parameters in the constraints
Right-hand-side values of the constraints
Addition/Deletion of constraints
3
What if questions!!!
Sensitivity analysis attempts to evaluate the
sensitivity of the optimal solution to changes in
the model parameters. Sensitivity analysis is
important to the manager who must operate in a
dynamic environment with imprecise estimates of
the parameters. Sensitivity analysis allows
him/her to ask certain what-if questions about
the problem.
4
Lets cover the class example that will be used!
What is Sensitivity Analysis? Examine the effect
on the optimal solution for changes
in Parameters of the objective function
Parameters in the constraints
Right-hand-side values of the constraints
Addition/Deletion of constraints
5

• Springfield Inc. has two products, A and B. The
  profit of A is 5 dollars per ton and the profit
  of B is 7 dollars per ton. The manager wants to
  maximize total profit.
• The maximum demand of A is 6 tons. Producing one
  unit of A consumes two units of material P.
  Producing one unit of B consumes three units of
  P. Springfield Inc. has nineteen tons of P.
  Producing one unit of A consumes one unit of
  material Q. Producing one unit of B consumes one
  unit of Q. Springfield Inc. has eight tons of Q.
• The manager needs to determine how many tons of A
  and B should be produced. Let x1 denotes the
  quantity of A produced and let x2 denotes the
  quantity of B produced. Please provide the
  model and solve.

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MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max Demand of
A 2x1 3x2 lt 19 Available P x1 x2 lt 8
Available Q x1,x2 gt0 Non-negativity
7
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2
2x1 3x2 lt 19
1
x1
1
2
3
4
5
6
7
8
9
10
8
Lets start with these changes KEEPING ALL ELSE
THE SAME!
What is Sensitivity Analysis? Examine the effect
on the optimal solution for changes in
Parameters of the objective function
Parameters in the constraints
Right-hand-side values of the constraints
Addition/Deletion of constraints
9
MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max Demand of
A 2x1 3x2 lt 19 Available P x1 x2 lt 8
Available Q x1,x2 gt0 Non-negativity
10
How do changes in the objective function
parameters affect the optimal solution? Range
of Optimality Simultaneous Changes
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• Feasible region WILL remain the same
• Slope of objective function IS changed
• Objective Function value WILL change
• Optimal Point MIGHT change

How do changes in the objective function
parameters affect the optimal solution?
12

• Feasible region WILL remain the same
• Slope of objective function IS changed
• Objective Function value WILL change
• Optimal Point MIGHT change

Feasible Region stays the same because there is
no change to the constraints!
13
x2
10
Max 5x17x2
9
x1 lt 6
What happens to the graph if the obj. function
changes to Max 6x17x2
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region stays the same because there is
no change to the constraints!
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
14
Max c1x1 c2x2 To graph set some arbitrary
c1x1 c2x2 A Standard Form of a line x2
-(c1/c2)x1 A/c2 Slope -c1/c2 If c1or c2
changes, so will the slope of the objective
function line!

• Feasible region WILL remain the same
• Slope of objective function IS changed
• Objective Function value WILL change
• Optimal Point MIGHT change

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x2
10
Max 5x17x2
9
x1 lt 6
What happens to the graph if the obj. function
changes to Max 6x17x2
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
16

• Feasible region WILL remain the same
• Slope of objective function IS changed
• Objective Function value WILL change
• Optimal Point MIGHT change

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x2
10
Max 5x17x2
9
x1 lt 6
What happens to the graph if the obj. function
changes to Max 6x17x2 6(5)7(3)51 Compare to
46!
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
18
What does MIGHT mean? That ishow much can you
change the parameters of the Objective Function
and have the same optimal point? This is called
the RANGE OF OPTIMALITY.

• Feasible region WILL remain the same
• Slope of objective function IS changed
• Objective Function value WILL change
• Optimal Point MIGHT change

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How do changes in the objective function
parameters affect the optimal solution? Range
of Optimality Simultaneous Changes
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The RANGE OF OPTIMALITY for each PARAMETER
provides the range of values over which the
current solution will remain optimal.
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The limits of the slope of the objective
function are the slopes of the binding
constraints. Thus, graphically, the limits of a
range of optimality are found by changing the
slope of the objective function line within the
limits of the slopes of the binding constraint
lines.
Binding constraints are the constraints that form
the optimal point!
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x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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x2
If the slope of the objective function is
changed, the optimal point will change when the
slope of the objective function line is parallel
to a binding constraint
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Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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This means that we are determining the range for
EACH parameter holding the other parameters
CONSTANT!

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiply both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

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Easy!...Just follow the steps! Lets start with
step 1!

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiply both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

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x2
What are the binding constraints?
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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x2
What are the binding constraints? 2x13x2 lt 19
x1x2 lt 8
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiply both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

Lets go over step 2 (and cover the Math Review!
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x2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes?
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Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
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7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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x2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
31

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiply both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

Lets go over step 3!
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x2
Given the objective function at the optimal
value, 5x17x246
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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x2
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246
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Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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x2
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246 Find slope
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
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x2
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246 Find
slope c1x17x246 7x2-c1x146 x2-c1x16.57
7 Slope-c1/7
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
36

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiply both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

Lets go over step 4!
37
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246 Find
slope c1x17x246 7x2-c1x146 x2-c1x16.57
7 Slope-c1/7
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
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Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246 Find
slope c1x17x246 7x2-c1x146 x2-c1x16.57
7 Slope-c1/7
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -c1/7 lt -2/3
39

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiple both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

Lets go over step 5!
40
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246 Find
slope c1x17x246 7x2-c1x146 x2-c1x16.57
7 Slope-c1/7
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -c1/7 lt -2/3
-1 lt -c1/7 lt -2/3 -7 -7 -7 7 gt c1 gt
4.67 -or- 4.67 lt c1 lt 7
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Holding c2 constant, c1 can change between 14/3
and 7 without affecting the optimal solution
(5,3). What happens to the value of the
objective function? It WILL change!
42
Holding c2 constant, c1 can change between 14/3
and 7 without affecting the optimal solution
(5,3). Q1 What is the optimal solution and
objective function value when c1 changes to
6?
42
43
Holding c2 constant, c1 can change between 14/3
and 7 without affecting the optimal solution
(5,3). Q1 What is the optimal solution and
objective function value when c1 changes to 6?
The optimal solution is the same (5,3) The new
objective function value is 6(5)7(3)51
43
44
Holding c2 constant, c1 can change between 14/3
and 7 without affecting the optimal solution
(5,3). Q1 What is the optimal solution and
objective function value when c1 changes to 6?
The optimal solution is the same (5,3) The new
objective function value is 6(5)7(3)51 Q2
What is the optimal solution and objective
function value when c1 changes to 4?
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45
Holding c2 constant, c1 can change between 14/3
and 7 without affecting the optimal solution
(5,3). Q1 What is the optimal solution and
objective function value when c1 changes to 6?
The optimal solution is the same (5,3) The new
objective function value is 6(5)7(3)51 Q2
What is the optimal solution and objective
function value when c1 changes to 4? The
optimal solution and objective function value
will change. Since c1 changes outside the range
of optimality, we do not know the new changes
without resolving the linear program.
45
46
More fun times!
Determine the range of optimality for c1 ,
Holding c2 constant
47
x2
Max 100x180x2
160
140
2x1 x2 lt 160
Optimal solution is x160 x240 Z9200
120
100
80
A
60
E
x1 x2 lt 100
Feasible Region
40
D
x1 2x2 lt 160
20
x1
C
B
20
40
60
80
100
120
140
160
48
(No Transcript)
49
Holding c2 constant, c1 can change between 80
and 160 without affecting the optimal solution
(60,40). Q1 What is the optimal solution and
objective function value when c1 changes to
75? Q2 What is the optimal solution and
objective function value when c1 changes to
150?
49
50
Holding c2 constant, c1 can change between 80
and 160 without affecting the optimal solution
(60,40). Q1 What is the optimal solution and
objective function value when c1 changes to 75?
The optimal solution and objective function
value will change. Since c1 changes outside the
range of optimality, we do not know the new
changes without re-solving the linear
program. Q2 What is the optimal solution and
objective function value when c1 changes to
150? The optimal solution is the same
(60,40) The new objective function value is
150(60)80(40)12,200
50
51
Now, lets determine the range of optimality for
c2, holding c1 constant!

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiply both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

52
What are the binding constraints?

53
What are the binding constraints? 2x13x2lt19 x1x2
lt8

54
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes?

55
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1

56
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1

57
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1

58
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find slope
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1

59
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find
slope 5x1c2x246 c2x2-5x146 x2-5x16.57
c2 Slope-5/c2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1

60
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find
slope 5x1c2x246 c2x2-5x146 x2-5x16.57
c2 Slope-5/c2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1

61
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find
slope 5x1c2x246 c2x2-5x146 x2-5x16.57
c2 Slope-5/c2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -5/c2 lt -2/3
62
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find
slope 5x1c2x246 c2x2-5x146 x2-5x16.57
c2 Slope-5/c2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -5/c2 lt -2/3 -1 gt -c2/5 gt -3/2
63
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find
slope 5x1c2x246 c2x2-5x146 x2-5x16.57
c2 Slope-5/c2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -5/c2 lt -2/3 -1 gt -c2/5 gt -3/2 -1 gt -c2/5
gt -3/2 -5 -5 -5
64
Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 5x1c2x246 Find
slope 5x1c2x246 c2x2-5x146 x2-5x16.57
c2 Slope-5/c2
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -5/c2 lt -2/3 -1 gt -c2/5 gt -3/2 -1 gt -c2/5
gt -3/2 -5 -5 -5 5 lt c2 lt 7.5
65
Holding c1 constant, c2 can change between 5 and
7.5 without affecting the optimal solution (5,3).
What happens to the value of the objective
function? It WILL change!
66
More fun times!
Determine the range of optimality for c2 ,
Holding c1 constant
67
Optimal solution is x160, x240, Z9200
68
How do changes in the objective function
parameters affect the optimal solution? Range
of Optimality Simultaneous Changes
69
How do changes in the objective function
parameters affect the optimal solution? Range
of Optimality Simultaneous Changes
70
100 Rule For all coefficients that are changed,
sum the percentages of allowable increases and
the allowable decreases represented by the
changes. If the sum of the percentage changes
does not exceed 100, the optimal solution will
not change.
Be aware of this rule, but you will not have to
know it for a test!
71
Lets make these changes KEEPING ALL ELSE THE
SAME!
What is Sensitivity Analysis? Examine the effect
on the optimal solution for changes in
Parameters of the objective function
Parameters in the constraints
Right-hand-side values of the constraints
Addition/Deletion of constraints
72
MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max
Demand of A 2x1 3x2 lt 19 Available P x1 x2
lt 8 Available Q x1,x2 gt0 Non-negativity
We will be examining changes to the PARAMETERS
(coefficients) of the CONSTRAINTS
73

• If not a redundant constraint, will change the
  feasible region.
• MAY change the optimum solution.
• Will effect the optimal value (from objective
  function) if the optimal point is changed.
• RESOLVE the new LP problem using the graphical
  method (you can answer most questions by
  RESOLVING!)

How do changes in the PARAMETERS (coefficients)
effect the optimal solution?
74
Lets make these changes KEEPING ALL ELSE THE
SAME!
What is Sensitivity Analysis? Examine the effect
on the optimal solution for changes in
Parameters of the objective function
Parameters in the constraints
Right-hand-side values of the constraints
Addition/Deletion of constraints
75
MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max
Demand of A 2x1 3x2 lt 19 Available P x1 x2
lt 8 Available Q x1,x2 gt0 Non-negativity
We will be examining changes to the RIGHT HAND
SIDE values of the CONSTRAINTS
76
How do changes in the RHS values of the
constraints affect the optimal solution? How to
Calculate the Shadow Price Range of Feasibility
of Shadow Price
77
The improvement in the value of the optimal
solution per unit increase in the right-hand side
is called the SHADOW PRICE. What does shadow
price mean? The shadow price of constraint i
(resource i) represents the price we should be
willing to pay to obtain one more unit of
resource i.
How do changes in the RHS values of the
constraints affect the optimal solution?
78
Changing a RHS value results in a parallel shift
of the changed constraint. This may affect both
the optimal solution and the optimal value of the
objective function.
How do changes in the RHS values of the
constraints affect the optimal solution?
79
The Right-hand-side (RHS) value usually
represents the available resource. A change in
RHS will result in changes in the intercepts
of the constraint function. Consequently, it may
result in changes in the area of feasible
solutions.
How do changes in the RHS values of the
constraints affect the optimal solution?
80
How do changes in the RHS values of the
constraints affect the optimal solution? How to
Calculate the Shadow Price Range of Feasibility
of Shadow Price
80
81

• The shadow price for a non-binding constraint is
  0
• Graphically, the shadow price of the RHS of a
  binding constraint is determined by
• Add 1 to the right hand side value of the
  constraint
• 2. Resolve for the optimal solution in terms of
  the same two binding constraints.
• 3. Shadow price new value of the objective
  function original value of the objective
  function
• Max problem
• shadow price znew zold
• Min problem
• shadow price zold znew

How do we compute the SHADOW PRICE ?
82
x2
Constraint 1 x1 lt 6
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
83

• The shadow price for a non-binding constraint is
  0. Why?
• Graphically, the shadow price of the RHS of a
  binding constraint is determined by
• Add 1 to the right hand side value of the
  constraint
• 2. Resolve for the optimal solution in terms of
  the same two binding constraints.
• 3. Shadow price new value of the objective
  function original value of the objective
  function
• shadow price znew zold

This is a non-binding constraint!
83
84
x2
Constraint 1 x1 lt 6
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
84
1
2
3
4
5
6
7
8
9
10
85
Constraint 1 Since x1 lt 6 is not a binding
constraint, its shadow price is 0. Springfield
Inc. does not want to pay more money to
increase one more unit of the demand of A (since
it is not even using what it has)
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
85
1
2
3
4
5
6
7
8
9
10
86
Constraint 2 2x13x2lt19
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
86
1
2
3
4
5
6
7
8
9
10
87

• The shadow price for a non-binding constraint is
  0
• Graphically, the shadow price of the RHS of a
  binding constraint is determined by
• Add 1 to the right hand side value of the
  constraint
• 2. Resolve for the optimal solution in terms of
  the same two binding constraints.
• 3. Shadow price new value of the objective
  function original value of the objective
  function
• shadow price znew zold

This is a binding constraint!
87
88
Constraint 2 (a binding constraint) Increase the
RHS value of the second constraint from 19 to 20
(by 1 unit) and resolve for the optimal point
determined by the two binding constraints 2x1
3x2 20 and x1 x2 8.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
89
Constraint 2 (a binding constraint) Increase the
RHS value of the second constraint from 19 to 20
(by 1 unit) and resolve for the optimal point
determined by the two binding constraints 2x1
3x2 20 and x1 x2 8. The solution is x1
4, x2 4, z 48.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
89
1
2
3
4
5
6
7
8
9
10
90
Constraint 2 (a binding constraint) Increase the
RHS value of the second constraint from 19 to 20
(by 1 unit) and resolve for the optimal point
determined by the two binding constraints 2x1
3x2 20 and x1 x2 8. The solution is x1
4, x2 4, z 48. Hence, the shadow price
znew-zold48-462 This means that 1 more unit of
material P will increase the total profit by 2.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
90
1
2
3
4
5
6
7
8
9
10
91
Constraint 2 Q If a supplier can provide 1 ton
of P at price 1.60, should Springfield Inc. buy
it?
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
92
Constraint 2 Q If a supplier can provide 1 ton
of P at price 1.60, should Springfield Inc. buy
it? A Yes, because Springfield Inc. can make 2
more profit if it has one more ton of material
P. Springfield Inc. will be willing to pay at
the most 2 for one ton of material P.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
93
Constraint 3 x1 x2 lt 8
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2lt19
2
1
x1
93
1
2
3
4
5
6
7
8
9
10
94
Constraint 3 Constraint 3 (a binding
constraint) Change the RHS value of the third
constraint from 8 to 9 and resolve for the
optimal point determined by the two
binding constraints 2x1 3x2 19 and x1 x2
9.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x13x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
95
Constraint 3 Constraint 3 (a binding
constraint) Change the RHS value of the third
constraint from 8 to 9 and resolve for the
optimal point determined by the two
binding constraints 2x1 3x2 19 and x1 x2
9. The solution is x1 8, x2 1, z 47.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x13x2 lt 19
2
1
x1
95
1
2
3
4
5
6
7
8
9
10
96
Constraint 3 Constraint 3 (a binding
constraint) Change the RHS value of the third
constraint from 8 to 9 and resolve for the
optimal point determined by the two
binding constraints 2x1 3x2 19 and x1 x2
9. The solution is x1 8, x2 1, z 47. The
shadow price is znew - zold 47 - 46 1. This
means that 1 more unit of material Q will
increase the total profit by 1.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x13x2 lt 19
2
1
x1
96
1
2
3
4
5
6
7
8
9
10
97
Constraint 3 Q If a supplier can provide 1 ton
of Q at price 1.60, should Springfield Inc. buy
it?
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
98
Constraint 3 Q If a supplier can provide 1 ton
of Q at price 1.60, should Springfield Inc. buy
it? A No, because Springfield Inc. can only
make 1 more profit if it has one more ton of
material Q. This cannot cover the additional
purchasing cost of 1.60. Springfield Inc. will
be willing to pay at the most 1 for one ton of
material Q.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
1
2
3
4
5
6
7
8
9
10
99
More fun times!
Determine the SHADOW PRICE for each of the
constraints.
100
x2
Max 100x180x2
160
140
2x1 x2 lt 160
Optimal solution is x160 x240 Z9200
120
100
80
A
60
E
x1 x2 lt 100
Feasible Region
40
D
x1 2x2 lt 160
20
x1
C
B
20
40
60
80
100
120
140
160
101
Question What is the shadow price of constraint
1?
Constraint 1 Change the RHS value of the first
constraint to 101 and resolve for the optimal
point determined by the binding constraints x1
x2 101 and 2x1 x2 160. The solution is
x1 59, x2 42, znew 9260. The shadow
price is znew - zold 9260 - 9200 60.
102
Question What is the shadow price of constraint
2?
Constraint 2 Change the RHS value of the first
constraint to 161 and resolve for the optimal
point determined by the binding constraints x1
x2 100 and 2x1 x2 161. The solution is
x1 61, x2 39, znew 9220 The shadow
price is znew - zold 9220 - 9200 20.
103
Question What is the shadow price of constraint
3?
Constraint 3 This is a non-binding constraint,
thus the shadow price is 0. Hence, we would
not pay more to increase the RHS of this
constraint because we are already not using all
of the constraint.
104
Remember that a positive shadow price means
an IMPROVEMENT which in a maximization problem
increases the obj. function value and decreases
the obj. function value in a minimization
problem!
105
If the shift in the constraint makes the feasible
region larger, the shadow price is positive or
zero. If the shift in the constraint makes the
feasible region smaller, the shadow price is
negative or zero.
106
This is a MAXIMIZATION problem This
means an IMPROVEMENT in the objective function
value Shadow price is POSITIVE
Feasible Region
Feasible Region
107
This is a MAXIMIZATION problem This
means a WORSE objective function value Shadow
price is NEGATIVE
Feasible Region
Feasible Region
108
This is a MINIMIZATION problem This
means a IMPROVEMENT objective function
value Shadow price is POSITIVE
Feasible Region
Feasible Region
109
This is a MINIMIZATION problem This
means a WORSE objective function value Shadow
price is NEGATIVE
Feasible Region
Feasible Region
110
In general the dual price and the shadow price
are the same for all MAX linear programs. In
MIN linear programs, the dual price is the
negative of the corresponding shadow price.
111
How do changes in the RHS values of the
constraints affect the optimal solution? How to
Calculate the Shadow Price Range of Feasibility
of Shadow Price
112
The range of values over which the shadow price
is applicable. The optimal solution is formed
by the same constraints. As the RHS increases,
other constraints will become binding and limit
the change in the value of the objective
function. Will be determined by computer
output, not graphically.
113
Constraint 2 (a binding constraint) Increase the
RHS value of the second constraint from 19 to 20
(by 1 unit) and resolve for the optimal point
determined by the two binding constraints 2x1
3x2 20 and x1 x2 8. The solution is x1
4, x2 4, z 48. Hence, the shadow price
znew-zold48-462 This means that 1 more unit of
material P will increase the total profit by 2.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3)
7
6
5
4
Feasible Region
3
2x1 3x2 lt 19
2
1
x1
113
1
2
3
4
5
6
7
8
9
10
114
x2
10
What happens if we increase the RHS of constraint
2 by 1 unit?
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x13x2lt19
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
115
x2
10
If we increase the RHS of constraint 2 by 1 unit,
it improves the optimal value by 2 (from 46 to
48) hence shadow price 2
Max 5x17x2
9
x1lt6
8
x1x2lt8
Optimal point (4,4) Optimal obj. function value
48
7
E
6
5
4
D
3
Feasible Region
2x13x2lt21
2
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
116
x2
10
If we increase the RHS of constraint 2 by 5
units, it improves the optimal value by 10 (from
46 to 56) hence shadow price 2
Max 5x17x2
9
x1lt6
8
D/E
x1x2lt8
Optimal point (0,8) Optimal obj. function value
56
7
6
5
4
Feasible Region
2x13x2lt25
3
2
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
117
x2
If we increase the RHS of constraint 2 by more
than 5 units it does not improve the objective
any more, hence shadow price is no longer
applicable
10
Max 5x17x2
D
9
x1lt6
8
E
x1x2lt8
Optimal point (0,8) Optimal obj. function value
56
7
6
5
4
Feasible Region
3
2x13x2lt30
2
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
118
x2
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x13x2lt19
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
119
x2
If we decrease the RHS of constraint 2 by 1 unit
it makes the objective function value worse by 2
units Hence shadow price 2
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
7
Optimal point (6,2) Optimal obj. function value
44
6
E
5
4
3
Feasible Region
2
2x13x2lt18
C/D
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
120
x2
If we decrease the RHS of constraint 2 by 3
units, it makes the objective function value
worse by (46-39.33) 6.67 According to the
shadow price, a 3 units decrease should have
resulted in 236 dollars change in the obj.
function value Hence shadow price is not
applicable because now the optimal solution is
formed by a set of new constraints
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
7
Optimal point (6,1.33) Optimal obj. function
value 39.33
6
E
5
4
3
Feasible Region
2
2x13x2lt16
C
D
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
121
Lets make these changes KEEPING ALL ELSE THE
SAME!
What is Sensitivity Analysis? Examine the effect
on the optimal solution for changes
in Parameters of the objective function
Parameters in the constraints
Right-hand-side values of the constraints
Addition/Deletion of Constraints
122

• Some Fact
• Deleting constraints leaves the feasible region
  either
• unchanged or enlarged
• Adding constraints leaves the feasible region
  either
• unchanged or smaller
• Adding constraints to a model will either
  impair the optimal
• value of the objective function or leave it
  unchanged
• Deleting constraints either improves the value
  of the
• objective function or leaves it unchanged
• A redundant constraint is one whose removal
  does not
• change the feasible

RESOLVE linear program!
123
End of Lecture 4
124

• Read 3.1-3.2
• Complete Homework from Book
• Chapter 3 5 (a-d only)
• Chapter 3 13 all
• Chapter 3 16 all

125

• You already developd the model, Solved using the
  Graphical Method, determined the feasible region,
  corner points, optimal solution and optimal
  objective function value. Also you graphed the
  objective function. Now, apply sensitivity
  analysis and determine shadow prices.
• HighTec Inc. assembles two different models of
  personal computers, Deskpro and Portable. The
  Deskpro generates a profit contribution of
  50/unit and the Portable generates a profit
  contribution of 40/unit. HighTec Inc. is
  currently interested in developing a weekly
  production schedule for both products. For next
  weeks production, a maximum of 150 hours of
  assembly time can be made available. Each unit
  of the Deskpro requires 3 hours of assembly time
  and each unit of the Portable requires 5 hours of
  assembly time. HighTec Inc. only has 20 Portable
  display components in inventory thus, no more
  than 20 units of the Portable may be assembled.
  Finally, only 300 square feet of warehouse space
  can be made available for new production.
  Assembly of each Deskpro requires 8 square feet
  of warehouse space similarly, each Portable
  requires 5 square feet.

126
MODEL Let d units Deskpro Let punits
Portable Max Profit 50d40p s.t. 3d5p lt 150
(Assembly Time) plt20 (Portable Display)
8d5p lt 300 (Warehouse
Capacity) d,pgt0 (Non-negativity) S
OLUTION Produce 30 Deskpro and 12 Portable for a
optimal value of 1980.
8d5p lt 300
p lt 20
A
E
3d5p lt 150
Feasible Region
D
B
C
d
126
127
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