b – 1 operations management module b – linear programming © 2006 prentice hall, inc. powerpoint...
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B – 1
Operations ManagementOperations ManagementModule B – Linear ProgrammingModule B – Linear Programming
© 2006 Prentice Hall, Inc.
PowerPoint presentation to accompanyPowerPoint presentation to accompany Heizer/Render Heizer/Render Principles of Operations Management, 6ePrinciples of Operations Management, 6eOperations Management, 8e Operations Management, 8e
B – 2
Lecture Outline
• Remind me what LP is
• What types of problems can we solve with LP?
• Formulating LP problems
• Example
• Sensitivity of the Answer
B – 3
Linear Programming
A model consisting of linear relationshipsrepresenting a firm’s objective and resource
constraints
LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called
constraints
B – 4
Common Elements to LP
• Decision variables– Should completely describe the decisions to be made
by the decision maker (DM)
• Objective Function (OF)– DM wants to maximize or minimize some function of
the decision variables
• Constraints– Restrictions on resources such as time, money,
labor, etc.
B – 5
LP Assumptions
• OF and constraints must be linear
• Proportionality– Contribution of each decision variable is
proportional to the value of the decision variable
• Additivity– Contribution of any variable is independent
of values of other decision variables
B – 6
LP Assumptions, cont’d.
• Divisibility– Allow both integer and non-integer (real)
numbers
• Certainty– All coefficients are known with certainty
– We are dealing with a deterministic world
B – 7
Types of Problems
• Module C– SCM: transportation models
• Chapter 13
• Aggregate Planning
• Module B– Product Mix
– Blending
– Scheduling (Production and Labor)
B – 8
Formulating LP Problems
The product-mix problem at Shader Electronics
Two products
1. Shader Walkman, a portable CD/DVD player
2. Shader Watch-TV, a wristwatch-size Internet-connected color TV
Determine the mix of products that will produce the maximum profit
B – 9
LP Model Formulation
• Data– Input to the model – given in the problem
• Decision variables– Mathematical symbols representing levels
of activity of an operation
– The quantities to be determined
B – 10
LP Model Formulation, cont’d.
• Objective function (OF)– The quantity to be optimized
– A linear relationship reflecting the objective of an operation
– Most frequent objective of business firms is to maximize profit
– Most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost
B – 11
LP Model Formulation, cont’d.
• Constraint– A linear relationship representing a
restriction on decision making
– Binding relationships
– Attach a word description to each set of constraints
– Include bounds on variables
B – 12
Formulating LP Problems
Walkman Watch-TVs Available HoursDepartment (X1) (X2) This Week
Hours Required to Produce 1 Unit
Electronic 44 33 240240
Assembly 22 11 100100
Profit per unit $7$7 $5$5
Decision Variables:XX11 = = number of Walkmans to be produced
XX22 = = number of Watch-TVs to be produced
Table B.1Table B.1
B – 13
Formulating LP Problems
Objective Function:
Maximize Profit = = $7$7XX11 + + $5$5XX22
There are three types of constraints Upper limits where the amount used is ≤ the
amount of a resource Lower limits where the amount used is ≥ the
amount of the resource Equalities where the amount used is = the
amount of the resource
B – 14
Formulating LP Problems
Second Constraint:
22XX11 + + 11XX22 ≤ 100 ≤ 100 (hours of assembly time)
Assemblytime available
Assemblytime used is ≤ ≤
First Constraint:
44XX11 + + 33XX22 ≤ 240 ≤ 240 (hours of electronic time)
Electronictime available
Electronictime used is ≤ ≤
B – 15
Formulating LP Problems
XX22 >> 0 0 (non-negativity)
NoneWatch-TVsProduced is >>
Third Constraint Set:
XX11 >> 0 0 (non-negativity)
NoneWalkmansProduced is >>
B – 16
Graphical Solution
Can be used when there are two decision variables
1. Plot the constraint equations at their limits by converting each equation to an equality
2. Identify the feasible solution space
3. Create an iso-profit line based on the objective function
4. Move this line outwards until the optimal point is identified
B – 17
Graphical Solution
100 –
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80 80 –
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60 60 –
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40 40 –
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20 20 –
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00 2020 4040 6060 8080 100100
Nu
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er o
f W
atch
-TV
sN
um
ber
of
Wat
ch-T
Vs
Number of WalkmansNumber of Walkmans
XX11
XX22
Assembly (constraint B)Assembly (constraint B)
Electronics (constraint A)Electronics (constraint A)Feasible region
Figure B.3Figure B.3
B – 18
Graphical Solution
100 –
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80 80 –
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60 60 –
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40 40 –
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00 2020 4040 6060 8080 100100
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atch
TV
sN
um
ber
of
Wat
ch T
Vs
Number of WalkmansNumber of Walkmans
XX11
XX22
Assembly (constraint B)Assembly (constraint B)
Electronics (constraint A)Electronics (constraint A)Feasible region
Figure B.3Figure B.3
Iso-Profit Line Solution Method
Choose a possible value for the objective function
$210 = 7X1 + 5X2
Solve for the axis intercepts of the function and plot the line
X2 = 42 X1 = 30
B – 19
Graphical Solution
100 –
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80 80 –
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00 2020 4040 6060 8080 100100
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-TV
sN
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ber
of
Wat
ch-T
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Number of WalkmansNumber of Walkmans
XX11
XX22
Figure B.4Figure B.4
(0, 42)
(30, 0)(30, 0)
$210 = $7$210 = $7XX11 + $5 + $5XX22
B – 20
Graphical Solution
100 –
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80 80 –
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60 60 –
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40 40 –
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20 20 –
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00 2020 4040 6060 8080 100100
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atch
-TV
sN
um
ber
of
Wat
ch-T
Vs
Number of WalkmansNumber of Walkmans
XX11
XX22
Figure B.5Figure B.5
$210 = $7$210 = $7XX11 + $5 + $5XX22
$350 = $7$350 = $7XX11 + $5 + $5XX22
$420 = $7$420 = $7XX11 + $5 + $5XX22
$280 = $7$280 = $7XX11 + $5 + $5XX22
B – 21
Graphical Solution
100 –
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80 80 –
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60 60 –
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40 40 –
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20 20 –
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00 2020 4040 6060 8080 100100
Nu
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atch
-TV
sN
um
ber
of
Wat
ch-T
Vs
Number of WalkmansNumber of Walkmans
XX11
XX22
Figure B.6Figure B.6
$410 = $7$410 = $7XX11 + $5 + $5XX22
Maximum profit lineMaximum profit line
Optimal solution pointOptimal solution point((XX11 = 30, = 30, XX22 = 40) = 40)
B – 22
Corner-Point Method
Figure B.7Figure B.7 1
2
3
100 –
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sN
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ch-T
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Number of WalkmansNumber of Walkmans
XX11
XX22
4
B – 23
Corner-Point Method The optimal value will always be at a
corner point
Find the objective function value at each corner point and choose the one with the highest profit
Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0
Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400
Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350
Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410
B – 24
Sensitivity Analysis
How sensitive the results are to parameter changes Change in the value of coefficients
Change in a right-hand-side value of a constraint
Trial-and-error approach
Analytic postoptimality method
B – 25
Sensitivity Report
Program B.1Program B.1
B – 26
Changes in Resources
The right-hand-side values of constraint equations may change as resource availability changes
The shadow price of a constraint is the change in the value of the objective function resulting from a one-unit change in the right-hand-side value of the constraint
B – 27
Changes in Resources
Shadow prices are often explained as answering the question “How much would you pay for one additional unit of a resource?”
Shadow prices are only valid over a particular range of changes in right-hand-side values
Sensitivity reports provide the upper and lower limits of this range
B – 28
Sensitivity Analysis
–
100 –
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60 60 –
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20 20 –
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00 2020 4040 6060 8080 100100 XX11
XX22
Figure B.8 (a)Figure B.8 (a)
Changed assembly constraint from Changed assembly constraint from 22XX11 + 1 + 1XX22 = 100 = 100
to to 22XX11 + 1 + 1XX22 = 110 = 110
Electronics constraint Electronics constraint is unchangedis unchanged
Corner point 3 is still optimal, but Corner point 3 is still optimal, but values at this point are now Xvalues at this point are now X11 = 45= 45, ,
XX22 = 20= 20, with a profit , with a profit = $415= $415
1
2
3
4
B – 29
Sensitivity Analysis
–
100 –
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80 80 –
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60 60 –
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40 40 –
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20 20 –
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00 2020 4040 6060 8080 100100 XX11
XX22
Figure B.8 (b)Figure B.8 (b)
Changed assembly constraint from Changed assembly constraint from 22XX11 + 1 + 1XX22 = 100 = 100
to to 22XX11 + 1 + 1XX22 = 90 = 90
Electronics constraint Electronics constraint is unchangedis unchanged
Corner point 3 is still optimal, but Corner point 3 is still optimal, but values at this point are now Xvalues at this point are now X11 = 15= 15, ,
XX22 = 60= 60, with a profit , with a profit = $405= $405
1
2
3
4
B – 30
Changes in the Objective Function
A change in the coefficients in the objective function may cause a different corner point to become the optimal solution
The sensitivity report shows how much objective function coefficients may change without changing the optimal solution point
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