modB.ppt modA.ppt operation management slide for business

© 2006 Prentice Hall, Inc. B – 1
Operations
Management
Module B –
Linear Programming
© 2006 Prentice Hall, Inc.
PowerPoint presentation to accompany
PowerPoint presentation to accompany
Heizer/Render
Heizer/Render
Principles of Operations Management, 6e
Principles of Operations Management, 6e
Operations Management, 8e
Operations Management, 8e

Outline
Outline
 Requirements Of A Linear
Requirements Of A Linear
Programming Problem
Programming Problem
 Formulating Linear Programming
Formulating Linear Programming
Problems
Problems
 Shader Electronics Example
Shader Electronics Example

Outline – Continued
 Graphical Solution To A Linear
Graphical Solution To A Linear
Programming Problem
Programming Problem
 Graphical Representation of
Graphical Representation of
Constraints
Constraints
 Iso-Profit Line Solution Method
Iso-Profit Line Solution Method
 Corner-Point Solution Method
Corner-Point Solution Method

 Sensitivity Analysis
Sensitivity Analysis
 Sensitivity Report
Sensitivity Report
 Change in the Resources of the
Change in the Resources of the
Right-Hand-Side Values
Right-Hand-Side Values
 Changes in the Objective Function
Changes in the Objective Function
Coefficient
Coefficient
 Solving Minimization Problems
Solving Minimization Problems

 Linear Programming Applications
Linear Programming Applications
 Production-Mix Example
Production-Mix Example
 Diet Problem Example
Diet Problem Example
 Production Scheduling Example
Production Scheduling Example
 Labor Scheduling Example
Labor Scheduling Example
 The Simplex Method Of LP
The Simplex Method Of LP

Learning Objectives
Learning Objectives
When you complete this module, you
should be able to:
should be able to:
Identify or Define:
Identify or Define:
 Objective function
Objective function
 Constraints
Constraints
 Feasible region
Feasible region
 Iso-profit/iso-cost methods
Iso-profit/iso-cost methods
 Corner-point solution
Corner-point solution
 Shadow price
Shadow price

Learning Objectives
Learning Objectives
should be able to:
should be able to:
Describe or Explain:
Describe or Explain:
 How to formulate linear models
How to formulate linear models
 Graphical method of linear
Graphical method of linear
programming
programming
 How to interpret sensitivity
How to interpret sensitivity
analysis
analysis

Linear Programming
Linear Programming
 A mathematical technique to
A mathematical technique to
help plan and make decisions
help plan and make decisions
relative to the trade-offs
relative to the trade-offs
necessary to allocate resources
necessary to allocate resources
 Will find the minimum or
Will find the minimum or
maximum value of the objective
maximum value of the objective
 Guarantees the optimal solution
Guarantees the optimal solution
to the model formulated
to the model formulated

LP Applications
LP Applications
1.
1. Scheduling school buses to minimize
Scheduling school buses to minimize
total distance traveled
total distance traveled
2.
2. Allocating police patrol units to high
Allocating police patrol units to high
crime areas in order to minimize
crime areas in order to minimize
response time to 911 calls
response time to 911 calls
3.
3. Scheduling tellers at banks so that
Scheduling tellers at banks so that
needs are met during each hour of the
needs are met during each hour of the
day while minimizing the total cost of
day while minimizing the total cost of
labor
labor

LP Applications
LP Applications
4.
4. Selecting the product mix in a factory
Selecting the product mix in a factory
to make best use of machine- and
to make best use of machine- and
labor-hours available while maximizing
labor-hours available while maximizing
the firm’s profit
the firm’s profit
5.
5. Picking blends of raw materials in feed
Picking blends of raw materials in feed
mills to produce finished feed
mills to produce finished feed
combinations at minimum costs
combinations at minimum costs
6.
6. Determining the distribution system
Determining the distribution system
that will minimize total shipping cost
that will minimize total shipping cost

LP Applications
LP Applications
7.
7. Developing a production schedule that
Developing a production schedule that
will satisfy future demands for a firm’s
will satisfy future demands for a firm’s
product and at the same time minimize
product and at the same time minimize
total production and inventory costs
total production and inventory costs
8.
8. Allocating space for a tenant mix in a
Allocating space for a tenant mix in a
new shopping mall so as to maximize
new shopping mall so as to maximize
revenues to the leasing company
revenues to the leasing company

Requirements of an
Requirements of an
LP Problem
LP Problem
1.
1. LP problems seek to maximize or
LP problems seek to maximize or
minimize some quantity (usually
minimize some quantity (usually
profit or cost) expressed as an
profit or cost) expressed as an
objective function
objective function
2.
2. The presence of restrictions, or
The presence of restrictions, or
constraints, limits the degree to
constraints, limits the degree to
which we can pursue our
which we can pursue our
objective
objective

Requirements of an
Requirements of an
LP Problem
LP Problem
3.
3. There must be alternative courses
There must be alternative courses
of action to choose from
of action to choose from
4.
4. The objective and constraints in
The objective and constraints in
linear programming problems
linear programming problems
must be expressed in terms of
must be expressed in terms of
linear equations or inequalities
linear equations or inequalities

Formulating LP Problems
The product-mix problem at Shader Electronics
The product-mix problem at Shader Electronics
 Two products
Two products
1.
1. Shader Walkman, a portable CD/DVD
Shader Walkman, a portable CD/DVD
player
player
2.
2. Shader Watch-TV, a wristwatch-size
Shader Watch-TV, a wristwatch-size
Internet-connected color TV
Internet-connected color TV
 Determine the mix of products that will
Determine the mix of products that will
produce the maximum profit
produce the maximum profit

Walkman
Walkman Watch-TVs
Watch-TVs Available Hours
Available Hours
Department
Department (
(X
X1
1)
) (
(X
X2
2)
) This Week
This Week
Hours Required
Hours Required
to Produce 1 Unit
to Produce 1 Unit
Electronic
Electronic 4
4 3
3 240
240
Assembly
Assembly 2
2 1
1 100
100
Profit per unit
Profit per unit $7
$7 $5
$5
Decision Variables:
Decision Variables:
X
X1
1 = number of Walkmans to be produced
= number of Walkmans to be produced
X
X2
2 = number of Watch-TVs to be produced
= number of Watch-TVs to be produced
Table B.1
Table B.1

Objective Function:
Objective Function:
Maximize Profit =
Maximize Profit = $7
$7X
X1
1 +
+ $5
$5X
X2
2
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

Second Constraint:
Second Constraint:
2
2X
X1
1 +
+ 1
1X
X2
2 ≤ 100
≤ 100 (hours of assembly time)
(hours of assembly time)
Assembly
Assembly
time available
time available
Assembly
Assembly
time used
time used is ≤
is ≤
First Constraint:
First Constraint:
4
4X
X1
1 +
+ 3
3X
X2
2 ≤ 240
≤ 240 (hours of electronic time)
(hours of electronic time)
Electronic
Electronic
time available
time available
Electronic
Electronic
time used
time used is ≤
is ≤

Graphical Solution
Graphical Solution
 Can be used when there are two
Can be used when there are two
decision variables
decision variables
1.
1. Plot the constraint equations at their
Plot the constraint equations at their
limits by converting each equation to
limits by converting each equation to
an equality
an equality
2.
2. Identify the feasible solution space
Identify the feasible solution space
3.
3. Create an iso-profit line based on the
Create an iso-profit line based on the
objective function
objective function
4.
4. Move this line outwards until the
Move this line outwards until the
optimal point is identified
optimal point is identified

Graphical Solution
Graphical Solution
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100
Number
of
Watch-TVs
Number
of
Watch-TVs
Number of Walkmans
Number of Walkmans
X
X1
1
X
X2
2
Assembly (constraint B)
Electronics (constraint A)
Feasible
region
Figure B.3
Figure B.3

Graphical Solution
Graphical Solution
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100
Number
of
Watch
TVs
Number
of
Watch
TVs
Number of Walkmans
Number of Walkmans
X
X1
1
X
X2
2
Feasible
region
Figure B.3
Figure 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

Graphical Solution
Graphical Solution
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100
Number
of
Watch-TVs
Number
of
Watch-TVs
Number of Walkmans
Number of Walkmans
X
X1
1
X
X2
2
Figure B.4
Figure B.4
(0, 42)
(30, 0)
(30, 0)
$210 = $7
$210 = $7X
X1
1 + $5
+ $5X
X2
2

Graphical Solution
Graphical Solution
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100
Number
of
Watch-TVs
Number
of
Watch-TVs
Number of Walkmans
Number of Walkmans
X
X1
1
X
X2
2
Figure B.5
Figure B.5
$210 = $7
$210 = $7X
X1
1 + $5
+ $5X
X2
2
$350 = $7
$350 = $7X
X1
1 + $5
+ $5X
X2
2
$420 = $7
$420 = $7X
X1
1 + $5
+ $5X
X2
2
$280 = $7
$280 = $7X
X1
1 + $5
+ $5X
X2
2

Graphical Solution
Graphical Solution
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100
Number
of
Watch-TVs
Number
of
Watch-TVs
Number of Walkmans
Number of Walkmans
X
X1
1
X
X2
2
Figure B.6
Figure B.6
$410 = $7
$410 = $7X
X1
1 + $5
+ $5X
X2
2
Maximum profit line
Maximum profit line
Optimal solution point
Optimal solution point
(
(X
X1
1 = 30,
= 30, X
X2
2 = 40)
= 40)

Corner-Point Method
Corner-Point Method
Figure B.7
Figure B.7 1
2
3
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100
Number
of
Watch-TVs
Number
of
Watch-TVs
Number of Walkmans
Number of Walkmans
X
X1
1
X
X2
2
4

Corner-Point Method
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

Corner-Point Method
Corner-Point Method
corner point
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
Solve for the intersection of two constraints
2X1 + 1X2 ≤ 100 (assembly time)
4X1 + 3X2 ≤ 240 (electronics time)
4X1 + 3X2 = 240
- 4X1 - 2X2 = -200
+ 1X2 = 40
4X1 + 3(40) = 240
4X1 + 120 = 240
X1 = 30

Corner-Point Method
Corner-Point Method
corner point
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

 How sensitive the results are to
How sensitive the results are to
parameter changes
parameter changes
 Change in the value of coefficients
Change in the value of coefficients
 Change in a right-hand-side value of a
Change in a right-hand-side value of a
constraint
constraint
 Trial-and-error approach
Trial-and-error approach
 Analytic postoptimality method
Analytic postoptimality method

Sensitivity Report
Sensitivity Report
Program B.1
Program B.1

Changes in Resources
 The right-hand-side values of
The right-hand-side values of
constraint equations may change
constraint equations may change
as resource availability changes
as resource availability changes
 The shadow price of a constraint is
The shadow price of a constraint is
the change in the value of the
the change in the value of the
objective function resulting from a
objective function resulting from a
one-unit change in the right-hand-
one-unit change in the right-hand-
side value of the constraint
side value of the constraint

 Shadow prices are often explained
Shadow prices are often explained
as answering the question “How
as answering the question “How
much would you pay for one
much would you pay for one
additional unit of a resource?”
additional unit of a resource?”
 Shadow prices are only valid over a
Shadow prices are only valid over a
particular range of changes in
particular range of changes in
right-hand-side values
right-hand-side values
 Sensitivity reports provide the
Sensitivity reports provide the
upper and lower limits of this range
upper and lower limits of this range

–
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100 X
X1
1
X
X2
2
Figure B.8 (a)
Figure B.8 (a)
Changed assembly constraint from
2
2X
X1
1 + 1
+ 1X
X2
2 = 100
= 100
to
to 2
2X
X1
1 + 1
+ 1X
X2
2 = 110
= 110
Electronics constraint
is unchanged
is unchanged
Corner point 3 is still optimal, but
values at this point are now X
values at this point are now X1
1 = 45
= 45,
,
X
X2
2 = 20
= 20, with a profit
, with a profit = $415
= $415
1
2
3
4

–
100 –
–
80
80 –
–
60
60 –
–
40
40 –
–
20
20 –
–
–
| | | | | | | | | | |
0
0 20
20 40
40 60
60 80
80 100
100 X
X1
1
X
X2
2
Figure B.8 (b)
Figure B.8 (b)
2
2X
X1
1 + 1
+ 1X
X2
2 = 100
= 100
to
to 2
2X
X1
1 + 1
+ 1X
X2
2 = 90
= 90
is unchanged
is unchanged
values at this point are now X
values at this point are now X1
1 = 15
= 15,
,
X
X2
2 = 60
= 60, with a profit
, with a profit = $405
= $405
1
2
3
4

Changes in the
Changes in the
Objective Function
Objective Function
 A change in the coefficients in the
A change in the coefficients in the
objective function may cause a
objective function may cause a
different corner point to become the
different corner point to become the
optimal solution
optimal solution
 The sensitivity report shows how
The sensitivity report shows how
much objective function coefficients
much objective function coefficients
may change without changing the
may change without changing the
optimal solution point
optimal solution point

Solving Minimization
Solving Minimization
Problems
Problems
 Formulated and solved in much the
Formulated and solved in much the
same way as maximization
same way as maximization
problems
problems
 In the graphical approach an iso-
In the graphical approach an iso-
cost line is used
cost line is used
 The objective is to move the iso-
The objective is to move the iso-
cost line inwards until it reaches the
cost line inwards until it reaches the
lowest cost corner point
lowest cost corner point

Minimization Example
X
X1
1 =
= number of tons of black-and-white chemical
number of tons of black-and-white chemical
produced
produced
X
X2
2 =
= number of tons of color picture chemical
number of tons of color picture chemical
produced
produced
Minimize total cost
Minimize total cost =
= 2,500
2,500X
X1
1 +
+ 3,000
3,000X
X2
2
X
X1
1 ≥ 30
≥ 30 tons of black-and-white chemical
tons of black-and-white chemical
X
X2
2 ≥ 20
≥ 20 tons of color chemical
tons of color chemical
X
X1
1 + X
+ X2
2 ≥ 60
≥ 60 tons total
tons total
X
X1
1,
, X
X2
2 ≥ $0
≥ $0 nonnegativity requirements
nonnegativity requirements

Table B.9
Table B.9
60
60 –
50 –
40
40 –
30 –
20
20 –
10 –
–
| | | | | | |
0
0 10
10 20
20 30
30 40
40 50
50 60
60
X
X1
1
X
X2
2
Feasible
region
X
X1
1 = 30
= 30
X
X2
2 = 20
= 20
X
X1
1 + X
+ X2
2 = 60
= 60
b
b
a
a

Total cost at a
Total cost at a =
= 2,500
2,500X
X1
1 +
+ 3,000
3,000X
X2
2
=
= 2,500 (40)
2,500 (40) +
+ 3,000(20)
3,000(20)
=
= $160,000
$160,000
Total cost at b
Total cost at b =
= 2,500
2,500X
X1
1 +
+ 3,000
3,000X
X2
2
=
= 2,500 (30)
2,500 (30) +
+ 3,000(30)
3,000(30)
=
= $165,000
$165,000
Lowest total cost is at point a
Lowest total cost is at point a

LP Applications
LP Applications
Department
Department
Product
Product Wiring
Wiring Drilling
Drilling Assembly
Assembly Inspection
Inspection Unit Profit
Unit Profit
XJ201
XJ201 .5
.5 3
3 2
2 .5
.5 $ 9
$ 9
XM897
XM897 1.5
1.5 1
1 4
4 1.0
1.0 $12
$12
TR29
TR29 1.5
1.5 2
2 1
1 .5
.5 $15
$15
BR788
BR788 1.0
1.0 3
3 2
2 .5
.5 $11
$11
Capacity
Capacity Minimum
Minimum
Department
Department (in hours)
(in hours) Product
Product Production Level
Production Level
Wiring
Wiring 1,500
1,500 XJ201
XJ201 150
150
Drilling
Drilling 2,350
2,350 XM897
XM897 100
100
Assembly
Assembly 2,600
2,600 TR29
TR29 300
300
Inspection
Inspection 1,200
1,200 BR788
BR788 400
400

LP Applications
LP Applications
X
X1
1 = number of units of XJ201 produced
= number of units of XJ201 produced
X
X2
2 = number of units of XM897 produced
= number of units of XM897 produced
X
X3
3 = number of units of TR29 produced
= number of units of TR29 produced
X
X4
4 = number of units of BR788 produced
= number of units of BR788 produced
Maximize profit
Maximize profit = 9
= 9X
X1
1 + 12
+ 12X
X2
2 + 15
+ 15X
X3
3 + 11
+ 11X
X4
4
subject to
subject to .5
.5X
X1
1 +
+ 1.5
1.5X
X2
2 +
+ 1.5
1.5X
X3
3 +
+ 1
1X
X4
4 ≤ 1,500
≤ 1,500 hours of wiring
hours of wiring
3
3X
X1
1 +
+ 1
1X
X2
2 +
+ 2
2X
X3
3 +
+ 3
3X
X4
4 ≤ 2,350
≤ 2,350 hours of drilling
hours of drilling
2
2X
X1
1 +
+ 4
4X
X2
2 +
+ 1
1X
X3
3 +
+ 2
2X
X4
4 ≤ 2,600
≤ 2,600 hours of assembly
hours of assembly
.5
.5X
X1
1 +
+ 1
1X
X2
2 +
+ .5
.5X
X3
3 +
+ .5
.5X
X4
4 ≤ 1,200
≤ 1,200 hours of inspection
hours of inspection
X
X1
1 ≥ 150
≥ 150 units of XJ201
units of XJ201
X
X2
2 ≥ 100
≥ 100 units of XM897
units of XM897
X
X3
3 ≥ 300
≥ 300 units of TR29
units of TR29
X
X4
4 ≥ 400
≥ 400 units of BR788
units of BR788

LP Applications
LP Applications
A
A 3
3 oz
oz 2
2 oz
oz 4
4 oz
oz
B
B 2
2 oz
oz 3
3 oz
oz 1
1 oz
oz
C
C 1
1 oz
oz 0
0 oz
oz 2
2 oz
oz
D
D 6
6 oz
oz 8
8 oz
oz 4
4 oz
oz
Feed
Feed
Product
Product Stock X
Stock X Stock Y
Stock Y Stock Z
Stock Z

LP Applications
LP Applications
X
X1
1 = number of pounds of stock X purchased per cow each month
= number of pounds of stock X purchased per cow each month
X
X2
2 = number of pounds of stock Y purchased per cow each month
= number of pounds of stock Y purchased per cow each month
X
X3
3 = number of pounds of stock Z purchased per cow each month
= number of pounds of stock Z purchased per cow each month
Minimize cost
Minimize cost = .02
= .02X
X1
1 + .04
+ .04X
X2
2 + .025
+ .025X
X3
3
Ingredient A requirement:
Ingredient A requirement: 3
3X
X1
1 +
+ 2
2X
X2
2 +
+ 4
4X
X3
3 ≥ 64
≥ 64
Ingredient B requirement:
Ingredient B requirement: 2
2X
X1
1 +
+ 3
3X
X2
2 +
+ 1
1X
X3
3 ≥ 80
≥ 80
Ingredient C requirement:
Ingredient C requirement: 1
1X
X1
1 +
+ 0
0X
X2
2 +
+ 2
2X
X3
3 ≥ 16
≥ 16
Ingredient D requirement:
Ingredient D requirement: 6
6X
X1
1 +
+ 8
8X
X2
2 +
+ 4
4X
X3
3 ≥ 128
≥ 128
Stock Z limitation:
Stock Z limitation: X
X3
3 ≤ 80
≤ 80
X
X1,
1, X
X2,
2, X
X3
3 ≥ 0
≥ 0
Cheapest solution is to purchase 40 pounds of grain X
Cheapest solution is to purchase 40 pounds of grain X
at a cost of $0.80 per cow
at a cost of $0.80 per cow

LP Applications
LP Applications
Manufacturing
Manufacturing Selling Price
Selling Price
Month
Month Cost
Cost (during month)
(during month)
July
July $60
$60 —
—
August
August $60
$60 $80
$80
September
September $50
$50 $60
$60
October
October $60
$60 $70
$70
November
November $70
$70 $80
$80
December
December —
— $90
$90
X
X1
1,
, X
X2
2,
, X
X3
3,
, X
X4
4,
, X
X5
5,
, X
X6
6 =
=number of units
number of units
manufactured during July (first month),
manufactured during July (first month),
August (second month), etc.
August (second month), etc.
Y
Y1
1,
, Y
Y2
2,
, Y
Y3
3,
, Y
Y4
4,
, Y
Y5
5,
, Y
Y6
6 =
=number of units sold during July, August,
number of units sold during July, August,
etc.
etc.

LP Applications
LP Applications
Maximize profit
Maximize profit =
= 80
80Y
Y2
2 + 60
+ 60Y
Y3
3 + 70
+ 70Y
Y4
4 + 80
+ 80Y
Y5
5 + 90
+ 90Y
Y6
6
- (60
- (60X
X1
1 + 60
+ 60X
X2
2 + 50
+ 50X
X3
3 + 60
+ 60X
X4
4 + 70
+ 70X
X5
5)
)
Inventory at
end of this
month
Inventory at
end of
previous month
Current
month’s
production
This month’s
sales
= + –
New decision variables: I1, I2, I3, I4, I5, I6
July:
July: I
I1
1 =
= X
X1
1
August:
August: I
I2
2 =
= I
I1
1 +
+ X
X2
2 -
- Y
Y2
2
September:
September: I
I3
3 =
= I
I2
2 +
+ X
X3
3 -
- Y
Y3
3
October:
October: I
I4
4 =
= I
I3
3 +
+ X
X4
4 -
- Y
Y4
4
November:
November: I
I5
5 =
= I
I4
4 +
+ X
X5
5 -
- Y
Y5
5
December:
December: I
I6
6 =
= I
I5
5 +
+ X
X6
6 -
- Y
Y6
6

LP Applications
LP Applications
Maximize profit
Maximize profit =
= 80
80Y
Y2
2 + 60
+ 60Y
Y3
3 + 70
+ 70Y
Y4
4 + 80
+ 80Y
Y5
5 + 90
+ 90Y
Y6
6
- (60
- (60X
X1
1 + 60
+ 60X
X2
2 + 50
+ 50X
X3
3 + 60
+ 60X
X4
4 + 70
+ 70X
X5
5)
)
July:
July: I
I1
1 =
= X
X1
1
August:
August: I
I2
2 =
= I
I1
1 +
+ X
X2
2 -
- Y
Y2
2
September:
September: I
I3
3 =
= I
I2
2 +
+ X
X3
3 -
- Y
Y3
3
October:
October: I
I4
4 =
= I
I3
3 +
+ X
X4
4 -
- Y
Y4
4
November:
November: I
I5
5 =
= I
I4
4 +
+ X
X5
5 -
- Y
Y5
5
December:
December: I
I6
6 =
= I
I5
5 +
+ X
X6
6 -
- Y
Y6
6
I1 ≤ 100, I2 ≤ 100 , I3 ≤ 100, I4 ≤ 100, I5 ≤ 100, I6 = 0
for all Y
for all Yi
i ≤ 300
≤ 300

LP Applications
LP Applications
Maximize profit
Maximize profit =
= 80
80Y
Y2
2 + 60
+ 60Y
Y3
3 + 70
+ 70Y
Y4
4 + 80
+ 80Y
Y5
5 + 90
+ 90Y
Y6
6
- (60
- (60X
X1
1 + 60
+ 60X
X2
2 + 50
+ 50X
X3
3 + 60
+ 60X
X4
4 + 70
+ 70X
X5
5)
)
July:
July: I
I1
1 =
= X
X1
1
August:
August: I
I2
2 =
= I
I1
1 +
+ X
X2
2 -
- Y
Y2
2
September:
September: I
I3
3 =
= I
I2
2 +
+ X
X3
3 -
- Y
Y3
3
October:
October: I
I4
4 =
= I
I3
3 +
+ X
X4
4 -
- Y
Y4
4
November:
November: I
I5
5 =
= I
I4
4 +
+ X
X5
5 -
- Y
Y5
5
December:
December: I
I6
6 =
= I
I5
5 +
+ X
X6
6 -
- Y
Y6
6
I1 ≤ 100, I2 ≤ 100 , I3 ≤ 100, I4 ≤ 100, I5 ≤ 100, I6 = 0
for all Y
for all Yi
i ≤ 300
≤ 300
X1 = 100, X2 = 200, X3 = 400,
X4 = 300, X5 = 300, X6 = 0
Y1 = 100, Y2 = 300, Y3 = 300,
Y4 = 300, Y5 = 300, Y6 = 100
I1 = 100, I2 = 0, I3 = 100,
I4 = 100, I5 = 100, I6 = 0
Final Solution
Profit = $19,000

LP Applications
LP Applications
Time
Time Number of
Number of Time
Time Number of
Number of
Period
Period Tellers Required
Tellers Required Period
Period Tellers Required
Tellers Required
9
9 AM
AM - 10
- 10 AM
AM 10
10 1
1 PM
PM - 2
- 2 PM
PM 18
18
10
10 AM
AM - 11
- 11 AM
AM 12
12 2
2 PM
PM - 3
- 3 PM
PM 17
17
11
11 AM
AM - Noon
- Noon 14
14 3
3 PM
PM - 4
- 4 PM
PM 15
15
Noon - 1
Noon - 1 PM
PM 16
16 4
4 PM
PM - 5
- 5 PM
PM 10
10
F
F = Full-time tellers
= Full-time tellers
P
P1
1 = Part-time tellers starting at 9
= Part-time tellers starting at 9 AM
AM (leaving at 1
(leaving at 1 PM
PM)
)
P
P2
AM (leaving at 2
(leaving at 2 PM
PM)
)
P
P3
AM (leaving at 3
(leaving at 3 PM
PM)
)
P
P4
4 = Part-time tellers starting at noon (leaving at 4
= Part-time tellers starting at noon (leaving at 4 PM
PM)
)
P
P5
= Part-time tellers starting at 1 PM
PM (leaving at 5
(leaving at 5 PM
PM)
)

LP Applications
LP Applications
= $75
= $75F
F + $24(
+ $24(P
P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5)
)
Minimize total daily
manpower cost
manpower cost
F
F + P
+ P1
1 ≥ 10
≥ 10 (
(9
9 AM
AM - 10
- 10 AM
AM needs
needs)
)
F
F + P
+ P1
1 + P
+ P2
2 ≥ 12
≥ 12 (
(10
10 AM
AM - 11
- 11 AM
AM needs
needs)
)
1/2 F
1/2 F + P
+ P1
1 + P
+ P2
2 + P
+ P3
3 ≥ 14
≥ 14 (
(11
11 AM
AM - 11
- 11 AM
AM needs
needs)
)
1/2 F
1/2 F + P
+ P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 ≥ 16
≥ 16 (
(noon - 1
noon - 1 PM
PM needs
needs)
)
F
F + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5 ≥ 18
≥ 18 (
(1
1 PM
PM - 2
- 2 PM
PM needs
needs)
)
F
F + P
+ P3
3 + P
+ P4
4 + P
+ P5
5 ≥ 1
≥ 17
7 (
(2
2 PM
PM - 3
- 3 PM
PM needs
needs)
)
F
F + P
+ P4
4 + P
+ P5
5 ≥ 15
≥ 15 (
(3
3 PM
PM - 7
- 7 PM
PM needs
needs)
)
F
F + P
+ P5
5 ≥ 10
≥ 10 (
(4
4 PM
PM - 5
- 5 PM
PM needs
needs)
)
F
F ≤ 12
≤ 12
4(
4(P
P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5) ≤ .50(10 + 12 + 14 + 16 + 18 + 17 + 15 + 10)
) ≤ .50(10 + 12 + 14 + 16 + 18 + 17 + 15 + 10)

LP Applications
LP Applications
= $75
= $75F
F + $24(
+ $24(P
P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5)
)
manpower cost
manpower cost
F
F + P
+ P1
1 ≥ 10
≥ 10 (
(9
9 AM
AM - 10
- 10 AM
AM needs
needs)
)
F
F + P
+ P1
1 + P
+ P2
2 ≥ 12
≥ 12 (
(10
10 AM
AM - 11
- 11 AM
AM needs
needs)
)
1/2 F
1/2 F + P
+ P1
1 + P
+ P2
2 + P
+ P3
3 ≥ 14
≥ 14 (
(11
11 AM
AM - 11
- 11 AM
AM needs
needs)
)
1/2 F
1/2 F + P
+ P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 ≥ 16
≥ 16 (
(noon - 1
noon - 1 PM
PM needs
needs)
)
F
F + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5 ≥ 18
≥ 18 (
(1
1 PM
PM - 2
- 2 PM
PM needs
needs)
)
F
F + P
+ P3
3 + P
+ P4
4 + P
+ P5
5 ≥ 1
≥ 17
7 (
(2
2 PM
PM - 3
- 3 PM
PM needs
needs)
)
F
F + P
+ P4
4 + P
+ P5
5 ≥ 15
≥ 15 (
(3
3 PM
PM - 7
- 7 PM
PM needs
needs)
)
F
F + P
+ P5
5 ≥ 10
≥ 10 (
(4
4 PM
PM - 5
- 5 PM
PM needs
needs)
)
F
F ≤ 12
≤ 12
4(
4(P
P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5)
) ≤ .50(112)
≤ .50(112)
F
F,
, P
P1
1,
, P
P2
2,
, P
P3
3,
, P
P4
4,
, P
P5
5 ≥ 0
≥ 0

LP Applications
LP Applications
= $75
= $75F
F + $24(
+ $24(P
P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5)
)
manpower cost
manpower cost
F
F + P
+ P1
1 ≥ 10
≥ 10 (
(9
9 AM
AM - 10
- 10 AM
AM needs
needs)
)
F
F + P
+ P1
1 + P
+ P2
2 ≥ 12
≥ 12 (
(10
10 AM
AM - 11
- 11 AM
AM needs
needs)
)
1/2 F
1/2 F + P
+ P1
1 + P
+ P2
2 + P
+ P3
3 ≥ 14
≥ 14 (
(11
11 AM
AM - 11
- 11 AM
AM needs
needs)
)
1/2 F
1/2 F + P
+ P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 ≥ 16
≥ 16 (
(noon - 1
noon - 1 PM
PM needs
needs)
)
F
F + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5 ≥ 18
≥ 18 (
(1
1 PM
PM - 2
- 2 PM
PM needs
needs)
)
F
F + P
+ P3
3 + P
+ P4
4 + P
+ P5
5 ≥ 1
≥ 17
7 (
(2
2 PM
PM - 3
- 3 PM
PM needs
needs)
)
F
F + P
+ P4
4 + P
+ P5
5 ≥ 15
≥ 15 (
(3
3 PM
PM - 7
- 7 PM
PM needs
needs)
)
F
F + P
+ P5
5 ≥ 10
≥ 10 (
(4
4 PM
PM - 5
- 5 PM
PM needs
needs)
)
F
F ≤ 12
≤ 12
4(
4(P
P1
1 + P
+ P2
2 + P
+ P3
3 + P
+ P4
4 + P
+ P5
5)
) ≤ .50(112)
≤ .50(112)
F
F,
, P
P1
1,
, P
P2
2,
, P
P3
3,
, P
P4
4,
, P
P5
5 ≥ 0
≥ 0
There are two alternate optimal solutions to this
problem but both will cost $1,086 per day
F = 10 F = 10
P1 = 0 P1 = 6
P2 = 7 P2 = 1
P3 = 2 P3 = 2
P4 = 2 P4 = 2
P5 = 3 P5 = 3
First Second
Solution Solution

The Simplex Method
The Simplex Method
 Real world problems are too
Real world problems are too
complex to be solved using the
complex to be solved using the
graphical method
graphical method
 The simplex method is an algorithm
The simplex method is an algorithm
for solving more complex problems
for solving more complex problems
 Developed by George Dantzig in the
Developed by George Dantzig in the
late 1940s
late 1940s
 Most computer-based LP packages
Most computer-based LP packages
use the simplex method
use the simplex method

modB.ppt modA.ppt operation management slide for business

More Related Content

Similar to modB.ppt modA.ppt operation management slide for business (20)

Recently uploaded (20)

modB.ppt modA.ppt operation management slide for business

Editor's Notes