Linear programming (Adapted from Chapter 13 Supplement, Operations and Management, 5Chapter 13 Supplement, Operations and Management, 5thth edition edition by by Roberta Russell & Bernard W. Taylor, III., Copyright 2006 John Wiley & Sons, Inc. This presentation also contains material of Pearson, Prentice hall
Dr. Arturo S. Leon, BSU (Spring 2010)
1© Arturo S. Leon, BSU, Spring
2010
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-2
Lecture Outline
� Model Formulation
� Graphical Solution Method
� Linear Programming Model
� Solution
� Solving Linear Programming Problems with Excel
� Sensitivity Analysis
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-3
A model consisting of linear relationshipsrepresenting a firm’s objective and resource constraints
Linear Programming (LP)
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
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-4
A model consisting of linear relationshipsrepresenting a firm’s objective and resource constraints
Linear Programming
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
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-5
http://video.google.com/videoplay?docid=754226565202103395#docid=8211036228894039768
Video of Linear Programming (LP)
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-6
LP Model Formulation
� Decision variables– mathematical symbols representing levels of activity of an
operation
� Objective function– 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
� Constraint– a linear relationship representing a restriction on decision
making
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-7
LP Model Formulation (cont.)
Max/min z = c1x1 + c2x2 + ... + cnxn
subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
:am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-8
LP Model: Example
LaborLabor ClayClay RevenueRevenuePRODUCTPRODUCT (hr/unit)(hr/unit) (lb/unit)(lb/unit) ($/unit)($/unit)
BowlBowl 11 44 4040
MugMug 22 33 5050
There are 40 hours of labor and 120 pounds of clay There are 40 hours of labor and 120 pounds of clay available each dayavailable each day
Decision variablesDecision variables
xx11 = number of bowls to produce= number of bowls to produce
xx22 = number of mugs to produce= number of mugs to produce
RESOURCE REQUIREMENTSRESOURCE REQUIREMENTS
LP Formulation: Example
Maximize Maximize ZZ = $40 = $40 xx11 + 50 + 50 xx22
Subject toSubject to
xx11 ++ 22xx22 ≤ ≤ 40 hr40 hr (labor constraint)(labor constraint)
44xx11 ++ 33xx22 ≤ ≤ 120 lb120 lb (clay constraint)(clay constraint)
xx1 1 , , xx22 ≥ ≥ 00
Quick solution with Excel solverQuick solution with Excel solver
Using solver for the Bowl/mug example
Decision variables
X1 -5 Maximize -250
x2 -1
constraint 1: -7 <= 40
constraint 2: -23 <= 120
constraint 3: -5 >= 0
constraint 4: -1 >= 0
Initial conditions for Solver
LP Example (Cont.)
Solution is Solution is xx11 = 24 bowls = 24 bowls xx2 2 = 8 mugs= 8 mugs
Revenue or benefit = Revenue or benefit = $1,360$1,360
Using solver for the Bowl/mug example
Decision variables
X1 24 Maximize 1360
x2 8
constraint 1: 40 <= 40
constraint 2: 120 <= 120
constraint 3: 24 >= 0
constraint 4: 8 >= 0
Solution
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-11
Graphical Solution Method
1.1. Plot model constraint on a set of coordinates Plot model constraint on a set of coordinates in a planein a plane
2.2. Identify the feasible solution space on the Identify the feasible solution space on the graph where all constraints are satisfied graph where all constraints are satisfied simultaneouslysimultaneously
3.3. Plot objective function to find the point on Plot objective function to find the point on boundary of this space that maximizes (or boundary of this space that maximizes (or minimizes) value of objective functionminimizes) value of objective function
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-12
LP Formulation: Example
Maximize Maximize ZZ = $40 = $40 xx11 + 50 + 50 xx22
Subject toSubject to
xx11 ++ 22xx22 ≤ ≤ 40 hr40 hr (labor constraint)(labor constraint)
44xx11 ++ 33xx22 ≤ ≤ 120 lb120 lb (clay constraint)(clay constraint)
xx1 1 , , xx22 ≥ ≥ 00
Solution is Solution is xx11 = 24 bowls = 24 bowls xx2 2 = 8 mugs= 8 mugs
Revenue = $1,360Revenue = $1,360
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-13
Graphical Solution: Example
4 4 xx11 + 3 + 3 xx2 2 ≤ ≤ 120 lb120 lb
xx11 + 2 + 2 xx2 2 ≤ ≤ 40 hr40 hr
Area common toArea common toboth constraintsboth constraints
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 – |1010
|6060
|5050
|2020
|3030
|4040 xx11
xx22
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-14
Computing Optimal Valuesxx11 ++ 22xx22 == 4040
44xx11 ++ 33xx22 == 120120
44xx11 ++ 88xx22 == 160160
--44xx11 -- 33xx22 == --120120
55xx22 == 4040
xx22 == 88
xx11 ++ 2(8)2(8) == 4040
xx11 == 2424
4 4 xx11 + 3 + 3 xx2 2 ≤ ≤ 120 lb120 lb
xx11 + 2 + 2 xx2 2 ≤ ≤ 40 hr40 hr
40 40 –
30 30 –
20 20 –
10 10 –
0 0 – |1010
|2020
|3030
|4040
xx11
xx22
ZZ = $50(24) + $50(8) = $1,360= $50(24) + $50(8) = $1,360
248
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-15
Extreme Corner Points
xx11 = 224 bowls= 224 bowls
xx2 2 == 8 mugs8 mugs
ZZ = $1,360= $1,360 xx11 = 30 bowls= 30 bowls
xx2 2 == 0 mugs0 mugs
ZZ = $1,200= $1,200
xx11 = 0 bowls= 0 bowls
xx2 2 == 20 mugs20 mugs
ZZ = $1,000= $1,000
AA
BB
CC|2020
|3030
|4040
|1010 xx11
xx22
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-16
44xx11 + 3+ 3xx2 2 ≤ ≤ 120 lb120 lb
xx11 + 2+ 2xx2 2 ≤ ≤ 40 hr40 hr
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –
BB
|1010
|2020
|3030
|4040 xx11
xx22
CC
AA
ZZ = 70= 70xx11 + 20+ 20xx22
Optimal point:Optimal point:
xx11 = 30 bowls= 30 bowls
xx2 2 == 0 mugs0 mugs
ZZ = $2,100= $2,100
Objective Function
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-17
Minimization Problem
CHEMICAL CONTRIBUTIONCHEMICAL CONTRIBUTION
BrandBrand Nitrogen (lb/bag)Nitrogen (lb/bag) Phosphate (lb/bag)Phosphate (lb/bag)
GroGro--plusplus 22 44
CropCrop--fastfast 44 33
Minimize Minimize ZZ = $6x= $6x11 + $3x+ $3x22
subject tosubject to
22xx11 ++ 44xx22 ≥≥ 16 lb of nitrogen16 lb of nitrogen
44xx11 ++ 33xx22 ≥≥ 24 lb of phosphate24 lb of phosphate
xx11, , xx22 ≥≥ 00
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-18
14 14 –
12 12 –
10 10 –
8 8 –
6 6 –
4 4 –
2 2 –
0 0 – |22
|44
|66
|88
|1010
|1212
|1414 xx11
xx22
A
B
C
Graphical Solution
x1 = 0 bags of Gro-plusx2 = 8 bags of Crop-fastZ = $24
Z = 6x1 + 3x2
Another example:
Max 7T + 5C (profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < 450 (max # chairs)
T > 100 (min # tables)
T, C > 0 (nonnegativity)
Graphical Solution
� Graphing an LP model helps provide insight into LP models and their solutions.
� While this can only be done in two dimensions, the same properties apply to all LP models and solutions.
Carpentry
Constraint Line
3T + 4C = 2400
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
0 800 T
C
600
0
Feasible
< 2400 hrs
Infeasible
> 2400 hrs
Painting
Constraint Line
2T + 1C = 1000
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
0 500 800 T
C
1000
600
0
0 100 200 300 400 500 T
C
500
400
300
200
100
0
Objective Function Line
7T + 5C = ProfitOptimal Point
(T = 320, C = 360)
0 100 200 300 400 500 T
C
500
400
300
200
100
0
Additional Constraint
Need at least 75 more chairs than tables
C > T + 75
Or
C – T > 75
T = 320
C = 360
No longer feasible
New optimal point
T = 300, C = 375
LP Characteristics
� Feasible Region: The set of points that satisfies all constraints
� Corner Point Property: An optimal solution must lie at one or more corner points
� Optimal Solution: The corner point with the best objective function value is optimal
Special Situation in LP
1. Redundant Constraints - do not affect the feasible region
Example: x < 10
x < 12
The second constraint is redundant because it is less restrictive.
Special Situation in LP
2. Infeasibility – when no feasible solution exists (there is no feasible region)
Example: x < 10
x > 15
Special Situation in LP3. Alternate Optimal Solutions – when there is
more than one optimal solution
Max 2T + 2CSubject to:
T + C < 10T < 5
C < 6T, C > 0
0 5 10 T
C
10
6
0
All points onRed segment are optimal
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