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ENGM 631 Optimization Ch. 3: Introduction to Linear Programming.
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Transcript of ENGM 631 Optimization Ch. 3: Introduction to Linear Programming.
ENGM 631 Optimization
Ch. 3: Introduction to Linear Programming
Sec. 3.1Alt. Prototype• K-Corp manufactures two primary products: a
small tractor suitable for an orchard and a tree shaker. K-Corp can make $3,000 per tractor and $5,000 per shaker. Currently, the demand for tractors is 8,000 per year and the demand for shakers is 6,000 per year. Currently, K-Corp is unable to meet total demand for both due to limitations in the assembly area. The assembly area has a capacity of 36,000 hours available. Each tractor takes 3 hours to assemble. Each shaker takes 4 hours.
Sec. 3.1Alt. Prototype
• Problem
Determine the appropriate number of tractors and shakers to make so as to maximize total profit.
• ConstraintsDo not exceed demand for either product
Do not exceed capacity of assembly area
Sec. 3.1Alt. Model
• VariablesX1 = number of tractors to manufacture
X2 = number of shakers to manufacture
• ProfitZ = 3X1 + 5X2 (in $1,000’s)
Sec. 3.1Alt. Model• Constraints
Demand for tractors
X1 < 8,000
Demand for shakers
X2 < 6,000
Capacity of Assembly
3X1 + 4X2 < 36,000
Non-negativity
X1 > 0 , X2 > 0
Sec. 3.1Alt. Model
Max Z = 3X1 + 5X2
s.t.
X1 < 8,000
X2 < 6,000
3X1 + 4X2 < 36,000
X1 > 0
X2 > 0
Sec. 3.1Alt. Rescale
• Let’s letX1 = number of tractors in 1,000’s
X2 = number of shakers in 1,000’s
Sec. 3.1Alt. ModelMax Z = 3X1 + 5X2
s.t.
X1 < 8
X2 < 6
3X1 + 4X2 < 36
X1 > 0
X2 > 0
where Z = profit (in $1,000,000’s)
Sec. 3.1Alt. Graphical Technique
2 4 6 8 10
8
6
4
2
Do not exceeddemand for tractors
X1 < 8
Sec. 3.1Alt. Graphical Technique
2 4 6 8 10
8
6
4
2
Do not exceeddemand for tractors, shakers
X2 < 6
X1 < 8
Sec. 3.1Alt. Graphical Technique
2 4 6 8 10 12
8
6
4
2
Do not exceeddemand for tractors, shakers, assembly capacity
3X1 + 4X2 < 36
X2 < 6
X1 < 8
Sec. 3.1Alt. Graphical Technique
2 4 6 8 10 12
8
6
4
2
Do not exceeddemand for tractors, shakers, assembly capacity,can’t make less thanzero
3X1 + 4X2 < 36
X2 < 6
X1 < 8
X1, X2 > 0
Sec. 3.1Alt. Feasible Region
2 4 6 8 10 12
8
6
4
2
3X1 + 4X2 < 36
X2 < 6
X1 < 8
X1, X2 > 0
Feasible Region
Sec. 3.1Alt. Problem
Out of the large (infinite)number of solutions available, which maximizesprofit?
2 4 6 8 10 12
8
6
4
2
Sec. 3.1Alt. Iso-Value Lines
2 4 6 8 10 12
8
6
4
2
Z = 3X1 + 5X2
Z = 15
Sec. 3.1Alt. Optimal Solution
2 4 6 8 10 12
8
6
4
2
Z = 3X1 + 5X2
= 3(4) + 5(6) = 42
(4,6)
Sec. 3.2 Some Terminology
2 4 6 8 10 12
8
6
4
2
FeasibleSolutions
Sec. 3.2 Some Terminology
2 4 6 8 10 12
8
6
4
2
Feasible Region
Sec. 3.2 Some Terminology
2 4 6 8 10 12
8
6
4
2
Boundary
Some Terminology
2 4 6 8 10 12
8
6
4
2
InfeasibleSolutions
Sec. 3.2 Aside
2 4 6 8 10 12
8
6
4
2
X1 < 8
X2 < 6
3X1 + 4X2 < 36 5X1 + 3X2 > 60
No Feasible Solutions
Sec. 3.2 Some Terminology
2 4 6 8 10 12
8
6
4
2
Corner PointFeasible Sol.(CPF)(0,6) (4,6)
(8,3)
(8,0)
(0,0)
Sec. 3.2 Relation: CPF & Optimal
2 4 6 8 10 12
8
6
4
2
(0,6) (4,6)
(8,3)
(8,0)
(0,0)
The best CPF solutionmust be an optimal solution.
Sec. 3.2 Relation: CPF & Optimal
2 4 6 8 10 12
8
6
4
2
(0,6) (4,6)
(8,3)
(8,0)
(0,0)
Z = 30 Z = 42
Z = 39
Z = 24Z = 0
The best CPF solutionmust be an optimal solution.
If a problem has multiple optimal solutions, at least twomust be CPF solutions.
Sec. 3.2 Relation: CPF & Optimal
2 4 6 8 10 12
8
6
4
2
(0,6) (4,6)
(8,3)
(8,0)
(0,0)
If a problem has multiple optimal solutions, at least twomust be CPF solutions.
Sec. 3.2 Optimal Solution
2 4 6 8 10 12
8
6
4
2
Recall,
Z = 3X1 + 5X2
(4,6)
3X1 + 4X2 < 36
Sec. 3.2 Optimal Solution
2 4 6 8 10 12
8
6
4
2
Suppose,
Z = 3X1 + 54X2
(4,6)
3X1 + 4X2 < 36
Sec. 3.2 Optimal Solution
2 4 6 8 10 12
8
6
4
2
Suppose,
Z = 3X1 + 54X2
(4,6)
3X1 + 4X2 < 36
Sec. 3.2 Optimal Solution
2 4 6 8 10 12
8
6
4
2
Suppose,
Z = 3X1 + 54X2
(4,6)
3X1 + 4X2 < 36
Multiple Optima
Standard Form for LP
Max Z c X c X c X
s t
a X a X a X b
a X a X a X b
a X a X a X b
X X X
n n
n n
n n
m m mn n m
n
1 1 2 2
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 20 0 0
. . .
. .
. . .
. . .
. . .
, , . . . ,
Sec. 3.2 Standard Form for LP
Max Z c X c X c X
s t
a X a X a X b
a X a X a X b
a X a X a X b
X X X
n n
n n
n n
m m mn n m
n
1 1 2 2
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 20 0 0
. . .
. .
. . .
. . .
. . .
, , . . . ,
DecisionVariables
Sec. 3.2 Standard Form for LP
Max Z c X c X c X
s t
a X a X a X b
a X a X a X b
a X a X a X b
X X X
n n
n n
n n
m m mn n m
n
1 1 2 2
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 20 0 0
. . .
. .
. . .
. . .
. . .
, , . . . ,
Objective
Sec. 3.2 Standard Form for LP
Max Z c X c X c X
s t
a X a X a X b
a X a X a X b
a X a X a X b
X X X
n n
n n
n n
m m mn n m
n
1 1 2 2
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 20 0 0
. . .
. .
. . .
. . .
. . .
, , . . . ,
Functional Constraints
Sec. 3.2 Standard Form for LP
Max Z c X c X c X
s t
a X a X a X b
a X a X a X b
a X a X a X b
X X X
n n
n n
n n
m m mn n m
n
1 1 2 2
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 20 0 0
. . .
. .
. . .
. . .
. . .
, , . . . , NonnegativityConstraints
Sec. 3.3 Assumptions
• Proportionality
The contribution of each activity to the value of the objective function is proportional to the level of activity.
Consider Z = 3X1 + . . .
Profit = R - C
= 8X1 - 5X1 = 3X1
Sec. 3.3 Assumptions• Proportionality
The contribution of each activity to the value of the objective function is proportional to the level of activity.
Consider Z = 3X1 + . . . X1 Z
0 01 32 63 94 125 15
Z
02468
10121416
0 1 2 3 4 5 6
X1
Co
ntr
ibu
tio
n t
o Z
Sec. 3.3 Assumptions• Proportionality
Now Suppose fixed costs = 3. Then
Profit = R - TC
= 8X1 - (3 + 5X1)
= 3X1 - 3
. X1 Z Z'
0 0 -31 3 02 6 33 9 64 12 95 15 12
Z
-5
0
5
10
15
20
0 1 2 3 4 5 6
X1
Co
ntr
ibu
tio
n t
o Z
Z
Z'
Sec. 3.3 Assumptions
• Proportionality
Now Suppose we have a decreasing marginal rate of return
Profit = R - TC
= (8-.025X1)X1 - 5X1
.
Sec. 3.3 Assumptions
• Additivity
Every function in the LP model is the sum of the individual contributions of the respective activities
Z = 3X1 + 5X2 + 2X1X2
Interaction terms not allowed
Sec. 3.3 Assumptions
• Divisibility
Decision variables, Xi i=1,2, . . . n, are allowed to have any values, including non-integer values, that satisfy the functional and nonnegativity constraints.
Sec. 3.3 Divisibility
2 4 6 8 10 12
8
6
4
2
2 4 6 8 10 12
8
6
4
2
For K-Corp, is it
This or this ?
Sec. 3.3 Assumptions• Certainty
The value assigned to each parameter of a linear programming model is assumed to be a known constant.Max Z = 3X1 + 5X2
s.t.
X1 < 8
X2 < 6
3X1 + 4X2 < 36
X1 > 0 , X2 > 0
How certain are we of this demand?
Sec. 3.4Alt Power Generation• A regional power system has three
generating stations: A, B, and C. Each station serves a local area. Three outlying areas are also served by the system: X, Y, and X. The power demand at areas X, Y, and Z is 25, 50, and 30 MW respectively. The maximum generating capacity beyond local requirements and the cost of generation at the three stations is shown below.
Sec. 3.4Alt Power Generation Station Excess Cap. Cost of Gen.
A 100 500
B 75 700
C 200 400
Power can be transmitted between any pair of generating stations, but 5% of the amount shipped will be lost. Power can be transmitted from some of the generating stations to outlying areas, but 10% of the amount shipped is lost.
Sec. 3.4Alt Power Generation Model
Lines exist from stations A & C to X, from B & C to Y, and from A & B to Z.
A
C
B
X
Z
Y
Formulate as a linear program!
Sec. 3.4Alt Power Generation Model
Problem:
Min Cost of satisfying outlying areas
s.t.
Do not exceed capacity at any plant
Exactly serve each outlying area
Sec. 3.4Alt Power Generation Model
• Decision VariablesXij = power shipped from plant i to outlying
area j
• Objective Function
Min Cost
Min Z = 500XAX + 500XAZ + 700XBY + 700XBZ
+ 400XCX + 400XCY
Sec. 3.4Alt Power Generation Model
• Do not Exceed Capacity
XAX + XAZ < 100
XBY + XBZ < 75
XCX + XCY < 200
Sec. 3.4Alt Power Generation Model• Shipped - Losses satisfy outlying areas
XAX - .1XAX + XCX - .1XCX = 25
XBY - .1XBY + XCY - .1XCY = 50
XAZ - .1XAZ + XBZ - .1XBZ = 30
or
.9XAX + .9XCX = 25
.9XBY + .9XCY = 50
.9XAZ + .9XBZ = 30
Sec. 3.4Alt Power Generation ModelMin Z = 500XAX + 500XAZ + 700XBY + 700XBZ
+ 400XCX + 400XCY
s.t.
XAX + XAZ < 100
XBY + XBZ < 75
XCX + XCY < 200
.9XAX + .9XCX = 25
.9XBY + .9XCY = 50
.9XAZ + .9XBZ = 30
Xi,j > 0 , i = A, B, C j = X, Y, Z
3.4 Additional Examples
• Design of Radiation therapy
• Regional Planning of Kibbutzim
• Controlling Air Pollution
• Reclaiming Solid waste
• Personnel Scheduling at Union Airways (United Airlines 24/7 40? $6m/yr saved)
• Distributing Goods in a Network (Like our Power Example)
3.5 Spreadsheet Solution
• Range name
• Data cells
• Target cell
• Output cells
• Changing cells
• Solver
3.5 Spreadsheet Fig. 3.14 Old
3.5 Spreadsheet Fig. 3.15
3.5 Spreadsheet Fig. 3.16
3.5 Spreadsheet Fig. 3.17 Old
3.5 Spreadsheet Fig. 3.18 Old
3.5 Spreadsheet Fig. 3.19 Old
3.5 Spreadsheet Fig. 3.20 Old
3.5 Spreadsheet Fig. 3.21 Old
3.5 Spreadsheet Fig. 3.22 Old
3.6 Large Models Modeling Languages
• Older edition Citgo (petroleum products distributor) had 3000 functional constraints and 15000 decision variables
• Need for a modeling language
• You can learn MPL and Lindo with links from the book’s website (not part of our course this year)
3.7 Conclusions
• LP is powerful
• LP is “standard”
• For allocating resources in any social organization
• Sometimes LP is not applicable; instead do other things
Cases
• 3.1 Auto assembly (Crusier &Thrillseeker)
• 3.2 (Online) Cutting Cafeteria costs
• 3.3 (Online) Staffing a Call Center at California Children’s Hospital
• 3.4 (Online) Promoting a Breakfast Cereal