Lecture 8: Optimization with a Min objective AGEC 352 Spring 2012 – February 8 R. Keeney.
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Transcript of Lecture 8: Optimization with a Min objective AGEC 352 Spring 2012 – February 8 R. Keeney.
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Lecture 8: Optimization with a Min objective
AGEC 352Spring 2012 – February 8
R. Keeney
![Page 2: Lecture 8: Optimization with a Min objective AGEC 352 Spring 2012 – February 8 R. Keeney.](https://reader036.fdocuments.us/reader036/viewer/2022082822/5697bffb1a28abf838cc0b41/html5/thumbnails/2.jpg)
UpcomingToday: Minimization lectureNext Week (Week 6)
◦2/13: Quiz on lecture 4-8; review HW 3◦2/14: Lab on Minimization◦2/15: Sensitivity
Week 7◦2/20: Exam Review in Class
(Assignment)◦2/21: No lab, Office Hrs from 9-12◦2/22: Exam I (50 mins., 30 ?’s, MC & T/F)
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Profits as the objectiveProfit has been used so far as our
objective variable and we assumed that this variable should be maximized
What if our problem has no revenue?◦Max P = R – C
R = 0 then Max P = -C
We could maximize profits by finding the maximum of the negative of costs. Or…
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Just treat Cost as the objective variableMax –C Min C
◦These two are equivalent. Any minimization problem can be rewritten as a maximization of the negative of the objective variable.
Given this relationship, we should expect everything about a basic minimization problem to be opposite (negative) of our basic maximization problem.
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Comparison of 2 variable problems
0;0:
200,1940100:
000,2010050:
320:
..
9060max
BCnegnon
BCstorage
BCcash
BCland
ts
BCP
;0;0:
500125100:
00.525.000.1:
00.125.010.0:
..
2.48.3min
BAnegNon
BACal
BANia
BAThi
ts
BAC
*Objective = max--better off with higher values of the objective variable*Constraints are upper limits--having higher resource levels increases the maximum making us better off
*Objective = min--better off with lower values of the objective variable*Constraints are lower limits--having higher requirement levels increases the minimum making us worse off
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Constraint AnalysisWe found the ‘most limiting
resource’ in maximization problems◦Divide RHS by coefficient for an
activity and choose the smallestNow we can find the ‘most
limiting requirement’◦Divide the RHS by coefficient for an
activity and choose the largestGraphically…
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Most limiting requirementCereal ARequirement
Supplied by Cereal A
Total required
Total/required
Thiamine 0.10 1.00 10
Niacin 1.00 5.00 5
Calories 100 500 5
If we were only going to eat cereal A, we would have to eat 10 ounces to fulfill the Thiamine requirement.
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Most limiting requirementCereal BRequirement
Supplied by Cereal B
Total required
Total/required
Thiamine 0.25 1.00 4
Niacin 0.25 5.00 20
Calories 125 500 4
If we were only going to eat cereal B, we would have to eat 20 ounces to fulfill the Niacin requirement.
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Some things we know just from the constraintsFeasible space graph
◦Cereal A axis intersects at A = 10◦Cereal B axis intersects at B = 20
The cereals have different most limiting requirements, a mixture of the two cereals is very likely to be cheaper than eating only one
Calories is not the most limiting resource for either food. It should not bind in the optimal solution…
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Feasible space of a maximization problem (see lecture for Jan. 25)
0 30 60 90 1201501802100
100
200
300
400
Corn constraintSugar constraint
Premium Finest
Sta
nd
ard
Stu
ff
Feasible space is inside of the lines.
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Feasible space of the cereal mix minimization problem
0
2
4
6
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10
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20
0 2 4 6 8 10
Cere
al B
Cereal A
Thiamine Niacin Calories
*Feasible space is everything northeast of the lines.*Completely defined by Thiamine and Niacin.*Three corner points:
A=0 & B=20A=10 & B=0A=4.4 & B=2.2
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The Objective (Level Curve)Price of Cereal A = 3.8Price of Cereal B = 4.2
C = 3.8A + 4.2 B◦Rewrite this isolating B (the y-axis
variable)B = C/4.2 – (3.8/4.2)A
◦Cost measured in units of Cereal B Intercept is determined by C, so we want the
lowest intercept Slope converts A to units of B
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Model with Level Curves
0
2
4
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10
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18
20
0 2 4 6 8 10
Cere
al B
Cereal A
Thiamine Niacin Calories
Cost (LC when C = 20) Cost (LC when C = 30)
*Lower intercepts indicate lower costs*Objective is improving when we lower the intercept due to minimization*Cost = 20 is not feasible (no point gives both enough thiamine and niacin)*Cost = 30 is feasible but is not optimal.
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Back to economicsThis is an expenditure
minimization model◦Similar to the utility problem faced by
consumers.◦What is the cheapest way to reach a
target level of satisfaction.◦Indifference curves and an
isoexpenditure line…Also the input-input model of
producer problems that use isoquants.
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Economic Model
X1
X2
U = target level
*Expenditures decrease in this direction but must be large enough to meet the target utility. *Optimum is the tangency point.
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Basic versus Non-basic modelsBasic models have only one type
of constraint◦Maximization
<= constraints
◦Minimization => constraints
More realistic models will have more options and may have both types of constraints…
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Calories in the cereal problemWhat if we cared about both
◦Eating enough calories and◦Not eating too many calories
How would we model this?
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Algebraic Model Extended
;0;0:
700125100:2
500125100:1
00.525.000.1:
00.125.010.0:
..
2.48.3min
BAnegNon
BACal
BACal
BANia
BAThi
ts
BAC
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New Feasible Space
0
1
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0 1 2 3 4 5 6
Cere
al B
Cereal A
Thiamine Niacin Cal Min Constraint Cal Max Constraint
Feasible Space is the triangle between Niacin, Thiamine, and Max Cal constraints.