Finance 30210: Managerial Economics Optimization.
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Transcript of Finance 30210: Managerial Economics Optimization.
Finance 30210: Managerial Economics
Optimization
Optimization deals with functions. A function is simply a mapping from one space to another. (that is, a set of instructions describing how to get from one location to another)
BAf :f
A
B
is a function
Is the range
Is the domain
A Bf
For example
For 53 xxf
Ax
50 x
5,0A ]20,5[B
3
14
Domain
205 y
Range
143 f
Function
y53 xy
For 50 x
0 5Domain
5
20
Ran
ge
X =3
Y =14
x
y 53max x50 x
0 5Domain
5
20R
ang
e
Optimization involves finding the maximum value for y over an allowable domain.
Here, the optimum occurs at x = 5 (y = 20)
x
y
x5
1max
100 x
What is the solution to this optimization problem?
105
There is no optimum because f(x) is discontinuous at x = 5
x
y x2max
60 x
0 6
12
What is the solution to this optimization problem?
There is no optimum because the domain is open (that is, the maximum occurs at x = 6, but x = 6 is NOT in the domain!)
x
y xmax
0x
0
12
What is the solution to this optimization problem?
There is no optimum because the domain is unbounded (x is allowed to become arbitrarily large)
x
The Weierstrass theorem provides sufficient conditions for an optimum to exist, the conditions are as follows:
x
y
x
x
The domain for is closed and bounded
is continuous
over the domain of
xf )(max xf
x
xfxxfxf
x
)()(lim'
0
xfFormally, the derivative of is defined as follows:
All you need to remember is the derivative represents a slope (a rate of change)
Actually, to be more accurate, the derivative represents a trajectory
Finding maxima/minima involves taking derivatives….
y
xxx x
Graphically…
x
xfxxfslope
)()(
xf
xxf
x
xfxxfxf
x
)()(lim'
0Now, let the change in x get arbitrarily small
y
xx
xf
x
xf
*x
0' xf
x
xf
*x
0' xf
0' xf 0' xf 0' xf 0' xf
First Order Necessary Conditions
If *x
)(max xfx
0*)(' xf
)(min xfx
is a solution to the optimization problem
then
or
Useful derivatives
Exponents1)(')( nn nAxxfAxxf
x
AxfxAxf )(')ln()(
LogarithmsLinear Functions
AxfAxxf )(')(
Example:
4'
4)(
xf
xxf Example:
x
xf
xxf
12'
ln12)(
Example:
4
5
15'
3)(
xxf
xxf
An Example
Suppose that your company owns a corporate jet. Your annual expenses are as follows:
You pay your flight crew (pilot, co-pilot, and navigator a combined annual salary of $500,000.
Annual insurance costs on the jet are $250,000
Fuel/Supplies cost $1,500 per flight hour
Per hour maintenance costs on the jet are proportional to the number of hours flown per year.
Maintenance costs (per flight hour) = 1.5(Annual Flight Hours)
If you would like to minimize the hourly cost of your jet, how many hours should you use it per year?
Let x = Number of Flight Hours
x
xx5.1$1500$
000,750$min
0
xx
CostHourly 5.1$1500$000,750$
hrsxx
7075.1
000,75005.1$
000,750$2
First Order Necessary Condition
An ExampleH
ou
rly
Co
st (
$)
Annual Flight Hours
0*)('' xf 0*)('' xf
Slope is increasingSlope is decreasing
x
xf
*x
0' xf
x
xf
*x
0' xf
0' xf 0' xf 0' xf 0' xf
How can we be sure we are at a maximum/minimum?
For a maximization… For a minimization…
Let x = Number of Flight Hours
x
xx5.1$1500$
000,750$min
hrsxx
7075.1
000,75005.1$
000,750$2
First Order Necessary Conditions
Second Order Necessary Conditions
0000,500,1$
3
xFor X>0
Suppose you know that demand for your product depends on the price that you set and the level of advertising expenditures.
22 5.8.4010000,5, pApAApApQ
Choose the level of advertising AND price to maximize sales
When you have functions of multiple variables, a partial derivative is the derivative with respect to one variable, holding everything else constant
First Order Necessary Conditions
010),( pAApQp
06.140),( ApApQA
22
0,05.8.4010000,5max pApAAp
Ap
10
010
pA
pA
010),( pAApQp
06.140),( ApApQA
50
40
06.24
0106.140
06.140
A
p
p
pp
Ap
With our two first order conditions, we have two variables and two unknowns
pAApQp 10),(
ApApQA 6.140),(
its generally sufficient to see if all the second derivatives are negative…
01),( ApQpp
06.1),( ApQAA
0*)('' xf
How can we be sure we are at a maximum?
Practice Questions
1) Suppose that profits are a function of quantity produced and can be
written as 222010 QQ
Find the quantity that maximizes profits
2) Suppose that costs are a function of two inputs and can be written as
2122
2121 32434 XXXXXXC
Find the quantities of the two inputs to minimize costs
Constrained optimizations attempt to maximize/minimize a function subject to a series of restrictions on the allowable domain
0
,max,
x,y gtosubject
yxfyx
To solve these types of problems, we set up the lagrangian
yxgyxfyx ,,),(
Function to be maximized Constraint(s)
Multiplier
To solve these types of problems, we set up the lagrangian
xgxfx )(
x
0
)(x
xf
*x
)(x xf
We know that at the maximum…
0)( xx
0xg
yxgyxfyx ,,),(
Once you have set up the lagrangian, take the derivatives and set them equal to zero
0,,),(
0,,),(
yxgyxfyx
yxgyxfyx
yyy
xxx
First Order Necessary Conditions
Now, we have the “Multiplier” conditions…
0, 0, 0 yxgyxg
Example: Suppose you sell two products ( X and Y ). Your profits as a function of sales of X and Y are as follows:
)(1.2010),( 22 yxyxyx
Your production capacity is equal to 100 total units. Choose X and Y to maximize profits subject to your capacity constraints.
100 yx
The key is to get the problem in the right format
0100
)(1.2010max 22
0,0
yxtosubject
yxyxyx
The first step is to create a Lagrangian
)100()(1.2010),( 22 yxyxyxyx
Objective FunctionConstraint
Multiplier
yxg ,
yxf ,
)100()(1.2010),( 22 yxyxyxyx
Now, take the derivative with respect to x and y
02.20),(
02.10),(
yyx
xyx
y
x
First Order Necessary Conditions
“Multiplier” conditions
0)100( 0100 0 yxyx
02.20
02.10
y
x
0 0100 yx
First, lets consider the possibility that lambda equals zero
0100 yx
50
02.10
x
x
100
02.20
y
y
05010050100 Nope! This can’t work!
02.20
02.10
y
x
The other possibility is that lambda is positive
0
0100 yx 0100 yx
x2.10 y2.20
50
50
2.202.10
xy
xy
yx
75
25
0250
050100
0100
y
x
x
xx
yx
yx 2.202.10
25
75
x
y05
Lambda indicates the marginal value of relaxing the constraint. In this case, suppose that our capacity increased to 101 units of total production.
Assuming we respond optimally, our profits increase by $5
Example: Postal regulations require that a package whose length plus girth exceeds 108 inches must be mailed at an oversize rate. What size package will maximize the volume while staying within the 108 inch limit?
X
Y
Z
Girth = 2x +2yVolume = x*y*z
10822
max0,0,0
zyxtosubject
xyzzyx
10822
max0,0,0
zyxtosubject
xyzzyx
First set up the lagrangian…
)22108(),,( zyxxyzzyx
Now, take derivatives…
0),,(
02),,(
02),,(
xyzyx
xzzyx
yzzyx
z
y
x
Lets assume lambda is positive 0
022108 zyx 022108 zyx
0
02
02
xy
xz
yz
xy
xz
xyyz
yz
2
02
02
yz
xyxz
xz
2
02
02
2
2
z x
z y
2 2 108
108
108
18
18
36
x y z
z z z
z
y
x
z
324xy
Suppose that you are able to produce output using capital (k) and labor (L) according to the following process:
5.5. lky
Labor costs $10 per hour and capital costs $40 per unit. You want to minimize the cost of producing 100 units of output.
klTC 4010
010
4010min
5..5
0,0
lktosubject
klkl
Minimizations need a minor adjustment…
yxgyxfyx ,,),(
A negative sign instead of a positive sign!!
)100(4010),( 5.5. lkllk
So, we set up the lagrangian again…now with a negative sign
Take derivatives…
05.40),(
05.10),(5.5.
5.5.
lklk
lklk
k
l
Lets again assume lambda is positive
05.40
05.105.5.
5.5.
lk
lk
01005.5. lk
5.5.
5.5.
20
05.10
kl
lk
5.5.
5.5.
80
05.40
lk
lk
kl
lkkl
4
8020 5.5.5.5.
200
50
1002
1004
1005.5.
5.5.
l
k
k
kk
lk
Suppose that you are choosing purchases of apples and bananas. Your total satisfaction as a function of your consumption of apples and bananas can be written as
6.4. BAU
Apples cost $4 each and bananas cost $5 each. You want to maximize your satisfaction given that you have $100 to spend
10054 BA
054100
max 6.4.
0,0
BAtosubject
BABA
First set up the lagrangian…
)54100(),( 6.4. BABABA
Now, take derivatives…
056.),(
044.),(4.4.
6.6.
BABA
BABA
B
A
054100
max 6.4.
0,0
BAtosubject
BABA
Lets again assume lambda is positive 0
AB
BABA
2.1
12.1. 4.4.6.6.
10054 BA
056.
044.4.4.
6.6.
BA
BA
6.1.
044.6.
6.6.
BA
BA
4.4.
4.4.
12.
056.
BA
BA
12
10
1002.154
10054
B
A
AA
BA
Suppose that you are able to produce output using capital (k) and labor (l) according to the following process:
5.5. lky
The prices of capital and labor are p and w respectively.Union agreements obligate you to use at least one unit of labor.
Assuming you need to produce y units of output, how wouldyou choose capital and labor to minimize costs?
1
min
5.5.
,
l
lky
tosubject
wlpklk
)1()(),( 5.5. lylkwlpklk
Just as in the previous problem, we set up the lagrangian. This time we have two constraints.
Doesn’t necessarily hold with equality
Will hold with equality
Non-Binding Constraints
.5 .5
.5 .5
.5 .5
( , ) .5 0
( , ) .5 0
0
1 0
k
l
k l p k l
k l w k l
y k l
l
First Order Necessary Conditions
01- 0 0)1( ll
)1()(),( 5.5. lylkwlpklk
0
05.
05.
5.5.
5.5.
5.5.
lky
lkw
lkp
First Order Necessary Conditions
Case #1: 1l
l
k
p
w
p
wll
p
wkly 2
0
1
w
pyl
2ypw
lp
wk
Constraint is non-binding
0
05.
05.
5.
5.
5.
ky
kw
kp
First Order Necessary Conditions
0Case #2: 1l
2yk
ypkp 22 22 ypypw
Constraint is binding
w
p
2ypw
w
pyl
p
wyk
Constraint is Non-Binding
Constraint is Binding
1
2
l
yk
Try this one…
You have the choice between buying apples and bananas. You utility (enjoyment) from eating apples and bananas can be written as:
BABAU 2, 5.
The prices of Apples and Bananas are given by BPAP and
Maximize your utility assuming that you have $100 available to spend
0 0
100$
2max 5.
,
BA
BPAP
tosubject
BA
BA
BA
BABPAPBABA BA 215. )100($2),(
(Income Constraint)
(Objective)
(You can’t eat negative apples/bananas!!)
ObjectiveNon-Negative Consumption Constraint
Income Constraint
BABPAPBABA BA 215. )100($2),(
0100$
02),(
05.),(
2
15.
BPAP
PBA
PABA
BA
BB
AA
First Order Necessary Conditions
We can eliminate some of the multiplier conditions with a little reasoning…
1. You will always spend all your income
2. You will always consume a positive amount of apples 01 0
0 0 0 22 BB
Case #1: Constraint is non-binding 0B 02
0100$
02),(
05.),( 5.
BPAP
PBA
PABA
BA
BB
AA
First Order Necessary Conditions
BA PP
A 25. 5.
B
A
P
APB
100$ AB PPB 400
Case #1: Constraint is binding 0B 02
0100$
02),(
05.),(
2
5.
AP
PBA
PABA
A
BB
AA
First Order Necessary Conditions
AA PA
P
A 100$
5. 5.
22 BPAB PP 4002
BP
AP
AB PP 40
Constraint is Non-Binding
Constraint is Binding
APA
B
100$
0
B
A
P
APB
100$2
25.
A
B
P
PA