C5Read.pdf 1 Chapter 5: The Economic Approach Applied to Single Variable Optimization A Brief Review of the Economic Approach ¥ The Economic Approach can be applied to optimization problems or equilibrium systems. Economic models are abstract, simplified descriptions of the optimization problem or equilibrium system. ¥ As applied to individual decision-makers, the Economic Approach says that agents act as if they optimize. In other words, they do the best they can under given conditions. ¥ The choices agents make are interpreted as being efficient (or optimal or best) from the perspective of the agent. ¥ The often seemingly ridiculous assumption of optimization (or rationality) is used because (1) it organizes variables in a coherent fashion, (2) agents really do face choices and optimization is one way to make decisions, (3) it helps us understand whats going on, and (4) it generates testable predictions. Did you know that lifeguards are very much aware of the problem they face? They are trained to neither jump right in the water nor to choose the path of least water! 1 ¥ Economists do not claim that individuals actually use sophisticated mathematics, rather, economists use mathematics to interpret observed variables with the aid of the optimization framework. Lifeguards are not aware of the mathematics involved in finding the optimal distance to run, then swim. They are conscious of their goal and the fact that either too little or too much running will raise the time it takes to get to the victim. They are trained to use rules of thumb, instinct, experience, and common sense to solve the problem. ¥ Assuming that individuals act as if they optimize enables accurate predictions to be made about the choices observed under varying conditions. It is the economist who imposes the optimization framework in order to make predictions about the lifeguards changes in behavior (in this case, the distance chosen to run on the sand) as an exogenous variable changes (say, their top running speed). ¥ Often, the optimization framework has provided good predictions of actual decisions. For many analysts, this is the acid test: to them, if a theory makes good predictions, then it doesnt matter to them how unrealistic are the assumptions of that theory. Having come up with a prediction, the economist would observe lifeguards in action in order to see if the predictions matched the empirical data. 1 In fact, Wabash College swim coach Gail Pebworth tells us that lifeguards must know their beach. Lifeguards are trained to constantly recalculate the optimal distance as the time of day and season changes on any particular beach.
Transcript
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Chapter 5: The Economic ApproachApplied to Single Variable Optimization
A Brief Review of the Economic Approach
¥ The Economic Approach can be applied to optimization problems or equilibrium systems.Economic models are abstract, simplified descriptions of the optimization problem orequilibrium system.
¥ As applied to individual decision-makers, the Economic Approach says that agents act as ifthey optimize. In other words, they do the best they can under given conditions.
¥ The choices agents make are interpreted as being efficient (or optimal or best) from theperspective of the agent.
¥ The often seemingly ridiculous assumption of optimization (or rationality) is used because(1) it organizes variables in a coherent fashion, (2) agents really do face choices andoptimization is one way to make decisions, (3) it helps us understand whatÕs going on, and (4)it generates testable predictions.
Did you know that lifeguards are very much aware of the problem they face? They aretrained to neither jump right in the water nor to choose the path of least water!1
¥ Economists do not claim that individuals actually use sophisticated mathematics, rather,economists use mathematics to interpret observed variables with the aid of the optimizationframework.
Lifeguards are not aware of the mathematics involved in finding the optimal distanceto run, then swim. They are conscious of their goal and the fact that either too little ortoo much running will raise the time it takes to get to the victim. They are trained touse rules of thumb, instinct, experience, and common sense to solve the problem.
¥ Assuming that individuals act as if they optimize enables accurate predictions to be madeabout the choices observed under varying conditions.
It is the economist who imposes the optimization framework in order to makepredictions about the lifeguardÕs changes in behavior (in this case, the distance chosento run on the sand) as an exogenous variable changes (say, their top running speed).
¥ Often, the optimization framework has provided good predictions of actual decisions. Formany analysts, this is the acid test: to them, if a theory makes good predictions, then it doesnÕtmatter to them how unrealistic are the assumptions of that theory.
Having come up with a prediction, the economist would observe lifeguards in actionin order to see if the predictions matched the empirical data.
1In fact, Wabash College swim coach Gail Pebworth tells us that lifeguards must Òknow their beach.ÓLifeguards are trained to constantly recalculate the optimal distance as the time of day and seasonchanges on any particular beach.
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The Structure of Optimization Problems
1) Setting Up the Problem
A) Objective function (e.g., utility or profit)
¥ Consumers are utility maximizers¥ Firms are profit maximizers¥ Students may be GPA maximizers¥ The lifeguard is a time minimizer
B) Endogenous variables are things the agent CAN choose;C) Exogenous variables are things the agent CANNOT control.
¥ Consumers choose goods; they cannot control prices¥ Firms choose output or (maybe) price, they cannot control a competitorÕs moves¥ Students choose the amount of time they want to study, not the content of the course¥ The lifeguard chooses the amount of time to run on the sand, not the location of thedrowning victim
Agent ObjectiveFunction
EndogenousVariable
ExogenousVariable
Consumer Max utilityType andnumber ofgoods
Prices,Income,Tastes andPreferences
CompetitiveFirm Max profit Output Price, cost
conditions
Monopoly Max profit Output, PriceMarketdemand, costconditions
Student Max GPATime SpentStudying ineach course
Ability,coursecontent,professor'sgrading scale
Lifeguard Min time Distance torun on sand
Top velocityon land andin water,location ofvictim
Sometimes agents have constraints on their choices. That is they face some restriction on thepossible values of the endogenous variables. For example, available income restricts therange of options in the consumerÕs choice problem. We will talk about this kind ofCONSTRAINED OPTIMIZATION problem later.
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2) Finding the Optimal Solution
An example of an optimization problem is the lifeguard problem you explored inC4Lab.xls. The optimal value of Distance_in_Sand (how far to run before jumping in andswimming) was found using ExcelÕs Solver. After you have Set Up the Problem, Solverdoes the work for you and determines the optimal value of the endogenous variable. Inearlier work (C3Lab.xls explored the profit maximization problem), we saw how tables andgraphs (using either the Direct Method or Method of Marginalism) could be used to find anoptimal solution.
Now we will see yet another way to solve optimization problems.
ASIDES:¥ Note how the steps for these various solution strategies are identical. Try to spot theconsistent patterns across the different solution strategies.
¥ There is no one best strategy. The idea is that by exploring these different ways to do theEconomic Approach, we learn and understand.
Solving the Lifeguard Problem with the Method of Marginalism via Calculus
Review of the Problem: You are a lifeguard who can run faster than you can swim. Youspot a drowning victim. What path should you take to minimize the time necessary toreach the victim? The essential trade-off is that running is faster than swimming, but if youswim all the way, the distance is shortest.
You are here
The drowning victim is here
SAND
What is the path of least time?
WATER
Nota t ion:
T = time to reach victim (in minutes)x = distance along sand the lifeguard runs (in yards)b = length of perpendicular connecting victim to beach (in yardsc = distance of victim from lifeguard along sand (in yards)vs = velocity of lifeguard running on sand (in yards/minute)vw = velocity of lifeguard swimming in water (in yards/minute)
x is a choice variable for thelifeguard. He or she can set it at x=0(which means jumping right in) orx=c (which means running down thebeach until he or she is directlyacross from the victim beforejumping in).
c
b
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Step 1: Setting Up the Problem
1) What is the Objective Function (Goal)?
minimize the time it takes to reach the victim (T)
¥ recall that dividing distance (in say, yards) by velocity (yd/min) gets you time (min)¥ total time to victim is the sum of running time and swimming time¥ once you choose how far to run on the sand, the swimming distance is also determined
because the lifeguard makes a bee-line (along the hypotenuse) for the victim
2) What are the Endogenous Variables?
distance to run along the sand (x) before jumping into the water
3) What are the Exogenous Variables?
velocity on sand (vs),velocity in water (vw),position of victim (that is, b and c)
Step 2: Finding the Optimal Solution
The problem can be stated mathematically in this way:
T =xvs
minx
+(c - x)2 + b2
vw
¥ The variable x, under the Òmin,Ó identifies the endogenous variables in the problem. It isclear that there is only one choice variable in this problem.
¥ The first term on the right-hand side (x/vs) is the time it takes to run on the sand; thesecond is the time it takes in the water.
¥ We divide distance (in yards or miles or feet) by velocity (in yards per minute or milesper hour or feet per second) to obtain travel time.
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Before we continue, letÕs quickly review the other methods of solving this problem:
A Review of the Direct Method (using Totals):
Two alternative ways of applying it:
(1) TABLE: For any set of values of the exogenous variables, have Excel calculate the valueof the function for different values of x. Choose the value of x that gives the lowest valuefor the function, total time to victim.
Solving the Lifeguard Problem Using the Direct Method
(2) GRAPH: Have Excel graph the value of the function for different levels of x, andchoose the x that corresponds to the lowest value of the function, total time.
A Totals Graph of the Lifeguard Problem
min
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
0 10 20 30 40 50 60 70 80 90 100
Time on Sand(minutes)
Time in Water(minutes)
Time to Victim(minutes)
Distance on Sand (yards)
Although the optimal distance on sand is about 65 yards, that U-shaped curve (or bowl) ispretty flat. That means that mistakes to the right or left (too much or too little running onsand) aren't that big a deal. Unless, of course, that half-second is the difference betweenlife and death!
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Finding the Optimal Solution with the Method of Marginalism viaCalculus
As usual, there are a series of steps that provide a framework for reaching the correctanswer. We provide the steps for you and encourage you to refer back to this pagewhenever you solve an optimization problem with the Method of Marginalism viaCalculus.
The steps assume that you have I) SET UP THE PROBLEM and are now ready for II)FINDING THE INITIAL SOLUTION
STEP (1) Set up the problem mathematically; that is, write out the objective function,indicate the choice variable, and indicate what the agent wishes to do (i.e., MAX or MIN).
STEP (2) Take the derivative of the objective function with respect to the choice variable.
STEP (3) Set the derivative of the objective function equal to zero. The value of theendogenous variable that satisfies this condition (called a Òfirst-order conditionÓ bymathematicians) is unique (for the simple problems we will be dealing with in thiscourse); we indicate this by sticking a star or asterisk (*) on the choice variable.
This is a crucial point. A choice variable, say X, can be set by the agent to manydifferent values. One (and only one) of these values is the optimal (or best orefficient) solution for the problem at hand. This special value, X*, is denoted by theasterisk convention.
STEP (4) Solve for the optimal value of the endogenous variable; that is, rearrange theequation (or first-order condition) in Step 3 so that you have X* by itself on the left-handside and only exogenous variables on the right. This is called a reduced form: it tells youthe optimal value of the endogenous variable for any set of values for the exogenousvariables.
STEP (5) Plug the optimal value back into the objective function. This will give you whatis called Òthe maximum value functionÓ. You can think of it as a sort of reduced form ofthe objective function: it gives the maximum value (or minimum value, in the case of, forexample, the lifeguard problem) of the objective function as a function of the parametersof the problem.
Let's put these steps to work in solving the lifeguard problem. We repeat that you shouldalways follow these steps carefully when solving an optimization problem with theMethod of Marginalism via Calculus.
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We have already Set Up the Problem , so we are prepared to apply the steps discussed onthe previous page.
STEP (1) Set up the problem mathematically; that is, write out the objective function,indicate the choice variable, and indicate what the agent wishes to do (i.e., MAX or MIN).
( )ws
x v
bxc
v
xT
22
min+−
+=
This can be rewritten in the following form which is easier to work with:
minx
s w
Tv
xv
c x b= + −( ) +[ ]1 1 2 21
2
STEP (2) Take the derivative of the objective function with respect to the choice variable.2
( )[ ] ( )( ))22
111 21
22xcbxc
vvdx
dT
ws
−−+−+=−
This derivative is a little complicated. We had to use the chain rule to get it. DonÕt worryabout that right now.
This expression can be evaluated at any value of x. The crucial idea is that thereÕs onespecial value of x, the optimal value, that can be found when the derived expression isequal to zero. Thus, the next step is to set the derivative equal to zero.
STEP (3) Set the derivative of the objective function equal to zero. The value of theendogenous variable that satisfies this condition (called a Òfirst-order conditionÓ bymathematicians) is unique; we indicate this by sticking a star or asterisk (*) on the choicevariable.
dT
dx v vc x b c x
s w
= + −( ) +[ ] − −( )( ) =−1 1 1
22 0
2 21
2* *)
Many students don't see the need to separate steps 2 and 3. After all, the only difference isthat Step 3 carries an "=0" and a few stars ("*"). But what a difference that is! Youcan take a derivative of any function at any time. Doing so will always report how thefunction changes. Setting the derivative equal to zero is a special step in the application ofthe derivative to the solving of optimization problems. The appendix at the end tries toemphasize this crucial point.
When you set the derivative of the objective function equal to zero and put asterisks on thechoice variables, you have made an important claim. By putting the Ò*Ó superscript on thechoice variable, you are saying that this is the optimal value. In essence, you have foundthe answer to the problem. All that remains is to present the solution in a nicer form.
2 We offer a helpful appendix at the end of this reading on the "derivative." Go to it now ifyou would like to learn more about the derivative before continuing through this example.
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STEP (4) Solve for the optimal value of the endogenous variable; that is, rearrange theequation (or first-order condition) in Step 3 so that you have the optimal value of the evariable by itself on the left-hand side and only exogenous variables on the right. This iscalled a reduced form: it tells you the optimal value of the endogenous variable for any setof values for the exogenous variables.
Applying this step to the equation in step 3 and solving for x* you obtain (after muchtedium):
bvv
vcx
ws
w
22
*
−−=
This is the answer. It tells you the optimal value of x* (distance on sand) for any value ofthe exogenous variables. All that remains is to substitute this optimal value into theobjective function in order to find the minimum time to the victim.
The expression above holds only for x* ≥ 0. If
bvv
vc
ws
w
22 −<
, then x* =
0. It doesnÕt make any sense for the lifeguard to run away from the
victim! One intuitive explanation for the corner solution is that if b is
much higher than c, then running is not at all worth it and the
lifeguard should jump right in.3
STEP (5) Plug the optimal value back into the objective function. This will give you whatis called Òthe maximum value functionÓ. You can think of it as a sort of reduced form ofthe objective function: it gives the maximum value (or minimum value, in the case of, forexample, the lifeguard problem) of the objective function as a function of the parametersof the problem.
T ∗ =
c − b vw
v 2 s − v 2
w
v s
+
�
� � � � � b vw
v 2 s − v 2
w
�
� � � � � 2
+ b2
v w
Now, most people would say, "Yecch! What a mess!!" or maybe even, "What on earth isthat thing?!?" But, actually, although admittedly seemingly complicated, it's merely anexpression that says, "The absolute minimum (i.e., fastest) time to the drowning victimdepends on b, c, vw, and vs." Given any configuration of those four exogenous variables,the expression above will immediately reveal T*, the minimum time to the victim. That'spowerful.
3 Mathematically, whatÕs happening is that minus x is being treated as a negative number in the Texpression, when, in fact, we should be using the absolute value of x. Minus x indicates travelingin the opposite direction, but such a move cannot possibly result in negative running time!Including |x| in the T expression doesnÕt seem worth the effort since we know the lifeguard willnever run away from the victim.
x*=0
c
b
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ASIDE ON MATH:We realize that cranking through a series of equations is probably not your idea of a greattime. That's why we hid a lot of the work. It turns out that the lifeguard problem isinteresting, but a bit cumbersome. Keeping the variables straight and not making anymistakes requires a little patience. WeÕre teaching you the Economic Approach, notmathematics.
LetÕs test out our answers to x* and T*:c = distance on sand = 100 yards
b = distance on water = 100 yardsvs = velocity on sand = 300 yards/minutevw = velocity on water = 100 yards/minute
x* =c -b⋅vw
(vs2 - vw
2)1 / 2
= − ∗−
≈100100 100
300 10064 6
2 2. yards
This is the same answer that Excel's Solver gave us, remember?
To find out what the minimum time, T*, is, we plug the values of the exogenous variablesinto our Step 5 maximum value function
T ∗ =
c − b vw
v 2 s − v 2
w
v s
+
�
� � � � � c −
�
� � � � � c −
b vw
v 2 s − v 2
w
�
� � � � � �
� � � � �
2
+ b2
v w b = 100
c = 100
v s = 300
v w
= 100
=
100 − 100• 100
3002 − 1002
300 +
�
� � � � � 100 − �
� � � � � 100 −
100• 100
3002 − 1002
�
� � � � � �
� � � � �
2
+ 1002
100
= 1 . 28 minutes
This, of course, perfectly agrees with our solution from the Solver (in C4Lab.xls).
To be sure, working with complicated expressions with several variables is not easy, butneither is it so fantastically complicated that it is beyond the reach of all but the mostbrilliant mathematicians. All we have done is found precisely where the objectivefunction has a flat spot (where the first derivative expression is equal to zero) and thenworked with the equation to isolate the optimal value of the endogenous variable, x*, onthe left-hand side. The power of the resulting reduced-form expression should beevidentÑany set of values for the four exogenous variables can be immediately pluggedin and x* can be calculated. The same holds true for the maximum value function, T*. Itmay be a messy expression, but once we have it, it is quite powerful.
The bar at theend tells you toevaluate theexpression at thevalues given.
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In fact, the ability to quickly find how x* and T* change given a shock (a change in a singleexogenous variable, holding all others constant), is the main reason why calculus is soprevalent in economics.
To see a demonstration of this, read on!
Step 3: Comparative Statics
In general, economists are less interested in finding a single answer to an optimizationproblem than in seeing how the optimal choice of an agent changes when the environmentin which she makes that choice changes. This is called comparative statics.
The exercise of comparative statics allows you to make predictions (either qualitative orquantitative) about the real world. These predictions can be tested to see if your model ofobserved behavior is any good. The test is simple: Is the prediction accurate?
Comparative statics takes the following form: InitialShockRecalculate NewCompare
Shock means change one of the exogenous variables while holding all of the othersconstant. Recalculate New means find the NEW optimal value (for example, by using theSolver).
TWO EXAMPLES:
Example 1: Suppose we solve the firmÕs profit maximization problem and determine itsoptimal output Q* as a function of the exogenous variables including output price. Wemay be interested in finding out how the optimal choice changes when the price of theproduct changes or when production inputs become more expensive.
Example 2: In the lifeguard problem, we may be interested in how the optimal choice ofdistance to run on sand changes when the location of the swimmer changes or when youget a lifeguard with a different top swimming or running speed.
Note: It is VERY important that you change only one exogenous variable at a time andobserve its effect on the optimal choice of the endogenous variable. If you change morethan one parameter (say, cost of labor and price in the profit max problem or lifeguardwater velocity and position of the victim in the lifeguard problem), it is extremely difficult todisentangle the effects in order to say how one variable alters the optimal choice. This is the heartof the difficulty facing the econometrician.
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TWO WAYS OF DOING COMPARATIVE STATICS:
The size of the shock, a finite distance (say a change in Price from $3/unit to $4/unit)versus an infinitesimally small shock, determines two alternative ways of doingcomparative statics.
As we saw in our study of the derivative in C3Lab.xls, as the size of the change getssmaller and smaller, the finite distance method approaches the method based on thecalculus. You will see what we mean when we work through an example.
Comparative Statics Method 1: Method of Actual Comparison
In this method, you simply redo the problem with the new value of the parameter inquestion. In the lifeguard problem, you were asked to run the Solver twice for differentrunning velocities of the lifeguard (300 yards/minute and 250 yards/minute). LetÕscompare your results:
X*(vw=100, c=b=100, vs=300) = 64.64 yards for a minimum time of 1.28 minutesX*(vw=100, c=b=100, vs=250) = 56.36 yards for a minimum time of 1.32 minutes
If you run Solver again with a running velocity of 200 yards/minute (an even slowerlifeguard) you will obtain:
X*(vw=100, c=b=100, vs=200) = 42.26 yards for a minimum time of 1.37 minutes
These data can be graphed in order to be understood more easily. The idea is that X*, theoptimal distance to run on the sand, depends upon the lifeguard's foot speed. Thus, thegraph is plotted with X* on the vertical axis and vs on the horizontal axis. It looks like this:
Top_Velocity_on_Sand Optimal_Distance_on_Sand
1 5 0 10.562 0 0 42.262 5 0 56.363 0 0 64.64
Comparative Statics Presentation Graph
0.00
20.00
40.00
60.00
80.00
0 100 200 300 400
Op
tima
l_D
ista
nce
_o
n_
Sa
nd
Top_Velocity_on_Sand
Notes:¥ the relationship between distance run on sand to sand velocity is a positive one: thefaster the lifeguard, the longer she should run on sand¥ the relationship appears to be non-linear: in other words, a one yard/minute increase inrunning speed does not uniformly increase optimal distance. It appears that the slope isdecreasingÑthe higher the lifeguard velocity on sand, the less a one yard/minute increasein speed will increase the time she should run on sand.
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Presentation Graphs:The graph of x*=Ä(vs|b, c, vw) (read "x star as a function of vs holding constant b, c, andvw") is called a Òpresentation graphÓ because it depicts how the optimal value of anendogenous variable changes as an exogenous variable changes, ceteris paribus.
The presentation graph on the previous page, reproduced below for ease of comparison,was constructed by TRACKING the optimal distance on sand for a given top velocity onsand as shown below:
A Totals Graph of the Lifeguard Problem
1.260
1.280
1.300
1.320
1.340
1.360
1.380
1.400
1.420
0 10 20 30 40 50 60 70 80 90 100
X (Distance on Sand)
vs = 30 0
vs = 25 0
64.6456.36
Comparat ive Stat ics Present at ion Graph
0.0 0
2 0.0 0
4 0.0 0
6 0.0 0
8 0.0 0
0 100 200 300 400
Opt imal_Dis tanc
Top_Velocity_on_Sand
6 4.6 4
5 6.3 6
How to Read a Presentation Graph:
One important question to ask in order to understand a presentation graph is, ÒWhat doesit mean to be off the curve?Ó
In the underlying graph, the lifeguard is certainly free to run other than 64.64 yards whentop velocity in sand is 300 yards/min. The point is that we will probably NOT observe thelifeguard doing this because points above or below the curve yield higher time to victim.
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In other words, this presentation graph depicts the connected minimum points of time tovictim given various values of top velocity in sand.
"Connected minimum points" is a difficult, but extremely important concept. The totalsgraph shows two alternative lifeguards and their optimal distance on sand values; whilethe presentation graph shows only the optimal solution for any lifeguard with a givenconfiguration of exogenous variables (vs, b, c, vw).
"Why is this so important?" Because there are essentially two kinds of graphs ineconomicsÑunderlying graphs and presentation graphs. Understand how presentationgraphs are created and read and you overcome a major hurdle to the understanding ofeconomics.
We could also have created a presentation graph of T*=Ä(velocity in sand). It would havesimilar properties to the x*=Ä(velocity of sand) presentation graph. Instead of trackingX*=Ä(vs|b, c, vw), we would plot T*=Ä(vs|b, c, vw). WeÕll examine this later.
Comparative Statics Method 2: Method of the Reduced Form
The Method of Actual Comparison works really well when you have access to Excel. TheSolver is quick, once youÕre accustomed to it, and you can very easily and quickly generatea list of the different values of x* when you vary one of the parameters. (WeÕll learn aneasy way to do this in the next lab using something called the Comparative Statics Wizard,which saves the values of x* and the different parameter values for you to work withlater).
Another method for generating comparative statics, i.e., to answer the question, ÒWhathappens when the environment changes?Ó is to use the mathematical expression for thereduced form. LetÕs talk about the reduced form in a Q&A format.
What is a reduced form?
The reduced form is an expression for the optimal value of the endogenous variable(s) interms of the exogenous variables.
How do I get a reduced form?
You get it by solving the optimization problem mathematically. For example, in thelifeguard problem, we took the derivative of the objective function with respect to x, set itequal to zero, and solved for x. The equation X*=Ä(vs, b, c, vw) is the reduced form. Whenwe substituted it into the objective function (step 5), we got a reduced form ofT*=Ä(exogenous variables), which has a special name, the maximum value function.
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What does the reduced form tell me?
It tells you how the optimal value of an endogenous variable is related to the exogenousvariables. From here, predictions are easy.
What is the advantage of the Method of the Reduced Form over the Method of ActualComparison (with Excel's Solver)?
Once you know the reduced form, you donÕt have to re-solve the problem: the reducedform tells you everything you want to know about how the optimal value of anendogenous variable changes when an exogenous variable changes.
So I should always use the Method of the Reduced Form over the Method of ActualComparison (with Excel's Solver)?
No. Sometimes the math to get the reduced form can be pretty cumbersome. It's prettyeasy to make algebra mistakes and other errors. In fact, sometimes, it can be impossible(and so the problem is said to have "no analytical solution" or "no closed form"). In thesetough cases, the Method of Actual Comparison is the only way to go.
Do the Method of the Reduced Form and the Method of Actual Comparison (with Excel'sSolver) always give the same answer?
Ouch! That's a tough question! The answer is "It depends." If the reduced form is linear inthe exogenous variable under consideration, then the Method of Actual Comparison (withExcel's Solver) and the Method of the Reduced Form give the exact same answer. But, fthe reduced form is non-linear in the exogenous variable under consideration, then theMethod of Actual Comparison (with Excel's Solver) and the Method of the Reduced Formgive different answers. We will explore this in much detail.
TWO EXAMPLES OF THE METHOD OF THE REDUCED FORM:
Example 1: We saw that the reduced form for x* was
x ∗ = c − b vw
v 2 s − v 2
w
Suppose that you are interested in seeing how x* varies when vs changes, other thingsequal. If you plug in numbers for all the OTHER exogenous variables, your reduced formbecomes
xvs
* = − ∗−
100100 100
1002 2
This tells you how x* varies for different values of vs, other things equal. You can see bylooking at the expression that x* is not a linear function of vs, just as the presentation graphwe drew above suggested. If you were really interested, you could take the derivative ofx* with respect to vs and find out exactly what the slope looks like and how it changeswhen vs changes.
Note how the presentation graph is a graph of the reduced form, x*=Ä(vs|b, c, vw).
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Example 2: Suppose that you were interested on how x* changed when the victim isfurther out from the beach (that is, when b increases)? Substituting for all of the otherparameters, we obtain:
xb
*.
= −1002 83
¥ Clearly x* is negatively related to b: the further out from the beach the victim is, the lesstime you should run on the sand.
¥ Furthermore, the relationship is linear: every yard further from the beach reducesoptimal distance on sand by 1/2.83 = 0.35 yard.
¥ You can plug in different values for b to get the corresponding x*. For example:When b = 100 (and vs = 300, vw = 100, c = 100), then x* = 64.64.When b = 200 (and vs = 300, vw = 100, c = 100), then x* = 29.29.
Thus, increasing b by 100 yards leads to approximately a 35 yard decrease (64.64 - 29.29 _35) in x*.
A presentation graph would be:
X* as a Function of b
0
2 0
4 0
6 0
8 0
5 0 150 250
b (yards)
x *
So, in this case, how are the Method of Actual Comparison and the Method of the ReducedForm related to each other?
In this case, they are the same because x* is linear in b. However, x* is non-linear in vs sothe two methods of comparative statics would give different answers. This is a reallydifficult idea and we will continue discussing it in the next few chapters.
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SUMMARY
Optimization problems have three steps:I. Setting up the problem (identifying the objective function and the exogenous andendogenous variables)II. Solving the problemIII. Generating comparative statics.
¥ The solution method, Step 2, can be either direct using totals (graphs or tables) or canrely on changes using marginals. Economists generally use marginals. ExcelÕs Solver is ayet another alternative.
¥ The Direct Method and Marginalism can be done using pencil and paper and your ownmath skills or by using Excel. Excel is much faster.
¥ Comparative statics, Step 3, involves changing one of the exogenous variables in theproblem and observing how the optimal choice changes. You can use either the Method ofDirect Comparison or the Method of the Reduced Form.
¥ If the reduced form is linear, the two methods of doing comparative statics give thesame answer; if not, they don't.
The material in this reading is difficult. Please do not panic. We will be using Excel to putinto practice the lessons stated here. You will have several opportunities to learn theseconcepts. Good luck!
Take the time to read the appendix and return to it as needed for help in understandingthe role of derivatives in optimization and comparative statics.
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Appendix: Some Information About Derivatives
What is a derivative?
A derivative is a mathematical expression that tells you how a function changes when theendogenous variable (or one of the endogenous variables, if there are several) changes a little bit.
Graphically, it is the slope of the function at that particular value of the endogenous variable.A function can change in two waysÑlinearly and non-linearly.
¥ Linear functions have a constant slope and, therefore, a constant value for the derivative. If the derivativeof a function is 0.3, for example, this tells you that every time the x variable goes up by 1, the y variable goesup by 0.3. This is true regardless of whether you increase x from 1 to 2 or from 3.000 to 3.001. An exampleof a linear function of x is y = 0.7 + 0.3x; its derivative with respect to x is 0.3. Notice how x does not appear inthe derivative. A mathematician would say, ÒIn this case, y is linear in x.Ó
y
x
dy/dx
x1 2 3
0.3
.71.01.31.6
This is a picture of a linear function of x
This is a picture of the derivative of a linear function of x
¥ Non-linear functions have a changing slope and, therefore, a derivative that takes on different values atdifferent values of x. If the derivative is increasing in x, that means that the slope is getting larger as x getslarger. If the derivative is decreasing in x, that means that the slope of the function is getting smaller as youmove out along the x axis. An example of a non-linear function of x is y = x2; its derivative, dy/dx, is 2x.Notice how x appears in the expression for the derivative. A mathematician would say, ÒIn this case, y is non-linear in x.Ó
y
x
dy/dx
x1 2 3
This is a picture of a non-linear function of x. Note that the slope is increasing as x increases.
This is a picture of the derivative of that non- linear function of x. Note that the value of the slope is increasing as x increases.
Keep clear in your mind the difference between the function itself and the slope of the function.
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Some Uses of Derivatives in Economics
(1) Solving Optimization Problems
An optimization problem typically requires you to find the value of an endogenous variable (or variables)that maximizes or minimizes a particular objective function. We can use derivatives to find that value (viathe Method of Marginalism using calculus) because of the following mathematical fact:
at the maximum or minimum value of functions that we will deal with in this course, the slope iszero (that is, the value of the derivative is zero). So all we need to do is find the derivative and thensolve for the value of the endogenous variable that makes the derivative equal to zero.
The equation that you make when you set the first derivative equal to zero is called the first-ordercondition. The first-order condition (FOC) is different from the derivative because the derivative by itself isnot equal to anythingÑyou can plug in any value of x and the derivative will pump out an answer that tells youwhether and by how much the function is rising or falling at that point.
When you solve the first -order condition for the optimal value of the endogenous variable, the expressionthat you get when you is called the reduced form.
Below are some graphs that show how the derivative can be used to solve an optimization problem:
x
y
x
yAt the maximum, the slope = dy/dx = 0
At the minimum, the slope = dy/dx = 0
x*x*Above are pictures of the functions; below are pictures of their derivatives.
y
x
dy/dx
xx*x*
To the left of x*the derivative is positive but decreasing; to theright of x* the derivative is negative.
To the left of x*the derivative is negative and falling (getting less negative); to the right of x*the derivative is positive and increasing.
The two graphs on the left show that if you want to maximize something by choosing x, find the value of xwhere the derivative is zero; the two graphs on the right show that if you want to minimize something bychoosing x, find the value of x where the derivative is zero.
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A Few Words About The Second DerivativeThe second derivative is simply the derivative of the first derivative. It tells you the slope of the
function that describes the slope of the function itself. For example, if a function has a constant slope, wesaw that its first derivative is a constant value (like 0.3 in the first example). The second derivative of thefunction tells you how the slope changes when x changes ... Well, since the slope is unchanging, the secondderivative would be zero.
Second derivatives are useful for the following reason: when you find the value of the endogenousvariable that makes the first derivative equal to zero, the point that you have located could be either amaximum or a minimum (see the set of pictures above). If you want to be sure which one you have found,you can check out the second derivative. If you plug in the proposed optimal value of x into the secondderivative and you find that the value of the second derivative is negative, this tells you that the slope of thefunction is declining when you move a little past x*. If the slope is declining after you leave x*, it mustmean that x* was the top of the function, i.e., that x* gave you the maximum value of the function. If on theother hand you plug x* into the second derivative and you find that its value is positive, this tells you that ifyou move a little past x* the slope starts increasing. This must mean that x* gave you the minimum value ofthe function. So, the second-order condition for a maximum is that the value of the second derivative at x* isnegative; the second-order condition for a minimum is that the value of the second derivative is positive atx*. (This is for single-variable optimization problems; the second-order conditions for multi-variableproblems are more complicated.)
The above is merely by way of explanation. Our primary focus is not mathematics and we will notwork any problems where the second-order conditions do not hold. You should understand, however, whata second derivative is and how it is used.
(2) Exploring Comparative Statics
Once you have used derivatives to solve for the reduced form of an endogenous variable, you have anexpression for the optimal or best value of the choice variable as a function solely of the exogenous (orenvironmental) variables. Economists are interested in how the optimal choice changes when one of thoseenvironmental variables changes. To explore this, you can take the derivative of the optimal value (x*) withrespect to the parameter you are interested in. Once you get that expression for the derivative, you donÕt setit equal to zero and solve for anything since you are not trying to find the optimal value of anything Ñthatpart is already done. What you are interested in is TELLING A STORY ABOUT HOW THE OPTIMALVALUE IS RELATED TO THE PARAMETERS OF THE PROBLEM. Basically, you should look at yourderivative and ask yourself two questions:
¥ Is the relationship between the exogenous variable and x* always positive, always negative, ordoes it depend?¥ Is the relationship between the exogenous variable and x* linear or non-linear? If the exogenousvariable does not appear in the expression for the derivative, then the relationship is LINEAR. If theexogenous variable DOES appear in the expression for the derivative, then the slope depends on thevalue of the exogenous variable and the relationship is NON-LINEAR.
ConclusionEconomists use derivatives for (1) Solving Optimization Problems and (2) Exploring the Comparative Staticsproperties of a reduced form. These are two different uses of the derivative and should not be confused. Inparticular, the derivative of a reduced form with respect to an exogenous variable is not a second derivative.It is merely another, different first derivative of a function. This common confusion can be avoided if youunderstand exactly where you are in the Economic Approach.
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Rules for Taking Derivatives
Let x be the variable; a, n, and e are constants (e = 2.7183....)General Rule Example of its Application
