Economists look at how costs and benefits change as there are small changes in actions.  We call this marginal analysis, and it is perhaps the key concept in economic analysis. It is an acknowledgement that people (should) make a decision based on the incremental gains and losses that result from that decision, and that sunk costs (money, time or other things of worth already expended and unredeemable) should not matter.

Marginal analysis, quite simply, balances the additional benefits from an action against the additional cost. In any case, be it a firm deciding whether or not to expand production, a student deciding if another beer is a good idea, or a professor choosing to give an extra exam, optimal performance requires that benefits and costs be equilibrated on the margin. What this means is that if the additional benefit exceeds the additional cost, take the action. Keep taking it as long as the benefit exceeds the cost, and to ensure that all excess benefits (those that exceed costs) are accrued, do it until for the last action, the benefits just equal the costs.

The benefits from the last action (such as unit of production or consumption) are termed marginal benefits, and the costs from that action are termed marginal costs. Economic assumptions, verified by much experience, shows that for most actions the benefits per unit are falling, while the costs are increasing. Thus, one measure of economic efficiency is that marginal benefits equal marginal costs. At that point, all the units for which benefits exceed costs are used. Too little, and some excess benefits are wasted; any more, and the costs for later units exceed the benefits.

Politicians, of course (economists love to beat up on politicians), often fail to ignore sunk costs, which economists usually refer to as fixed costs. In 1990, the National Aeronautics and Space Administration (NASA) launched into orbit the Hubble Space Telescope, a new orbiting telescope that by being in space avoided atmospheric distortions from astronomical observations. Astronomers expected vast new gains and insights in their scientific explorations.

Unfortunately, someone goofed. While being tested after being in orbit, scientists and engineers at NASA discovered some flaws in the mirrors used in the telescope that significantly diminished the ability of the telescope to gather signals from deep space. The scientists were devastated, but immediately set upon ways to rectify the problem. Of course the solution was costly.

Politicians were outraged, some calling the Hubble Space Telescope a $2 billion boondoggle and debacle. There was a strong movement to deny any more funds to the project, since NASA had not gotten it right initially. But marginal analysis reveals a much different perspective.

The $2 billion or so dollars already spent on the Hubble Telescope did not matter. What was relevant at that point was what were the gains and losses from fixing the problem or leaving the telescope as it was. In its flawed state, the Hubble Telescope could still perform many useful and interesting scientific functions. Corrected, it could perform more. The only relevant question at that point was whether or not the additional scientific discoveries that would come from fixing the problems would be worth the cost of the repairs. The initial expenditure to build and launch the Hubble Space Telescope did not matter anymore.

People seem to violate this maxim all the time, counting sunk or fixed costs in decision making. Politicians and bureaucrats might do it because the cost of not doing so may include being accused of wastefulness, not being careful guardians of the publics fiscal well-being. And individuals often go to shows or take trips that they claim they do not want to because they have already bought the tickets. But that is the key. Since the tickets are already bought, the marginal cost of attending is probably very small, just some time and exertion. And the potential gains, the enjoyment from the activity, must be high. We know this because when buying the tickets the cost was high (it included the dollars spent) so the expected benefits must have been high as well. Now, although the expected benefits might be diminished somewhat by weariness, the costs are significantly lower, only the time spent in the activity, so the expected net benefits are still positive.
 

The Mathematics of Marginal Analysis

We can also have a function of more than one variable.  For example, the variable y can be a function of variables x, z and w.  These are called multivariate functions.  Common examples are:

y= x+z+w

y=2x2+6wz

y=log(x)+z+w2

y=ax2+bx+cw+dxw+fw2 where a, b, c, d and f are parameters

y=1/x+1/w

A very important functional form in economics is called the Cobb-Douglas function, and it has the form y=aX^aY^b where a, a and b are parameters.  We will use this function extensively when we do mathematical analysis in this class.

When we do marginal analysis we are just looking for the slope of the line described by the function.  If you were to graph a function, say y=3x, we can find the slope by seeing how much y changes as x changes.  The notation we use for the slope is Dy/Dx.  Try graphing this function, using several values for x, starting at 1 and getting bigger.  You'll see that for every 1 unit that x increases, y increases 3 units.  This is called a linear function, and the rate of change in y as x grows larger is constant.  The slope, which is just this rate of change, is 3 (the slope is always constant for the linear function).  When the function is nonlinear -- that is there is an exponent for other function, like a logarithm -- the slope changes at different points on the line.  Try graphing y=4x².  You'll see that the line gets steeper and steeper as x gets larger.  If you figure the slope, say from x=4 to x=5, and then from x=6 to x=7, you'll see that the slope is increasing.  If you graph y=1/x you'll find the opposite -- as x gets larger and larger the slope of the line gets smaller and smaller.

Differential calculus is the tool we use for finding the slope of a function.  As long as you keep thinking of derivatives as the slope of the line you would find if you graph the function, you'll know what you need to succeed in the MBA program.  There are lots of rules for finding derivatives, but most of what you need to know is summarized in the table found at:

http://web.mit.edu/wwmath/calculus/differentiation/table.html

We probably won't use anything beyond the first three rules shown in the table, and nothing more than the quotient rule.  The notation used for derivatives is dy/dx or df(x)/dx or f '(x).  The d are the notation meaning to take a derivative, as is the f ', and both mean take the derivative of the function [which is y or f(x)] with respect to x.  The function we will use most in this class is aXⁿ where X is the variable (argument) and a and n are constants.  If f(X)=aXⁿ then the derivative of the function, noted f''(X)=anX^n-1.  For example, if a=3 and n=5, then f''(X)=15X⁴.  The other common function used in business is g(X)=ln(X) where ln is the natural logarithm (sometimes it is written log rather than ln).  The derivative of this function with respect to X is 1/X, that is, g'(X)=1/X.

It may seem that if there are more than one variable on the right-hand-side, like the second set of examples given, that finding a derivative is impossible.  But the rules are really the same.  Just treat the extra variables as constants.  So if we have the function y=3xz and we take the derivative with respect to x, we get the value 3z.  that is, ¶y/¶x=3z.  Similarly, ¶y/¶w=3x.  Notice, we change the symbol used to denote a derivative.  These are called partial derivatives.
 

Optimization

When y is a function of more than one variable, for example when y=-x²-w²+10x=20w, we find the optimum by taking each partial derivative separately, setting them equal to zero, and solving the equations together.  So we have ¶y/¶x=-2x+10 and ¶y/¶w=-2w+20.  Setting each equal to zero we have -2x+10=0, or x=5, and -2w+20=0 so w=10. Take the second derivatives (¶²y/¶x² =-2 and ¶²y/¶w²=-2 ).  Since these are negative, if means we have a maximum.  Y reaches its highest value if x=5 and w=10.

Constrained Optimization

Constrained optimization is more difficult.  It is most common to run into constrained optimization with two variables.  Think of consumption -- you choose lots of goods, but must spend within your budget.  So you pick the best combination of goods you can afford.  This is constrained optimization.  We try to simplify things, so we'll use only 2 goods, and one budget.  So you are choosing among, say, food (F) and clothes (C), and have a budget of 1000.  Food costs 5 per unit and clothes 15.  Thus your budget constraint (the most you can spend) is 1000=5F+15C, and you choose F and C to get the most satisfaction out of that budget.

We use a function to represent the satisfaction.  Let's choose a simple one, like U=F*C, where U is utility (the amount of satisfaction you get).  Now, we want to maximize U, but there is a problem -- because we must spend within our budget.

The way to do this is called constrained optimization, and uses what is called the method of Lagrange.  We form a new function, call it L, where

L=F*C+m(1000-5F-15C)

where m is called the lagrange multiplier.  Any symbol can be used, although traditionally the Greek letter lambda is used.  For convenience, however, we'll just use m.  It is a new variable, which we create.  Notice, the term in parentheses is just the budget constraint.

Now, maximize L as you would any function -- which means set the partial derivatives with respect to the variables, F, C and  m equal to 0.

So, the partial derivative of L with respect to F is C - 5m = 0, with respect to C is F-15m = 0, and with respect to m is 1000-5F- 15C=0  (which is just the budget constraint).

We have three equations, C - 5m= 0, F-15m = 0, and 1000-5F- 15C=0 and three variables (unknowns) F, C and  m.  Solve these three equations, and you have the answer.

C = 5m

F=15m

1000=5F+ 15C

From the first two, F/C=3 so F=3C.  Use this in the third, 1000=30C, or C=1000/30.  then solve for F, and you have the best combination of F and C given your budget constraint.