Edit made on January 04, 2009 by ColinWright at 21:14:34
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HEADERS_END
Optimal pricing is a difficult subject, with many unquantifiable issues to take into account.
However, we can analyse a simple case, and build from there.
[[[>50 This model is pretty simplistic, but the principle is sound.
In particular, even if the price-sales function is more complex, each
part of it is, when you zoom in, an approximation to a straight-line.
The calculations still work too, but the function can't almost certainly
can't be described by a simple function, so it must be done numerically.
Additionally, the robustness of the solution should be considered. It
may be that small changes to the price/sales response curve causes the
solution to "leap" to another part of the curve, giving a completely
different "optimal" solution.
Finally, "optimal" is actually a relative term. Maximum profits might
not be the objective of a business in a complex market. Preventing
competition might be more important.
]]]
Suppose we have identical items to sell, each of which costs /c/ to make. We have a market
saturation at /N,/ which is as many as we can dispose of even charging nothing, and that the
sales drop off linearly as the price increases. That means our sales, /S,/ are given by a
formula:
EQN:S=N-\alpha{p}
where EQN:\alpha is the "drop-off factor" and is greater than zero, and /p/ is the price charged.
Then for a given price /p/ the profits are:
EQN:P=S(p-c)=(N-\alpha{p})(p-c)=Np-Nc-\alpha{p^2}+\alpha{c} EQN:\begin{eqnarray}P&=&S(p-c)\\&=&Np-Nc-\alpha{p^2}+\alpha{p}{c}\\&=&-\alpha{p^2}+(N+{\alpha}c)p-cN\end{eqnarray}
This is a downwards facing parabola with a maximum for some value of /p./ As a quick sanity
check, if /p=0/ then we lose /cN/ because we dispose of /N,/ items, each costing /c/ to provide.
Substituting EQN:p=N/\alpha we get 0, which is right because we sell none, hence make none, hence
no money changes hands anywhere.
We want to maximise this profit by adjusting /p/ to get the largest /P,/ which is not necessarily
obtained by getting the maximum sales!
We can use calculus to maximise /P./ We differentiate with respect to the thing we can change, /p,/
and we get this:
EQN:\frac{dP}{dp}=N-2{\alpha}p
We get a maximum (in this case - because the EQN:p^2 term is negative) when EQN:\frac{dP}{dp}=0,
and that happens when EQN:p=N/(2\alpha).
Exercises:
* Compute the profit when /p/ is this optimal value,
* Show that that varying the value of /p/ slightly up or down makes /P/ smaller
** ... thereby showing that the profit is a local maximum
* Show that the sales achieved are less than half the market saturation /N./
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