       Newbie with simple questions (take 2)

• To: mathgroup at smc.vnet.net
• Subject: [mg62322] Newbie with simple questions (take 2)
• From: misha <iamisha1 at comcast.net>
• Date: Sat, 19 Nov 2005 23:19:16 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```I am a new user (errr..purchaser) of Mathematica, but I have not been
able to find answers to these (probably) simple questions with
Mathematica?s help browser.  I am trying to use Mathematica to solve a
simple system of simultaneous equations.  I suppose I could use it to
solve the first order conditions (FOCs), but I?m having enough problems
as it is.  I have more ambitious goals than this, but I thought this
would be an easy place to start.

By the way, can anyone recommend a book heavy in examples for a
beginning user such as myself?

Here is the complete problem:

P(Q) = a - Q (inverse demand curve)

Q = q1 + q2 (Cournot) duopoly

C1(q1) = c*q1 (firm 1?s commonly known cost function, with constant
marginal cost, c)

C2 =  cL*q2 with probability t
cH*q2 with probability 1 - t

(firm 2?s cost functions for constant marginal costs cL < cH, known to
firm 2 but unknown with certainty to firm 1)

If Firm 2 has constant marginal cost cH, firm 2 chooses q2 to solve

max{[(a - q1* - q2) - cH]*q2}

If Firm 2 has constant marginal cost cL, then firm 2 chooses q2 to solve

max{[(a - q1* - q2) - cL]*q2}

The resulting FOCs are:

<<the asterisk denotes ?optimal? and the cH in parentheses denotes that
it is a function of cH>>

q2*(cH) = (1/2)*(a - q1* - cH)

q2*(cL) = (1/2)*(a - q1* - cL)

Similarly, Firm 1 chooses q1 to solve

max{t[(a - q1 - q2*(cH)) - c]*q1 + (1 - t)[(a - q1 - q2*(cL)) - c]*q1},

which yields FOC:

q1* = (1/2)*[t(a - q2*(cH) - c) + (1 - t)(a - q2*(cL) - c)]

So I want to use Mathematica to do the tedious algebra to get me the
following:

q2*(cH) = (a - 2cH + c)/3 + (1 - t)(cH - cL)/6

q2*(cL) = (a - 2cL + c)/3 - t(cH - cL)/6

q1* = (a - 2c + tcH + (1 - t)cL)/3

```

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