Very good. The second approach is what I am doing now--you are right. I agree with your concerns though, and I have to frequently quit to avoid spurious results.

BTY, why not write a book on Mathematica applications (if you haven't done so already)? I am sure it would be very helpful!

BTW, The second approach is what you are essentially doing now so that might be what you want.

You are right Neil. I think I made some mistakes the first time and got an error message. Now it works fine. Many thanks.

At the risk of abusing your generosity, I have one more question.

If I compute the minimization problem (or solve simultaneous equations problem), I get a set of results for the specified variables, e.g.

{p[1] -> 1.71304, p[2] -> 1.66576, p[3] -> 1.58799, 
 alpha[1, 2] -> 0.273479, alpha[1, 3] -> 0.386724, 
 alpha[2, 1] -> 0.331958, alpha[2, 3] -> 0.613276, 
 alpha[3, 1] -> 0.668042, alpha[3, 2] -> 0.726521}

Now, I need to substitute for these results, i.e. set

p[1] = 1.7130407333027033`
p[2] = 1.6657564453986469`
p[3] = 1.5879949741053119`
alpha[1, 2] = 0.273478839300758`
alpha[1, 3] = 0.3867235962022603`
alpha[2, 1] = 0.3319579814272776`
alpha[2, 3] = 0.6132764037035208`
alpha[3, 1] = 0.6680420181563871`
alpha[3, 2] = 0.7265211604700595`

I am doing it now manually. Is there a way to automate this process?

Thanks in advance for your help.

Best

One more question if our good friends don't mind.

How can I use "eqns" as a set of constraints in a minimization problem? Recall that:

eqns = Table[DemandEquation[i] == SupplyEquation[i], {i, 1, n}]

where n could be 2 or 3 or more. A constraint requires "And" or "&&" between the equations, but I could not put these equations together using &&. Any suggestions?

Best

Sami,

I would suggest generating your equations with:

eqns = Table[DemandEquation[i] == SupplyEquation[i], {i, 1, 2}]

or

eqns = MapThread[Equal, {DemandFunctions, SupplyFunctions}]

You can get the list of variables as the number of equations grows:

vars = Table[p[n], {n, 1, 2}]

And solve with

Solve[eqns, vars]

Neil: Here is how it worked for me.

Assume 2 markets, each with one supply and one demand function. Total of four functions. Of course, the number of markets, n, can be adjusted easily.

In[1]:= SupplyEquation[x_] := s1[x]*p[x]^s2[x]

In[2]:= SupplyFunctions = Table[SupplyEquation[n], {n, 1, 2}]

Out[2]= {p[1]^s2[1] s1[1], p[2]^s2[2] s1[2]}

In[3]:= DemandEquation[x_] := d1[x]*p[x]^d2[x]

In[4]:= DemandFunctions = Table[DemandEquation[n], {n, 1, 2}]

Out[4]= {d1[1] p[1]^d2[1], d1[2] p[2]^d2[2]}

In[5]:= Solve[DemandFunctions[[1]] == SupplyFunctions[[1]] && 
  DemandFunctions[[2]] == SupplyFunctions[[2]], {p[1], p[2]}]

Out[5]= {{p[1] -> E^((-Log[d1[1]] + Log[s1[1]])/(d2[1] - s2[1])), 
  p[2] -> E^((-Log[d1[2]] + Log[s1[2]])/(d2[2] - s2[2]))}}

Sami,

Is this what you want?

supplyEqn[x_] := supply[x] == a[x] + p[x]^b[x]

Get a list of equations with

In[3]:= eqns = Table[supplyEqn[n], {n, 1, 3}]

Out[3]= {supply[1] == a[1] + p[1]^b[1], supply[2] == a[2] + p[2]^b[2],
  supply[3] == a[3] + p[3]^b[3]}

Now you can use Solve, Minimize, etc. on those equations.

I hope this helped.

Regards,

Neil

Hi Sami,

I don't really understand why you need different functions if they all have the same definition. Anyway, here is a way to do that.

ToExpression@
    StringTemplate[
      "supply`i`[p`i`_, a`i`_, b`i`_] := a`i` + p`i`^b`i`"][<|"i" -> #|>] & /@ Range[10];

Definition@supply1
(* supply1[p1_, a1_, b1_] := a1 + p1^b1 *)

Hi Sami,

If all of the functions have the same form why are different functions needed?

supply[p_, a_, b_] := a + p^b
supply[2, 3, 4]
(* 19 *)

19 = 3 + 2^4