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# [SOLVED] Solve for Iterative Variable, 'What If --> 'Goal Seek' in Excel

Posted 10 years ago
 Hi, How could I solve for a variable that generates another set value? I'll write a specific example with a known solution as an example. "Solving for:" unknownVariable = ? "Setup:" beginningValue = 1; endingValue = 0.03; increments = 9; intermediateValue1 = beginningValue * unknownVariable; intermediateValue2 = intermediateValue1 * unknownVariable; "..." endingValue = intermediateValue8 * unknownVariable; "Solution to be solved for:" unknownVariable = 0.68  Thanks, Greg
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Posted 10 years ago
 One could recast this in Mathematica as below. f[1] = 1; f[n_] := f[n - 1]*x Solve[f[10] == .03 && x > 0, x] (* Out[234]= {{x -> 0.677315868387}} *) Or just take (.3)^(1/9).
Posted 10 years ago
 That's exactly what I needed- thanks!
Posted 8 years ago
 Hello, Kind of a related question. But what if I am trying to Repeat Calculations/Iterations without a specific function?I am trying to repeat an initial condition many times such that I get a desired final output. ONE final output depends on many subsequent calculations from ONE initial condition. But I want to get a specific final output. I have many given parameters Clear["Global*"] tid = 50; taid = 20; tcid = 30; rho = 0.975; pe1 = 4.37; pe2 = 4.4509; qedom = 1021167; qeim = 5700; qcid = 529.59; el = 0.655; did = 4.22; dtot = 430932.474; iad = 445002; cad = 688723.8554; qetot = qedom + qeim; uel = 1 - el; sigid = 0.20568; rid = 0.02958; upid = N[Exp[Sqrt[1]*sigid]]; dwid = N[1/upid]; rf = N[Exp[rid*1]]; pid = N[(rf - dwid)/(upid - dwid)]; qid = N[1 - pid]; prf2 = 0.001*pe2*qetot-0.01*cad; npva = ((rho - rho^(taid + 1))/(1 - rho))*prf2 - 0.01*iad; I would encourage you NOT TO CHANGE the values of the parameters(you are free to change the names if you wish); these are from real data and heavily influence the results. The following parameter "cprd" is the INITIAL CONDITION (given the other parameters) that I want to change many times exch=63.56; cprd = 200; pin = exch*cprd;(*63.56 is a FIXED constant and "pin" is the product exch*cprd*) prf1[p_] = 0.001*pe1*qedom*el - 0.001*p*qcid - 0.001*pe1*qeim; As you see, ALL THE FOLLOWING calculations depends on ONE chosen value of "pin" and if I change "cprd" all the results would change. stsp = Table[pin*(upid^i), {i, -tid, tid}]; Dimensions[stsp] (*checking this gives a column vector*) {101} lensp = Length[stsp] 101 prf1[stsp]; per[i_] := prf1[stsp][[i]]; up[n1_, sig_, T_] := N[Exp[Sqrt[T/n1] sig]]; down[n1_, sig_, T_] := N[1/up[n1, sig, T]]; int[n1_, Rf_, T_] := N[Exp[Rf*(T/n1)]]; P[up_, down_, al_] := N[(al - down)/(up - down)]; Q[up_, down_, al_] := N[1 - P[up, down, al]]; mean[l_List] := Apply[Plus, l]/Length[l]; AmericanOption[p0_, n1_, sig_, T_, Rf_, exercise_Function] := Module[{u = up[n1, sig, T], d = down[n1, sig, T], al = int[n1, Rf, T], p, q, OpRecurse, res}, p = P[u, d, al]; q = Q[u, d, al]; OpRecurse[node_, level_] := OpRecurse[node, level] = If[level == n1, exercise[p0*d^node u^(level - node)], rho*{p, q}.{OpRecurse[node, level + 1], OpRecurse[node + 1, level + 1]} + exercise[p0*d^node u^(level - node)]]; res = OpRecurse[0, 0]; Clear[OpRecurse]; res]; AmericanPut[p0_, n1_, sig_, T_, Rf_] := AmericanOption[p0, n1, sig, T, Rf, # &] AmericanPut[per[1], tcid, sigid, tcid, rid] (*just checking*) 95832.1 tabnpvc =Table[AmericanPut[per[i], tcid, sigid, tcid, rid], {i, 1, lensp}]; e1 = 0; newp = Table[e1, {i, lensp}, {j, lensp}]; Dimensions[newp] {101, 101} Table[newp[[i, Min[i + 1, lensp]]] = pid, {i, lensp}]; Table[newp[[i, Max[i - 1, 1]]] = qid, {i, lensp}]; valtern = Max[#, npva] & /@ tabnpvc; The matrix "val" is the MOST IMPORTANT. I want to REPEAT THE ITERATIONS (by changing "cprd") such that difference between the 1st and the 50th column val[[All, 1]]-val[[All, 50]] is first positive and then zero (vector contains positive numbers followed by zeros). val = Table[e1, {i, lensp}, {j, tid + 1}]; val[[All,tid + 1]] = tabnpvc; Do[Table[val[[All, i]] = Max[#, npva] & /@ (prf1[stsp] + rho*newp.val[[All, i + 1]]), {i, 1, tid}], {tid}]; val // MatrixForm; Dimensions[val]; diff = val[[All, 1]] - val[[All, 50]]; (*this vector should ideally contain positive values FOLLOWED BY zeros but NOT negative*) pos = Position[diff, _?(# < 0 &)] (*this is not desired*) {{36}, {37}, {38}, {39}, {40}, {41}, {42}, {43}, {44}, {45}, {46},{47}} Finally, just checking with figures valplot[i_, j_] := {stsp[[i]], val[[All, j]][[i]]}; val1 = ListLinePlot[Table[valplot[i, 1], {i, 1, lensp}], PlotRange -> All]; val2 = ListLinePlot[Table[valplot[i, 50], {i, 1, lensp}], PlotRange -> All]; Show[val1, val2]; (*this should be such that the curves don't ever cross but converge to the value of "npva"*) So MY DESIRED FINAL OUTPUT is that the vector "diff" should contain BOTH positive numbers and zeros. It should first be positive and then zeros (maybe "pos" should equal the null vector {}). As you see, you can keep changing the value of "cprd" until you get the desired resultI have no idea on how to do incorporate all this in a "While" or "Do" function. Please help. Thank You.
Posted 10 years ago
 I don't know much about Excel, but the simplest type of thing would be to use Solve or Reduce e.g. Solve[ 24*x==96, x] Note some differences in syntax, especially use "==" for Equal as "=" is reserved for Set. Note that you can use natural language inputs, e.g. in Wolfram Alpha, if you find this easier.Wolfram Language is pretty powerful in solving equations, and on the other hand there are more choices for how to solve something. Sometimes there is no answer (like a real x with x^2==-1) and you can then use minimization to get the closest possibility. Since solving equations can become arbitrarily hard I would guess that solving is better done in Wolfram Language than in Excel.The idea of searching for a value is an important one in Wolfram's way of doing things. For example, find the smallest elementary cellular automaton that displays random behavior from a single black cell? Sounds hard. You don't need to understand what this is, just make a table of graphics and pick out the answer. Table[ArrayPlot[CellularAutomaton[rule, {{1},0},{100,All}],PlotLabel->rule],{rule,0,255}] `(See Stephen Wolfram's A New Kind of Science )