I got x as a result of the symbolic equation solver!

Hi all, When I run the below code I got the following answer, and I don't know what does x mean:

ans =

struct with fields:

p1: x

parameters: x

conditions: x < -((lambda + v)*(delta - 2))/2 & (12*B*delta + 24*delta*x + 4*B*delta^3 + 8*delta^2*lambda…

This a part of my whole code, but in this part I am trying to find the optimal p1 which maximize the totalprofit. Obviously the function totalprofit is not quadratic but still I couldn't get the meaning of x. Can someone please help?

Thanks in advance.

syms n p1 lambda delta v N F B 

n=(delta - B*delta + 2*delta*v + (4*B*delta - 4*delta - 8*delta*lambda + 8*delta*p1 - 8*delta*v - 2*B*delta^2 + 4*delta^2*lambda + 4*delta^2*v + delta^2 + B^2*delta^2 + 4)^(1/2) - 2)/(2*delta) assume(0<n & n<v & v-B/2+(B^2/4+B/2+lambda-v+1/4)^(1/2)-1/2<=n & B>0 & lambda>0 & 0<delta & delta<1) assume((delta - 1)*(B/2 - v - (B^2/4 + B/2 + lambda - v + 1/4)^(1/2) + lambda/(delta - 1) + 1/2)<p1 & p1<-((lambda + v)*(delta - 2))/2)
p2 =simplify(lambda/2 + n/2 - (n - v)^2/2 - (B*(n - v))/2)
D1=simplify(v-n)
D2=simplify(n-p2+lambda-D1^2+B*D1)
profit1=simplify(p1*(D1)-F*(lambda)^2)
profit2=simplify(p2*(D2))

totalprofit=simplify(profit1+profit2)
firstdifofp1_totalprofit=simplify(diff(totalprofit,p1)) 
solve(firstdifofp1_totalprofit,p1, ReturnConditions=true)

[Aquatris]

You might want to check here. I tried it but it returned empty. Maybe there is no solution to the problem i dont know :P

[melda]

I couldn't relate that one with mine.. You may right there might be no solution but in that case doesn't matlab return empty ? When I remove assumptions it give me z1 instead of x. So, I am thinking maybe it is because of the lenght of my solution?

[Aquatris]

As @John D'Errico explained, you have a condition under which you have a solution. The link I had in my comment solves the condition and finds the solution that satisfies the condition. However in your example it returns empty saying it cannot find explicit solution. Maybe I was not clear when I said 'no solution'. What I meant was 'no unique solution'.

Answer 1

[John D'Errico]

syms n p1 lambda delta v N F B
n=(delta - B*delta + 2*delta*v + (4*B*delta - 4*delta - 8*delta*lambda + 8*delta*p1 - 8*delta*v - 2*B*delta^2 + 4*delta^2*lambda + 4*delta^2*v + delta^2 + B^2*delta^2 + 4)^(1/2) - 2)/(2*delta);
assume(0<n & n<v & v-B/2+(B^2/4+B/2+lambda-v+1/4)^(1/2)-1/2<=n & B>0 & lambda>0 & 0<delta & delta<1)
assume((delta - 1)*(B/2 - v - (B^2/4 + B/2 + lambda - v + 1/4)^(1/2) + lambda/(delta - 1) + 1/2)<p1 & p1<-((lambda + v)*(delta - 2))/2)
p2 =simplify(lambda/2 + n/2 - (n - v)^2/2 - (B*(n - v))/2);
D1=simplify(v-n);
D2=simplify(n-p2+lambda-D1^2+B*D1);
profit1=simplify(p1*(D1)-F*(lambda)^2);
profit2=simplify(p2*(D2));
totalprofit=simplify(profit1+profit2);
firstdifofp1_totalprofit=simplify(diff(totalprofit,p1));
solve(firstdifofp1_totalprofit,p1, ReturnConditions=true)

ans = struct with fields:
            p1: x
    parameters: x
    conditions: x < -((lambda + v)*(delta - 2))/2 & (12*B*delta + 24*delta*x + 4*B*delta^3 + 8*delta^2*lambda + 4*delta^4*lambda + 8*delta^2*v + 4*delta^4*v + 12*delta^3*x + 6*delta^2 + delta^4 +…

Simple enough.

The result is ANY number x that satisfies those bounds.

Let me create a much simpler problem as an example.

syms X
Xsol = solve((X^2 + 4*X + 4) <= 4,'returnconditions',true)

Xsol = struct with fields:
             X: [2×1 sym]
    parameters: x
    conditions: [2×1 sym]

I'll plot the function

fplot((X^2 + 4*X + 4))
yline(4)

So that is true when x >= -4 AND x<=0. Do you agree? And ANY number in that interval is a valid solution. Now look back at the conditions in Xsol.

Xsol.conditions

ans = 

So much less complicated than what you got, but much the same too. In your case, the solution is any number x that satisfies those inequalities. There is no unique solution.

[melda]

I got it thank you. If that is the case having no unique solution is not good for my model at all.