Subs Error while using solve function

Hello!

I have question about using solve function.

I'm trying to derive yield to maturity using forward contract prices.

During the process, I tried to use solve function as shown below, which gives me error about

not able to transform data into double and recommending usage of 'subs' function.

I think i am doing this in the worst way possible, so Could you please give me the solution about this error?

Thanks.

T = rows(Data_full);
yield = zeros(T,1);
syms R P
for t = 1:T
    eqn = (5/2)/(1+(R/2))+(5/2)/(1+(R/2)^2)+(5/2)/(1+(R/2)^3)+(5/2)/(1+(R/2)^4)+(5/2)/(1+(R/2)^5)...
        +(5/2)/(1+(R/2)^6)+100/(1+(R/2)^6) == P;
    P = Data_full(t,1);
    yield(t) = solve(eqn,R);
end
yield

Answer 1

[Dyuman Joshi]

Allocate the value to P before defining the equation solve().

T = rows(Data_full);
yield = zeros(T,1);
syms R P

for t = 1:T
    
    %Define the value for P
    P = Data_full(t,1);
    
    %Equation to solve for 
    eqn = (5/2)/(1+(R/2))+(5/2)/(1+(R/2)^2)+(5/2)/(1+(R/2)^3)+(5/2)/(1+(R/2)^4)+(5/2)/(1+(R/2)^5)...
    +(5/2)/(1+(R/2)^6)+100/(1+(R/2)^6) == P;
    
    %Solution
    yield(t) = solve(eqn,R);
end
yield

Or you can directly use the value, no need to define P -

T = rows(Data_full);
yield = zeros(T,1);
syms R

for t = 1:T
    
    %Equation to solve for 
    eqn = (5/2)/(1+(R/2))+(5/2)/(1+(R/2)^2)+(5/2)/(1+(R/2)^3)+(5/2)/(1+(R/2)^4)+(5/2)/(1+(R/2)^5)...
    +(5/2)/(1+(R/2)^6)+100/(1+(R/2)^6) == Data_full(t,1);
    
    %Solution
    yield(t) = solve(eqn,R);
end
yield

[Dyuman Joshi]

I think fzero or vpasolve will be a better fit here.

Answer 2

[Torsten]

You will have to find criteria to sort out the solution you want. E.g. for real solutions only you could use

syms R
P = 10;
eqn = (5/2)/(1+(R/2))+(5/2)/(1+(R/2)^2)+(5/2)/(1+(R/2)^3)+(5/2)/(1+(R/2)^4)+(5/2)/(1+(R/2)^5)...
      +(5/2)/(1+(R/2)^6)+100/(1+(R/2)^6) == P;
SOL = double(solve(eqn,R))

SOL = 
   1.6229 - 1.1718i
   1.6229 + 1.1718i
  -1.4316 - 1.4125i
  -1.4316 + 1.4125i
   1.4324 - 1.4129i
   1.4324 + 1.4129i
  -0.6115 - 1.9172i
  -0.6115 + 1.9172i
   1.0190 - 1.7650i
   1.0190 + 1.7650i
  -2.3857 - 0.0221i
  -2.3857 + 0.0221i
   1.4932 - 2.4597i
   1.4932 + 2.4597i
  -1.4304 - 2.5567i
  -1.4304 + 2.5567i
   3.0832 + 0.0000i

SOL = SOL(abs(imag(SOL))<1e-5)

SOL = 3.0832