Why I can't get my output? What's wrong with line 13?

function [call,put,calldif,putdif]=bs(S,r,T,sigma,K,q)
  % S is current stock price
  % r is annualized risk free rate
  % T is time to expiration (in years)
  % sigma is annualized stock return standard deviation/volatility
  % K is strike price
  % q is annualized dividend rate
  temp1=(log(S/K)+(r-q+sigma^2/2)*T)/(sigma*T^0.5);
  temp2=temp1-sigma*T^0.5;
  call=S*exp(-q*T)*normcdf(temp1)-K*exp(-r*T)*normcdf(temp2);
  put=-S*exp(-q*T)*normcdf(-temp1)+K*exp(-r*T)*normcdf(-temp2);
  [a,b]=blsprice(S,K,r,T,sigma,q);
  [calldif,putdif]=[call,put]-[a,b]
  end

[the cyclist]

For future reference, it's better to post code, rather than an image of code, so that we can paste it into a file to test.

[Shawn Miller]

Sorry for this, the code is as follows,

function [call,put,calldif,putdif]=bs(S,r,T,sigma,K,q)
  % S is current stock price
  % r is annualized risk free rate
  % T is time to expiration (in years)
  % sigma is annualized stock return standard deviation/volatility
  % K is strike price
  % q is annualized dividend rate
  temp1=(log(S/K)+(r-q+sigma^2/2)*T)/(sigma*T^0.5);
  temp2=temp1-sigma*T^0.5;
  call=S*exp(-q*T)*normcdf(temp1)-K*exp(-r*T)*normcdf(temp2);
  put=-S*exp(-q*T)*normcdf(-temp1)+K*exp(-r*T)*normcdf(-temp2);
  [a,b]=blsprice(S,K,r,T,sigma,q);
  [calldif,putdif]=[call,put]-[a,b]
  end

[Star Strider]

Is your code throwing an error? If so, please copy and paste the entire red text from your Command Window and paste it to a Comment here.

What are the sizes of ‘call’, ‘put’, ‘a’ and ‘b’?

[Shawn Miller]

>> bs(100, 0.1, 0.25, 0.5, 95, 0)
Error using  - 
Too many output arguments.

Error in bs (line 17)
[calldif,putdif]=[call,put]-[a,b]

call, put, a, b are all scalers. call and put are what I attempt to compute using my defined formula, and a and b are the values computed using blsprice function in MATLAB. Basically, what I am doing here is to compare the two results.

[Joseph Cheng]

shawn read the cyclist's answer to remedy your comment above.

Answer 1

[the cyclist]

[calldif, putdif] = [call,put] - [a,b]

is not valid MATLAB syntax. MATLAB can't distinguish how the right-hand variables should be sorted into the left-hand variables. The right-hand side is a 1x2 vector, and MATLAB doesn't "know" how you want that parceled out between the two output variables. It is similar to the (invalid) syntax

 [x,y] = [[1 2 3],[4 5]] - [[5 6 7],[8 9]]

In that case, the right-hand side is a vector of length 5, and MATLAB would not know how to split it up between x and y.

Instead, you could do

callputdiff = [call,put] - [a,b]
calldiff = callputdiff(1);
putdiff = callputdiff(2);

Alternatively, you might not need to separate them at all, just keeping them in the vector while you operate on it. Similarly, perhaps call and put could just have been in a vector themselves.

[Shawn Miller]

Many thanks, I think I'll just change my output to a vector called "callputdiff".