How to use the solve function with a changing parameter?

Hi all,

I am working on an allocation problem that changes over time. I am hoping to find quantities a,b,c,d given this system of 4 equations. My problem is, I need to add a for loop where t changes symbolic of years. How can I input changing parameters to this solve function? When I run as it is it says "explicit solution cannot be found". I know this is not true because when I manually enter the numbers it works. I need t to ultimately be t=1:100 so that is why I do not want to do this without a loop.

 for T=1:5

       t=double(T);

     % 532K  
     syms a b c d t;
    [a, b, c, d]=solve('(a/(1.01^t)*1.499424825623625e+06)^(-1/.3)-977.55=(b/(1.01^t)*1.658846640009515e+07)^(- 1/0.75)-977.55','(a/(1.01^t)*1.499424825623625e+06)^(-1/.3)-977.55=(c/2.415456478597783e+07)^(- 1/1.5)-30','(a/(1.01^t)*1.499424825623625e+06)^(-1/.3)-977.55=(d/5.477225575051662e+05)^(-1/0.5)- 30','a+b+c+d = 532000');
    MI_in1=double(a);
    MI_out1=double(b);
    Irr_crop1=double(c);
    Irr_veg1=double(d);
    Quant1=[MI_in1,MI_out1,Irr_crop1,Irr_veg1];

 end

Answer 1

[Roger Stafford]

There is an important difference between solving your equations symbolically in terms of t, as compared with solving them with t as a given single numerical value. Apparently 'solve' is able to do the latter but not the former. (I would be very surprised if it could do the former with your equations.)

Change your for-loop so that 't' is not a symbolic quantity but assumes each of the numerical values t = 1:100, one at a time. For each value of 't' store the corresponding values of a, b, c, and d in appropriate arrays as you solve them.

An alternate method would be to solve algebraically for 'b', 'c', and 'd' in terms of 'a', in the first three equations and after substituting the solutions into the fourth equation, you have a single equation in a single unknown, 'a'. Then call on 'fzero' for each value of 't' to find a corresponding 'a'. Having found it, then you can easily determine 'b', 'c', and 'd'.