Solve function is unable to return the results of System of equations in the desired single number format instead of fraction ? help

% In am interested in obtaining Steady state solution of 6 state ergodic

% continuous time markov chain where the long term probabilies in each

% state is given by Vector P = [ p1 p2 p3 p4 p5 p6] and

% State Transition rate matrix A wehre

% A =[-0.00074 0.00062 0.00012 0 0 0;

% 30 -30.00077 0.00031 0.00046 0 0;

% 0 0 -0.00012 0.00012 0 0;

% 0 0 0 - 0.00063 0.0004 0.00023;

% 2 0 0 0 -2 0 ;

% 0.2 0 0 0 0 -0.2]

% We obtain p1, p2 ...p6 by solving Equation 1 & 2 below:

% P*A = 0 .... Equation 1

% p1+p2+p3+4+p5+p6 = 1 ..... Equation 2

%

format long

syms eqn1 eqn2 eqn3 eqn4 eqn5 eqn6 eqn7 eqns sol p1sol p2sol p3sol p4sol p5sol p6sol p1 p2 p3 p4 p5 p6

eqn1 = -p1*0.00074+p2*30+ p5*2 + p6*0.2 == 0

eqn2 = p1*0.00062 - p2*30.00077 == 0;

eqn3 = p1*0.00012 +0.00031*p2 - p3*0.00012 == 0;

eqn4 = p2*0.00046 +p3*0.00012- p4*0.00063 == 0;

% eqn5 = p4*0.0004 - p5*2 == 0; we drop this equation and use eqn7 below

% for obatining p1, p2 ...p6

eqn6 = p4*0.00023 - p5*0.2 == 0;

eqn7 = p1+p2+p3+p4+p5+p6==1;

eqns = [eqn1, eqn2, eqn3, eqn4, eqn6, eqn7]

sol = solve([eqns],[p1,p2,p3,p4,p5,p6]);

format short

p1sol = sol.p1

p2sol = sol.p2

p3sol = sol.p3

p4so4 = sol.p4

p5so5 = sol.p5

p6so6 = sol.p6

p1sol = sol.p1

p2sol = sol.p2

p3sol = sol.p3

p4so4 = sol.p4

p5so5 = sol.p5

p6so6 = sol.p6

>> Solve_ergodic_ctmc

eqn1 =

30*p2 - (37*p1)/50000 + 2*p5 + p6/5 == 0

eqns =

[30*p2 - (37*p1)/50000 + 2*p5 + p6/5 == 0, (31*p1)/50000 - (8444466037051747*p2)/281474976710656 == 0, (3*p1)/25000 + (31*p2)/100000 - (3*p3)/25000 == 0, (23*p2)/50000 + (3*p3)/25000 - (63*p4)/100000 == 0, (23*p4)/100000 - p5/5 == 0, p1 + p2 + p3 + p4 + p5 + p6 == 1]

p1sol =

16625042510445626906250000/36395607773662890434476703

p2sol =

343575393447444480000/36395607773662890434476703

p3sol =

16625930080212032804490000/36395607773662890434476703

p4so4 =

3167094689375285334920000/36395607773662890434476703

p5so5 =

3642158892781578135158/36395607773662890434476703

p6so6 =

-26445240656283633798455/36395607773662890434476703

Answer 1

[John D'Errico]

Sure it is. You just need to convert the result from a fraction, into a floating point number. So either use double to create a double precision result, or use vpa to create a symbolic floating point number. Thus...

double(p1sol)

or

vpa(p1sol)

Note that setting format short did nohing, since the result that was displayed was in symbolic form.

Or, you could have used vpasolve instead of solve. vpasolve would return a floating point symbolic result directly.