Only seven lines of code, the program runs continuously without any results when solving the equation

This is my code

syms I A
B = 0.02;%blocking probability
N = 800;%number of channels
FN = factorial(N);
ACell = solve((A^N)/FN == B*symsum(A^I/factorial(I),I,0,N),A);
A = double(ACell);
A = A(end)

[VBBV]

syms I A
B = 0.02;%blocking probability
N = 8;%number of channels very huge
FN = factorial(N);
ACell = solve((A^N)/FN == B*symsum(A^I/factorial(I),I,0,N),A);
A = double(ACell);
A = A(end)

A = 3.6271

Very huge value of N is chosen. Select a smaller number for N

[VBBV]

Factorial of 800!! is huge. Matlab interprets it as INF

[VBBV]

%% A huge number when N = 800
factorial(800)

ans = Inf

syms I A
B = 0.02;%blocking probability
N = 80;%number of channels very huge
FN = factorial(N);
ACell = solve((A^N)/FN == B*symsum(A^I/factorial(I),I,0,N),A);
A = double(ACell);
A = A(end)

A = 68.6881

[wei zeng]

Thank u for your answering!800 is indeed too large. I will improve my algorithm to calculate the corresponding answer for 800

[VBBV]

The sym method did not solve your actual problem equation, Walter modified the equation by applying log on both sides and then solved it. That's simplified the problem

[Walter Roberson]

The log makes it much much much faster !

Just one of those "tricks of the trade"

Answer 1

[Walter Roberson]

syms I A
B = sym(0.02);%blocking probability
N = sym(800);%number of channels very huge
FN = factorial(N);
eqn = (A^N)/FN == B*symsum(A^I/factorial(I),I,0,N);
ACell = vpasolve(log(lhs(eqn))==log(rhs(eqn)),A);
A = double(ACell)

A = 789.3325

[wei zeng]

Thank you for your answer. But I want to know why the sym method can solve the problem of the large number 800?

[Walter Roberson]

factorial(sym(800))

ans = 

771053011335386004144639397775028360595556401816010239163410994033970851827093069367090769795539033092647861224230677444659785152639745401480184653174909762504470638274259120173309701702610875092918816846985842150593623718603861642063078834117234098513725265045402523056575658860621238870412640219629971024686826624713383660963127048195572279707711688352620259869140994901287895747290410722496106151954257267396322405556727354786893725785838732404646243357335918597747405776328924775897564519583591354080898117023132762250714057271344110948164029940588827847780442314473200479525138318208302427727803133219305210952507605948994314345449325259594876385922128494560437296428386002940601874072732488897504223793518377180605441783116649708269946061380230531018291930510748665577803014523251797790388615033756544830374909440162270182952303329091720438210637097105616258387051884030288933650309756289188364568672104084185529365727646234588306683493594765274559497543759651733699820639731702116912963247441294200297800087061725868223880865243583365623482704395893652711840735418799773763054887588219943984673401051362280384187818611005035187862707840912942753454646054674870155072495767509778534059298038364204076299048072934501046255175378323008217670731649519955699084482330798811049166276249251326544312580289357812924825898217462848297648349400838815410152872456707653654424335818651136964880049831580548028614922852377435001511377656015730959254647171290930517340367287657007606177675483830521499707873449016844402390203746633086969747680671468541687265823637922007413849118593487710272883164905548707198762911703545119701275432473548172544699118836274377270607420652133092686282081777383674487881628800801928103015832821021286322120460874941697199487758769730544922012389694504960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

When you use factorial() of a symbolic number, you can get a full precision answer up to about 999; and functions can reason about the value of factorial() of larger integers even if larger values have delayed calculation.