# Please ans this....how to write the equation into Matlab Code

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Kunwar Pal Singh on 26 Apr 2021
Commented: Kunwar Pal Singh on 27 Apr 2021
Walter Roberson on 26 Apr 2021
To search for a maximum, minimize the negative of the result equation that I posted.
You can use lower bound and upper bound to apply those constraints. I suspect that t_1 should have a lower bound of 0.

Walter Roberson on 26 Apr 2021
Edited: Walter Roberson on 27 Apr 2021
%these variables must be defined in a way appropriate for your situation
S_N = rand() * 10
S_N = 5.6100
theta = randn() * 2 * pi
theta = -9.9489
l = randi([2 10])
l = 4
b_1 = rand()
b_1 = 0.4189
c_11 = rand()
c_11 = 0.7008
t_year = randi([1950 2049])
t_year = 1960
d_11 = rand()
d_11 = 0.3502
t_1 = rand()
t_1 = 0.1989
t_x = t_1 + rand()
t_x = 0.5704
lambda_a = randi([500 579])
lambda_a = 551
LOTF_a = rand()
LOTF_a = 0.1140
P = rand()
P = 0.7418
K_l = rand()
K_l = 0.7342
k_0 = rand()
k_0 = 0.2623
t_tau = randi(10)
t_tau = 7
overhaulcost_a = 1000 + rand()*100
overhaulcost_a = 1.0864e+03
%the work
syms t
part1 = int(S_N .* cos(theta) .* l .* b_1 .* t_year .* d_11, t, t_1, t_x);
part2 = int(lambda_a .* LOTF_a, t, t_1, t_x);
part3 = int(P*K_l .* t_year + P .* k_0 .* l .* l .* t_tau, t, t_1, t_x);
part4 = overhaulcost_a ;
result = part1 - part2 - part3 - part4;
result
result =
Kunwar Pal Singh on 27 Apr 2021
ok thanx

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