Trying to solve a reliability problem, to find lamda and mu, is there a way to
get a numerical result? I tried a couple different variations of solve, but
never got it in terms of just lambda or mu.
Thanks,
-d
lamda = var('lamda')
mu = var('mu')
a2 = .5*(3*lamda + mu) - (.5*sqrt((lamda^2 + 6*lamda*mu + mu^2)))
a3 = .5*(3*lamda + mu) + (.5*sqrt((lamda^2 + 6*lamda*mu + mu^2)))
k2 = (2*lamda^2)/(-a2*(a3 - a2))
k3 = (2*lamda^2)/(+a3*(a3 - a2))
R(t) = -k2*exp(-a2*t) - k3*exp(-a3*t)
Z = integral(-t*R(t), t, 0, oo)
print Z
#print solve([a2,a3,k2,k3,R,Z], lamda)
#solve(Z,lamda)
2*integrate((lamda^2*e^((-1.5*lamda - 0.5*mu + 0.5*sqrt(lamda^2 + 6*lamda*mu +
mu^2))*t)/(sqrt(lamda^2 + 6*lamda*mu + mu^2)*(-1.5*lamda - 0.5*mu +
0.5*sqrt(lamda^2 + 6*lamda*mu + mu^2))) + lamda^2*e^((-1.5*lamda - 0.5*mu -
0.5*sqrt(lamda^2 + 6*lamda*mu + mu^2))*t)/(sqrt(lamda^2 + 6*lamda*mu +
mu^2)*(1.5*lamda + 0.5*mu + 0.5*sqrt(lamda^2 + 6*lamda*mu + mu^2))))*t, t, 0,
+Infinity)
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