written 2.8 years ago by | modified 20 months ago by |
The values of λ and μ are 4 per hour and 6 per hour, respectively.Compute the steady-state parameters and the probabilities for zero, one, two three, and four or more customers in the shop.
written 2.8 years ago by | modified 20 months ago by |
The values of λ and μ are 4 per hour and 6 per hour, respectively.Compute the steady-state parameters and the probabilities for zero, one, two three, and four or more customers in the shop.
written 2.8 years ago by | • modified 2.8 years ago |
Solution:
Values are: $\lambda = 4$, $\mu = 6$
$\rho = \frac{\lambda}{\mu} = \frac{4}{6}=\frac{2}{3}$
$P_{0}-1-p=\frac{1}{3}$
$P_{1}=\frac{1}{3}\left(\frac{2}{3}\right)=\frac{2}{9}$
$P_{2}=\frac{1}{3}\left(\frac{2}{3}\right)^{2}=\frac{4}{27}$
$P_{\geqslant 4}=1-\sum_{n=0}^{3} P_{n}=\frac{16}{81}$
Steady state parameters are:
$L=\frac{\lambda}{\mu-\lambda}=\frac{4}{6-4}=\frac{4}{x}=2 \text{ customers} $
$\omega=\frac{L}{\lambda}=\frac{2}{4}=0.5$ hour
$w_{k}=w-\frac{1}{\mu}=1-\frac{1}{6}=\frac{5}{6}$ hour
$L_{\theta_1}=\frac{x}{ \mu ( \mu -\lambda)}=\frac{16}{6(6-4)}=\frac{16}{12} = \frac{4}{3}$ customers
$L= L_{\theta}+\frac{\lambda}{\mu}=\frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2$ customers