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The interarrival times as well as service time at a single chair unisex barbershop have been shown to be exponentially distributed.

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.

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Solution:

Values are: $\lambda = 4$, $\mu = 6$

$\rho = \frac{\lambda}{\mu} = \frac{4}{6}=\frac{2}{3}$

$P_{0}-1-p=\frac{1}{3}$

$P_{1}=\frac{1}{3}\left(\frac{2}{3}\right)=\frac{2}{9}$

$P_{2}=\frac{1}{3}\left(\frac{2}{3}\right)^{2}=\frac{4}{27}$

$P_{\geqslant 4}=1-\sum_{n=0}^{3} P_{n}=\frac{16}{81}$

$L=\frac{\lambda}{\mu-\lambda}=\frac{4}{6-4}=\frac{4}{x}=2 \text{ customers}$
$\omega=\frac{L}{\lambda}=\frac{2}{4}=0.5$ 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