Round answers to THREE decimal PLACES as stated
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Givenarrival rate =18 customers per hour\therefore \lambda=18a) Now, in this single server operationService rate =28 customers per hour\therefore \mu=28Liq or Average waiting time spent by Customer\omega_{q}=\frac{\lambda}{\mu(\mu-\lambda)} \Rightarrow(1Puting values of \lambda_{\text {ste }} in equation (T)\begin{aligned}\Rightarrow \frac{18}{28(10)} & =0.0642 \text { hours } \\& =3.857 \text { minutes }\end{aligned}b) Now for multi server model, we will first calculate the po\text { Here, } \lambda=18, \mu=20Important formula needed are\begin{array}{l}P=\frac{\lambda}{s \mu}(s \text { is no. of server } \\P_{0}=\left[\sum_{n=0}^{s-1}\left(\frac{\lambda}{n !}\right)^{n}+\left(\frac{\lambda}{n !}\right)_{s !}^{s}\left(\frac{1}{1-p}\right)\right]^{-1} \\L_{\theta}=\frac{P_{0} \times\left(\frac{\lambda}{u}\right)^{s} \times p}{\frac{s}{n}(1-p)^{2}} \quad \text { and } \omega_{q}=\frac{1 q}{\lambda}\end{array}Now calculating p=\frac{18}{2 \times 20}=\frac{9}{20}\begin{array}{l}\omega_{q}=\frac{L q}{\lambda}=\frac{0.228532}{18}=0.01269 \text { hour } \\ =0.762 \text { minutes } \\\end{array} ...