Round answers to THREE decimal PLACES as stated

Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

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} ...