Correct Answer - Option 2 : 30 minutes, 22.5 minutes and 2.25 technicians
Concept:
Arrival rate = λ /hr
Service rate = μ /hr
\(\rho=\frac{λ}{μ}\)
Expected(avg) number of customers in queuing system or Average number of customers in the system:
\({L_s} = \frac{{{\rho }}}{{1 - \rho }} \)
Expected waiting time in system (includes service time) for each individual customer or time a customer spends in the system:
\( \Rightarrow {W_s} =\frac{1}{μ-λ }= \frac{{{L_s}}}{λ } \)
Expected (avg) queue length (excludes customers being served) or Average number of customers in the queue:
\(L_q=\frac{\rho^2}{1-\rho}\)
Waiting time in queue (excludes service time) for each individual customer or Expected time a customer spends in a queue:
\(W_q=\frac{L_q}{λ}=\frac{λ}{μ(μ-λ)}\)
Calculation:
Arrival Rate, \(λ=6/hr\)
Service Rate, \(μ=8/hr\)
\(\rho=\frac{λ}{μ}=\frac{6}{8}= 0.75\)
Expected(avg) number of customers in queuing system or Average number of customers in the system:
\({L_s} = \frac{{{\rho }}}{{1 - \rho }} = \frac{{{0.75 }}}{{1 - 0.75 }}= 3\) persons
Expected waiting time in system (includes service time) for each individual customer or time a customer spends in the system:
\( \Rightarrow {W_s} =\frac{1}{μ-λ }= \frac{{{L_s}}}{λ } = \frac{{3}}{{6}} = \frac{1}{{2}}\) hr = 30 min
Expected (avg) queue length (excludes customers being served) or Average number of customers in the queue:
\(L_q=\frac{\rho^2}{1-\rho}=\frac{0.75^2}{1-0.75}= 2.25\)
Waiting time in queue (excludes service time) for each individual customer or Expected time a customer spends in a queue:
\(W_q=\frac{L_q}{λ}=\frac{2.25}{6}\)hr = 22.5 minutes
