Respuesta :
Answer:
The answer is "Option F".
Explanation:
The arrival rate[tex](\lambda) = 10 \ / \ hour[/tex]
The service rate [tex](\mu) = 1 \ in \ 5 \ minutes = 12\ /\ hour[/tex]
The time of waiting at queue
[tex]\frac{M}{\frac{M}{1}} \ queue, \ W_q, \\\\ \frac{M}{\frac{M}{1}} = \frac{\lambda}{ \mu \times (\mu - \lambda)}[/tex]
[tex]= \frac{10}{(12 \times (12 - 10))} \\\\ = 0.4167 \ hours \\\\ = 25 \ minutes[/tex]
It was not a queue for [tex]\frac{M}{\frac{M}{1}}[/tex]. This is a [tex]\frac{G}{\frac{G}{1}}[/tex] queue so because intercom or time intervals differ coefficient is given.
[tex]\to c_a = 2.58\\\\\to c_e = 0.75[/tex]
[tex]\to W_q , \frac{G}{\frac{G}{1}} = \frac{(c_{a}^2 + c_{e}^2)}{2} \times W_q \frac{M}{\frac{M}{1}}[/tex]
[tex]= \frac{(2.58^2 + 0.75^2)}{2} \times 25 \\\\ = \frac{(6.6564 + 0.5625)}{2} \times 25 \\\\ = 3.60945 \times 25\\\\ = 90.23625\\\\ = 90 \ minutes[/tex]
Therefore, the entire waiting time even before order is issued,
[tex]\to W_s , \frac{G}{\frac{G}{1}} = W_q \frac{G}{\frac{G}{1}} + \text{Avg service time}[/tex]
[tex]=90+5\\\\=95 \ minutes[/tex]
