Question

You are planning employees for a bank. You plan for six tellers. Tellers take 15 minutes per customer with a standard deviation of seven minutes. Customers will arrive one every three minutes according to an exponential distribution (recall that the standard deviation is equal to the mean). Every customer who arrives eventually gets serviced.

a. On average, how many customers would be waiting in line?

b. On average, how long would a customer spend in the bank?

c. If a customer arrived, saw the line, and decided not to get in line, that customer has ________.

d. A customer who enters the line but decides to leave the line before getting service is said to have ________.

Short answer

Mean arrival time,$\overline{{\text{X}}_{\text{a}}}\text{=\hspace{0.17em}3\hspace{0.17em}min}$

Exponential distribution, mean and standard deviation are equal i.e.${\text{S}}_{\text{a}}\text{\hspace{0.17em}=\hspace{0.17em}3min}$

Mean service time, $\overline{{\text{X}}_{\text{a}}}\text{=\hspace{0.17em}}15\text{\hspace{0.17em}min}$

Standard deviation,$S_s=7\min$

No. of tellers, S=6

Expected arrival rate

$\begin{array}{l}\left(\text{λ}\right)=\frac{1}{{\text{X}}_a}\\ =\frac{1}{3}\times 60\\ =20\text{\hspace{0.17em}customer\hspace{0.17em}per\hspace{0.17em}hour}\end{array}$

Service rate per server

$\begin{array}{l}\left(\text{μ}\right)=\frac{1}{{\text{X}}_{\text{s}}}\\ =\frac{1}{15}\times 60\\ =4\text{\hspace{0.17em}customer\hspace{0.17em}per\hspace{0.17em}hour}\end{array}$

Coefficient of variation for arrival

$\begin{array}{l}\left(C_a\right)=\frac{S_a}{X_a}\\ =\frac{3}{3}\\ =1\end{array}$

Coefficients of variation for service

$\begin{array}{l}\left(C_s\right)=\frac{S_s}{X_s}\\ =\frac{7}{15}\\ =0.4667\end{array}$

Expected server utilization is

Expected server utilization

$\begin{array}{l}\left(\text{ρ}\right)\text{=\hspace{0.17em}}\frac{\text{λ}}{\text{S×\hspace{0.17em}μ}}\\ \end{array}$

Obtain the following results

$\begin{array}{l}\left(\text{ρ}\right)\text{=}\frac{\text{20}}{\text{6×4}}\\ =0.833\end{array}$

The expected server utilization is 0.833

The expected number of people waiting with the help of people waiting is

Expected number of people waiting

$\begin{array}{l}{\text{L}}_{\text{q}}=\frac{{\text{ρ}}^{\sqrt{\text{2}(\text{s}+\text{1})}}}{\text{1}-\text{ρ}}\text{×}\frac{{\text{C}}_{\text{a}}^{\text{2}}+{\text{C}}_{\text{s}}^{\text{2}}}{\text{2}}\\ \\ {\text{\hspace{0.17em}L}}_{\text{q}}=\frac{{0.833}^{\sqrt{\text{2}(6+\text{1})}}}{\text{1}-0.833}\text{×}\frac{1^2+{\left(0.4667\right)}^2}{\text{2}}\\ \\ =1.8468\end{array}$

The expected number of people waiting is 1.8468.

Step-by-step solution

Step by Step Solution Step 1: (a) Average number of people waiting in line is

$L_q=1.8468$

(b) Average number of people waiting in the system is

${\text{L}}_{\text{s}}{\text{=\hspace{0.17em}L}}_{\text{q}}\text{\hspace{0.17em}+\hspace{0.17em}S×ρ}$

${\text{L}}_{\text{s}}=\text{\hspace{0.17em}1.8468}+\text{6}\times \left(\text{0.8448}\right)\text{\hspace{0.17em}customers}$

Therefore, the average time waiting in the system

$\begin{array}{l}{\text{W}}_{\text{s}}=\frac{{\text{L}}_{\text{s}}}{\text{λ}}\\ {\text{W}}_{\text{s}}=\frac{\text{6.8448}}{\text{20}}\times \text{60}\\ {\text{W}}_{\text{s}}=\text{\hspace{0.17em}20.53\hspace{0.17em}min}\end{array}$

The average time waiting in the system is 20.53 min.

(c)Customer arrived, saw the line, and decided not to get in line, that customer is

If a consumer arrives, sees the queue, and decides not to join it, he is said to have balking.

(d) Enters the line but decides to leave the line before getting service is said to have is

Reneging service occurs when a customer leaves the line before the service is delivered.