**Problem 11-13 (Algorithmic)**

Pete's Market is a small local grocery store with only one
checkout counter. Assume that shoppers arrive at the checkout lane
according to a Poisson probability distribution, with an arrival
rate of **18** customers per hour. The checkout
service times follow an exponential probability distribution, with
a service rate of **20** customers per hour. The
manager’s service goal is to limit the waiting time prior to
beginning the checkout process to no more than **five
minutes**. Also the manager of Pete's Market wants to
consider one of the following alternatives for improving service.
Calculate the value of *W _{q}* for each
alternative.

a. Hire a second person to bag the groceries while the
cash register operator is entering the cost data and collecting
money from the customer. With this improved single-server
operation, the service rate could be increased to
**28** customers per hour. Round your answer to three
decimal places. Do not round intermediate calculations.

*W _{q}* =_______ minutes

b. Hire a second person to operate a second checkout
counter. The two-server operation would have a service rate of
**20** customers per hour for each
server. *Note*:
Use *P*_{0} values from Table
11.4 to answer this question. Round your answer to three
decimal places. Do not round intermediate calculations.

*W _{q}* = ____________ minutes

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For tue poison arrillal rate and tece Exponentiou seulice time the average time a customa spends waiting inline for tee seaccice in a sinal salua quaing sustemis \log =\frac{\lambda}{\text { N(al- } \lambda)}\begin{array}{l}\begin{array}{l}\lambda=18 \\\mu=20\end{array} \quad \omega q=\frac{18}{20(20-18)} \\=\frac{18}{20(q)}=\varnothing \% 99 \times 60 \approx 5 \cdot 88 \text { orinttes } \\=0.45 * 60=27 \text { minuty } \\\omega q=24 \text { minutes } \\\end{array}b)\omega q=1 q(\lambdawhere.\begin{array}{l}L q=\cdot \frac{(\lambda / \mu)^{s} \times \lambda \mu}{(s-1) !(5 \mu-\lambda)^{2}} P_{0}=\frac{\left(18(20)^{2}+18 \times 20 \times 0.5094\right.}{(2-1) !(2(20)-181)^{\gamma}} \\=\frac{0.81 \times 183.384}{484} \\=0.30690297521 \\\approx 0.3069 \quad \omega a=\mid 9 / \lambda \\=\frac{0.3069}{18} \\\omega q=0.0170 \text { mintus } \\=0.01705016529 \\\end{array}ple like if it help's me ...