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 Wq 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.
Wq =_______ 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 P0 values from Table
11.4 to answer this question. Round your answer to three
decimal places. Do not round intermediate calculations.
Wq = ____________ minutes
Solved 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
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 ...