A restaurant\'s reservation agent takes reservations for dinner by telephone. If

Question

A restaurant's reservation agent takes reservations for dinner by telephone. If he is already on the phone when a patron calls to make a reservation, the incoming call is answered automatically, and the customer is asked to wait for the agent. As soon as the agent is free, the patron who has been on hold the longest is transferred to the agent to be served. Calls for reservations come in at a rate of about 17 per hour. The agent is able to make a reservation in an average of three minutes. The agent is paid $13 per hour. The restaurant estimates that that the restaurant loses $20 due to goodwill and dissatisfaction for every hour customers are waiting to talk to an agent. a) What is the hourly system cost for the situation where the restaurant has only one agent to take calls? (Enter two decimal places) b) The restaurant is considering adding a second agent who would also be paid $13 per hour. What is the hourly system cost for the situation where the restaurant has two agents to take calls? (Enter two decimal places) c) Should the restaurant hire the second agent?

Explanation / Answer

Part (a)

This is a case of M/M/1 queue system with arrival rate = 17 per hour and

service rate µ = 20 per hour [3 minutes per reservation => 60/3 = 20 reservations per hour.]

Average waiting time, E(W) = /{µ(µ - )} = 17/(20 x 3) = 17/60 hours = 17 minutes.

Since on an average 17 customers arrive per hour, total waiting time = 17 x 17 = 289 minutes.

Given loss-of-goodwill cost due to customers waiting = 20 per hour of waiting, total cost due to waiting

= (289 x 20)/60 = 96.67.

Cost of service = payment to the agent = 13 per hour.

So, per hour total system cost = 96.67 + 13 = $109.66 ANSWER

Part (b)

This is a case of M/M/c queue system with arrival rate = 17 per hour and

service rate µ = 20 per hour and number of service channels (agnts), c = 2.

Requisite formulae:

E(w) = E(m)/()   E(m) = P0{(µ)(/µ)c}/{(c - 1)!(cµ - )2}

P0 = 1/(S1 + S2), where S1 = [0,c - 1](n/n!) and S2 = (c)/[c!{1 - (/c)}]

Preparatory calculations:

= µ = 17/20 = 0.85; S1 = (0/0!) + (/1!) [because c = 2 and hence c- 1 = 1]

So, S1 = 1 + 0.85 = 1.85

S2 = (0.852)/[2!{1 - (0.85/2)}] = 0.6283

So, P0 = 1/(1.85 + 0.6283) = 0.4035

Average number of customers waiting in the queue = E(m)

= 0.4035{(17 x 20)(17/20)2}/(1)!{(2 x 20) – 17}2 = 0.1874 hours.

Since on an average 17 customers arrive per hour, total waiting time = 17 x 0.1874 = 3.1853 hours.

Given loss-of-goodwill cost due to customers waiting = 20 per hour of waiting, total cost due to waiting = 3.1853 x 20 = 63.716.

Cost of service = payment to two agents = 26 per hour.

So, per hour total system cost = 63.716 + 26 = $89.72 ANSWER

Part (c)

Since Part (b) cost < Part (a) cost, additional agent is recommended. ANSWER

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