Question

A firm has a single-channel service station with the following empirical data available to its management.
i) The mean arround rate is 6.2 minutes
ii) The mean service time is 5.5 minutes
iii) The arrival and service time probability distributions are as follows:

Arrivals ( Minutes)         Probability              Service Time (Minutes)                Probability
3-4                                  .05                           3-4                                          .10
4-5                                  .20                           4-5                                          .20
5-6                                  .35                           5-6                                          .40
6-7                                  .25                           6-7                                          .20
7-8                                  .10                           7-8                                          .10
8-9                                  .05                           8-9                                          .00
Total                               1.00    1.00          

The queuing process begins at 10:00 am proceeds for nearly 2 hours. An arrival goes to the service facility immediately; if it is empty otherwise it will wait in a queue. The queue. The queue discipline is first come first served. If the attendant's wages are K10,000 per hour and the customer's waiting time cost K11,000 per hour, would it be an economical proposition to engage second attendant?                                                           

Answer 1

Interarrival time (minutes)	Midpoint (Li)	Probability (pi)	pi*(Li - L)^2
3-4	3.7	0.05	0.3125
4-5	4.7	0.20	0.4500
5-6	5.7	0.35	0.0875
6-7	6.7	0.25	0.0625
7-8	7.7	0.10	0.2250
8-9	8.7	0.05	0.3125
Average (L)	6.2	Variance	1.4500
		SD =	1.2042

Note that the midpoints are not exactly average of the ranges. They have been strategically chosen in order to match the given average value of 6.2

Similarly, for the service time distribution,

Interarrival time (minutes)	Midpoint (Mi)	Probability (pi)	pi*(Mi - M)^2
3-4	3	0.1	0.6250
4-5	4	0.2	0.4500
5-6	5	0.4	0.1000
6-7	6	0.2	0.0500
7-8	7	0.1	0.2250
8-9	8	0	0.0000
Average (M)	5.5	Variance	1.4500
		SD =	1.2042

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Ca = coefficient of variation for interarrival times = 1.2042/6.2 = 0.194
Cs = coefficient of variation for interarrival times = 1.2042/5.5 = 0.219

M = average service time = 5.5 mintues
L = average interarrival time = 6.2 minutes

u = tilization = M/L = 5.5/6.2 = 0.887

For a G/G/1 queue,

Lq = average number of people waiting in the queue is given by -

So, Lq = ((0.194^2 + 0.219^2)/2)*(0.887^2)/(1-0.887) = 0.30 persons

Cost of server = $10,000 per hour
Cost of waiting = Lq x $11,000 = $3,300 per hour

So, total cost = $13,300 per hour

For a G/G/2 queue,

Lq for M/M/2 will come as 0.2173

Sp. Lq = 0.2173 x (0.194^2 + 0.219^2)/2) = 0.0093

Cost of server = $20,000 per hour
Cost of waiting = Lq x $11,000 = $102 per hour

So, total cost = $20,102

Therefore, it makes no real sense to employ another server and incur most cost per hour.