April
1999 QUESTION 5 Total Marks: 20 Marks |
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questions
GRADE A
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| Irwin Transport Company is setting up an
office to handle phone reservations. During the busy period it is expected that phone
calls will come in at a rate of 1 every 4 minutes. From past experience management knows
that, on the average, it takes about 3 minutes to process a call. When a customer calls in
and all the agents are busy, the caller listens to music until an agent is free.
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[2] | |
| (a) | Assume Irwin Transport Company will staff the office with one reservation agent. | |
| (i) What percentage of the time will customer calls be answered immediately? | [3] | |
| mean arrival rate, l = 60/4 = 15 calls/hr mean service rate, m = 60/3 = 20 calls/hr P = l / m Percentage of time calls will be
answered immediately
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| (ii) How long will the average customer need to wait before an agent comes on the line? | [2] | |
| mean waiting time in
queue = l / [m (m - l)] =15 / [20(20-15)] = 0.15 hrs » 9 mins
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| (iii) What is the call intensity? | [1] | |
| Call intensity is
0.75.
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| (iv) At any given time, how many customers will be in the system? | [2] | |
| mean number of
customers in the system = P / 1-P = 0.75 / (1-0.75) = 3 customers
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| (b) | If Irwin Transport Company adds a second reservation agent then calculate the following: | |
| (i) What percentage of the time will a customers call be answered immediately? | [5] | |
| P = l
/ cm = 15/ 2(20) = 0.375 Percentage
calls will be answered immediately
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| (ii) How long will the average customer need to wait before an agent comes on the line? | [4] | |
| mean waiting time in
queue =[ (0.375(2))2 / 2!(1- 0.375)2(2)(20) ]* 0.4545 = 0.5625/31.25 * 0.4545 = 0.008181 hrs » 0.49 mins
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| (iii) At any given time, how many customers will be waiting? | [3] | |
| mean number of
customers will be waiting =[ ( 0.375(0.375 * 2)2 / 2!(1- 0.375)2 ) * 0.4545 + 0.372(2) = 0.2109/0.781 * 0.4545 + 0.75 = 0.873
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