Question

During the early morning hours, customers arrive at a branch post office at an average of 1.5 minutes per customer (Poisson),

while clerks can handle transactions in an average time (exponential) of 3 minutes each. Please find the following:

(Please keep five decimal points).


a. The average time of customers waiting for service if three clerks are used (provide your answer in minutes).

b. If four clerks are used, how much will customer waiting time be reduced (in minutes)?

Answer 1

This is M/M/s system with following parameters

Arrival rate, $\lambda$ = 1/1.5 customers per minute

Service rate, $\mu$ = 1/3 per minute

Number of clerks, M = 3

$\lambda$/$\mu$ = (1/1.5)/(1/3) = 2

In the following table, look for $\lambda$ /$\mu$ = 2 and M = 3

Nu M TABLE 18.4 Infinite-source values for Lg and A given /p and M La Po p M La P% λμ M 4 Po 0.15 1 0.026 1.1 2.4 3 2 0.001 2 3 4 0.20 2 1.2 0.25 1 2 2.5 .850 .860 .800 .818 .750 .778 .700 .739 .650 .702 .600 .667 0.30 2 1.3 3 4 5 2 3 4 5 0.35 2 0.40 2.6 2 1.4 AN .056 .083 .089 .090 .091 .045 .074 ,080 .082 .082 .035 .065 .072 .074 .074 .025 .057 .065 .067 .067 .016 .050 .058 0.45 1 .550 3 4 5 0.50 2 3 1 2 3 1.5 2 2.7 0.55 1 2 1.6 3 2.8 0.60 1 AN 1.7 .060 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 4 5 6 7 8 4 5 6 7 8 4 5 6 0.65 1 .633 .637 .500 .600 .606 .450 .569 .576 .400 .538 .548 350 .509 .521 .300 .481 .495 -250 .455 .471 200 .429 .447 .150 0.050 0.002 0.083 0.004 0.129 0.007 0.188 0.011 0.267 0.017 0.368 0.024 0.002 0.500 0.033 0.003 0.672 0.045 0.004 0.900 0.059 0.006 1.207 0.077 0.008 1.633 0.098 0.011 2.250 0.123 0.015 3.200 0.152 0.019 4.817 0.187 0.024 0.003 8.100 0.229 0.030 0.004 18.050 0.277 0.037 0.005 0.333 0.045 0.007 3 4 5 2 3 4 5 2 3 4 5 2 3 4 5 2 3 4 5 2 3 2.9 0.477 0.066 0.011 0.675 0.094 0.016 0.003 0.951 0.130 0.023 0.004 1.345 0.177 0.032 0.006 1.929 0.237 0.045 0.009 2844 0.313 0.060 0.012 4.426 0.409 0.080 0.017 7.674 0.532 0.105 0.023 17.587 0.688 0.136 0.030 0.007 0.889 0.174 0.040 0.009 1.149 0.220 0.052 0.012 1.491 0.277 0.066 0.016 1.951 0.346 0.084 0.021 290 327 332 250 294 300 301 212 .264 271 272 ..176 236 .245 .246 .143 211 221 223 .111 .187 199 .201 .081 .166 .180 .182 .053 .146 .162 .165 .026 .128 .145 .149 149 .111 .130 .134 .135 .096 .117 .121 .122 .081 .105 .109 .111 .068 .093 .099 .100 2.589 0.431 0.105 0.027 0.007 3.511 0.583 0.130 0.034 0.009 4.963 0.658 0.161 0.043 0.011 7.354 0.811 0.198 0.053 0.014 12.273 1.000 0.241 0.066 0.018 27.193 1.234 0.295 0.081 0.023 1.528 0.354 0.099 0.028 0.008 1.902 0.427 0.120 0.035 0.010 2.386 0.513 0.145 0.043 0.012 3.027 0.615 0.174 0.052 0.015 0.70 1 1.8 2 3 0.75 3.0 1.9 0.80 2 3 1 2 3 .061 .008 .044 .052 .054 .055 .038 .047 .049 .050 .050 .032 .042 .044 .045 .045 .027 .037 .040 .040 .041 3.1 0.85 2.0 .404 2 3 4 0.90 1 2.1 3.2 2 3 4 3 4 5 6 3 4 5 6 3 4 5 6 3 4 5 7 .425 .427 .100 379 .403 .406 .050 356 383 .386 333 364 .367 0.95 2.2 3.3 .023 2 3 4 8 4 5 6 7 8 .033 .036 .037 .037 1.0 2.3 3 4 (continued)

The corresponding value of Lq = 0.889

Lq is the average number of customers waiting in line

Average time of customers waiting for service, Wq = Lq/$\lambda$

= 0.889/(1/1.5)

= 1.333 minutes

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In table 18.4, look for $\lambda$ /$\mu$ =2 and M = 4

The corresponding value of Lq = 0.174

Average time of customers waiting for service, Wq = Lq/$\lambda$

= 0.174/(1/1.5)

= 0.261 minutes

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Reduction in customer waiting time = 1.333 - 0.261

= 1.0723 minutes

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