| M / M / 1 EXAMPLE |
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7 |
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| A car wash can serve 50 vehicles in an hour. |
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| During the peak
Saturday hour, 40 vehicles arrive. |
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| Assume both
arrivals and service times are random. |
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| A. How much vehicle
storage space do we need to handle the average queue |
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| assuming an average
vehicle length of 18 feet and an average stationary gap of 7 feet? |
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Car Wash |
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| l = |
40 |
vph |
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m = |
50 |
vph |
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| l = |
0.0111 |
vps |
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m = |
0.0139 |
vps = |
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Service Time = |
1/m = |
72 |
sec/veh = |
1.2 |
min/veh |
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r = l/m = |
40 vph/50 vph = |
0.80 |
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Avg Q |
= r2/(1-r) = |
3.2 |
veh |
(round up to 4 vehicles) |
4 |
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| Storage Space
Needed = 4 veh x (18 ft + 7 ft)/veh = |
100 |
feet |
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| B. How long does
the average customer that
arrives during the |
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| Saturday rush hour
have to wait before entering the
car wash? |
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Avg W |
= r/(m-l) = |
288 |
sec = |
4.8 |
min |
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(or |
Avg W |
= r/(m-l) = |
0.08 |
hrs = |
4.8 |
min |
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| C. How much time
should this person allocate for the trip to the car wash? |
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Avg t |
= 1/(m-l) = |
360 |
sec = |
6.0 |
min |
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| Time Being Washed =
Service Time = |
Avg t - Avg W = |
1.2 |
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