Queueing Theory Help M/M/3
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Comtex plc employs three people in its mail room to sort and despatch mail going through its internal mail system. Letters arrive at an average rate of 150 an hour and each employee can deal with 60 letters an hour. Assuming you can treat this as an M/M/3/∞/FIFO system, calculate:
- The probability that there are no letters waiting to be sorted
- The proportion of the time each sorter is busy
- The probability that there are more than 4 letters in the sorting office
I have attempted these questions... please can someone tell me if I am doing it right
service rate is now S$mu$ so traffic intensity is now $lambda$/$mu$S where S is 3 since there are 3 people. So $rho$ = 150/180 = 0.833333333 and 1-$rho$=0.16666 = $p_0$
the proportion of the time each sorter is busy
$rho$= 0.833333333 so 83.3333% of the time? not sure if this is rightsince question is asking if there are more than 4 letters in the sorting office now $n>S$ so
$p_4$ = $2.5^4$/4! * 0.166666 = 0.27126...
Is this all correct? thanks
queueing-theory operations-research
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add a comment |
$begingroup$
Comtex plc employs three people in its mail room to sort and despatch mail going through its internal mail system. Letters arrive at an average rate of 150 an hour and each employee can deal with 60 letters an hour. Assuming you can treat this as an M/M/3/∞/FIFO system, calculate:
- The probability that there are no letters waiting to be sorted
- The proportion of the time each sorter is busy
- The probability that there are more than 4 letters in the sorting office
I have attempted these questions... please can someone tell me if I am doing it right
service rate is now S$mu$ so traffic intensity is now $lambda$/$mu$S where S is 3 since there are 3 people. So $rho$ = 150/180 = 0.833333333 and 1-$rho$=0.16666 = $p_0$
the proportion of the time each sorter is busy
$rho$= 0.833333333 so 83.3333% of the time? not sure if this is rightsince question is asking if there are more than 4 letters in the sorting office now $n>S$ so
$p_4$ = $2.5^4$/4! * 0.166666 = 0.27126...
Is this all correct? thanks
queueing-theory operations-research
$endgroup$
$begingroup$
@callculus not sure if you can help?
$endgroup$
– Charlotte Sacks
Dec 6 '18 at 23:01
add a comment |
$begingroup$
Comtex plc employs three people in its mail room to sort and despatch mail going through its internal mail system. Letters arrive at an average rate of 150 an hour and each employee can deal with 60 letters an hour. Assuming you can treat this as an M/M/3/∞/FIFO system, calculate:
- The probability that there are no letters waiting to be sorted
- The proportion of the time each sorter is busy
- The probability that there are more than 4 letters in the sorting office
I have attempted these questions... please can someone tell me if I am doing it right
service rate is now S$mu$ so traffic intensity is now $lambda$/$mu$S where S is 3 since there are 3 people. So $rho$ = 150/180 = 0.833333333 and 1-$rho$=0.16666 = $p_0$
the proportion of the time each sorter is busy
$rho$= 0.833333333 so 83.3333% of the time? not sure if this is rightsince question is asking if there are more than 4 letters in the sorting office now $n>S$ so
$p_4$ = $2.5^4$/4! * 0.166666 = 0.27126...
Is this all correct? thanks
queueing-theory operations-research
$endgroup$
Comtex plc employs three people in its mail room to sort and despatch mail going through its internal mail system. Letters arrive at an average rate of 150 an hour and each employee can deal with 60 letters an hour. Assuming you can treat this as an M/M/3/∞/FIFO system, calculate:
- The probability that there are no letters waiting to be sorted
- The proportion of the time each sorter is busy
- The probability that there are more than 4 letters in the sorting office
I have attempted these questions... please can someone tell me if I am doing it right
service rate is now S$mu$ so traffic intensity is now $lambda$/$mu$S where S is 3 since there are 3 people. So $rho$ = 150/180 = 0.833333333 and 1-$rho$=0.16666 = $p_0$
the proportion of the time each sorter is busy
$rho$= 0.833333333 so 83.3333% of the time? not sure if this is rightsince question is asking if there are more than 4 letters in the sorting office now $n>S$ so
$p_4$ = $2.5^4$/4! * 0.166666 = 0.27126...
Is this all correct? thanks
queueing-theory operations-research
queueing-theory operations-research
asked Dec 6 '18 at 22:45
Charlotte SacksCharlotte Sacks
132
132
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@callculus not sure if you can help?
$endgroup$
– Charlotte Sacks
Dec 6 '18 at 23:01
add a comment |
$begingroup$
@callculus not sure if you can help?
$endgroup$
– Charlotte Sacks
Dec 6 '18 at 23:01
$begingroup$
@callculus not sure if you can help?
$endgroup$
– Charlotte Sacks
Dec 6 '18 at 23:01
$begingroup$
@callculus not sure if you can help?
$endgroup$
– Charlotte Sacks
Dec 6 '18 at 23:01
add a comment |
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@callculus not sure if you can help?
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– Charlotte Sacks
Dec 6 '18 at 23:01