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Urban OR: Quiz 2 Solutions (2003)
Problem 1:
(a)
W
o
=
ƒ
2
λ
i
E
[
S
i
2
]
=
12
⋅
1
+
12
⋅
9
=
1 min
2
60 2
60 2
i
=
1
W
q
1
=
W
o
=
1
=
1.25 min
(1
−
ρ
1
)
(1
−
12
)
60
W
q
2
=
(1
−
1
)(1
W
−
o
1
−
2
)
=
(1
−
12
)(1
1
−
12
−
12
)
=
6.25 min
60
60
20
W
q
,
priority
=
(1.25
+
6.25) / 2
=
3.75 min
⋅
[
1
2
+
σ
S
2
]
24
⋅
[2
2
+
1]
W
q
,
FCFS
=
W
q
,
M
/
G
/1
=
µ
=
60
=
5 min
2(1
−
)
2
⋅
(1
−
24
)
30
The expected waiting time for the FCFS system is greater than the expected waiting time
for the priority system, as expected. When we assign priority to the customers with the
shortest expected service time, we minimize the expected waiting time (see Corollary on
page 239 of the text).
(b) Use expression (4.107a) – note that
1
=20/60=1/3 and
2
=20/20=1 and
1
+
2
= 4/3 (>1).
E
[
S
1
2
]
E
[
S
2
2
]
1
1
2
9
1
⋅
+
(1
−
1
)
⋅
+
⋅
2
⋅
E
[
S
1
]
2
⋅
E
[
S
2
]
3
2
⋅
1
3
2
⋅
3
W
q
,1
=
(1
−
1
)
=
(
)
=
1.75 min
2
3
and
W
q
,2
= ∞
(c) Now, since
=1, neither class of customers attains steady state. Both classes
experience infinite expected waiting time in the long run.
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View Full DocumentProblem 2.
(a) The ambulances are almost always available. The system is either in state 000 or
(very rarely) in state 001 or 010 or 100. Thus, whenever an ambulance is
dispatched, it is dispatched to a random location in its own district or sector (i.e., the
set of points within one mile of the ambulance's home location).
Thus the response
distance is a r.v. uniform over [0,1]. The mean response distance is 1/2 mile.
(b) The workload or utilization factor is
ρ
=
λ
/(3
µ
) < 1. Since
µ
= 1,
max
= 3 calls/hour.
(c) Here all three servers are almost always busy. That is, with probability 1 
epsilon
,
there is always a queue of waiting 'customers.' FCFS means that the location of the
next queued customer is uniform over the entire triangle. The triangle circumference
is 3 miles. The ambulance takes a shortest route from its home location to the
customer. The customer is in one of two 3milelength segments emanating from the
ambulance’s home location, and conditioned on the segment he or she is in, is at a
location that is uniformly distributed over that segment. Thus the mean travel
distance is 3/2 = 1.5 mi.
(d) The ambulance will travel more than one mile if it is serving an interdistrict
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 Fall '06
 hansman

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