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首页Data networks--Solutions 数据网络的部分习题解答
Data networks--Solutions 数据网络的部分习题解答
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Data Networks (2nd Edition) by Dimitri P. Bertsekas (Author), Gallager (Author)
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1
TCOM 501 Homework 1
Solutions
Problem 3.1
A customer that carries out the order (eat in the restaurant) stays for 5 minutes (25
minutes). Therefore the average customer time in the system is
15255.05.05.0
=
×
+
×
=
T minutes
Applying Little’s Theorem the average number in the system is
75155
=
×
=
×
=
TN
λ
minutes
Problem 3.3
The system can be represented as shown in the figure below. In particular, once a
machine breaks down, it goes into repair if a repairperson is available at that time,
otherwise it waits in a queue for a repairperson to become free. Note that if m = 1 this
system will be identical to Example 3.7.
Let λ be the throughput of the system and let Q be the average time a broken down
machine waits for a repairperson to become free. Applying Little’s Theorem to the
system, we then obtain
(
)
NPQR =++λ , from which
(
)
NPR ≤+λ .
Since the number of machines waiting require can be at most
(
)
mN − , the average
waiting time λQ is at most the average time to repair
(
)
mN − machines, which is
(
)
m
P
mN − . To see this, we can consider a system that a machine fails right after it is
repaired. Then the repair personnel are always busy and the queue always full. Applying
Little’s Theorem to the queue, we obtain
m
N
Q
−
=
′
′
λ
By applying Little’s Theorem to the always-busy service personnel, we also have
Working
Machines
Machines
Waiting
for
Repair
1
2
m
![](https://csdnimg.cn/release/download_crawler_static/1099537/bg2.jpg)
2
mP
=
′
λ
Eliminating
λ
′
we get
(
)
QQ
m
PmN
Q ≥
′
−
=
′
,
Applying Little’s Theorem to the repairperson, we have mP
≤
λ
. So we have the
following bound for the throughput
+
≤≤
+ PR
N
P
m
R
N
m
NP
,minλ
Note that these bounds generalize the ones obtained in Example 3.7 (see Eq. (3.9)). By
using the equation
λ
N
T = for the average time between repairs, we have
m
NP
RTPR
m
NP
+≤≤
+,min
Problem 3.4
(a) If λ is the throughput of the system, Little’s theorem gives TN
λ
=
, so from the
relation
2
NT βα += , we obtain
22
TT βλα += , or
2
T
T
β
α
λ
−
=
This relation between λ and T is plotted below.
αβ
λ
1
=
′
λ
λ
α
2
=
′
T
T
![](https://csdnimg.cn/release/download_crawler_static/1099537/bg3.jpg)
3
The maximum value of λ is attained for the value T’ for which the derivative of
(
)
2
T
T
β
α
−
is zero. This is T’ = 2α and from the above equation, the corresponding
maximal throughput value
αβ
λ
1
=
′
.
(b) When λ < λ’, there are two corresponding values of T: a low value corresponding to
an uncontested system where N is relatively low, and a high value corresponding to a
congested system where N is relatively high. This assumes that the system reaches a
steady-state. However, it can be argued that when the system is congested a small
increase in the number of cars in the system due to statistical fluctuations will cause an
increase in the time in the system, which will tend to decrease the rate of departure of
cars from the system. This will cause a further increase in the number in the system and a
further increase in the time in the system, etc. In other words, when we are operating on
the right side of the curve of the figure, there is a tendency for instability in the system,
whereby a steady-state is never reached: the system tends to drift towards a traffic jam
where the car departure rate from the system tends towards zero and the time a car spends
in the system tends towards infinity. Phenomena of this type are analyzed in the context
of the Aloha multi-access system in Chapter 4.
Problem 3.6
(a) The probability that the person will be the last to leave is ¼ because the exponential
distribution is memoryless, and all customers have identical service time distribution. In
particular, at the instant the customer enters service, the remaining service time of each of
the other three customers has the same distribution as the service time of that customer.
(b) The average time in the bank is 1 (the average service time) plus the expected time for
the first customer to finish service. The latter time is ¼ since the departure process is
statistically identical to that of a single server facility with 4 times larger service rate.
More precisely, we have
P { no customer departs in the next t mins}
= P { 1
st
does not depart in next t mins} P{ 2
nd
does not depart in next t mins }
P { 3
rd
does not depart in next t mins} P{ 4
th
does not depart in next t mins }
=
(
)
tt
ee
4
4
−−
=
Therefore
P{ the first departure occurs with in the next t mins} =
t
e
4
1
−
−
And the expected time to the next departure is ¼. So the answer is 1.25 minutes.
(c) The answer will not change because the situation at the instant when the customer
begins service will be the same under the conditions for (a) and the conditions for (c).
![](https://csdnimg.cn/release/download_crawler_static/1099537/bg4.jpg)
4
Problem 3.7
In the statistical multiplexing case the packets of at most one of the streams will wait
upon arrival for a packet of the other stream to finish transmission. Let W be the waiting
time, and not that 0 = W = T/2. W have that one half of the packets have system time T/2
+ W and waiting time in queue W. Therefore
Average System Time =
222
1
22
1 WT
W
TT +
=
++
Average Waiting Time in Queue = W/2
Variance of Waiting Time =
422
1
22
1
2
22
WWW
=
+
So the average system time is between T/2 and 3T/4, and the variance of waiting time is
between 0 and
16
2
T
.
Problem 3.10
(a) Let
n
t be the nth arrival, and
nnn
tt −=
+1
τ . We have for s = 0
{
}
(
)
(
)
{
}
s
nnn
etAstAPsP
λ
τ
−
==−+=> 0
(by the Poisson distribution of arrivals in an interval). So
{
}
s
n
esP
λ
τ
−
−=≤ 1
which is (3.11).
To show that
n
τττ ,...,,
21
are independent, note that (using the independence of the
numbers of arrivals in disjoint intervals)
{
}
==> τττ
1
|sP
n
P{ 0 arrival in
(
]
ττττ =+
1
|, s }
= P { 0 arrival in
(
]
s+ττ , } =
{
}
sPe
s
>=
−
2
τ
λ
Therefore,
21
,ττ are independent.
To verify (3.12), we observe that
(
)
(
)
{
}
λδ
δ
−
==−+ etAtAP 0
![](https://csdnimg.cn/release/download_crawler_static/1099537/bg5.jpg)
5
So (3.12) will be shown as (using L’Hospital’s rule)
( )
0lim
1
lim
00
=+−=
+−
−
→
−
→
λλ
δ
λδ
λδ
δ
λδ
δ
e
e
To verify (3.13) we note that
(
)
(
)
{
}
λδ
λδδ
−
==−+ etAtAP 1
so (3.13) will be shown if 0lim
0
=
−
−
→
δ
λδλδ
λδ
δ
e
This is equivalent to
(
)
0lim
0
=−
−
→
λλ
λδ
δ
e , which is clearly true.
To verify (3.14) we note that
(
)
(
)
{
}
(
)
(
)
{
}
(
)
(
)
{
}
( )( ) ( )( ) ( )
δδλδδλδ
δ
δ
δ
ooo
tAtAPtAtAPtAtAP
=+−+−−=
=
−
+
−
=
−
+
−
=
≥
−
+
11
1012
(b) Let
21
,NN be the number of arrivals in two disjoint intervals of lengths
21
,ττ . Then
{
}
{
}
{
}
{
}
( ) ( )
( )
( )
( )
!!!
,
21
0
21
0
21
0
2121
2121
n
e
kn
e
k
e
knNPkNPknNkNPnNNP
n
n
k
knk
n
k
n
k
λτλτλτλτ
ττλλτλτ
+
=
−
=
−===−====+
+−
=
−
−−
==
∑
∑∑
(c) The number of arrivals of the combined process in disjoint intervals is clearly
independent, so we need to show that the number of arrivals in an interval is Poisson
distributed, i.e.
( ) ( ) ( ) ( ){ }
( )
(
)
!
...
......
1
...
11
1
n
entAtAtAtAP
n
k
kk
k
τλτλ
ττ
τλλ
++
==−−−++++
++−
For simplicity let k = 2. A similar proof applies for k > 2. Then
(
)
(
)
(
)
(
)
{
}
( ) ( ) ( ) ( ){ }
( ) ( ){ } ( ) ( ){
}
∑
∑
=
=
−=−+=−+=
−=−+=−+=
=−−+++
n
k
n
k
kntAtAPktAtAP
kntAtAktAtAP
ntAtAtAtAP
0
2211
0
2211
2121
,
ττ
ττ
ττ
and the calculation continues as in part (b). Also
P { 1 arrival from A
1
prior to t | 1 occurred }
= P { 1 arrival from A
1
, 0 from A
2
} / P { 1 occurred }
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