Download to read offline
![4 Chapter 1. Packet-Switched Networks
(c) Data: (10 B)×8 b
100 Mb/s
+ tp = 0.79 μs + 0.087 s
Ack: (2.5 B)×8 b
100 Mb/s
+ tp = 0.19 μs + 0.087 s
Total time ≈ 1μs (transfer) + 0.173 s (prop.)
3. Assuming the speed of transmission at 2.3 × 108:
(a) Total Delay: D = [np + (nh − 2)]tf + (nh − 1)tp + nhtr
(b) tp1 =
50 miles×1600 m/miles
2.3×108 m/s
= 0.35 ms
tp2 =
400 miles×1600 m/miles
2.3×108 m/s
= 2.8 ms
Number of packets = np = 200MB
10KB = 20, 000
tf =
10,040 B/pockets×8 b/B
100 Mb/s
= 0.8 ms/pockets
D = [20, 000+(5−2)]0.8+[(3−1)0.35+(3−1)2.8]+5×0.2×103
≈ 16.6 s
4. Dp = [np + (nh − 2)]tf1 + nhtr1 + (nh − 1)tp
Dc = 3 ([1 + (nh − 2)]tf2 + nhtr2 + (nh − 1)tp)
Dt = Dp + Dc = (np + nh − 2)tf1 + 3(nh − 1)tf2 + nh(tr1 + 3tr2) +
4(nh − 1)tp
5. Number of packets = np = 200MB
10KB = 20, 000
Dt = Dp + Dc
Dc = dconn-req + dconn-accep + dconn-release
(a) Here, the problem askes that tr be defined as the processing time
for each packet. Therefore, tr = 20, 000× 0.2 = 4,000 s](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-12-320.jpg)
![5
Dc = dconn-req = dconn-accep = dconn-release
= [np + (nh − 2)]tf + nhtr + (nh − 1)tp
= [1 + (5 − 2)]
500 b/packet
100 mb/s
+ 3 × 4, 000 s + 4.84 ms
≈ 12, 000 s
(b) Same as Part (a).
(c) Dt = Dp + Dc = 17 + 3 × 12, 000 = 36, 017 s
6. s = 109 b/s
nh = 10 nodes
tr1 = tr2 = 0.1 s = tr
Data forms two packets:
(9960 + 40) bytes for packet1
(2040 + 40) bytes for packet2
tf1−packet1 =
10,000 B×8 b/B
109 b/s
= 8 × 10−5 s
tf1−packet2 =
2,080 B×8 b/B
109 b/s
= 16.64 × 10−6 s
tf2 = transfer times for control packets = 500 b
109 b/s
= 5 × 10−7 s
tp =
c = 500 miles×1.61×103 m
2.3×108 m/s
= 3.5 × 10−3 s
(a) request + accept time:
t1 + t2 = 2 ([np + (nh − 1)]tf2 + (nh − 1)tp + nhtr]) = 2.06 s
(b) t3 = 1
2 (t1 + t2) = 1.03 s
(c) Dt = Dp + Dc
Dp = Dp−packet1 + Dp−packet2
= [np + (nh − 2)]tf1−packet1 + nhtr1 + (nh − 1)tp + [np + (nh −](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-13-320.jpg)
![6 Chapter 1. Packet-Switched Networks
2)]tf2−packet1 + nhtr1 + (nh − 1)tp
Dc = t1 + t2 + t3
Dt = 4.1 s
7. d + h = 10, 000 b,
ρd = 72%,
h
d = 0.04,
s = 100 Mb/s,
(a) h = 0.04d then: d
d+h = 0.96.
Since: ρd = ρ d
h+d then: 0.72 = ρ0.96
⇒ ρ = 0.74
(b) μ = s
h+d =
100×106 b/s
10×103 b = 10 × 103 = 10, 000 packets/sec
(c) λ = ρμ = 0.74 × 10, 000 = 74, 000
D̄ = 1
10,000−7,400 = 0.38 ms
(d) D̄opt = h
s
√
ρd
1−
√
ρd
2
d + h = 10, 000h/d = 0.04
⇒ h = 384 b
D̄opt = 0.12
8. s = 100 Mb/s
ρ = 80%
(a) ρ = λ
μ
⇒ 0.8 = 8000
μ
⇒ μ = 10, 000 packets/s](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-14-320.jpg)
![7
t
t
t
t
Connection
Request
Connection
Accept
Connection
Release
Data
Node A
Node C
Node B
Node D
Figure 1.1: Signaling delay in connection-oriented packet-switched environ-
ment.
μ = s
h+d ⇒ 10, 000 = 100×106
h+d
⇒ h + d = 10, 000 b
(b) ρh = 0.008 and ρh = ρ h
d+h
⇒ d + h = 100h
⇒ h = 100 b, and d = 9900 b.
(c) ρh = 0.008
ρd = ρ − ρh = 0.8 − 0.008
⇒ ρd = 0.792
dopt = h
√
ρd
1−
√
ρd
= 809 bits
(d + h)opt = h + dopt = 100 + 809
⇒ (d + h)opt = 909 b
(d) D̄opt = h
s
1
1−
√
ρd
2
⇒ D̄opt = 8.2 × 10−5 s
9. See Figure 1.1.
10. D = d+h
s[1−ρd/d(d+h)]](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-15-320.jpg)
![16 Chapter 3. Networking Devices
digital channels, and one control channel:
Bits in each frame = (8 + 4 + 1)×5 b/channel + 1 guard-bit=
66 b/frame
Frame rate =
160×103 b/s
66 b/frame
= 2,424 frames/s
3. (a) Pulse stuffing = 4800 b/s × 0.03 = 144 b/s
Number of characters are = 1 (sync) + 99 (data)= 100
Synchronization bit rate =
4800 b/s
100 ≈ 50 b/s
Number of 150 b/s terminals
=
4800−(2×600+5×300+50+144) b/s
150 b/s
= 12.7 ≈ 12 terminals
(b) The number of characters for synchronization is proportional to
bit rates. For example, since we need 12 characters for 150 b/s
terminals, therefore we need 3 characters for synchronization.
Frame format in terms of bits is:
2 × (12 char × 10 b/char) + 5 × (6 char × 10 b/char) + 12 ×
(3 char × 10 b/char) + 3 × 10 b/char + 1 × 10 b/char
= 940 b/frame
4. (a) #bits/frame (total) = 2 Mb/s × 26 μs/frame = 52 bits/frame
#bits/frame (data) = 52 bits/frame − 10 = 42 bits/frame
#channels = n = 42 bits/frame 1
6 bits/ch
= 7 ch/frame
(b) P[clipping] =
m−1
i=n
m − i
i
pi(1 − p)m−i−1
m = 10, n = 7, p = 0.9
P[clipping] = 0.947](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-24-320.jpg)
![17
5. (a) ρ = ta
tc+d = 2/8 = 25%
(b) Pj=3 =
m
j
(
ti
ta
)j
j
i=0
m
j
(ta
td
)i
=
8
3
(2
6
)3
3
j=0
8
1
(2
6
)i
≈ 21%
(c) B = Pj=n=4 =
m
n
(1/3)n
n
i=0
m
i
(1/3)i
=
8
7
(1/3)4
4
i=0
8
i
(1/3)i
= 70/81
1+8/3+28/9+56/27+70/81 ≈ 9%
(d) E[C] =
4
j=1 jpj = 1(0.275) + 2(0.32) + 3(0.213) + 4(0.0889) ≈
1.94
6. (a) m = 4
n = 2
Prob[clipping] = Pc =
3
i=2
3
i
ρi(1 − ρ)3−i
= 10.4% for ρ = 0.2
= 35.2% for ρ = 0.4
= 64.8% for ρ = 0.6
= 89.6% for ρ = 0.8
(b) m = 4
n = 3
Prob[clipping]Pc =
3
i=3
3
3
ρ3(1 − ρ)3−3
= 0.0% for ρ = 0.2
= 6.4% for ρ = 0.4
= 21.6% for ρ = 0.6
= 51.2% for ρ = 0.8
(c) n = 4
P4 = 100%](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-25-320.jpg)
![33
pi = pc(1 − pc)i = 0.387(1 − 0.387)7 = 0.0116
(e) na = 4
E[c] = 1−pc
pc
= 1−0.387
0.387 = 1.52
Listen to Medium Transmit data
Wait for 512g
Bit Time
Random
# g
Medium not
available
Medium
available
Transmission
success
Collision
Listen to Medium
Transmit with
Prob P for Max
tp
Medium not
available
Medium
available
Collision
Fixed P
Transmission
success
a)
b)
Figure 5.2: Answer to exercise.
8. See Figure 5.3.](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-41-320.jpg)
![49
1
R
2
R
3
R
4
R
7
R
6
R
5
R
1
R
2
R
3
R
4
R
7
R
6
R
5
R
1
R
2
R
3
R
4
R
7
R
6
R
5
R
1
R
2
R
3
R
4
R
7
R
6
R
5
R
Figure 7.8: Answer to exercise.
PCE = (0.3)(0.3)(0.6) = 0.054
PCDF = 1 − (1 − 0.5)(1 − 0.8) = 0.9
PEF = 0.2
PCF = [1 − (1 − PCE)(1 − PEF )]PCDF (0.3)(0.3) = 0.0197
PAC = 1 − (1 − PAB)(1 − PBC) = 0.4126
PAF = 1 − (1 − PAC)(1 − PCF ) = 0.424](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-57-320.jpg)
![71
taking the eth root.
(b) The values m = 0 or m = 1 always produce ciphertexts equal to
0 or 1 respectively, due to the properties of exponentiation. Thus,
the keys containing of all 0’s or all 1’s can be easily recovered by
the attacker. An example could be {x = 3, y = 7}.
9. To overcome the vulnerability in the above combination, practical RSA
implementations typically embed some form of structured, randomized
padding into the value m before encrypting it. This padding ensures
that m does not fall into the range of insecure plaintexts, and that a
given message, once padded, encrypts to one of a large number of dif-
ferent possible ciphertexts. The latter property can increase the cost
of a dictionary attack beyond the capabilities of a reasonable attacker.
Modern constructions use secure techniques such as optimal asymmet-
ric encryption padding (OAEP) to protect messages.
The intuitive solution to this problem is that user 1 must select a larger
random number for RSA encryption. In this case, both users 1 and 2
use this number to create key, k.
A second solution is that user 1 pads k with random bits so that the
message has almost the same number bits as x does.
10. Suppose that user 1 chooses a prime number a, a random number x1,
and a generator g and creates y1. We can say:
k1 = yx1
2 mod a = (gx2 mod a)x1 mod a = [(gx2 )x1
mod a] mod a =
[(gx1 )x2
mod a] mod a = (gx1 mod a)x2 mod a = yx2
1 mod a = k2](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-79-320.jpg)
![77
(c) Percentage of time that the buffer is not empty can be realized
from Figure 11.1. If we include all the activities of the queue as
described in Part (a), the queue activity stops at around 680 μs.
Thus, the percentage of time that the buffer is not empty
= Time That the Buffer Is Empty
Duration of System Processing Time
= 20 μs
680 μs = 0.029
Percentage of time the buffer is not empty = 1 - 0.029 = 0.97
2. (a) E[Kq(t)] = λE[Tq] = ρ2
1−ρ = 0.92
1−0.9 = 8.1
(b) E[Tq] = E[T] − E[Ts]
E[T] = 1
μ(1−ρ)
E[Ts] = 1
μ
⇒ E[Tq] = ρ
1
μ−λ
λ
μ = ρ, ⇒ μ = λ
ρ = 44.44
E[Tq] = 0.9
1
44.44−40
= 0.2 s
(c) P0 = 1 − ρ = 1 − 0.9 = 0.1
λ = 40 packets/s
μ = 44.44 packets/s
3. (a) T = min(T1, T2, ..., Ti)
(b) P[T t] = P[time until next packet departures]
= P[min(T1, T2, ..., Ti)]
= P[T1 t1, T2 t2, ..., Ti ti]
= P[T1 t]P[T2 t]...P[Ti t]
= e−μte−μt...e−μt
= e−iμt](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-85-320.jpg)
![78 Chapter 11. Packet Queues and Delay Analysis
4. (a) P[K(t) k] = 1 − P[K(t) ≥ k]
= 1 −
∞
i=k Pi = 1 −
∞
i=k ρi(1 − ρ) = 1 − (1−ρ)ρk
1−ρ = 1 − ρk
(b) P[K(t) 20] = 0.9904 ⇒ k = 20
⇒ 0.9904 = 1 − ρ20
⇒ ρ = 0.7927
ρ = λ
μ = 0.7927 = 300
μ
⇒ μ = 378.45 packet/s
5. (a) P[K(t) ≥ k] = (1 − ρ)
∞
j=k ρj = (1 − ρ) ρk
1−ρ = ρk
(b) P[K(t) ≥ 60] = ρ60 = 0.01
⇒ ρ ≈ 0.92
⇒ λ ≈ 0.92μ
6. (a) The Markov chain is similar to a regular M/M/1 except the ar-
raival rate to any state i is λi = λ
i
(b) For State 0: p0λ = p1μ
⇒ p1 = λ
μp0 ⇒ p1 =ρp0
For State 1: p1
λ
2 + p1μ = p0 + p2μ
⇒ p1
λ
2 = p2μ ⇒ p2 = λ
2μp1 ⇒ p2 = 1
2 ρp1 = 1
2ρ2p0
Continuing this trend for next states, a generic form can be devel-
oped as:
pi−1
λ
i = piμ ⇒ pi = λ
iμpi−1 = 1
i!ρip0
Since
∞
i=0 pi = 1⇒ p0 = e−ρ
⇒ pi = 1
i! ρie−ρ](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-86-320.jpg)
![79
(c) When i → ∞ while ρ 1, we have pi → 0, thus, the system is in
a steady state.
(d) The utilization (of the server) is:
ρi = λi
μ =
1
i
λ
μ = λ
iμ
(e) E[K(t)] =
∞
i=0 iP[K(t) = i] =
∞
i=0 i
ρk
i!
e−ρ
Solving this equation using
∞
i=0
ρe
e! = eρ will result in:
E[K(t)] = ρ
(f) The mean system delay considering any State i is obtained using
Little’s law:
E[T]i = E[K(t)]
λi
= ρ
λ
i
= iρ
λ
However, since the arrival rate is different in each state, we need
to compute the mean over all states:
E[T] =
∞
i=0 piE[T]i =
∞
i=0 pi
iρ
λ
= ρ
∞
i=0
1
i!λρie−ρ
i
λ
= ρ2
λ
Since E[Ts] = 1
μ, then:
E[Tq] = E[T] - E[Ts] = ρ2
λ − 1
μ
7. a = 2
1
μ = 100 ms/packet
λ = 18 packets/s
(a) Prob[blocking a packet]=Prob[all servers are busy]
ρ1 = λ
μ = (18)(100 × 10−3) = 1.8
ρ = λ
aμ = 18/2.1 = 0.9
P[i 2] = Pa
1−ρ
Pa = P2 =
ρ2
1
2! P0
P0 = 1
(1+ρ1)+
ρ2
1
2!
1
1−ρ
= 1
(1+1.8)+ 1.82
2!
1
1−0.9
= 0.052](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-87-320.jpg)
![80 Chapter 11. Packet Queues and Delay Analysis
ρ = 18/20 = 0.9
P[K(t) = 2] = Pa =
ρa
2
a! ρ0 = 1.82
2 (0.0526) = 0.0853
⇒ Prob[Waiting] = P[K(t) ≥ 2] =
∞
i=a Pi = Pa
1−ρ = 0.0853
1−0.9 =
0.853
(b) E[K(t)] = Paρ
(1−ρ)2 + ρ1
= (0.0853)(0.9)
(1−0.9)2 + 1.8 = 9.474
(c) E[T] = E[Tq] + E[Ts]
= Pa
aμ(1−ρ)2 + 1
μ = 0.421 + 0.1 = 0.521 s
(d) P[K(t) 50] =
∞
i=50 ρi
=
∞
i=50 ρi−aPa = Pa
ρ2
∞
i=50 ρi
= Pa
ρ2 [
∞
i=50 ρi −
i=50 ρi]
= Pa
ρ2 [ 1
1−ρ − 1−ρ50+1
1−ρ ]
= Pa
ρ2 [ ρ51
1−ρ] = 0.0853
0.92
0.951
0.1
P[K(t) 50] = 0.00488
8. (a) Prob[blocking a packet]=Prob[all servers are busy]
λ = 100 packets/s
mean service rate:μ = 20 packets/s; ρ1 = 100/20 = 5
Use Erlang-B:
Pa =
ρa
1
a!
1
a
i=0
ρi
1
i!
= 56
6!
1
6
i=0
5i
i!
= 0.1935
(b) Pa = 0.1935
2 = 0.0967
By plugging numbers in the equation, we have:
P7 = 0.121
P8 = 0.075
We need two switches to lower the blocking probability below
0.0967, this is 4 more switches then the setup in Part (a).](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-88-320.jpg)
![86 Chapter 11. Packet Queues and Delay Analysis
ρ3 = λ3
μ3
= 0.666
ρ4 = λ4
μ4
= 0.666
E[K1(t)] = ρ1
1−ρ1
= 0.5 packets
E[K2(t)] = 0.133
1−0.133 = 0.15 packets
E[K3(t)] = 0.666
1−0.666 = 2 packets
E[K4(t)] = 0.666
1−0.666 = 2 packets
(c) E[T] =
4
i=1
E[Ki(t)]
α = 0.5+0.15+2+2
20 = 0.232 ms
12. N/A
13. (a) Queuing unit 1: λ1 = 3λ
Queuing unit 2: λ2 = 6λ + 6λ + 3λ + λ = 20λ
Queuing unit 3: λ3 = 6λ
Queuing unit 4: λ4 = 6λ
(b) E[K1] = ρ1
1−ρ1
; ρ1 = 3λ
μ1
E[K2] = ρ2
1−ρ2
; ρ2 = 20λ
μ2
E[K3] = ρ3
1−ρ3
; ρ3 = 6λ
μ3
E[K4] = ρ4
1−ρ4
; ρ4 = 6λ
μ4
(c) E[T] = E[K1]
3λ + E[K2]
20λ + E[K3]
6λ + E[K4]
6λ
14. α = 200 packets/ms
μ0 = 100 packets/ms
μi = 10 packets/ms
(a) λ = 0.4λ + α ⇒ λ = α/0.6 = 333.33 packets/s
Thus, the arrival rate to each of the queuing units in parallel is:
λi = 0.4/5 × 333.33 = 26.67 packets/s](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-94-320.jpg)
![87
(b) E[K0(t)] = ρ0
1−ρ0
Since ρ0 = λ
μ0
, then:
E[K0(t)] =
λ
μ0
1− λ
μ0
= λ
μ0−λ =
333.33 packets/s
100×103 packets−333.33 packets/s
=
3.34 × 10−3 packets
E[Ki(t)] = ρi
1−ρi
= λi
μi−λi
=
26.67 packets/s
10×103 packets−26.67 packets/s
= 2.67×
10−3 (for 1 ≤ i ≤ 5)
(c) E[T0] = E[K0(t)]
λ = 3.34×10−3 packets
333.33 packets/s
= 10−5 s
E[Ti] = E[K0(t)]+E[E[Ki(t)]]
α =
E[K0(t)]+
5
i=1
PiE[Ki(t)]
α = 2.67×10−3+5(0.2E[Ki(t)])
200
= 30 ms
15. (a) λ3 = 0.3λ − 1
λ2 = α + 0.3λ1
λ1 = λ2 + λ3 + λ4
λ4 = 0.3λ1
λ1 = λ2 + λ3 + λ4 = α + 0.3λ1 + 0.3λ1 + 0.3λ1 = α + 0.9λ1
⇒ λ1 = 8 packets/ms
λ2 = α + 0.3λ1 = α + 0.3 × 10α = 4α = 3.2 packets/ms
λ3 = 0.3 × 80 = 2.4 packets/ms
λ4 = 0.3λ1 = 0.3 × 80 packets/ms
(b) ρ1 = 8
10 ; E[K1(t)] = ρ1
1−ρ1
= 0.8
1−0.8 = 4
ρ2 = 3.2
12 ; E[K2(t)] = ρ2
1−ρ2
= 0.267
1−0.267 = 0.364
ρ3 = 2.4
14 ; E[K3(t)] = ρ1
1−ρ1
= 0.171
1−0.171 = 0.206
ρ4 = 2.4
16 ; E[K4(t)] = ρ1
1−ρ1
= 0.15
1−0.15 = 0.176
(c) E[T] = E[K(t)]
α = 4+0.364+0.206+0.176
0.8 = 5.93 ms](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-95-320.jpg)
![Chapter 12
Quality-of-Service and
Resource Allocation
1. (a) P0 = PX (0)P1 + [PX (0) + PX (1)]P0
P1 = PX (2)P0 + PX(1)P1 + PX (0)P2
P2 = PX (3)P0 + PX(2)P1 + PX (1)P2 + PX (0)P3
P3 = PX (4)P0 + PX(3)P1 + PX (2)P2 + PX (1)P3 + PX (0)P4
P4 = PX (5)P0+PX (4)P1+PX (3)P2+PX (2)P3+PX (1)P4+PX (0)P5
(b) See Figure 12.1.
0 1 2 3 4
Px(5)
Px(1)
Px(1)
Px(1) Px(1)
Px(0)
Px(0) + Px(1)
Px(0) Px(0) Px(0)
Px(2)
Px(2) Px(2) Px(2)
Px(3) Px(3) Px(3)
Px(4) Px(4)
Figure 12.1: Markov chain.
89](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-97-320.jpg)
![90 Chapter 12. Quality-of-Service and Resource Allocation
2. (a) PX (k) = 1
4 for k = 0, 1, 2, 3
PX (k) = 0 for k = 0, 1, 2, 3
Also for k = 0, 1, 2, 3 : PX(0) = PX(1) = PX (2) = PX (3) = 1
4
(b) P0 = [PX (0) + PX (1)]P0 + PX(0)P1 = (1
4 + 1
4)P0 + 1
4P1
= 1
2P0 + 1
4P1 ⇒ 1
2P0 = 1
4P1 ⇒ 2P0 = p1
P1 = PX (2)P0 + PX (1)P1 + PX (0)P2 = 1
4P0 + 1
4P1 + 1
4P2
= 1
4P1 + 1
4 × 1
2P1 + 1
4P2
⇒ 5
2 P1 = P2, P2 = 5
2 × 2P0 = 5P0
P2 = PX (3)P0 + Px(2)P1 + Px(1)P2 + PX(0)P3
= 1
4(P0 + P1 + P2 + P3)
= 1
4(1
5 P2 + 2
5P2 + P2 + P3)
⇒ P3 = 12
5 P2, P3 = 12P0
P0 = 0.008
P1 = 2P0 = 0.016
(c) See Figure 12.2.
0 1 2 3
¼
¼
¼
¼
¼
¼
¼
½
Figure 12.2: Markov chain.
P00 = PX (0) + PX (1) = 1
4 + 1
4 = 1
2
P01 = PX (2) = 1
4
P02 = PX (3) = 1
4
P11 = PX (1) = 1
4
P10 = PX (0) = 1
4
P12 = PX (2) = 1
4](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-98-320.jpg)
![91
P20 = 0
P21 = PX (0) = 1
4
P22 = PX (1) = 1
4
3. (a) For Poisson distribution:
PX (x) = (λt)xe−λt
x! .
For t = 1
g
PX (x) =
(λ
g
)xe
− λ
g
x! .
With λ = 20 packets/s and g = 30 packets/s
PX (k) =
(2
3
)ke− 2
3
k! .
(b) P0 = [PX (0) + PX (1)]P0 + PX (0)P1
⇒ P1 = [1−PX(0)−PX (1)]P0
PX (0)
We know P0 = 0.007. From Part (a), we also know:
PX (0) = e− 2
3
1 = 0.513
PX (1) =
2
3
e− 2
3
1 = 0.34, thus:
P1 = 0.00197
P1 = PX (2)P0 + PX(1)]P1 + PX(0)P2
⇒ P2 = [1−PX(1)]P1−PX (2)P0
PX(0)
We know PX(2) =
(2
3 )
2
e− 2
3
2! = 0.114, thus:
P2 = 0.249
P2 = PX (3)P0 + PX(2)]P1 + PX(1)P2 + PX (0)P3
⇒ P3 = [1−PX(1)]P2−PX (3)P0−PX (2)P1
PX (0)
We know PX(3) =
(2
3 )
3
e− 2
3
3! = 0.025, thus:
P3 = 0.318
(c) Transition probabilities:](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-99-320.jpg)
![93
5. N/A
6. Solving this problem requires a great deal of ellaborations in complex
mathematical background. We try to summarize this background. In
general, for a continuous random variable X ≥ 0, with mean E[X,
second moment E[X2], and PDF fX(x), the Laplace-Steiljes transform
(LST) of FX (x) is defined by
ˆ
fX(δ) = ∞
0 e−δxfX(x)dx
Hence, for the residual time distribution, Rj(t), LST is given by
r̂j(δ) = 1− ˆ
fX (δ)
E[X]δ
Now, we can derive the mean residual time, rj, can be derived as
rj = −limδ→0
dr̂j (δ)
dδ
= E[X2]
2E[X]
7. (a) For non-preemtive scheduler:
E[T1] = 0.37 s, E[T2] = 0.62 s, E[T3] = 0.25 s
(b) For preemtive scheduler:
E[T1] = 0.12 s, E[T2] = 0.66 s, E[T3] = 1.91 s
(c) The waiting time E[Ti] for class i packet is lower for preemtive
scheduler as long as i is low and it will be reversed for higher is.
8. We compare the impact of an increased number of inputs on total delay
in priority scheduler with: three flows (n=3), and four flows (n=4).
λi = λ = 0.2 packets/ms](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-101-320.jpg)
![94 Chapter 12. Quality-of-Service and Resource Allocation
1
μi
= 1
μ = 1ms
ri = r = 0.5 ms
(a) For a non-preemptive scheduler, we know:
E[Tq,i] = Wx + E[Tq,i]2 + E[Tq,i]3
=
n
j=1 ρjrj +
i
j=1 ρjE[Tq,j] + E[Tq,i]
i−1
j=1 ρj
n = 3, non-preemptive scheduler:
Wx =
n
j=1 ρjrj =
3
j=1 ρjrj = (3)(0.2)(0.5) = 0.3
For i = 1:
E[Tq,1] = Wx +
1
j=1 ρjE[Tq,j] + 0 = 0.3 + ρ1E[Tq,1]
= 0.3 + 0.2E[Tq,1]
E[Tq,1] = 0.375
For i = 2:
E[Tq,2] = Wx +
2
j=1 ρjE[Tq,j] + E[Tq,2]
1
j=1 ρj
= 0.3 + ρ1E[Tq,1] + ρ2E[Tq,2] + E[Tq,2]ρ1
= 0.3 + 0.2 × 0.375 + 0.2E[Tq,2] + 0.2E[Tq,2]
E[Tq,2] = 0.625 ms
For i = 3:
E[Tq,3] = Wx +
3
j=1 ρjE[Tq,j] + E[Tq,3]
2
j=1 ρj
= 0.3 + ρ1E[Tq,1] + ρ2E[Tq,2] + ρ3E[Tq,3] + ρ1E[Tq,3] + ρ2E[Tq,3]
= 0.3 + 0.2 × 0.375 + 0.2 × 0.625 + 0.2E[Tq,3]
E[Tq,3] = 1.25 ms
E[Ti] = E[Tq,i] + 1
μi
For i = 3 ⇒
E[T3] = E[Tq,i] + 1
μ3
= 1.25 + 1 = 2.25 ms](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-102-320.jpg)
![95
n = 4, non-preemptive scheduler:
Wx =
n
j=1 ρjrj =
4
j=1 ρjrj = (4)(0.2)(0.5) = 0.4
For i = 1:
E[Tq,1] = Wx +
1
j=1 ρjE[Tq,j] + 0 = 0.4 + 0.2E[Tq,1]
E[Tq,1] = 0.5
For i = 2:
E[Tq,2] =0.4 + ρ1E[Tq,1] + ρ2E[Tq,2] + E[Tq,2]ρ1
= 0.4 + 0.2 × 0.5 + 0.4E[Tq,2] + 0.2E[Tq,2]
E[Tq,2] = 0.833 ms
For i = 3:
E[Tq,3] =0.4 + ρ1E[Tq,1] + ρ2E[Tq,2] + ρ3E[Tq,3] + 2ρ1E[Tq,3]
= 0.4 + 0.2 × 0.5 + 0.2 × 0.833 + 0.6E[Tq,3]
E[Tq,3] = 1.667 ms
E[Ti] = E[Tq,i] + 1
μi
For i = 3 ⇒
E[T3] = E[Tq,3] + 1
μ3
= 1.667 + 1 = 2.667 ms
(b) For a preemptive scheduler, we know:
E[Ti] = E[Tq,i] + θi
where:
θi = 1
μi
1−
i−1
j=1
ρj
n = j = 3:
From non-preemtive case: E[Tq,3] = 1.25 ms
θ3 = 1
μ3
1−
2
j=1
ρj
= 1.667
E[T3] = 1.25 + 1.667 = 2.92 ms](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-103-320.jpg)
![96 Chapter 12. Quality-of-Service and Resource Allocation
n = j = 4:
From non-preemtive case: E[Tq,3] = 1.667 ms
θ3 = 1
μ3
1−
3
j=1
ρj
= 2.5
E[T3] = 2.5 + 1.667 = 4.167 ms
(c) The total delay obtained in the non-preemtive scheduler where
n = 3 and n = 4 are close.
In a preemtive scheduler, the difference in delay when n = 3 and
n = 4 is very large.
This is very obvious as when the numbers of flows increases in the
case of preemtive scheduler, the waiting time becomes large for a
low priority packet to be procesed. However, in a non-preemtive
scheduler, this has a littile impact. This is because in an non-
preemptive scheduler, lower priority packets cannot be interrupted
immidiately upon the arriaval of higher priority packets.
9. (a) 2.1,2.1,4.1,1.1,3.2,4.2,1.2,4.3,1.3,2.2,4.4,1.4,3.3,1.5,2.3,3.4,2.4,3.6,4.5,2.5
(b) 2.1,3.1,4.1,4.2,4.3,4.4,1.1,2.2,2.3,3.2,3.3,3.4,4.5,1.3,2.4,2.5,1.5
10. N/A
11. N/A](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-104-320.jpg)
![113
9. (a) N/A
(b) The complexity of this network is estimated to be
(nc/n) = d(h + logd n)
(nL/n) = 1 + h + logd n
(c) For the extended the Delta network DE
n,d,h,the blocking probability
is estimated as follows
Pp(n, d, 0) = 1 − (1 − p)k−1
Pp(n, d, 0) = [1 − (1 − p)2(1 − Pp(n/d, d, h − 1))]d
10. (a) Architecture 1: choose d = 2
k ≥ 2d − 1 ⇒ k = 3
complexity = 4(2 × 3) + 3(4 × 4) + 4(3 × 2) = 96
This is optimal in terms of complexity:Dopt =
n
2 = 2
⇒ k = 3
See Figure 13.12
Figure 13.12: Clos network, d = 2 and k = 3.
Architecture 2: choose d = 4
k ≥ 2d − 1 ⇒ k = 7
Complexity = 2(4 × 7) + 7(2 × 2) + 2(7 × 4) = 140](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-121-320.jpg)
![116 Chapter 13. Networks in Switch Fabrics
(c) C6,3,5 has lower probability of blocking. But it yields a higher
complexity and higher cost due to more crossbars is used. So it
really depends on the need of the network to decide which is better.
14. Five-stage Clos network with n = 8, d = 2.
(a) See Figure 13.15.
(b) See Figure 13.16.
(c) Total blocking probability with p = 0.2.
The probability of blocking for the middle stage is (2p−p2)3. Thus
B = (1 − (1 − p) × (1 − p) × [1 − (2p − p2)3])3 = 0.059
15. (a) See Figure 13.17.
(b) See Figure 13.18.
BXY = [1−(1−p)2(1−(1−p)2)] = [1−(1−0.2)(1−0.2)(1−0.059)] =
0.0629
BZW = [1 − (1 − 0.2)(1 − 0.0629)(1 − 0.2)]3 = 0.02
16. Consider a five-stage Clos network (similar to problem 13.14) whose
stages use the following crossbar dimensions: 1st stage: d × k, 2nd
stage: e × j, 3rd stage: n
de × n
de , 4th stage: j × e, and 1st stage: k × d.
(a)
(b) We find k, d, j, e in terms of n. Non blocking conditions: j ≥ 2e−1
k ≥ 2d − 1
Xc = dkn
d + k[ej n/d
e + n/d
e × n/d
e × j + ej n/d
e ] + dkn
d
= kn + k(jn
d + jn2
d2e2 + jn
d ) + kn](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-124-320.jpg)
![118 Chapter 13. Networks in Switch Fabrics
Figure 13.16: Lee’s model for the five-stage Clos network with n = 8, d = 2.
Xc = 2kn + k(2jn
d + jn2
d2e2 )
k = 2d − 1
j = 2e − 1
Xnb
c = 2n(2d − 1) + (2d − 1)[2n(2e−1)
e + n2(2e−1)
d2e2 ]
= 4nd − 6n + 8ne + 4n2
de2 − 2n2
de2 − 4ne
d − 2n
d − 2n2
d2e − n2
d2e2
To optimize:
dXnb
c
de = 8n − 4n2
de2 + 4n2
de3 − 4n
d + 2n2
d2e2 + 2n2
d2e3 = 0
⇒ eopt ≈
−4n2d+2n2
−8nd2+4nd
We know: dopt ≈
n
2 ⇒ eopt ≈
−4n2
√n
2
+2n2
−4n2+4n
√n
2
(c) Plug dopt and eopt into Xnb
c .](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-126-320.jpg)
![122 Chapter 14. Optical Switches and Networks, and WDM
6. (a) E[T] = 1
μi,j −Li,j
(b) E[Ts] = 1
μi,j −si,jLi,j
(c) E[Tn] = 1
n(n−1)
i,j
si,j
μi,j −si,jΛ
7. N/A
8. N/A
9. N/A
10. N/A](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-130-320.jpg)
![Chapter 17
Compression of Digital Voice
and Video
1. See Figure 17.1.
g(t) = sin(200πt)
s(t) =
+∞
n=−∞
[t−nTs
τ ] =
+∞
n=−∞
[2000t − 2n]
gs(t) = g(t) × s(t) =
+2
n=−2
[2000t − 2n] sin(200πt)
G(f) = j[δ(f − 100) + δ(f + 100)]
S(f) = 1
2
+∞
n=−∞ sinc[5 × 10−4(f − 1000n)]
Gs(f) = G(f) × S(f)
= 1
2
+2
n=−2 sinc[5 × 10−4(f − 1100n)] + sinc[5 × 10−4(f − 900n)]
2. N/A
3. Mean: E[X] = 0
Rate Capacity: R = 4 b/sample
Variance: V [X] = 2
(a) For this source, we know Db = V [X]2−2R. With R = 4, we obtain
135](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-143-320.jpg)
![136 Chapter 17. Compression of Digital Voice and Video
0
1 2
-1
-2
-2 -1 0 1 2
1
-2 -1
0 1 2
-100 100 -
1000 1000
-1100
- 100 1100
t t
f f f
G(f) S(f) G(f)
s
g(t) s(t) g(t)
s
t
Figure 17.1: Solution to exercise. Sampling process in time and frequency
domains.
Db=0.0078
(b) If tolerable distortion becomes Db = 0.05 by using the same for-
mula we obtain
R = 1
2 log2
1
Db
= 2.66
Thus, the required transmission capacity=2.66 b/sample
4. Variance = σ2 = V [X] = 10
N = 12](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-144-320.jpg)
![137
(a) Δ
√
V [X]
= 0.4238
Δ = 0.4238(
√
10) = 1.34
(b) ai = −aN−i = −(N
2 − i)Δ
a1 = −a11 = −(6 − 1)1.34 = −6.7
a2 = −a10 = −(6 − 2)1.34 = −5.36
a3 = −a9 = −(6 − 3)1.34 = −4.02
a4 = −a8 = −(6 − 4)1.34 = −2.68
a5 = −a7 = −(6 − 5)1.34 = −1.34
a6 = 0
(c) x̂i = −x̂N+1−i = −(N
2 − i + 1
2)Δ
x̂1 = −x̂12 = −(6 − 1 + 1
2)1.34 = −7.37
x̂2 = −x̂10 = −(6 − 2 + 1
2)1.34 = −6.03
x̂3 = −x̂9 = −(6 − 3 + 1
2)1.34 = −4.69
x̂4 = −x̂8 = −(6 − 4 + 1
2)1.34 = −3.35
x̂5 = −x̂7 = −(6 − 5 + 1
2)1.34 = −2.01
x̂6 = −x̂6 = −(6 − 6 + 1
2)1.34 = −0.67
(d) D
V [X] = 0.0.1885
D = 0.1885
(e) For the source, we have Db = V [X]2−2R. Here, we are asked to
find Db given R.
N = 12
R = log2 N ≈ 4
Db = V [X]2−2R = 0.039
5. Note to instructors: this problem needs to be corrected to “· · ·
repeat Problem 4, this time using a 16-level optimal uniform quantizer
· · ·”](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-145-320.jpg)
![138 Chapter 17. Compression of Digital Voice and Video
Variance = σ2 = V [X] = 10
N = 16
(a) Δ
√
V [X]
= 0.4908
Δ = 0.4908(
√
10) = 1.06
(b) ai = −aN−i = −(N
2 − i)Δ
a1 = −a15 = −(8 − 1)1.06 = −7.42
a2 = −a14 = −(8 − 1)1.06 = −6.36
a3 = −a13 = −(8 − 1)1.06 = −5.3
a4 = −a12 = −(8 − 1)1.06 = −4.24
a5 = −a11 = −(8 − 1)1.06 = −3.18
a6 = −a10 = −(8 − 1)1.06 = −2.12
a7 = −a9 = −(8 − 2)1.06 = −1.06
a8 = 0
(c) x̂i = −x̂N+1−i = −(N
2 − i + 1
2 )Δ
x̂1 = −x̂16 = −(8 − 1 + 1
2)1.06 = −7.95
x̂2 = −x̂15 = −(8 − 2 + 1
2)1.06 = −6.69
x̂3 = −x̂14 = −(8 − 3 + 1
2)1.06 = −5.83
x̂4 = −x̂13 = −(8 − 4 + 1
2)1.06 = −4.77
x̂5 = −x̂12 = −(8 − 5 + 1
2)1.06 = −3.71
x̂6 = −x̂11 = −(8 − 6 + 1
2)1.06 = −2.65
x̂7 = −x̂10 = −(8 − 6 + 1
2)1.06 = −1.59
x̂8 = −x̂9 = −(8 − 6 + 1
2)1.06 = −0.53
(d) D
V [X] = 0.01154
D = 0.1154
6. N/A](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-146-320.jpg)
![142 Chapter 17. Compression of Digital Voice and Video
513 -46 0 -2 0 -1 0 0
-46 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0
-2 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0
-1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Figure 17.3: Quantization of a still image to produce Matrix Q[i][j], and
the order of matrix elements for transmission.
Entropy of a uniformly distributed source is more compared to the above
entropy of a non-uniformly distributed suorce.
13. (a) PX (1) = PXY (1, 1) + PXY (1, 2) + PXY (1, 3) = 0.1+ 0.2+ 0.4 = 0.7
PX (2) = PXY (2, 1) + PXY (2, 2) + PXY (2, 3) = 0.1 + 0 + 0.2 = 0.3
PX (3) = PXY (3, 1) + PXY (3, 2) + PXY (3, 3) = 0
H(X) = −(0.7 log 0.7 + 0.3 log 0.3) = 0.881
PY (1) =
3
x=1 PXY (x, 1) = 0.1 + 0.1 = 0.2
PY (2) =
3
x=1 PXY (x, 2) = 0.2
PY (3) =
3
x=1 PXY (x, 3) = 0.4 + 0.2 = 0.6
H(Y ) = −(0.2 log 0.2 + 0.2 log 0.2 + 0.6 log 0.6) = 1.371
(b) The marginal entropy shows the average information that we re-
cieve from one source if we know the other one.
(c) H(X, Y ) = −(0.1 log 0.1 + 0.2 log 0.2 + 0.2 log 0.2 + 0.1 log 0.1 +](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-150-320.jpg)
![144 Chapter 17. Compression of Digital Voice and Video
(a) H(x) = −
N
k=1 Pk log2 Pk
= −(0.15 log2 0.15 + 0.2 log2 0.2 + 0.3 log2 0.3 + 0.35 log2 0.35)
= 1.926
Number of typical sequences = 2nH(x) = 2100(1.926) = 9.6 × 1057
(b) Total number of sequences = Nn = 4100 = 1.607 × 1060
Total number of non-typical sequences = 1.607 × 1060 − 9.9 × 1057
≈ 1.597 × 1060
Number of typical sequences
Number of non-typical sequences
= 9.6×1057
1.597×1060 = 0.006
(c) P[X̄] = 2−nH(x) = 2−100(1.926) = 1.04 × 10−58
(d) Number of bits to represent typical sequence= nH(x) = 100(1.926) =
193 b
(e) Most probable sequence is: {a4, a4, · · · , a4} with probability Pn
4 =
(0.35)100 = 2.55 × 10−46
17. (a) See Figure 17.4. The compressed codes are:
a0 = 0, a1 = 110, a2 = 10100, a3 = 100, a4 = 1011, a5 = 111,
a6 = 10101.
(b) The entropy of the source is computed as
H(X) = −
6
k=0 Pk log2 Pk = 2.07
The average code length is:
R̄ =
6
i=0 Pii = 1(0.55) + 3(0.1) + 5(0.05) + 3(0.14) + 4(0.06) +
3(0.08) + 5(0.02) = 2.1
As it is expected: H(X) ≤ R̄ ≤ H(X) + 1](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-152-320.jpg)
![150 Chapter 18. VoIP and Multimedia Networking
+ 12B (UDP header) + 20B (total RTP headers)]= 1,448 B
(b) Total combined two source data rates = 2×31 Kb/s ×(1/8) B/b=
7.75 KB/s
Time required for one packet = 1,448 B
7.75 KB/s
= 186.8 ms
Number of packets generated in 5 minutes = 5×60
186.8 ms = 1,605
packets.
3. N/A
4. 1,280×1,024 pixel blocks
1 packet = 0.1 row
1 chunk = 1 pixel cblock
(a) Sample sapce ⇒ 77 samples ⇒ each sample (pixel) = 7 b
A chunk = 1 pixel block = (8×8 pixels) × 7 b = 448 b
A chunk + header = 448 b + (4 B) × 8 (b/B) = 480 b
(b) 1 packet = 0.1 row = (0.1)(1,280) = 128 pixel blocks = 128 chunks
= 128 chunks × 480 b/chunk = 61,440 b
A packet + header = 61,440 b + 12 × 8 (b/B) = 61,536 b
(c) Video clip = 4 minutes
1 sec = 30 images
Total number of images/min = (4 min) × 60 (sec/min) × 30 =
7,200 images
Total number of packets/image = (10 packets/row) × 1,024 (rows)
= 10,240 packets/image
Total number of packets/4 min = (7,200) × (10,240) = 73,728,000](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-158-320.jpg)
![151
Total number of bits/4 min = (73,728,000) × (61,440 b/packets)
= 4.53 Tb
Bandwidth = 4.53Tb
4 min = 18.8 Gb/s
5. Data of Chunk: 1 pixel block × 10 phrases × 5 bits/phrase = 50 b
Header of Chunk: = 4 B
Chunk = 50 + 4 × 8 = 82 bits
(a) Row data = 1280 pixel block × 82 b/pixel block
SCTP header = 12 bytes
Packet = 1280×82
8 + 12 = 13, 132 byte ≈ 12.8 kbyte
(b) One frame = 12.8 kbyte/row × 1024 rows = 12.8 Mbyte
Required B.W. = 12.8 Mbyte/frame × 30 frames/s × 8 b/B =
3 Gb/s
(c) Total data = 3 Gb/s × 2 × 60 × 60 = 2, 700 GB/h
6. (a) FY (t)(y) = P[Y (t) ≤ y] = P[X(t) + 2t ≤ y] = P[X(t) − 2t]
= FX(t)(y − 2t) ⇒ fY (t)(y) = FX(y − 2t) = 1
√
2παt
e−(y−2t)2/2αt
(b) FY (t)Y (t+1)(y1, y2) = P[X(t) + 2t ≤ y1, X(t + 1) + 2(t + 1) ≤ y2]
= FX(t),X(t+1)(y1 − 2t, y2 − 2(t + 1))
⇒ fY (t)Y (t+s)(y1, y2) = fX(t),X(t+1)(y1 − 2t, y2 − 2(t + 1))
= fX(t)(y1 − 2t)fX(t)(y2 − y1 − 2)
= 3−(y1−2t))2/2αt
√
2παt
× e−(y2−y1−2)2/2α
√
2πα
7. 70 cycles/min
6 pulses/cycle
1 pulse/chunk](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-159-320.jpg)
![152 Chapter 18. VoIP and Multimedia Networking
Q,R,S → 4 samples
P,T,U → 1 sample
(a) We use variable size chunk.
Assume each sample is encoded by 8 bits.
A packet consists of: 12B SCTP header + 6 chunks each with 16B
chunk header.
Total bits of the 3 chunks made by P,T,U pulses
= 3(16B (chunk header) × 8 b/B + 1 sample/pulse × 8 b/sample)
= 408 b
Total bits of the 3 chunks made by Q,R,S pulses
= 3(16B (chunk header) × 8 b/B+ 4 samples/pulse × 8 b/sample)
= 480 b
Total SCTP packet size = (408b + 480b) + [20B (IP header) +
12B (SCTP header)] × 8 b/B = 1144 b/(SCTP packet)
Bandwidth = 70 cycles/min =
70×1144 b/packet
60s = 1335 b/s
(b) H = Max. Number of Human’s Heartbeat Cycles
G = Max. number of Patients
L = Max. Link Bandwidth
We can choose the above parameters as long as we must can sat-
isfy:
L = H×G×1144
60s
(c) The second option reduces the transmission overhead (packet over-
head) but requires a larger bandwidth.
8. N/A](https://image.slidesharecdn.com/solutionsmanual-231111145522-78984253/85/SOLUTIONS-MANUAL-pdf-160-320.jpg)
SOLUTIONS MANUAL.pdf
- 1. SOLUTIONS MANUAL FOR COMPUTERAND COMMUNICATION NETWORKS Nader F. Mir Upper Saddle River, NJ . Boston . Indianapolis . San Francisco . New York Toronto . Montreal . London . Munich . Paris . Madrid . Capetown Sydney . Tokyo . Singapore . Mexico City
- 2. The author andpublisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein. Visit us on the Web: www.prenhallprofessional.com Copyright © 2007 by Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work(including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. ISBN 0-13-234570-6 First release, March 2007
- 3. Contents Preface ii 0.1 Howto Obtain Errata of Text-Book . . . . . . . . . . . . . . iii 0.2 Errors in This Solution Manual . . . . . . . . . . . . . . . . . iii 0.3 How to Obtain an Updated Solution Manual . . . . . . . . . iii 0.4 How to Contact Author . . . . . . . . . . . . . . . . . . . . . iv About the Author v I Fundamental Concepts 1 1 Packet-Switched Networks 3 2 Foundation of Networking Protocols 9 3 Networking Devices 15 4 Data Links and Transmission 21 5 Local-Area Networks and Networks of LANs 29 6 Wireless Networks and Mobile IP 35 7 Routing and Inter-Networking 41 8 Transport and End-to-End Protocols 51 i
- 4. ii CONTENTS 9 Applicationsand Network Management 57 10 Network Security 67 II Advanced Concepts 73 11 Packet Queues and Delay Analysis 75 12 Quality-of-Service and Resource Allocation 89 13 Networks in Switch Fabrics 101 14 Optical Switches and Networks, and WDM 121 15 Multicasting Techniques and Protocols 123 16 VPNs, Tunneling, and Overlay Networks 133 17 Compression of Digital Voice and Video 135 18 VoIP and Multimedia Networking 149 19 Mobile Ad-Hoc Networks 155 20 Wireless Sensor Networks 157
- 5. Preface An updated versionof the solution manual is hereby provided to instructors. For any problem marked with “N/A,” the solution will be provided in the upcoming versions. Please check with the publisher or the author time- to-time to obtain the latest verion of the solution manual. For effective educational learning purposes, instructors are kindly requested not to allow any student to access this solution manual. 0.1 How to Obtain Errata of Text-Book The Errata of the text-book, Edition 1, is now available. Please contact the author directly at nader.mir@sjsu.edu for copy. 0.2 Errors in This Solution Manual If you find any error in this solution manual, please directly inform the author at nader.mir@sjsu.edu . 0.3 How to Obtain an Updated Solution Manual Please check time to time the web page of the text-book in the Prentice- Hall site and click on “Instructors’ link” to obtain the latest version of this solution manual. iii
- 6. iv CONTENTS 0.4 Howto Contact Author Please feel free to send any feedback on the text book to Department of Elec- trical Engineering, San Jose State University, San Jose, CA 95192, U.S.A, or via email at nader.mir@sjsu.edu. The preparation of this version of the solution manual took years and the manual may contain some errors. Please also feel free to send me any feedback on this solution manual. I would love to hear from you especially if you have suggestions for im- proving this book further for its next editions. I will carefully read all review comments. You can find out more about me at: http://www.engr.sjsu.edu/nmir I hope that you enjoy the text and that you receive a little of my liking for the computer communications from it. Nader F. Mir San Jose, California
- 7. About the Author NaderF. Mir received the B.Sc. degree (with honors) in electrical & computer engineering in 1985 and the M.Sc. and Ph.D. degrees both in electrical engineering from Washington University in St. Louis, MO, in 1990 and 1994 respectively. He is currently a professor and department associate chairman of Elec- trical Engineering at San Jose State University, California. He is also the director of MSE Program in Optical Sensors and Networks for Lockheed Martin Space Systems. Previously, he was an associate professor at this school and assistant professor at the University of Kentucky in Lexington. From 1994 to 1996, he was a research scientist at the Advanced Telecommu- nications Institute, Stevens Institute of Technology, in New Jersey, working on the design of advanced telecommunication networks. From 1990 to 1994, he was with the Computer and Communications Research Center at Wash- ington University in St. Louis and worked as a research assistant on design and analysis of high-speed switching-systems projects. His research interests are: analysis of computer communication networks, design and analysis of switching systems, network design for wireless ad- hoc and sensor systems, and applications of digital integrated circuits in computer communications. He is a senior member of the IEEE and has served as the member of Technical Program Committee and Steering Committee of a number of major IEEE networking conferences such as WCNC, GLOBECOM, and ICC. Dr. v
- 8. vi CONTENTS Mir haspublished numerous refereed technical journal and conference papers all in the field of communications and networking. He has published a book in video communication engineering, and another text-book published by Prentice Hall Publishing Co. entitled “Computer & Communication Networks, Design and Analysis”. Dr. Mir has received a number of prestigious national and university awards including the university teaching recognition award and research excellence award. He is also the recipient of the 2004 IASTED Outstanding Tutorial Presentation award. Currently, he has several journal editorial positions such as: the Edito- rial Board Member of the International Journal of Internet Technology and Secured Transactions, the Editor of Journal of Computing and Information Technology, and the Associate Editor of IEEE Communication Magazine.
- 9.
- 11. Chapter 1 Packet-Switched Networks 1.Total distance = = 2( 3, 0002 + 10, 0002) = 20,880.61 km. Speed = c = 2.3 ×108 m/s. (a) proagation delay = tp = c = 20,880.61 km 2.3×108 m/s = 90.8 ms (b) Number of bits in transit during the propagation delay = (90.8 ms) × (100 × 106 b/s) = 9.08 Mb (c) 10 bytes = 80 bits 2.5 bytes = 20 bits, then: total length = 80 + 20 = 100 bits T = 100 b 100×106 b/s = 1 μs 2. Total distance = = 2( (30/1000)2 + 10, 0002) ≈ 20,000 km. Speed = c = 2.3 ×108 m/s. (a) tp = c = 20,000 km 2.3×108 m/s = 87 ms (b) 100 Mb/s × 0.087 s = 8.7 Mb 3
- 12. 4 Chapter 1.Packet-Switched Networks (c) Data: (10 B)×8 b 100 Mb/s + tp = 0.79 μs + 0.087 s Ack: (2.5 B)×8 b 100 Mb/s + tp = 0.19 μs + 0.087 s Total time ≈ 1μs (transfer) + 0.173 s (prop.) 3. Assuming the speed of transmission at 2.3 × 108: (a) Total Delay: D = [np + (nh − 2)]tf + (nh − 1)tp + nhtr (b) tp1 = 50 miles×1600 m/miles 2.3×108 m/s = 0.35 ms tp2 = 400 miles×1600 m/miles 2.3×108 m/s = 2.8 ms Number of packets = np = 200MB 10KB = 20, 000 tf = 10,040 B/pockets×8 b/B 100 Mb/s = 0.8 ms/pockets D = [20, 000+(5−2)]0.8+[(3−1)0.35+(3−1)2.8]+5×0.2×103 ≈ 16.6 s 4. Dp = [np + (nh − 2)]tf1 + nhtr1 + (nh − 1)tp Dc = 3 ([1 + (nh − 2)]tf2 + nhtr2 + (nh − 1)tp) Dt = Dp + Dc = (np + nh − 2)tf1 + 3(nh − 1)tf2 + nh(tr1 + 3tr2) + 4(nh − 1)tp 5. Number of packets = np = 200MB 10KB = 20, 000 Dt = Dp + Dc Dc = dconn-req + dconn-accep + dconn-release (a) Here, the problem askes that tr be defined as the processing time for each packet. Therefore, tr = 20, 000× 0.2 = 4,000 s
- 13. 5 Dc = dconn-req= dconn-accep = dconn-release = [np + (nh − 2)]tf + nhtr + (nh − 1)tp = [1 + (5 − 2)] 500 b/packet 100 mb/s + 3 × 4, 000 s + 4.84 ms ≈ 12, 000 s (b) Same as Part (a). (c) Dt = Dp + Dc = 17 + 3 × 12, 000 = 36, 017 s 6. s = 109 b/s nh = 10 nodes tr1 = tr2 = 0.1 s = tr Data forms two packets: (9960 + 40) bytes for packet1 (2040 + 40) bytes for packet2 tf1−packet1 = 10,000 B×8 b/B 109 b/s = 8 × 10−5 s tf1−packet2 = 2,080 B×8 b/B 109 b/s = 16.64 × 10−6 s tf2 = transfer times for control packets = 500 b 109 b/s = 5 × 10−7 s tp = c = 500 miles×1.61×103 m 2.3×108 m/s = 3.5 × 10−3 s (a) request + accept time: t1 + t2 = 2 ([np + (nh − 1)]tf2 + (nh − 1)tp + nhtr]) = 2.06 s (b) t3 = 1 2 (t1 + t2) = 1.03 s (c) Dt = Dp + Dc Dp = Dp−packet1 + Dp−packet2 = [np + (nh − 2)]tf1−packet1 + nhtr1 + (nh − 1)tp + [np + (nh −
- 14. 6 Chapter 1.Packet-Switched Networks 2)]tf2−packet1 + nhtr1 + (nh − 1)tp Dc = t1 + t2 + t3 Dt = 4.1 s 7. d + h = 10, 000 b, ρd = 72%, h d = 0.04, s = 100 Mb/s, (a) h = 0.04d then: d d+h = 0.96. Since: ρd = ρ d h+d then: 0.72 = ρ0.96 ⇒ ρ = 0.74 (b) μ = s h+d = 100×106 b/s 10×103 b = 10 × 103 = 10, 000 packets/sec (c) λ = ρμ = 0.74 × 10, 000 = 74, 000 D̄ = 1 10,000−7,400 = 0.38 ms (d) D̄opt = h s √ ρd 1− √ ρd 2 d + h = 10, 000h/d = 0.04 ⇒ h = 384 b D̄opt = 0.12 8. s = 100 Mb/s ρ = 80% (a) ρ = λ μ ⇒ 0.8 = 8000 μ ⇒ μ = 10, 000 packets/s
- 15. 7 t t t t Connection Request Connection Accept Connection Release Data Node A Node C NodeB Node D Figure 1.1: Signaling delay in connection-oriented packet-switched environ- ment. μ = s h+d ⇒ 10, 000 = 100×106 h+d ⇒ h + d = 10, 000 b (b) ρh = 0.008 and ρh = ρ h d+h ⇒ d + h = 100h ⇒ h = 100 b, and d = 9900 b. (c) ρh = 0.008 ρd = ρ − ρh = 0.8 − 0.008 ⇒ ρd = 0.792 dopt = h √ ρd 1− √ ρd = 809 bits (d + h)opt = h + dopt = 100 + 809 ⇒ (d + h)opt = 909 b (d) D̄opt = h s 1 1− √ ρd 2 ⇒ D̄opt = 8.2 × 10−5 s 9. See Figure 1.1. 10. D = d+h s[1−ρd/d(d+h)]
- 16. 8 Chapter 1.Packet-Switched Networks (a) ∂D ∂h = 0 Thus: hopt = d(1−2ρd) 2ρd (b) Queueing delay
- 17. Chapter 2 Foundation ofNetworking Protocols 1. (a) Address: 127.156.28.31 = 0111 1111 . 1001 1100 . 0001 1100 . 0001 1111 Mask: 255.255.255.0 = 1111 1111 . 1111 1111 . 1111 1111 . 0000 0000 Class A Subnet ID: 1001 1100 0001 1100=39964 (b) Address: 150.156.23.14 = 1001 0110 . 1001 1100 . 0001 0111 .0000 1110 Mask: 255.255.255.128 = 1111 1111 . 1111 1111 . 1111 1111 . 1000 0000 Class B Subnet ID: 000101110 = 46 (c) Address: 150.18.23.101 = 9
- 18. 10 Chapter 2.Foundation of Networking Protocols 1001 0110 . 0001 0010 . 0001 0111 . 0110 0101 Mask: 255.255.255.128 = 1111 1111 . 1111 1111 . 1111 1111 . 1000 0000 Class B Subnet ID: 000101110 = 46 2. (a) IP: 1010 1101 . 1010 1000 . 0001 1100 . 0010 1101 Mask: 1111 1111 . 1111 1111 . 1111 1111 . 0000 0000 Class B Subnet ID=00011100=28 (b) A packet with IP address 188.145.23.1 using mask pattern 255.255.255.128 IP: 1011 1100 . 1001 0001 . 0001 0111 . 0000 0001 Mask: 1111 1111 . 1111 1111 . 1111 1111 . 1000 0000 Class B Subnet ID=000101110=46 (c) A packet with IP address 139.189.91.190 using a mask pattern 255.255.255.128 IP: 1000 1011 . 1011 1101 . 0101 1011 . 1011 1110 Mask: 1111 1111 . 1111 1111 . 1111 1111 . 1000 0000 Class B Subnet ID=010110111=183 3. IP1: 1001 0110 . 0110 0001 . 0001 1100 . 0000 0000 IP2: 1001 0110 . 0110 0001 . 0001 1101 . 0000 0000 IP3: 1001 0110 . 0110 0001 . 0001 1110 . 0000 0000 New IP (CIDR): 150.97.28.0/22
- 19. 11 4. Address: 141.33.11.0/22 =1000 1101 . 0010 0001 . 0000 1011 . 0000 0000 141.33.12.0/22 = 1000 1101 . 0010 0001 . 0000 1100 . 0000 0000 141.33.13.0/22 = 1000 1101 . 0010 0001 . 0000 1101 . 0000 0000 141.33.8.0/21 5. (a) 191.168.6.0 1011 1111 . 1010 1000 . 0000 0110 . 0000 0000 1111 1111 . 1111 1111 . 1111 1110 . 0000 0000 Result: 1011 1111 . 1010 1000 . 0000 0110 . 0000 0000 = 191.168.6.0/23 (b) 173.168.28.45 1010 1101 . 1010 1000 . 0001 1100 . 0010 1101 1111 1111 . 1111 1111 . 1111 1110 . 0000 0000 Result: 1010 1101 . 1010 1000 . 0001 1100 . 0000 0000 = 173.108.28.0/23 (c) 139.189.91.190 1000 1011 . 1011 1101 . 0101 1011 . 1011 1110 1111 1111 . 1111 1111 . 1111 1110 . 0000 0000 Result: 1000 1011 . 1011 1101 . 0101 1010 . 0000 0000 = 139.189.90.0/23 6. 180.19.18.3: 1011 0100 . 0001 0011 . 0001 0010 . 0000 0011
- 20. 12 Chapter 2.Foundation of Networking Protocols (a) 180.19.0.0/18: 1011 0100 . 0001 0011 . 0000 0000 . 0000 0000 180.19.3.0/22: 1011 0100 . 0001 0011 . 0000 0011 . 0000 0000 180.19.16.0/20: 1011 0100 . 0001 0011 . 0001 0000 . 0000 0000 (b) The longest match is 180.19.16.0/20. 7. (a) N1 L11: 1100 0011 . 0001 1001 . 0000 0000 . 0000 0000 N2 L13: 1000 0111 . 0000 1011 . 0000 0010 . 0000 0000 N3 L21: 1100 0011 . 0001 1001 . 0001 1000 . 0000 0000 N4 L23: 1100 0011 . 0001 1001 . 0000 1000 . 0000 0000 N5 L31: 0110 1111 . 0000 0101 . 0000 0000 . 0000 0000 N6 L33: 1100 0011 . 0001 1001 . 0001 0000 . 0000 0000 (b) Packet: 1100 0011 . 0001 1001 . 0001 0001 . 0000 0011 L11: 1100 0011 . 0001 1001 . 0000 0000 . 0000 0000 L12: 1100 0011 . 0001 1001 . 0001 0000 . 0000 0000 L12: 1100 0011 . 0001 1001 . 0000 1000 . 0000 0000 L13: 1000 0111 . 0000 1011 . 0000 0010 . 0000 0000 L21: 1100 0011 . 0001 1001 . 0001 1000 . 0000 0000 L22: 1100 0011 . 0001 1001 . 0001 0000 . 0000 0000 L23: 1100 0011 . 0001 1001 . 0000 1000 . 0000 0000 L31: 0110 1111 . 0000 0101 . 0000 0000 . 0000 0000 L33: 1100 0011 . 0001 1001 . 0001 0000 . 0000 0000 Answer=N6 (c) 32-21=11 b So, 211 = 2,048 users
- 21. 13 8. (a) IPv4address field has 32 bits Total number of IP addersses available = 232 Number of IP addresses per person for 620 million people = 232 620×106 = 6.9 ≈ 7 (b) Number of bits required to serve 620 million people is 30 since 229 ≈ 536 mil 620 mil 230 ≈ 1, 074 mil Thus, CIDR can only have 32-30 = 2 bit as Network ID field as x.x.x.x/2 (c) IPv6 address field has 128 bits: Total number of IPaddresses available = 2128 Number of IP addresses per person for 620 million people = 2128 620×106 = 5.49 × 1029 9. (a) 1111:2A52:111:73C2:A123:56F4:1B3C binary form: 0001 0001 0001 0001 : 0010 1010 0101 0010 : 1010 0001 0010 0011 : 0000 0001 0001 0001 : 0111 0011 1100 0010 : 1010 0001 0010 0011 : 0101 0110 0111 0100 : 0001 1011 0011 1100 (b) 2532::::FB58:909A:ABCD:0010 binary form: 0010 0101 0011 0010 : : : : 1111 1011 0101 1000 : 10001 0000 1001 1010 : 1010 1011 1100 1101 : 0000 0000 0001 0000 (c) 2222:3333:AB01:1010:CD78:290B::1111 binary form: 0010 0010 0010 0010 : 0011 0011 0011 0011 : 1010 1011 0000 0001: 0001 0000 0001 0000 : 1100 0000 0111 1000 : 0010 1001 0000 1011 : : 0001 0001 0001 0001
- 22. 14 Chapter 2.Foundation of Networking Protocols 10. N/A 11. Connection set up can be greatly simlpified because the VP is already selected. Only the VC has to be chosen with 216 = 64K choices. 12. Retain features: • conection-oriented network. Lost Features: • shorter process time • lower header/data ratio • harder to multiplex
- 23. Chapter 3 Networking Devices 1.(a) Number of channels= 12 × 5 × 10 = 600 (b) Capacity= 600 × 4 KHZ = 2.4 MHZ 2. (First, please make a correction: 170 Kb/s must change to 160 Kb/s) Total output bit-rate of the multiplexer is 160 Kb/s (a) Total bit-rate from analog links = (5 KHz + 2 KHz + 1 KHz) × 2 samples/cycle × 5 bits/sample = 80 Kb/s (b) Total bit-rate from digital links = 160 Kb/s - 80 Kb/s = 80 Kb/s Total bit-rate per digital line = 80 Kb/s 4 lines = 20 Kb/s Pulse stuffing per digital line = 20 Kb/s - 8 Kb/s = 12 Kb/s (c) The number of channels of the frame dedicated to each line is proportional to the data rate of that line. Let’s consider one channel when upto 10 Kb/s of data rate is present. Using a proportional channel assignment , we need a total of 5 + 2 + 1 = 8 analog channels. As each digital line requires 8 Kb/s, we can also assign one channel per digital line. Therefore we need 4 15
- 24. 16 Chapter 3.Networking Devices digital channels, and one control channel: Bits in each frame = (8 + 4 + 1)×5 b/channel + 1 guard-bit= 66 b/frame Frame rate = 160×103 b/s 66 b/frame = 2,424 frames/s 3. (a) Pulse stuffing = 4800 b/s × 0.03 = 144 b/s Number of characters are = 1 (sync) + 99 (data)= 100 Synchronization bit rate = 4800 b/s 100 ≈ 50 b/s Number of 150 b/s terminals = 4800−(2×600+5×300+50+144) b/s 150 b/s = 12.7 ≈ 12 terminals (b) The number of characters for synchronization is proportional to bit rates. For example, since we need 12 characters for 150 b/s terminals, therefore we need 3 characters for synchronization. Frame format in terms of bits is: 2 × (12 char × 10 b/char) + 5 × (6 char × 10 b/char) + 12 × (3 char × 10 b/char) + 3 × 10 b/char + 1 × 10 b/char = 940 b/frame 4. (a) #bits/frame (total) = 2 Mb/s × 26 μs/frame = 52 bits/frame #bits/frame (data) = 52 bits/frame − 10 = 42 bits/frame #channels = n = 42 bits/frame 1 6 bits/ch = 7 ch/frame (b) P[clipping] = m−1 i=n m − i i pi(1 − p)m−i−1 m = 10, n = 7, p = 0.9 P[clipping] = 0.947
- 25. 17 5. (a) ρ= ta tc+d = 2/8 = 25% (b) Pj=3 = m j ( ti ta )j j i=0 m j (ta td )i = 8 3 (2 6 )3 3 j=0 8 1 (2 6 )i ≈ 21% (c) B = Pj=n=4 = m n (1/3)n n i=0 m i (1/3)i = 8 7 (1/3)4 4 i=0 8 i (1/3)i = 70/81 1+8/3+28/9+56/27+70/81 ≈ 9% (d) E[C] = 4 j=1 jpj = 1(0.275) + 2(0.32) + 3(0.213) + 4(0.0889) ≈ 1.94 6. (a) m = 4 n = 2 Prob[clipping] = Pc = 3 i=2 3 i ρi(1 − ρ)3−i = 10.4% for ρ = 0.2 = 35.2% for ρ = 0.4 = 64.8% for ρ = 0.6 = 89.6% for ρ = 0.8 (b) m = 4 n = 3 Prob[clipping]Pc = 3 i=3 3 3 ρ3(1 − ρ)3−3 = 0.0% for ρ = 0.2 = 6.4% for ρ = 0.4 = 21.6% for ρ = 0.6 = 51.2% for ρ = 0.8 (c) n = 4 P4 = 100%
- 26. 18 Chapter 3.Networking Devices 7. ρ = ta ta+td = 0.9 for m = 11 and n = 10 : Pc = the clipping prabability = m−1 i=n m − 1 i ρi(1−ρ)m−1−i = 10 10 ρ10(1−ρ)0 = ρ10 =≈ 0.35 8. 1 μ 1 − η ρ m 9. See Figure 3.1. 1 0 1 1 1 1 1 0 0 Natural NRZ Polar NRZ Manchester Figure 3.1: Line coding.
- 27. 19 10. See Figure3.2. ASK FSK PSK Figure 3.2: Modulation techniques. 11. (a) Assume a packet incoming at input port of IPP has length of L bits. Then, T = d+50 r ⇒ ∂T2 ∂d∂r = 0 ⇒ dopt, ropt Therefore ways to optimize the transmission delay T are follow- ings: • Increase transmission rate (r) by reducing clock cycle time of CPU. • Define value of d to be equal to highest-probability packet length(L).
- 28. 20 Chapter 3.Networking Devices (b) For example, if the switch fabric has 5 stages of routing in its internal network, the processing delay mostly depends on AND gate switch time of gates on a fabric. Assume applying CMOS transistors,which are slowest technology for switching transistors, for this switch fabric. Assuming 50-80 ns switch time for CMOS AND gate, the total propagation delay in this switch fabric = 80 ns ×5 stages=0.4 μs. On the other hand, the delay in IPP (D) mostly depends on packet fragmentation and encapsulation delay time. Typical value of this delay time is about tens or hundreds of milliseconds for a 512-bytes packet. Therefore processing delay in the switch fabric is not significant compared to delay in IPP. 12. N/A
- 29. Chapter 4 Data Linksand Transmission 1. Tprop = 5000×103 m 3×108 m/s = 16.7 m/s Total size=(500 page)(1000 char/page)(8 bits/char)=4 Mb (a) T = 16.7 ms + 4 Mb/(64 kb/s)=62.51 s (b) T = 16.7 ms + 4 Mb/(620 mb/s)=23.15 ms (c) With two million volumes of books: Total size = 4Mb ×2 × 106 = 8000 Gb i. T = 16.7 ms + 8000 Gb/(64 kb/s) = 1.25 × 108 s ≈ 4 years ii. T = 16.7 ms+8000 Gb/(620 mb/s) = 12903.2167 s ≈ 3.6 hours 2. N/A 3. (a) See Figure 4.1 (a). CRC-12: X12 + X11 + X3 + X2 + X + 1 Rule of hardware: For each existing term except the first term (in this case X12) assign an EXOR followed by a 1-bit register. 21
- 30. 22 Chapter 4.Data Links and Transmission (a) (b) 14 3 2 1 0 11 10 4 3 2 1 0 15 Figure 4.1: Answer to exercise. For each non-existing term assign a 1-bit register. Once all bits of data (D, 0) moves in completely, the content of registers show the remainder of the division process. (b) See Figure 4.1 (b). CRC-16: X16 + X15 + X2 + 1 4. (a) Dividend = X10 + X8 + X6 + X5 + X4 Divisor = X4 + X (b) If dividend = X10 + X8 + X6 + X5 + X4, and divisor = X4 + X, then, quotient = X6 + X4 − X3 + X2 + 2 and, remainder = −X3 − 2X
- 31. 23 5. The hardwareis shown in figure 4.2. 1-bit Shift Register dividend 10101110000 Starting bit to enter 0 1 2 3 Power of x: Figure 4.2: Contents of the four shift registers. If we sift in D, 0 = 1010111,0000 G = 10010 ⇒ X4 + X The final contents of shift registers as the step-by-step implementation of D,0 G |2 shows CRC = 0 0 0 1 (MSB at right): Bits of D, 0 left to shift in Shift registers’ contents 1010111,0000 0 0 0 0 010111,0000 1 0 0 0 10111,0000 0 1 0 0 0111,0000 1 0 1 0 111,0000 0 1 0 1 11,0000 1 1 1 0 1,0000 1 1 1 1 0000 1 0 1 1 000 0 0 0 1 00 0 1 0 0 0 0 0 1 0 - 0 0 0 1 If we sift in D, CRC = 1010111,1000 G = 10010 ⇒ X4 + X The final contents of shift registers as the step-by-step implementation of D,CRC G |2 shows 0 0 0 0 indicating no errors:
- 32. 24 Chapter 4.Data Links and Transmission Bits of D, CRC left to shift in Shift registers’ contents 1010111,1000 0 0 0 0 010111,1000 1 0 0 0 10111,1000 0 1 0 0 0111,1000 1 0 1 0 111,1000 0 1 0 1 11,1000 1 1 1 0 1,1000 1 1 1 1 1000 1 0 1 1 000 1 0 0 1 00 0 0 0 0 0 0 0 0 0 - 0 0 0 0 6. (a) D=1010 1101 0101 111 G=1110 10 g=6, then, g-1=5 D,0=1010 1101 0101 111,0000 0 CRC=D,0 G |2 Dividend = 101011010101111,00000 Divisor = 111010 Quotient = 111011111000010 Remainder = 10100 CRC=10100 D,CRC=1010 1101 0101 111,10100 (b) D,CRC = 1010 1101 0101 111,10100 G=1110 10 D,CRC G |2 Dividend = 101011010101111,10100 Divisor = 111010 Quotient = 111011111000010
- 33. 25 Remainder = 0 Thedata is correct. 7. v = 3 × 108 m/sec = 80 b/frame r = 10 × 106 bits/sec, tp/tf = 10 (a) E = tf t = tf tf +2tp = 1 1+2tptf = 1 1+2(10) = 0.04762 or 4.76% (b) Assuming the speed of light to be 3 × 108 in the cable: 10 = tp tf = d v r = d 3×108 80 10×106 ⇒ d(10×106) 80(3×108) = 10 d = 24 km (c) See Figure 4.3. tp = d/v = 24 km/3 × 108 ⇒ tp = 8 × 10−5 s (d) 8 : E = 1 1+2(8) = 0.0588 = 5.88% tp = 6.4 × 10−5 sec, d = 19.2 km 6 : E = 1 1+2(6) = 0.0769 = 7.69% tp = 4.8 × 10−5 sec, d = 14.4 km 4 : E = 1 1+2(4) = 0.111 = 11.11% tp = 3.2 × 10−5 sec, d = 9.6 km 2 : E = 1 1+2(2) = 0.2 = 20% tp = 1.6 × 10−5 sec, d = 4.8 km 8. tp = 0.2 s r = 2 Mb/s f = 800 b tf = f r = 800 2×106 = 4 × 10−4 s
- 34. 26 Chapter 4.Data Links and Transmission 0.2 0.1111 2 0.0476 0.0588 0.0769 4 6 8 10 E tp tf Figure 4.3: Answer to exercise. The efficiency trend. (a) Stop-and-Wait protocol: E = tf t = tf tf +2tp = 4×10−4 4×10−4+2(.2) = 0.0010 ≈ 0.1% (b) Sliding window protocol, w = 6: E = w w+2 tp tf = 6 6+2 0.2 4×10−4 = 0.00596 ≈ 0.6% 9. Frame = 5,000 b tp = 1 μs/km (a) Rate on R2-R3 = 1 Gb/s Required condition on R3-R4: Link R3-R4 must transfer slower
- 35. 27 shuch that: Rate onR3-R4 = ER3−R4 ER2−R3 × 1 Gb/s (b) tp = 1,800 km × 1 μs/km = 1,800 μs tf = 5,000 b/frame 1 Gb/s = 5 μs ER2−R3 = w w+2 tp tf = 5 5+2 1,800 μs 5 μs = 6.8×10−3. (c) tp = 800 km × 1 μs/km = 800 μs tf = 5,000 b/frame ER3−R4 ER2−R3 ×1 Gb/s = 5 ER2−R3 ER3−R4 = 5 6.8×10−3 ER3−R4 = 0.034 ER3−R4 ER3−R4 = 1 1+2 tp tf = 1 1+2 800 0.034 ER3−R4 ⇒ ER3−R4 = 4.6 × 10−3.
- 36. 28 Chapter 4.Data Links and Transmission
- 37. Chapter 5 Local-Area Networksand Networks of LANs 1. (a) 88 Bit Packet ⇒ Data Part=88-80 Prop. Speed= 200 m/μs One cycle time = (transmission time+propagation time) for data packet+ (transmission time+propagation time) for ack packet = 256 b 106 b/s + 103 m 200×10−6 + 88 b 106 b/s + 103 m 200×106 = 354 × 10−6 s Total time = (one cycle time) × (total bits)/data size of packet = (354 × 10−6 s) × (8 b/ch × 106 ch)/176 b/packet = 16 s/packet (b) One cycle time = (transmission time + propagation time)for data packet+ (transmission time + propagation time)for ack packet = 256 b+(100/2)×1 b 106 b/s + 103 m 200×106 m/s + 50 nodes×1 b/node 106 b/s + 103 m 200×106 = 366×10−6 sTotal time = (one cycle time)×(total bits)/data size of packet = (366 × 10−6 s) × (8 b/ch × 106 ch)/176 b/packet = 16.54 s/packet 29
- 38. 30 Chapter 5.Local-Area Networks and Networks of LANs Ground Floor 3 5 LAN 5 4th Floor 3rd Floor 2nd Floor Figure 5.1: Answer to exercise. The LAN overview of connections in a building. 2. Assuming that the computers and phones are placed at the corners of rooms, the overview of the LAN connections in a building is shown in Figure 5.1. (a) 2nd floor: d=5+3+5=13 m 3rd floor: d=5+3+3+5=16 m 4th floor: d=5+3+3+3+5=19 m (b) VoIP rate per office=64 × 2 = 128 Kb/s Web rate per office=(22 KB/page/s × 1 60 )(2 min ) ×8 b/B =5.86 Kb/s LAN rate=(128+5.86) × 12 offices=1.6 Mb/s 3. Please make the following correction: 100 m to be 1 km. Also combine
- 39. 31 Parts (a) and(b) to be Part (a) and thus Part (c) to become Part (b) Data rate=100 × 106 b/s Speed = 200 m/μs, Frame = 1000 b, (a) Mean distance = 0.375 km total time/frame = (transmission time) + (propagation time) = 103 b/frame 100×106 b/s + 0.375 km 200×106 m/s = 11.87 μs (b) Time is seconds = to sense a collision in the midpoint of two users’ distance = total time to send a frame up to the midpoint leading to a collision, and then sense back the collision = 0.5 (11.87 μs) + 0.5 (11.87 μs) = 11.87 μs Time in bits = 11.87 μs × 107 b/s = 1, 187 b 4. 100 Mb/s 96 bit time to clear tp=180 b (a) g=2 Retransmission time = (96 + 512 × 2) × 10−8 = 1.12 × 10−5 s (b) g=1 Retransmission time = (96 × 512) × 10−8 = 6.08 × 10−6 s (c) tp = l c = 180 b 100 Mb/s = 1.8 × 10−6 s 5. (a) α = tp T β = λT
- 40. 32 Chapter 5.Local-Area Networks and Networks of LANs R = pt tB +t = e−λtp T+2tp− 1−e−p λ + 1 λ = −λtp λ(T+2tp)+e−p Un = −αβ λ(T+2αβ)+e−αβ = β T e−αβ β+2αβ+e−αβ (b) Rn is in terms of frames/time slot due to the throughput R be- ing normalized. Rn makes it easier to use for estimation of the system. β is called ”offered load” since β is equal to λ ( is the av- erage arrival rate) multiplied by T (is a frame duration in seconds) resulting in “offered load”. (c) α = {0.001, 0.01, 0.1, 1.0} Un1 = β T e−0.001β/β + 2(0.001β) + e−0.001β Un2 = β T e−0.01β/β + 2(0.01β) + e−0.01β Un3 = β T e−0.1β/β + 2(0.1β) + e−0.1β Un4 = β T e−1β/β + 2(1β) + e−1β 6. N/A 7. na = 4 = 10 n = 10 f = 1500 byte × 8 b/1 byte = 12, 000 b (a) tp = c = 10 3×108 = 3.33 × 10−8 s tr = f r = 12,000 bits 10×109 b/s = 1.2 × 10−6 s (b) u = tr tr+tp = 1.2×10−6 s 1.2×10−6 s+3.33×10−8 s = 0.973 s (c) pc = na−1 na na−1 = 4−1 4 4−1 = 0.422 (d) na = 7, i = 7 pc = na−1 na na−1 = 7−1 7 7−1 = 0.387
- 41. 33 pi = pc(1− pc)i = 0.387(1 − 0.387)7 = 0.0116 (e) na = 4 E[c] = 1−pc pc = 1−0.387 0.387 = 1.52 Listen to Medium Transmit data Wait for 512g Bit Time Random # g Medium not available Medium available Transmission success Collision Listen to Medium Transmit with Prob P for Max tp Medium not available Medium available Collision Fixed P Transmission success a) b) Figure 5.2: Answer to exercise. 8. See Figure 5.3.
- 42. 34 Chapter 5.Local-Area Networks and Networks of LANs Hub Network Analyzer Bridge To Internet 1 R To Other Buildings Hub Repeater Figure 5.3: Answer to exercise.
- 43. Chapter 6 Wireless Networksand Mobile IP 1. N/A 2. N/A 3. N/A 4. N/A 5. N/A 6. (a) The probability of reaching a cell boundary or the probability of requiring a handoff as a function of db is shown in Figures 6.1 and 6.2. Suppose that a vehicle initiates a call in a cell with 10 miles radius. The vehicle speed is chosen to be 45 m/h (within a city) 35
- 44. 36 Chapter 6.Wireless Networks and Mobile IP Table 6.1: Probability of having a handoff for Case 4 db Handoff Probability (%) α01 (m) k = 35 m/h k = 60 m/h 0 50 50 5 37 43 1 10 29 36 20 17 26 0 50 50 5 12 22 5 10 4 9 20 1 2 0 50 50 5 4 9 10 10 1 2 20 0 0 and 75 m/h (on highways). In case 1, since a vehicle is resting all the time with an average speed of 0 m/h, the probability of reaching a cell boundary is clearly 0 percent. In contrast to Case 1, for a vehicle is moving with an average speed in Case 2, the chance of reaching a cell boundary is always 100 percent. Thus, when a vehicle is either at rest or moving with some speed, the probability of requiring a handoff is independent on db. From the figure, we see that as α01 increases, the chance of reaching a cell boundary is lower. Also, with a fixed db, the handoff probability in highway is much higher than in the city area. This is because the higher the speed limit, the higher the probability of reaching a cell. Table 6.1 summarizes the results. Assume α01 = 1 and in the city area where k = 45 m/h: if db is 5 miles the only chance of reaching a cell is 87 percent, or the chance that a handoff occurs for the cell is 87 percent. If db is 10 miles, the probability of reaching a cell boundary is 76 percent. As db increases, the probability of
- 45. 37 0 2 46 8 10 12 14 16 18 20 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 db, miles Probability of Requirement For Handoff α01 = 1 α01 = 5 α01 = 10 Figure 6.1: The probability of reaching a cell boundary for Case 4: (a) within a city 0 2 4 6 8 10 12 14 16 18 20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 db, miles Probability of Requirement For Handoff α01 = 1 α01 = 5 α01 = 10 Figure 6.2: The probability of reaching a cell boundary for Case 4: (b) in highway.
- 46. 38 Chapter 6.Wireless Networks and Mobile IP 25 30 35 40 45 50 55 60 65 70 75 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Average Speed of A Vehicle, mph Probability of Requirement For Handoffs α 01 = 1 α01 = 5 α01 = 10 Figure 6.3: The probability of reaching a cell boundary in terms of a vehicle’s speed for Case 4. reaching the call holding time decreases. As the cell size increases, the probability of reaching a boundary decreases in an exponential manner. The only difference between Case 3 and Case 4 is that the change of handoff probability for the latter is between 0 percent to 50 percent while the former has a change of probability between 0 percent to 100 percent. This is mainly due to the difference between the two initial state probabilities for both case. (b) The relationship between a vehicle’s speed and the chance of reach- ing a cell boundary is shown in Figure 6.3 As shown earlier, for Case 1 and Case 2, the probability of requiring a handoff is inde- pendent on the call holding time and db and therefore, it is also independent on the vehicle’s speed. For Case 1 in which a vehicle is resting all the time, the vehicle will never reach a cell boundary. For Case 2 in which a vehicle is moving all the time with some speed, the chance of reaching a cell boundary is always 100 per- cent. The probability of reaching a cell boundary is proportional to the vehicle’s speed. This is because the increase of the speed of a vehicle increases the chance of reaching a cell boundary. As α01
- 47. 39 increases, the probabilityof requiring a handoff decreases. 7. N/A
- 48. 40 Chapter 6.Wireless Networks and Mobile IP
- 49. Chapter 7 Routing and Inter-Networking 1.(a) See Figure 7.1 (a). min=2 max=2 H = min+max 2 = 2 (b) See Figure 7.1 (b). min = 3 For max: n=4, max=4, n=5, max=9/2, n=6, max=5 in general, for n, the max is n 2 + 2 H = min+max 2 = 3+(n/2+2) 2 (c) See Figure 7.1 (c). H=3 (d) See Figure 7.1 (d). min = 3 max = 4 H = 2(3)+(n−4)4 n−2 41
- 50. 42 Chapter 7.Routing and Inter-Networking (a) (d) (c) (b) Figure 7.1: Answer to exercise. Four different network topologies to connect two users.
- 51. 43 2. Using Dijkstra’sAlgorithm Table 7.1: Solution to problem. (a) k βA,C βA,D βA,F βA,E βA,B {A} AC(5) × AF(9) × × {A,F} AC(5) AFD(12) ACF(8) AFE(10) AFB(14) {A,F,C} AC(5) ACD(9) ACF(8) ACE(7) AFB(14) {A,F,C,D} AC(5) ACD(9) ACF(8) ACE(7) AFB(14) {A,F,C,D,E} AC(5) ACD(9) ACF(8) ACE(7) ACEB(9) {A,F,C,D,E,B} AC(5) ACD(9) ACF(8) ACE(7) ACEB(9) (b) See Figure 7.2 3. Using Bellman-Ford Algorithm (a) βA,C βA,D βA,F βA,E βA,B 1 AC(5) × AF(9) × × 2 AC(5) ACD(9) ACF(8) ACE(7) AFB(14) 3 AC(5) ACD(9) ACF(8) ACE(7) ACEB(9) 4 AC(5) ACD(9) ACF(8) ACE(7) ACEB(9) (b) See Figure 7.3. 4. Using Dijkstra’s Algorithm (b) See Figure 7.4. 5. Using Dijkstra’s Algorithm
- 52. 44 Chapter 7.Routing and Inter-Networking Figure 7.2: Answer to exercise. Figure 7.3: Answer to exercise.
- 53. 45 Figure 7.4: Answerto exercise. Figure 7.5: Answer to exercise.
- 54. 46 Chapter 7.Routing and Inter-Networking (a) k βA,C βA,D βA,F βA,E βA,B {A} AC(5) × AF(9) × × {A,F} AC(5) AFD(12) ACF(8) AFE(10) AFB(14) {A,F,C} AC(5) ACD(9) ACF(8) ACE(7) ACFB(13) {A,F,C,D} AC(5) ACD(9) ACF(8) ACE(7) ACFB(13) {A,F,C,D,E} AC(5) ACED(8) ACF(8) ACE(7) ACEB(9) {A,F,C,D,E,B} AC(5) ACED(8) ACF(8) ACE(7) ACEB(9) (a) k β1,2 β1,3 β1,4 β1,5 β1,6 β1,7 {1} 1,2(3) 1,3(3) × × 1,6(9) × {1,2} 1,2(3) 1,3(3) 1,2,4(11) 1,2,5(15) 1,6(9) × {1,2,3} 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,6(9) × {1,2,3,4} 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,6(9) × {1,2,3,4,5} 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,3,5,6(8) 1,3,5,7(20) {1,2,3,4,5,6} 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,3,5,6(8) 1,3,5,6,7(16) (b) See Figure 7.5. 6. Using Bellman-Ford Algorithm (a) β1,2 β1,3 β1,4 β1,5 β1,6 β1,7 1 1,2(3) 1,3(3) × × 1,6(9) × 2 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,6(9) 1,6,7(17) 3 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,3,5,6(8) 1,6,7(17) 4 1,2(3) 1,3(3) 1,3,4(7) 1,3,5(7) 1,3,5,6(8) 1,3,5,6,7(16) (b) See Figure 7.6. 7. From R1 to R4 using Dijkstra’s Algorithm (b) See Figure 7.7.
- 55. 47 Figure 7.6: Answerto exercise. 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R Figure 7.7: Answer to exercise.
- 56. 48 Chapter 7.Routing and Inter-Networking (a) k β1,2 β1,3 β1,4 β1,5 β1,6 β1,7 {1} 1,2(2) × × × 1,6(2) 1,7(8) {1,6} 1,2(2) 1,6,3(7) × 1,6,5(7) 1,6(2) 1,6,7(3) {1,6,5} 1,2(2) 1,6,3(7) 1,6,5,4(11) 1,6,5(7) 1,6(2) 1,6,7(3) {1,6,5,4} 1,2(2) 1,6,3(7) 1,6,5,4(11) 1,6,5(7) 1,6(2) 1,6,7(3) {1,6,5,4} 1,2(2) 1,6,3(7) 1,6,5,4(11) 1,6,5(7) 1,6(2) 1,6,7(3) {1,6,5,4,3} 1,2(2) 1,6,3(7) 1,6,3,4(9) 1,6,5(7) 1,6(2) 1,6,7(3) {1,6,5,4,3,2} 1,2(2) 1,2,3(6) 1,2,3,4(8) 1,6,5(7) 1,6(2) 1,6,7(3) {1,6,5,4,3,2,7} 1,2(2) 1,2,3(6) 1,6,7,4(5) 1,6,5(7) 1,6(2) 1,6,7(3) 8. From R1 to R4 using Bellman-Ford Algorithm (a) β1,2() β1,3() β1,4() β1,5() β1,6() β1,7() 1 1,2(2) × × × 1,6(2) 1,7(8) 2 1,2(2) 1,2,3(6) 1,7,4(10) 1,6,5(7) 1,6(2) 1,6,7(3) 3 1,2(2) 1,2,3(6) 1,6,7,4(5) 1,6,7,5(4) 1,6(2) 1,6,7(3) 4 1,2(2) 1,2,3(6) 1,6,7,4(5) 1,6,7,5(4) 1,6(2) 1,6,7(3) (b) See Figure 7.8. 9. PBC = (0.3)(0.1)(0.7) = 0.021 PCE = (0.3)(0.6) = 0.18 PCDF = 1 − (1 − 0.5)(1 − 0.8) = 0.9 PCEF = 1 − (1 − 0.18)(1 − 0.2) = 0.344 PCF = (0.9)(0.344) = 0.31 PBCF = 1 − (1 − 0.021)(1 − 0.31) = 0.324 PBF = (0.3)(0.324) = 0.097 PAF = 1 − (1 − 0.4)(1 − 0.097) = 0.458 10. PAB = 0.4 PBC = (0.3)(0.1)(0.7) = 0.021
- 57. 49 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R 1 R 2 R 3 R 4 R 7 R 6 R 5 R Figure 7.8: Answerto exercise. PCE = (0.3)(0.3)(0.6) = 0.054 PCDF = 1 − (1 − 0.5)(1 − 0.8) = 0.9 PEF = 0.2 PCF = [1 − (1 − PCE)(1 − PEF )]PCDF (0.3)(0.3) = 0.0197 PAC = 1 − (1 − PAB)(1 − PBC) = 0.4126 PAF = 1 − (1 − PAC)(1 − PCF ) = 0.424
- 58. 50 Chapter 7.Routing and Inter-Networking
- 59. Chapter 8 Transport andEnd-to-End Protocols 1. Figure 8.1 shows the operation. Each packet size is 1000 bytes since that’s the MSS of Host B. For Host A: MSS = 2000, ISN = 2000, File Size = 200 Kb = 25 KB For Host B: MSS = 1000, ISN = 4000, Packet Size = 100 Kb = 25 KB where data per packet is 960 bytes Three stages in file transfer: Connection establishment, Segment trans- fer, and Connection termination. 2. (a) TCP sequence number field includes 4 B = 32 b. Thus: • Maximum number of bytes to be identified in a connection = 232 • We consume one sequence number for connection setup, seq(i), and one sequence number for connection termination, seq(k). As each byte of data is identified by a unique sequence number, therefore, the maximum number of data bytes that can be identified for a connection and we can transfer = f = 232 −2 = 4, 294, 967, 294 B 51
- 60. 52 Chapter 8.Transport and End-to-End Protocols HOST B HOST A 2-WAY CONNECTION ESTABILSHED PACKET TRANSMISSION STARTS A gap in transmission to allow Host A not tot wait for ACK and keep transmitting 2-WAY CONNECTION TERMINATED LAST 40 BYTES Figure 8.1: Answer to exercise. (b) Total size of each segment = 2,000 B. Also, each segment has the following headers: 20 B Link + 20 B IP + 20 B TCP = 60 B. Thus: • Maximum size of data in each segment = 2,000 B - 60 B = 1,940 B. • Maximum number of segments to be produced in a connection = 4,294,967,294 B 1,940 B ≈ 2,213,901 • Total size of all segment headers = 2,213,901 × (60 B) = 132,834,060 B • Total size of all segment headers and data = 4,294,967,294 B + 132,834,060 B = 4,427,801,354 B • Total time it takes to transfer all segments = 4,427,801,354 B×8 b/B 100×106 b/s
- 61. 53 = 354.21 s≈ 5.9 minutes 3. (a) Slow start congestion control: Since the number of packets transmitted doubles every time, the number of round trips to reach n is log2 n − 1. (b) Additive increase congestion control: Since the number of packets transmitted increases by 1 every time, the number of round trips to reach n is n − 1. 4. This Problem belongs to Chapter 12 please see Problem 12.16. 5. N/A 6. = 1.2 Gb/s RTT = 3.3 ms File size f = 2 MB packet size = 1 KB Hence the total number of packets needed to be transmitted = 2MB / 1 KB = 2000 (a) With an additive increase/multiplative decrease protocol, the win- dow size increases by one all the times until a congestion when the window size is divided by two. Therefore, the window size starting at wg = 1 KB changes its value as follows: wg = 1 KB, 2 KB, 3 KB, 4 KB, · · ·, n KB Therefore: 1 + 2 + 3 + · · · + n = 2000
- 62. 54 Chapter 8.Transport and End-to-End Protocols Thus: n(n+1) 2 = 2000 where we can obtain n = 62.74 ≈ 63 Since, the congestion window size of 500 KB is never reached, no multiplicative decrease takes place. Thus The total time = 63 × 3.3 ms = 207 ms Clearly, the window size takes a total of 500×RTT = 500×3.3=1.65 seconds to reach 500 KB. (b) With a slow start protocol, the window size is doubled all the times until a congestion. Therefore, the window size starting at wg = 1 KB changes its value as follows: wg = 1 KB, 2 KB, 4 KB, 8 KB, · · ·, approximatley 1,024 KB Therefore, we will have to make 11 roundtrips to transmit the file: 1+2+4+8+16+32+64+128+256+512+1024 = 2047 Thus the total time = 11 × RTT = 11 × 3.3 ms = 36.3 ms The window size takes a total of 10× (c) With the additive increase/multiplative decrease protocol, it takes 63 RTTs to transfer the 2 MB file. Therefore, Δ = 63 × 3.3 ms ≈ 208 ms r = f Δ = 2 MB 208 ms = 76.9 Mb/s (d) With the additive increase/multiplative decrease protocol, ρu = r B = 76.9 Mb/s 1.2 Gb/s = 64 × 10−3 7. Round trip time = 0.5 s Packet transmitted every 50 ms Let’s assume packet (segment) P-11 is lost. The first acknowledgement, ACK-10, is received when P-21 is about to be transmitted. No ACK is received before P-22 as P-11 is lost. We receive the fourth ACK-10 after P-24 is sent. Segment loss is detected. P-11 is transmitted instead
- 63. 55 Source Destination 0.5 sec LOSS OFSEGMENT DETECTED 50ms Figure 8.2: Answer to exercise. of P-25. See Figure 8.2. (a) In this case, P-11 is transmitted and after 50 ms, P-25 is trans- mitted, and the cycle continues. Hence, we lose only 50 ms. See Figure 8.2. (b) In this case, the sender waits for the acknowledgment of retrans- mitted P-11. Thus, it has to wait for the complete round trip. Hence, the time lost here is 0.5 s.
- 64. 56 Chapter 8.Transport and End-to-End Protocols
- 65. Chapter 9 Applications andNetwork Management 1. The command for is “nslookup.” In the command prompt in that window we put the name of the website in the format: www.name.com. When entering this command in the command prompt of windows, it can be seen that we get the server name, IP address, and Aliases. The command ns-lookup works both ways, that is, if we give the name we get the IP addresses and also vice versa. The snapshot in the Figure 9.1 shows the IP addresses of some of the most frequently used websites with their server names. 2. (a) To obtain the file name in a remote machine, the DNS server re- quests the local DNS server if it is not the local DNS server to contact the remote machine. These requests are carried out by the server either recursively or iteratively. On the other hand, if the server wants to obtain the file name from another DNS server, depending upon the type of information and the file location, the DNS server either requests the root DNS or another local DNS. (b) When we take the domain name from the DNS server, our query 57
- 66. 58 Chapter 9.Applications and Network Management Figure 9.1: Solution to exercise.
- 67. 59 gives us aresult which includes all the possible aliases of the par- ticular domain names and their corresponding IP addresses. On the other hand, if the query is done using the IP address, then we get only the particular alias that corresponds to that IP address in response. This is illustrated by the Figure 9.2 using the example of gmail.com. (c) As seen in Figure 9.2, all hosts from the same subnet need not be identified by the same DNS server as we can assign different subnets with different IP addresses. This is done for various reasons like traffic sharing, having different names (alias) for same website, etc. 3. (a) SSH provides a far better security of transmission compared with TELNET. (b) The functionality given by Rlogin implementation in Telnet are: It passes terminal type It bypasses the need for username/password to be entered. No newline etc processing is applied to data transferred It has better out-of-band data handling It has better flow-control handling It has window-size negotiation 4. (a) No, FTP does not compute checksum any checksum for its file transfer. It relies on the underlying TCP layer for error control. TCP layer uses checksum for error control. (b) If the TCP connection is shut down, the browser tries to set up the connection once. If this attempt fails then the browser quits the file transfer. (c) Following are the list of commands that may be set for FTP clients.
- 68. 60 Chapter 9.Applications and Network Management Figure 9.2: Solution to exercise.
- 69. 61 Command Explanation ABOR Abortan active file transfer. ACCT Account information. ALLO Allocate sufficient disk space to receive a file. APPE Append CDUP Change to Parent Directory. CLNT Send FTP Client Name to server CWD Change working directory. DELE Delete file. EPSV Enter extended passive mode. EPRT Specifies an extended address and port to which the server should connect. FEAT Get the feature list implemented by the server. GET Use to download a file from remote HELP Returns usage documentation on a command if specified, else a general help document is returned. LIST Returns information of a file or directory if specified, else information of the current working directory is returned. LPSV Enter long passive mode. LPRT Specifies a long address and port to which the server should connect. MDTM Return the last-modified time of a specified file. MGET Use to download multiple files from remote MKD Make directory (folder). MODE Sets the transfer mode. MPUT Use to upload multiple files to remote NLST Returns a list of filenames in a specified directory. NOOP No operation (dummy packet; used mostly on keep alive). OPTS Select options for a feature. PASS Authentication password.
- 70. 62 Chapter 9.Applications and Network Management PASV Enter passive mode. PORT Specifies an address and port to which the server should connect. PUT Use to upload a file to remote PWD Print working directory. Returns the current directory of the host. QUIT Disconnect. REIN Re initializes the connection. REST Restart transfer from the specified point. RETR Retrieve a remote file. RMD Remove a directory. RNFR Rename from RNTO Rename to. SITE Sends site specific commands to remote server. SIZE Return the size of a file. SMNT Mount file structure. STAT Returns the current status. STOR Store a file. STOU Store a file uniquely. STRU Set file transfer structure. SYST Return system type. TYPE Sets the transfer mode (ASCII/Binary). USER Authentication username. 5. The total file transfer delay is: (a) On both directions when the network is in its best state of traffic, the average file transfer delay is 3.5 ms. (b) On both directions when the network is in its worst state of traffic, the average file transfer delay is 9 ms.
- 71. 63 (c) On onedirection when we try FTP from one computer to itself: The average file transfer is 7.5 ms. 6. All characters of the URL must be from the following: A-Z, a-z, 0-9 . / ∼ % - + # ? ! = () @ If a URL contains a different character it should be converted; for example,ˆmust be written as %5e, the hexadecimal ASCII value with a percent sign in front. A blank space can also be converted into an underscore. 7. (a) The purpose of the GET command in the HTTP is to request a representation of the specified resource. The GET method re- trieves whatever information (in the form of an entity) identified by the Request-URI. If the Request-URI refers to a data-producing process, it is the produced data is returned as the entity in the re- sponse and not the source text of the process, unless that text happens to be the output of the process. (b) The purpose of the PUT command in HTTP is to request the enclosed entity to be stored under the supplied Request-URI. Thus it basically uploads a representation of the specified resource. (c) The GET command needs to use the name of the contacted server when it is applied as HTTP is a stateless protocol. This means that it keeps the state information and live connections to remote clients. Thus, we connect to the server, get the info we need, and then disconnect. Therefore, we need to give the name of the contacted server.
- 72. 64 Chapter 9.Applications and Network Management 8. (a) The role of ASN.1 on the 7 layer OSI model is shown in Figure 9.3. The ASN.1 notation is used in the application layer as a notation. (b) The impact of constructing a grand set of unique global ASN.1 names for MIB systems is that the network management can iden- tify an object by a sequence of names or numbers from the root to that object. This enables designers to produce specifications without undue consideration to the encoding issues. (c) A US based organization must register under the following: Root : ISO : company name : dod : internet : MIB Figure 9.3: Solution to exercise. 9. (a) The SNMP protocol has the function wherein the network man- ager can use this protocol to find the location of fault. The task of SNMP is to transport MIB information among all the manag- ing centers and agents executing on its behalf. The most efficient method for the above functions is unarguably UDP as it will be faster and efficient. (b) The pros of letting all the managing centers access the MIB is bet- ter connection ability and better communication. It would greatly
- 73. 65 help in thedevelopment and servicing of the MIB. All these things would result in the better utilization of the network and also more efficiency. On the other hand, the price to pay for this kind of flexibility is having a huge impact on the security of the network. Also, even if the security aspect is taken care of, letting everyone access to the MIB variables increases the complexity of the MIB design and maintenance. (c) MIB is the information storage medium that contains managed objects reflecting the current status of the network. Now, if the MIB variables are located in the router memory, it would greatly improve the efficiency and the speed of the process. However, it brings with it the problem of updating the MIB. In the scenario where router B is not involved in, any communication must also be notified about the change in order to update the MIB if the communication is limited between routers A and C. This would create unnecessary overheads and wastage of network bandwidth. This is of course on top of increasing the router complexity, buffer size, and host of other problems. Thus, MIB variables should not be organized in the local router memory.
- 74. 66 Chapter 9.Applications and Network Management
- 75. Chapter 10 Network Security 1.L4 = 4de5635d, R4 = 3412a90e, k5 = be11427e6ac2. L4 = 0100;1110;1111;0101;0110;0011;0101;1110. R4 = 0011;0100;0001;0010;1010;0101;0000;1110. After the expansion stage the right half will become: R4 = 000110;101000;000010;100101;010100;001010;100001;011100. k5 = 101111;110001;000101;000010;011111;110110;101011;000010. R4 Xor k5 gives us: 101001;011001;000111;100111;001011;111100;001010;011110. Now passing it through the S-Box: R4 = 0100;0110;1001;0110;0111;1011;0000;0111. L4 = 0100;1110;1111;0101;0110;0011;0101;1110. Xor with the left half: R4 = 0000;1000;0110;0011;0001;1000;0101;1001 After permutation: R5 = 1011;1010;0100;1001;0010;1000;0000;0100 = ba492804 L5 = 0011;0100;0001;0010;1010;0101;0000;1110 = 3412a50e 67
- 76. 68 Chapter 10.Network Security 2. Key generation: The key is 010101. . . .01 and is 56 bit long. Thus, the parity bits have already been discarded. The key is first divided into two blocks of 28 bits using the standard permutation block provided by the DES algorithm: the left block say C0 = 0000000; 0111111; 1100000; 0001111. the right block say D0= 0000000; 0111111; 1100000; 0001111. Now, we shift left both C0 and D0 by 1 thus we get C1 and D1 as follows: C1 = 0000000; 1111111; 1000000; 0011110. D1 = 0000000; 1111111; 1000000; 0011110. ki(left) = 101100;001001;001011;001010. ki(right) = 010101;010000;001001;010100. ki = ki(left);ki(right). Message generation: The message is all ones: 111. . . .111 (64). With left half (32) 11. . . .11 and right half(32) 11. . . 11. (Here the initial permutation has no effect.) Converting the 32 message into 48 by passing through the mangler (1): 111. . . 11(32 bit) = 111. . . 111(48bits). Xoring with the key ki: 010011; 110110; 110100; 110101; 101010; 101111; 110110; 101011. Now, passing it through the S-Box: 0110;0110;0010;0101;1101;1011;1000;1010. Xor with left half: 1001;1001;1101;1010;0010;0100;0111;0101. After permutation of the right half we get: 0000;0110;1101;1001;0100;1101;1110;1010 (R1) 1111;1111;1111;1111;1111;1111;1111;1111 (L1).
- 77. 69 3. N/A 4. N/A 5.From the text book: c = mx mod n and m = cy mod n. Note that x and y are mod inverse of each other. Thus, c = ((cy)x) mod n. Since x and y are inverse of each other, we then get c = c mod n = c. 6. M = 1010. The two four bit primes are a = 5 and b = 11. Also x = 3. To find the keys, we have n = ab = (5)(11) = 55 q = (a − 1)(b − 1) = (4)(10) = 40 Thus, xy mod (a − 1)(b − 1) = 1 resulting in 3y mod 40 = 1 which implies that y = 27, since (3)(27) = 81 and 81 mod 40 = 1. Therefore the keys are: The public key = {3, 55} The private key = {27, 55}. Thus: the cipher text from the message 1010 (10 in decimal) is 103 mod 55 = 1000 mod 55 = 10 mod 55. Therefore, the cipher text is 10. 7. m = 13 a = 5
- 78. 70 Chapter 10.Network Security b = 11 x = 7 (a) Encryption: The public key = {7, 55} C = 137 mod 55 = 62748517 mod 55 = (55)(1140882) + 7 mod 55 = 7 mod 55. C = 7. (b) The corresponding y is given as follows: n = ab = (5)(11) = 55 q = (a − 1)(b − 1) = (4)(10) = 40 Also x = 7 Thus, xy mod (a − 1)(b − 1) = 1 7y mod 40 = 1 Which implies y = 23 (since (7)(23) = 161 and 161 mod 40 = 1) The private key = {23, 55}. (c) The decryption is 723 mod 55 = 13. 8. (a) When encrypting with small values of the m, the (non-modular) result of me may be strictly less than the modulus n. In this case, ciphertexts may be easily decrypted by taking the the root of the ciphertext with regardless of the modulus. For systems that conventionally use small values of e, such as 3, the AES key of 256 bits using this scheme would be insecure since the largest m would have a value of 2563, and 2553 is less than any reasonable modulus. Such plaintexts could be recovered by simply taking the cube root of the ciphertext. Thus, the 256-bit AES key, k, chosen by user 1 is too small to encrypt securely with RSA having a public key as {x, 5} since ke x. Thus, ke mod x = ke and an intruder only recovers k by
- 79. 71 taking the ethroot. (b) The values m = 0 or m = 1 always produce ciphertexts equal to 0 or 1 respectively, due to the properties of exponentiation. Thus, the keys containing of all 0’s or all 1’s can be easily recovered by the attacker. An example could be {x = 3, y = 7}. 9. To overcome the vulnerability in the above combination, practical RSA implementations typically embed some form of structured, randomized padding into the value m before encrypting it. This padding ensures that m does not fall into the range of insecure plaintexts, and that a given message, once padded, encrypts to one of a large number of dif- ferent possible ciphertexts. The latter property can increase the cost of a dictionary attack beyond the capabilities of a reasonable attacker. Modern constructions use secure techniques such as optimal asymmet- ric encryption padding (OAEP) to protect messages. The intuitive solution to this problem is that user 1 must select a larger random number for RSA encryption. In this case, both users 1 and 2 use this number to create key, k. A second solution is that user 1 pads k with random bits so that the message has almost the same number bits as x does. 10. Suppose that user 1 chooses a prime number a, a random number x1, and a generator g and creates y1. We can say: k1 = yx1 2 mod a = (gx2 mod a)x1 mod a = [(gx2 )x1 mod a] mod a = [(gx1 )x2 mod a] mod a = (gx1 mod a)x2 mod a = yx2 1 mod a = k2
- 80. 72 Chapter 10.Network Security
- 81.
- 83. Chapter 11 Packet Queuesand Delay Analysis 1. Note to Instructors: This problem requires a good understanding of queueing theory without any use of formulas. Advance explanation of objectives of this problem to students will be very helpful. We can summarize the queing situation as follows: • Interarrival time = 20 μs • Service time = 0 μs, if there is no packet misordering = 10 + 30×(n=number of misorderings) μs, if there are n packet misorderings in a block. (a) Packet block arrival and departure activities are shown in Table 11.1. If we consider one packet block between arrival time 20 μs, and departure time 90 μs (for the duration of 70 μs), and then one packet block between arrival time 40 μs, and departure time 170 μs (for the duration of 130 μs), and continue this trend, the queuing behaviour can be shown in Figure 11.1. (b) Mean number of packet blocks = (Service Times)×(1 packet block) Duration of System Processing Time = 1,200 blocks 680 μs = 1.76 75
- 84. 76 Chapter 11.Packet Queues and Delay Analysis Table 11.1: Packet block arrival and departure activities. Packet Block Number Number of Misor- derings Arrival Time in μs Service Time in μs Departure Time in μs 1 2 20 70 90 2 4 40 130 170 3 0 60 10 70 4 0 80 10 90 5 1 100 40 140 6 4 120 130 250 7 3 140 100 280 8 5 160 160 320 9 2 180 70 250 20 4 200 130 330 11 0 220 10 230 12 2 240 70 310 13 5 260 160 420 14 2 280 70 350 15 1 300 40 340 0 1 2 3 4 5 6 7 20 40 600 640 680 Processing Time (micro-sec) Number of Waiting Packet-Blocks in Queue 90 … … Figure 11.1: Solution to exercise. The trend of packets accumulated in the queue over time.
- 85. 77 (c) Percentage oftime that the buffer is not empty can be realized from Figure 11.1. If we include all the activities of the queue as described in Part (a), the queue activity stops at around 680 μs. Thus, the percentage of time that the buffer is not empty = Time That the Buffer Is Empty Duration of System Processing Time = 20 μs 680 μs = 0.029 Percentage of time the buffer is not empty = 1 - 0.029 = 0.97 2. (a) E[Kq(t)] = λE[Tq] = ρ2 1−ρ = 0.92 1−0.9 = 8.1 (b) E[Tq] = E[T] − E[Ts] E[T] = 1 μ(1−ρ) E[Ts] = 1 μ ⇒ E[Tq] = ρ 1 μ−λ λ μ = ρ, ⇒ μ = λ ρ = 44.44 E[Tq] = 0.9 1 44.44−40 = 0.2 s (c) P0 = 1 − ρ = 1 − 0.9 = 0.1 λ = 40 packets/s μ = 44.44 packets/s 3. (a) T = min(T1, T2, ..., Ti) (b) P[T t] = P[time until next packet departures] = P[min(T1, T2, ..., Ti)] = P[T1 t1, T2 t2, ..., Ti ti] = P[T1 t]P[T2 t]...P[Ti t] = e−μte−μt...e−μt = e−iμt
- 86. 78 Chapter 11.Packet Queues and Delay Analysis 4. (a) P[K(t) k] = 1 − P[K(t) ≥ k] = 1 − ∞ i=k Pi = 1 − ∞ i=k ρi(1 − ρ) = 1 − (1−ρ)ρk 1−ρ = 1 − ρk (b) P[K(t) 20] = 0.9904 ⇒ k = 20 ⇒ 0.9904 = 1 − ρ20 ⇒ ρ = 0.7927 ρ = λ μ = 0.7927 = 300 μ ⇒ μ = 378.45 packet/s 5. (a) P[K(t) ≥ k] = (1 − ρ) ∞ j=k ρj = (1 − ρ) ρk 1−ρ = ρk (b) P[K(t) ≥ 60] = ρ60 = 0.01 ⇒ ρ ≈ 0.92 ⇒ λ ≈ 0.92μ 6. (a) The Markov chain is similar to a regular M/M/1 except the ar- raival rate to any state i is λi = λ i (b) For State 0: p0λ = p1μ ⇒ p1 = λ μp0 ⇒ p1 =ρp0 For State 1: p1 λ 2 + p1μ = p0 + p2μ ⇒ p1 λ 2 = p2μ ⇒ p2 = λ 2μp1 ⇒ p2 = 1 2 ρp1 = 1 2ρ2p0 Continuing this trend for next states, a generic form can be devel- oped as: pi−1 λ i = piμ ⇒ pi = λ iμpi−1 = 1 i!ρip0 Since ∞ i=0 pi = 1⇒ p0 = e−ρ ⇒ pi = 1 i! ρie−ρ
- 87. 79 (c) When i→ ∞ while ρ 1, we have pi → 0, thus, the system is in a steady state. (d) The utilization (of the server) is: ρi = λi μ = 1 i λ μ = λ iμ (e) E[K(t)] = ∞ i=0 iP[K(t) = i] = ∞ i=0 i ρk i! e−ρ Solving this equation using ∞ i=0 ρe e! = eρ will result in: E[K(t)] = ρ (f) The mean system delay considering any State i is obtained using Little’s law: E[T]i = E[K(t)] λi = ρ λ i = iρ λ However, since the arrival rate is different in each state, we need to compute the mean over all states: E[T] = ∞ i=0 piE[T]i = ∞ i=0 pi iρ λ = ρ ∞ i=0 1 i!λρie−ρ i λ = ρ2 λ Since E[Ts] = 1 μ, then: E[Tq] = E[T] - E[Ts] = ρ2 λ − 1 μ 7. a = 2 1 μ = 100 ms/packet λ = 18 packets/s (a) Prob[blocking a packet]=Prob[all servers are busy] ρ1 = λ μ = (18)(100 × 10−3) = 1.8 ρ = λ aμ = 18/2.1 = 0.9 P[i 2] = Pa 1−ρ Pa = P2 = ρ2 1 2! P0 P0 = 1 (1+ρ1)+ ρ2 1 2! 1 1−ρ = 1 (1+1.8)+ 1.82 2! 1 1−0.9 = 0.052
- 88. 80 Chapter 11.Packet Queues and Delay Analysis ρ = 18/20 = 0.9 P[K(t) = 2] = Pa = ρa 2 a! ρ0 = 1.82 2 (0.0526) = 0.0853 ⇒ Prob[Waiting] = P[K(t) ≥ 2] = ∞ i=a Pi = Pa 1−ρ = 0.0853 1−0.9 = 0.853 (b) E[K(t)] = Paρ (1−ρ)2 + ρ1 = (0.0853)(0.9) (1−0.9)2 + 1.8 = 9.474 (c) E[T] = E[Tq] + E[Ts] = Pa aμ(1−ρ)2 + 1 μ = 0.421 + 0.1 = 0.521 s (d) P[K(t) 50] = ∞ i=50 ρi = ∞ i=50 ρi−aPa = Pa ρ2 ∞ i=50 ρi = Pa ρ2 [ ∞ i=50 ρi − i=50 ρi] = Pa ρ2 [ 1 1−ρ − 1−ρ50+1 1−ρ ] = Pa ρ2 [ ρ51 1−ρ] = 0.0853 0.92 0.951 0.1 P[K(t) 50] = 0.00488 8. (a) Prob[blocking a packet]=Prob[all servers are busy] λ = 100 packets/s mean service rate:μ = 20 packets/s; ρ1 = 100/20 = 5 Use Erlang-B: Pa = ρa 1 a! 1 a i=0 ρi 1 i! = 56 6! 1 6 i=0 5i i! = 0.1935 (b) Pa = 0.1935 2 = 0.0967 By plugging numbers in the equation, we have: P7 = 0.121 P8 = 0.075 We need two switches to lower the blocking probability below 0.0967, this is 4 more switches then the setup in Part (a).
- 89. 81 i 0 1 2j i i i i i i j i (j+1) i i i λ λ λ λ λ λ i c c i i μ 2μ 3μ μ μ μ Figure 11.2: Solution to exercise, Markov chain for the M/M/c/c system. 9. (a) The handoff process is modeled using M/M/c/c Markovian system in which there is no queueing line with random interarrival call time and exponential service time (channel holding time) with c servers (c channels) as well as c sources (c handoff calls). The Markovian chain of this system is illustrated in Figure 11.2 where in our case: λi = Handoff request rate for traffic type i ∈ {0, 1, ..., k} following a Poisson process. 1/μi = Mean holding time of a channel or mean channel exchange time for traffic type i with an exponential distribution. Let j be channels that are busy. Thus, handoff calls depart at rate jμi. (b) When the number of requested channels reaches the total number of available channels (ci), ie. j = ci, then, all ci channels are in use and the channel exchange rate is ciμi. In this case, any new arriving handoff calls are blocked since there is no queueing. The global balance equations are: λiP0 = μiP1 for j = 0 (11.1) λiPj−1 = jμiPj for 0 j ≤ ci (11.2) where, P0 is the probability that no channel exchange is requested for traffic type i, and Pj is the probability that j channel exchanges
- 90. 82 Chapter 11.Packet Queues and Delay Analysis are requested for traffic type i. It then follows that P1 = ρiP0 (11.3) Pj = ρiPj−1 j (11.4) where, ρi = λi/μi is the offer load of the system. In Equation (11.4), let j = 2 and 3, then: P2 = ρiP1 2 = ρi 2P0 2 × 1 = ρi 2P0 2! (11.5) P3 = ρiP2 3 = ρi 3P0 3 × 2! = ρi 3P0 3! (11.6) By induction from Equations 11.5 and 11.6 Pj = ρi jP0 j! (11.7) Knowing that the sum of the probabilities must be one 1 = ci j=0 ρi jP0 j! ⇒ P0 = 1 ci j=0 ρi j j! (11.8) Equations (11.7) and (11.8) can be combined as: Pj = ρi j j! 1 ci j=0 ρi j j! (11.9) When j = ci, all the channels are busy and any handoff call gets blocked. The handoff blocking probability denoted as Pci are ex- pressed by Pci = ρi ci ci! 1 ci j=0 ρi j j! (11.10)
- 91. 83 0 1000 20003000 4000 5000 6000 7000 8000 9000 10000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Handoff Request Rate (calls/sec) Handoff Blocking Probablity (%) ← 1/μ i = 10 ms ← 1/μi = 20 ms 1/μ i = 30 ms → c i = 50 Figure 11.3: Hanfoff Blocking Probability (ci = 50). 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Handoff Request Rate (calls/sec) Handoff Blocking Probablity (%) 1/μ i = 10 ms ← 1/μ i = 20 ms 1/μ i = 30 ms → ↓ ci = 100 Figure 11.4: Hanfoff Blocking Probability (ci = 100).
- 92. 84 Chapter 11.Packet Queues and Delay Analysis 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Handoff Request Rate (calls/sec) Handoff Blocking Probablity (%) ← ci = 100 ci = 50 → ← ci = 10 1/μ i = 30 ms Figure 11.5: Hanfoff Blocking Probability (ci = 50 and 100, 1/μi = 30 ms). 0 50 100 150 200 250 300 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Number of Channels Handoff Blocking Probablity (%) ← 1/μi = 30 ms −−− 1/μi = 20 ms −.−. 1/μi = 10 ms Figure 11.6: Hanfoff Blocking Probability (ci ranging from 0 to 300).
- 93. 85 10. (a) Thehandoff blocking probability of the selected handoffs as a func- tion of system offered load are shown in Figures 11.3, 11.4, 11.5, and 11.6. (b) In Figure 11.4, we assume the total available channels (ci) to be 50 and 100 respectively. The handoff blocking probabilities with a choice of three different mean holding times of 15, 20, and 30 ms are plotted. The times shown in the plot are estimates of the switched or exchanged channel latencies. The figures show that the blocking probability is directly proportional to the mean channel exchange time. As the mean holding time increases, the performance reaches to its ideal value. In Figure 11.5, the handoff blocking probability drops when the numbers of available channel increases. At last, we plot the graph of blocking probability versus the number of channels for different values of 1/μi in Figure 11.6. The handoff call request rate is fixed while we vary the number of channels from 0 to 300. For 1/μi = 30 ms, the blocking probability is dramatically decreased when the number of channels are less than 150. After exceeding 150, the behavior of decreasing is not obviously seen. 11. (a) λ1 = α + λ2 λ2 = λ3 λ3 = 0.41 λ4 = 0.6λ1 λ1 = α + λ3 = α + 0.4λ1 = α 0.6 = 33 packets/ms λ2 = λ3 = 0.4λ1 = 0.4 × 33 = 13.33 packet/ms λ4 = 0.6 × 33 = 20 packet/ms (b) ρ1 = λ1 μ1 = 0.33 ρ2 = λ2 μ2 = 0.133
- 94. 86 Chapter 11.Packet Queues and Delay Analysis ρ3 = λ3 μ3 = 0.666 ρ4 = λ4 μ4 = 0.666 E[K1(t)] = ρ1 1−ρ1 = 0.5 packets E[K2(t)] = 0.133 1−0.133 = 0.15 packets E[K3(t)] = 0.666 1−0.666 = 2 packets E[K4(t)] = 0.666 1−0.666 = 2 packets (c) E[T] = 4 i=1 E[Ki(t)] α = 0.5+0.15+2+2 20 = 0.232 ms 12. N/A 13. (a) Queuing unit 1: λ1 = 3λ Queuing unit 2: λ2 = 6λ + 6λ + 3λ + λ = 20λ Queuing unit 3: λ3 = 6λ Queuing unit 4: λ4 = 6λ (b) E[K1] = ρ1 1−ρ1 ; ρ1 = 3λ μ1 E[K2] = ρ2 1−ρ2 ; ρ2 = 20λ μ2 E[K3] = ρ3 1−ρ3 ; ρ3 = 6λ μ3 E[K4] = ρ4 1−ρ4 ; ρ4 = 6λ μ4 (c) E[T] = E[K1] 3λ + E[K2] 20λ + E[K3] 6λ + E[K4] 6λ 14. α = 200 packets/ms μ0 = 100 packets/ms μi = 10 packets/ms (a) λ = 0.4λ + α ⇒ λ = α/0.6 = 333.33 packets/s Thus, the arrival rate to each of the queuing units in parallel is: λi = 0.4/5 × 333.33 = 26.67 packets/s
- 95. 87 (b) E[K0(t)] =ρ0 1−ρ0 Since ρ0 = λ μ0 , then: E[K0(t)] = λ μ0 1− λ μ0 = λ μ0−λ = 333.33 packets/s 100×103 packets−333.33 packets/s = 3.34 × 10−3 packets E[Ki(t)] = ρi 1−ρi = λi μi−λi = 26.67 packets/s 10×103 packets−26.67 packets/s = 2.67× 10−3 (for 1 ≤ i ≤ 5) (c) E[T0] = E[K0(t)] λ = 3.34×10−3 packets 333.33 packets/s = 10−5 s E[Ti] = E[K0(t)]+E[E[Ki(t)]] α = E[K0(t)]+ 5 i=1 PiE[Ki(t)] α = 2.67×10−3+5(0.2E[Ki(t)]) 200 = 30 ms 15. (a) λ3 = 0.3λ − 1 λ2 = α + 0.3λ1 λ1 = λ2 + λ3 + λ4 λ4 = 0.3λ1 λ1 = λ2 + λ3 + λ4 = α + 0.3λ1 + 0.3λ1 + 0.3λ1 = α + 0.9λ1 ⇒ λ1 = 8 packets/ms λ2 = α + 0.3λ1 = α + 0.3 × 10α = 4α = 3.2 packets/ms λ3 = 0.3 × 80 = 2.4 packets/ms λ4 = 0.3λ1 = 0.3 × 80 packets/ms (b) ρ1 = 8 10 ; E[K1(t)] = ρ1 1−ρ1 = 0.8 1−0.8 = 4 ρ2 = 3.2 12 ; E[K2(t)] = ρ2 1−ρ2 = 0.267 1−0.267 = 0.364 ρ3 = 2.4 14 ; E[K3(t)] = ρ1 1−ρ1 = 0.171 1−0.171 = 0.206 ρ4 = 2.4 16 ; E[K4(t)] = ρ1 1−ρ1 = 0.15 1−0.15 = 0.176 (c) E[T] = E[K(t)] α = 4+0.364+0.206+0.176 0.8 = 5.93 ms
- 96. 88 Chapter 11.Packet Queues and Delay Analysis
- 97. Chapter 12 Quality-of-Service and ResourceAllocation 1. (a) P0 = PX (0)P1 + [PX (0) + PX (1)]P0 P1 = PX (2)P0 + PX(1)P1 + PX (0)P2 P2 = PX (3)P0 + PX(2)P1 + PX (1)P2 + PX (0)P3 P3 = PX (4)P0 + PX(3)P1 + PX (2)P2 + PX (1)P3 + PX (0)P4 P4 = PX (5)P0+PX (4)P1+PX (3)P2+PX (2)P3+PX (1)P4+PX (0)P5 (b) See Figure 12.1. 0 1 2 3 4 Px(5) Px(1) Px(1) Px(1) Px(1) Px(0) Px(0) + Px(1) Px(0) Px(0) Px(0) Px(2) Px(2) Px(2) Px(2) Px(3) Px(3) Px(3) Px(4) Px(4) Figure 12.1: Markov chain. 89
- 98. 90 Chapter 12.Quality-of-Service and Resource Allocation 2. (a) PX (k) = 1 4 for k = 0, 1, 2, 3 PX (k) = 0 for k = 0, 1, 2, 3 Also for k = 0, 1, 2, 3 : PX(0) = PX(1) = PX (2) = PX (3) = 1 4 (b) P0 = [PX (0) + PX (1)]P0 + PX(0)P1 = (1 4 + 1 4)P0 + 1 4P1 = 1 2P0 + 1 4P1 ⇒ 1 2P0 = 1 4P1 ⇒ 2P0 = p1 P1 = PX (2)P0 + PX (1)P1 + PX (0)P2 = 1 4P0 + 1 4P1 + 1 4P2 = 1 4P1 + 1 4 × 1 2P1 + 1 4P2 ⇒ 5 2 P1 = P2, P2 = 5 2 × 2P0 = 5P0 P2 = PX (3)P0 + Px(2)P1 + Px(1)P2 + PX(0)P3 = 1 4(P0 + P1 + P2 + P3) = 1 4(1 5 P2 + 2 5P2 + P2 + P3) ⇒ P3 = 12 5 P2, P3 = 12P0 P0 = 0.008 P1 = 2P0 = 0.016 (c) See Figure 12.2. 0 1 2 3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ½ Figure 12.2: Markov chain. P00 = PX (0) + PX (1) = 1 4 + 1 4 = 1 2 P01 = PX (2) = 1 4 P02 = PX (3) = 1 4 P11 = PX (1) = 1 4 P10 = PX (0) = 1 4 P12 = PX (2) = 1 4
- 99. 91 P20 = 0 P21= PX (0) = 1 4 P22 = PX (1) = 1 4 3. (a) For Poisson distribution: PX (x) = (λt)xe−λt x! . For t = 1 g PX (x) = (λ g )xe − λ g x! . With λ = 20 packets/s and g = 30 packets/s PX (k) = (2 3 )ke− 2 3 k! . (b) P0 = [PX (0) + PX (1)]P0 + PX (0)P1 ⇒ P1 = [1−PX(0)−PX (1)]P0 PX (0) We know P0 = 0.007. From Part (a), we also know: PX (0) = e− 2 3 1 = 0.513 PX (1) = 2 3 e− 2 3 1 = 0.34, thus: P1 = 0.00197 P1 = PX (2)P0 + PX(1)]P1 + PX(0)P2 ⇒ P2 = [1−PX(1)]P1−PX (2)P0 PX(0) We know PX(2) = (2 3 ) 2 e− 2 3 2! = 0.114, thus: P2 = 0.249 P2 = PX (3)P0 + PX(2)]P1 + PX(1)P2 + PX (0)P3 ⇒ P3 = [1−PX(1)]P2−PX (3)P0−PX (2)P1 PX (0) We know PX(3) = (2 3 ) 3 e− 2 3 3! = 0.025, thus: P3 = 0.318 (c) Transition probabilities:
- 100. 92 Chapter 12.Quality-of-Service and Resource Allocation P00 = PX (0) + PX (1) = 0.855 P01 = PX (2) = 0.114 P02 = PX (3) = 0.025 P03 = PX (4) = 0.004 P01 = PX (2) = 0.114 P10 = PX (0) = 0.513 P11 = PX (1) = 0.342 P12 = PX (2) = 0.114 P13 = PX (3) = 0.025 P20 = 0 P21 = PX (0) = 0.513 P22 = PX (1) = 0.342 P23 = PX (2) = 0.114 P30 = 0 P31 = 0 P32 = PX (0) = 0.513 P33 = PX (1) = 0.342 (d) Sketch from Part (c). 4. (a) b + vTb = zTb Tb = b z−v (b) b = 0.5 Mb z = 100 Mb/s v = 10 Mb/s Tb = 0.5 Mb 100 Mb/s−10 Mb/s = 5.56 ms
- 101. 93 5. N/A 6. Solvingthis problem requires a great deal of ellaborations in complex mathematical background. We try to summarize this background. In general, for a continuous random variable X ≥ 0, with mean E[X, second moment E[X2], and PDF fX(x), the Laplace-Steiljes transform (LST) of FX (x) is defined by ˆ fX(δ) = ∞ 0 e−δxfX(x)dx Hence, for the residual time distribution, Rj(t), LST is given by r̂j(δ) = 1− ˆ fX (δ) E[X]δ Now, we can derive the mean residual time, rj, can be derived as rj = −limδ→0 dr̂j (δ) dδ = E[X2] 2E[X] 7. (a) For non-preemtive scheduler: E[T1] = 0.37 s, E[T2] = 0.62 s, E[T3] = 0.25 s (b) For preemtive scheduler: E[T1] = 0.12 s, E[T2] = 0.66 s, E[T3] = 1.91 s (c) The waiting time E[Ti] for class i packet is lower for preemtive scheduler as long as i is low and it will be reversed for higher is. 8. We compare the impact of an increased number of inputs on total delay in priority scheduler with: three flows (n=3), and four flows (n=4). λi = λ = 0.2 packets/ms
- 102. 94 Chapter 12.Quality-of-Service and Resource Allocation 1 μi = 1 μ = 1ms ri = r = 0.5 ms (a) For a non-preemptive scheduler, we know: E[Tq,i] = Wx + E[Tq,i]2 + E[Tq,i]3 = n j=1 ρjrj + i j=1 ρjE[Tq,j] + E[Tq,i] i−1 j=1 ρj n = 3, non-preemptive scheduler: Wx = n j=1 ρjrj = 3 j=1 ρjrj = (3)(0.2)(0.5) = 0.3 For i = 1: E[Tq,1] = Wx + 1 j=1 ρjE[Tq,j] + 0 = 0.3 + ρ1E[Tq,1] = 0.3 + 0.2E[Tq,1] E[Tq,1] = 0.375 For i = 2: E[Tq,2] = Wx + 2 j=1 ρjE[Tq,j] + E[Tq,2] 1 j=1 ρj = 0.3 + ρ1E[Tq,1] + ρ2E[Tq,2] + E[Tq,2]ρ1 = 0.3 + 0.2 × 0.375 + 0.2E[Tq,2] + 0.2E[Tq,2] E[Tq,2] = 0.625 ms For i = 3: E[Tq,3] = Wx + 3 j=1 ρjE[Tq,j] + E[Tq,3] 2 j=1 ρj = 0.3 + ρ1E[Tq,1] + ρ2E[Tq,2] + ρ3E[Tq,3] + ρ1E[Tq,3] + ρ2E[Tq,3] = 0.3 + 0.2 × 0.375 + 0.2 × 0.625 + 0.2E[Tq,3] E[Tq,3] = 1.25 ms E[Ti] = E[Tq,i] + 1 μi For i = 3 ⇒ E[T3] = E[Tq,i] + 1 μ3 = 1.25 + 1 = 2.25 ms
- 103. 95 n = 4,non-preemptive scheduler: Wx = n j=1 ρjrj = 4 j=1 ρjrj = (4)(0.2)(0.5) = 0.4 For i = 1: E[Tq,1] = Wx + 1 j=1 ρjE[Tq,j] + 0 = 0.4 + 0.2E[Tq,1] E[Tq,1] = 0.5 For i = 2: E[Tq,2] =0.4 + ρ1E[Tq,1] + ρ2E[Tq,2] + E[Tq,2]ρ1 = 0.4 + 0.2 × 0.5 + 0.4E[Tq,2] + 0.2E[Tq,2] E[Tq,2] = 0.833 ms For i = 3: E[Tq,3] =0.4 + ρ1E[Tq,1] + ρ2E[Tq,2] + ρ3E[Tq,3] + 2ρ1E[Tq,3] = 0.4 + 0.2 × 0.5 + 0.2 × 0.833 + 0.6E[Tq,3] E[Tq,3] = 1.667 ms E[Ti] = E[Tq,i] + 1 μi For i = 3 ⇒ E[T3] = E[Tq,3] + 1 μ3 = 1.667 + 1 = 2.667 ms (b) For a preemptive scheduler, we know: E[Ti] = E[Tq,i] + θi where: θi = 1 μi 1− i−1 j=1 ρj n = j = 3: From non-preemtive case: E[Tq,3] = 1.25 ms θ3 = 1 μ3 1− 2 j=1 ρj = 1.667 E[T3] = 1.25 + 1.667 = 2.92 ms
- 104. 96 Chapter 12.Quality-of-Service and Resource Allocation n = j = 4: From non-preemtive case: E[Tq,3] = 1.667 ms θ3 = 1 μ3 1− 3 j=1 ρj = 2.5 E[T3] = 2.5 + 1.667 = 4.167 ms (c) The total delay obtained in the non-preemtive scheduler where n = 3 and n = 4 are close. In a preemtive scheduler, the difference in delay when n = 3 and n = 4 is very large. This is very obvious as when the numbers of flows increases in the case of preemtive scheduler, the waiting time becomes large for a low priority packet to be procesed. However, in a non-preemtive scheduler, this has a littile impact. This is because in an non- preemptive scheduler, lower priority packets cannot be interrupted immidiately upon the arriaval of higher priority packets. 9. (a) 2.1,2.1,4.1,1.1,3.2,4.2,1.2,4.3,1.3,2.2,4.4,1.4,3.3,1.5,2.3,3.4,2.4,3.6,4.5,2.5 (b) 2.1,3.1,4.1,4.2,4.3,4.4,1.1,2.2,2.3,3.2,3.3,3.4,4.5,1.3,2.4,2.5,1.5 10. N/A 11. N/A
- 105. 97 Packet No. SizeFlow Fi (FQ) Fi (WQ) 1 110 1 110 1100 2 110 1 220 2200 3 110 1 330 3300 4 100 1 430 4300 5 100 1 530 5300 6 100 1 630 6300 7 100 2 100 500 8 200 2 300 1500 9 200 3 200 666.6 10 240 3 440 1466.6 11 240 3 680 2266.6 12 240 4 240 600 12. Priority Queueing: (a) With 10% of the bandwidth,the low-priority will at least be able to transmit,without the guarantee bandwidth, a low-priority might never transmit. (b) The high-priority flows will lose 10% bandwidth and this 10% hit will be distributed evenly throughout all the high-priority flows, therefore the performance hit will not be noticeable. 13. See the table for the following arrangement: Flow1 110,110,110,100,100,100,(Flow2) 100,200, (Flow3) 200,240,240, (Flow4) 240 (a) Packets in fair queueing: 7,1,9,2,12,8,3,4,10,5,6,11 (b) Packet in weighted queueing: 7,12,9,1,10,8,2,11,3,4,5,6 14. (a) Priority Q.: A1,B1,B2,B3,A2,C1,C2,C3,A3,A4,B4,B5,C4,D1,D2,D3,A5,C5,D4,D5
- 106. 98 Chapter 12.Quality-of-Service and Resource Allocation (b) Fair Q.: A1,B1,D1,B2,C1,D2,A2,B3,C2,D3,A3,B4,C3,D4,A4,B5,C4,D5,A5,C5 (c) Weighted: A1,B1,B2,C1,C2,C3,D1,D2,D3,D4,A2,B3,B4,C4,D5,A3,C5,B5,A4,A5 15. Please make a correction: Part (c), flow D is 30 percent. (a) Priority Q.: B1,–,B2,A1,A2,B3,C1,B4,A3,A4,B5,C2,C3,C4,C5,D1,D2,D3,A5,D4,D5 (b) Fair Q.: B1,–,B2,C1,D1,A1,B3,C2,D2,A2,B4,C3,D3,A3,B5,C4,D4,,A4,C5,,D5,A5 (c) Weighted: B1,–,B2,C1,C2,C3,C4,D1,D2,A1,A2,A3,B3,C5,D3,D4,A4,B4,D5,A5,B5 16. This Problem is a result of moving Problem 8.4 to here as Problem 12.16. (a) Fairness index of B1, B2, B3 is given by σ = ( n i=1 fi) 2 n n i=1 f2 i = (f1+f2+f3)2 3(f2 1 +f2 2 +f2 3 ) = (1+1+1)2 3(12+12+12) = 1 (b) For equal throughput rates, the fairness index is σ = (f+f+f)2 3(f2+f2+f2) = (3f)2 3(3f2) = 1 We know: 0 is for the worst and 1 is for the best allocation of resource allocation. Thus from result of Part (a) we can say that we can have the best resource allocation when the throughput rates are equal. (c) Fairness index of B1, through B5 is given by σ = (f1+f2+f3+f4+f5)2 5(f2 1 +f2 2 +f2 3 +f2 4 +f2 5 ) = (1+1+1+1.2+16)2 5(1+1+1+1.44+256) = 0.313
- 107. 99 (d) The resultof Part (c) shows us that the resource allocation is not the best when the throughput rates are different. This is because the network cannot offer the fair amount of resource to each flow.
- 108. 100 Chapter 12.Quality-of-Service and Resource Allocation
- 109. Chapter 13 Networks inSwitch Fabrics 1. XCrossbar = n2 XDelta = nd logd n n Complexity of crossbar Complexity of Delta network 22 16 16 24 256 128 25 1024 320 The complexity of crossbar increases dramatically compared to Delta network as n increases. See Figures 13.1. 2. (a) See Figure 13.2. (b) Complexity: X(D16,2) = n d × d2 logd n = nd logd n = (16)(2) log2 16 = 128 X(D16,4) = 2 n d × d2 = (2)(16)(2) = 128 They have the same complexity 101
- 110. 102 Chapter 13.Networks in Switch Fabrics 5 10 15 20 25 30 500 1000 Crossbar Delta network n Complexity Figure 13.1: Comparison on the complexity of crossbar and Delta network. Communication Delay: The delay for D16,2 is higher than D16,4 because it has to go through more stages. However, D16,4 is nonblocking while D16,2 is blocking. 3. (a) See Figures 13.3 and 13.4. (b) Complexity: X(Ω16,2) = n d × d2 logd n = nd logd n = (16)(2) log2 16 = 128 X(Ω16,4) = 2 n d × d2 = (2)(16)(2) = 128 They have the same complexity Communication Delay: The delay for Ω16,2 is higher than Ω16,4 because it has to go through more stages. However, Ω16,4 is nonblocking while Ω16,2 is blocking.
- 111. 103 (a) (b) Figure 13.2:(a) D16,2 switch fabric, (b) a D16,4 switch fabric. 4. B = 1 − (1 − P)3 B = P 5. (a) See Figures 13.5 and 13.6. (b) B16,2number of stages → 2 log2 16 − 1 = 7 B16,4number of stages → 2 log4 16 − 1 = 3 Compare in terms of complexity: In general, the comlexity of Bn,d is : nd(r logd n − 1) B16,2 : 32 × 7 = 224 B16,4) : 64 × 3 = 192
- 112. 104 Chapter 13.Networks in Switch Fabrics Figure 13.3: Ω16,2 switch fabric.
- 113. 105 Figure 13.4: Ω16,4switch fabric.
- 114. 106 Chapter 13.Networks in Switch Fabrics 11 21 31 41 13 23 33 43 25 35 17 27 37 47 11 21 31 23 33 15 25 35 45 17 27 37 Figure 13.5: B16,2 switch fabric.
- 115. 107 Figure 13.6: B16,4switch fabric.
- 116. 108 Chapter 13.Networks in Switch Fabrics Compare in terms of communication Delay : B16,2 : 7 × delay per stage B16,4 : 3 × delay per stage Thus, B16,2 has higher complexity and delay. 6. See Figures 13.7 and 13.8. B16,2 : P2 = 1 − (1 − P1)2 P3 = (1 − (1 − P1)2)2 P4 = 1 − (1 − P1)2(1 − P3) P5 = P4 P6 = 1 − (1 − P1)2(1 − P5) Pblock = P2 6 = (1 − (1 − P1)2(1 − P5))2 = (1 − (1 − P1)2(1 − P2 4 ))2 = (1 − (1 − P1)2(1 − (1 − (1 − P1)2(1 − (1 − (1 − P1)2)2)2))2 B16,4 : Pblock = (1 − (1 − P1)2)4 Pblock(B16,2) Pblock(B16,4) 7. (a) See Figures 13.9 and 13.10. (b) The Banyan network is similar to Delta network. Thus, the routing rule of the Delta network can be applied to a Banyan network. 8. (a) See Figure 13.11. (b) B9,3 : XB = nd(2 logd n − 1) = 9(3)(2 log3 9 − 1) = 81 crosspoints s = 2 logd n − 1 = 2 log2 9 − 1 = 3 stages
- 117. 109 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 Figure 13.7: Lee’s blocking model for B16,2 switch fabric. P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 Figure 13.8: Lee’s blocking model for B16,4 switch fabric.
- 118. 110 Chapter 13.Networks in Switch Fabrics 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Figure 13.9: Y16,2 switch fabric.
- 119. 111 Figure 13.10: Y16,4switch fabric.
- 120. 112 Chapter 13.Networks in Switch Fabrics (b) (a) Figure 13.11: Comparing two switching networks: (a) B9,3 and (b) Ω9,3.
- 121. 113 9. (a) N/A (b)The complexity of this network is estimated to be (nc/n) = d(h + logd n) (nL/n) = 1 + h + logd n (c) For the extended the Delta network DE n,d,h,the blocking probability is estimated as follows Pp(n, d, 0) = 1 − (1 − p)k−1 Pp(n, d, 0) = [1 − (1 − p)2(1 − Pp(n/d, d, h − 1))]d 10. (a) Architecture 1: choose d = 2 k ≥ 2d − 1 ⇒ k = 3 complexity = 4(2 × 3) + 3(4 × 4) + 4(3 × 2) = 96 This is optimal in terms of complexity:Dopt = n 2 = 2 ⇒ k = 3 See Figure 13.12 Figure 13.12: Clos network, d = 2 and k = 3. Architecture 2: choose d = 4 k ≥ 2d − 1 ⇒ k = 7 Complexity = 2(4 × 7) + 7(2 × 2) + 2(7 × 4) = 140
- 122. 114 Chapter 13.Networks in Switch Fabrics See Figure 13.13 k=7 d =4 Figure 13.13: Clos network, d = 2 and k = 3. (b) For reliability and fault talerance. 11. N/A 12. The Clos network is designed to be non-blocking with extra link in the middle stages, however the Lee’s model referred to the possible conges- tion for each link, there for the number calculated values using Lee’s method referred to the conceptual blocking of each link between notes. Summary: With Lee’s method, we consider only one path. But for the analysis of blocking k ≥ 2d − 1 we consider all paths.
- 123. 115 (a) (b) Figure 13.14: Comparingtwo Clos networks: (a) C6,2,3 and C6,3,5 switching networks (b) Comparing Lee’s models. 13. (a) See Figure 13.14. d = 6/2 ⇒ d = 2 k ≥ 2d − 1 = 2 × 3 − 1 ⇒ k = 3 k = 2 × 3 − 1 = 5 (b) Lee’s model, total blocking probability. Assuming the probability of deley for all links are p, C6,2,3 ⇒ B = (1 − (1 − p)2)3 = (2p − p2)3 C6,3,5 ⇒ B = (1 − (1 − p)2)5 = (2p − p2)5
- 124. 116 Chapter 13.Networks in Switch Fabrics (c) C6,3,5 has lower probability of blocking. But it yields a higher complexity and higher cost due to more crossbars is used. So it really depends on the need of the network to decide which is better. 14. Five-stage Clos network with n = 8, d = 2. (a) See Figure 13.15. (b) See Figure 13.16. (c) Total blocking probability with p = 0.2. The probability of blocking for the middle stage is (2p−p2)3. Thus B = (1 − (1 − p) × (1 − p) × [1 − (2p − p2)3])3 = 0.059 15. (a) See Figure 13.17. (b) See Figure 13.18. BXY = [1−(1−p)2(1−(1−p)2)] = [1−(1−0.2)(1−0.2)(1−0.059)] = 0.0629 BZW = [1 − (1 − 0.2)(1 − 0.0629)(1 − 0.2)]3 = 0.02 16. Consider a five-stage Clos network (similar to problem 13.14) whose stages use the following crossbar dimensions: 1st stage: d × k, 2nd stage: e × j, 3rd stage: n de × n de , 4th stage: j × e, and 1st stage: k × d. (a) (b) We find k, d, j, e in terms of n. Non blocking conditions: j ≥ 2e−1 k ≥ 2d − 1 Xc = dkn d + k[ej n/d e + n/d e × n/d e × j + ej n/d e ] + dkn d = kn + k(jn d + jn2 d2e2 + jn d ) + kn
- 125. 117 D= 2 K=2 D= 2 K=3 D=2 K=3 Figure 13.15: Five-stageClos network with n = 8, d = 2.
- 126. 118 Chapter 13.Networks in Switch Fabrics Figure 13.16: Lee’s model for the five-stage Clos network with n = 8, d = 2. Xc = 2kn + k(2jn d + jn2 d2e2 ) k = 2d − 1 j = 2e − 1 Xnb c = 2n(2d − 1) + (2d − 1)[2n(2e−1) e + n2(2e−1) d2e2 ] = 4nd − 6n + 8ne + 4n2 de2 − 2n2 de2 − 4ne d − 2n d − 2n2 d2e − n2 d2e2 To optimize: dXnb c de = 8n − 4n2 de2 + 4n2 de3 − 4n d + 2n2 d2e2 + 2n2 d2e3 = 0 ⇒ eopt ≈ −4n2d+2n2 −8nd2+4nd We know: dopt ≈ n 2 ⇒ eopt ≈ −4n2 √n 2 +2n2 −4n2+4n √n 2 (c) Plug dopt and eopt into Xnb c .
- 127. 119 n=8 n Multiplexer 5-Stage Clos Network Figure 13.17:A Cantor network with three parallel switching planes. 17. (a) m = 512 bytes × 16 = 8192 bytes (b) rows = 8192×8 B/b 32 = 2048 rows C = log2 2048 = 11 bits (c) RAM segment process time: 518× 8× 1 32 × (2 ns + 2 ns + 1 ns) = 640 ns/segment Transmission time of a segment = 0.4μs 16 = 25 ns Bit rate = 512×8 bits 640+25 = 6.15 × 109 = 6.15 Gb/s (d) 6.15 Gb/s 512×8 = 1.5 Msegment/sec/port
- 128. 120 Chapter 13.Networks in Switch Fabrics Z X Y W Figure 13.18: Lee’s blocking probability model for the Cantor network with three parallel switching planes.
- 129. Chapter 14 Optical Switchesand Networks, and WDM 1. n ≤ ≤ 2n − 1 8 ≤ ≤ 15 (a) Minimum loss = 8 × 40 = 32 dB Maximum loss = 15 × 40 = 600 dB (b) ave ≈ n+(2n−1) 2 ≈ 3n−1 2 ≈ 12 Average loss = 12 × 40 = 240 dB 2. N/A 3. N/A 4. N/A 5. N/A 121
- 130. 122 Chapter 14.Optical Switches and Networks, and WDM 6. (a) E[T] = 1 μi,j −Li,j (b) E[Ts] = 1 μi,j −si,jLi,j (c) E[Tn] = 1 n(n−1) i,j si,j μi,j −si,jΛ 7. N/A 8. N/A 9. N/A 10. N/A
- 131. Chapter 15 Multicasting Techniquesand Protocols 1. Multicast Connection. A multicast protocol such as MOSPF, PIM, and CBT helps keep the traffic down by requiring the source to trans- mit only one packet. In the down side, a multicast protocol causes some computational delay associated with the implementation of the protocol in a router. Additionally, a special type router is needed to implement the multicast protocol. Another major drwback with the multicast protocol is that if the source packet is lost on its way before being copied, no multicast group members receive a copy of the packet. Multicast Connection. In contrast, sending the packet separately to each group would increase the traffic substantially. But it is more convenient from the network management stand-point. 2. In the sparse-mode algorithm, a shared-tree technique is used, and a relatively low cost path is selected. As a result, the sparse-mode ap- proach intruduces extra delay than the dense-mode approach. 123
- 132. 124 Chapter 15.Multicasting Techniques and Protocols Multicast Group 2 1 R 2 R LAN 1 LAN 2 LAN 3 LAN 5 LAN 4 3 R 4 R 7 R 8 R 6 R 5 R Figure 15.1: MOSPF protocol. A rendesvous point (RP) router is selected as a shared root of dis- tribution sub-tree. RP router is used to coordinate forwarding packets and prevent initial flooding of datagram. However, RP router can a hot spot for multicast traffic congestion and a possible point of failure in routing. Also, as finding a low cost path is not necessary in the sparse-mode, less hardware complexity meay be needed. However, the sparse-mode approach lowers the efficiancy of MOSPF as this protocol employs OSPF unicast routing that requires that each router in a network be aware of all available links. If the sparse mode is chosen, each router using a shared-tree may cause some involving routers select longers paths than they would normally select in a dense- mode. 3. (a) See Figure 15.1.
- 133. 125 (b) k β3,2 β3,4β3,5 β3,6 β3,7 β3,8 {3} 3-2(5) 3-4(7) × × 3-7(10) 3-8(11) {2,3} 3-2(5) 3-4(7) 3-2-5(12) × 3-7(10) 3-8(11) {2,3,4} 3-2(5) 3-4(7) 3-2-5(12) × 3-7(10) 3-8(11) {2,3,4,7} 3-2(5) 3-4(7) 3-2-5(12) 3-7-6(23) 3-7(10) 3-8(11) {2,3,4,7,8} 3-2(5) 3-4(7) 3-2-5(12) 3-7-6(23) 3-7(10) 3-8(11) {2,3,4,5,7,8} 3-2(5) 3-4(7) 3-2-5(12) 3-7-6(23) 3-7(10) 3-8(11) {2,3,4,5,6,7,8} 3-2(5) 3-4(7) 3-2-5(12) 3-7-6(23) 3-7(10) 3-8(11) The copying router is R5. 4. N/A 5. (a) We choose R2 as a rendezvous point as this node is at the network edge and has the least cost to multicast group. (b) Form a least-cost tree for the multicast action. For LAN 1: R3 → R2 → R5 → LAN 4, and R3 → R2 → R5 → R6 → LAN 3. The copying router is R5, and The total cost = 26. 6. N/A 7. (a) In the MOSPF deployment, the multicast tree is: R3 → R8 → LAN2, and R8 → R6 → LAN3. The total cost is 2.
- 134. 126 Chapter 15.Multicasting Techniques and Protocols (b) In the PIM deployment, we choose R4 as the rendezvous router. The multicast tree is: R3 → R7 → R4 → R7 → R6 → LAN3, and R7 → R8 → LAN2. The total cost is 5. 8. See Figure 15.2. For the copy network with F = 7, d = 2, and k = 7: Divide the stages into two halves. The first half has 7 2 = 3 stages. For Stages 1, 2, and 3: Initialize: F1 = 7, f1 = 1 ⇒ route the packet randomly then at each stage: The second half of the network has 7 2 = 4 stages. Stages 4, 5, 6, and 7: Stage 4: F4 = 7 1 = 7 and f4 = 7 27−4 = 7 8 = 1 ⇒ make one copy at this stage Stage 5: F4 = 7 1 = 7 and f4 = 7 27−5 = 7 4 = 2 ⇒ make two copy at this stage Stage 6: F4 = 7 2 = 4 and f4 = 7 27−6 = 4 2 = 2 ⇒ make two copy at this stage Stage 7: F4 = 4 2 = 2 and f4 = 7 27−7 = 2 1 = 2 ⇒ make two copy at this stage At the end of stage 7, we have 8 copies in total, since we only need 7, we’ll discard one of them and send the remaining 7 copies to the interface. 9. (a) We prefer a copy node not to be at the edge of a network:
- 135. 127 11 21 3141 13 23 33 43 25 35 17 27 37 47 11 21 31 23 33 15 25 35 45 17 27 37 Figure 15.2: Multicasting with F = 7 copies in a copy network with d = 2 and k = 7.
- 136. 128 Chapter 15.Multicasting Techniques and Protocols R12 → R13 → R10, and R12 → R17 → R15 → R16 → LAN 1, and R12 → R17 → R15 → R14 → LAN 2. Total multicast cost is 10. (b) R12 → R13 → R10 → R20 → R28 → R27 → R26 → R24 → R25 → R23 → LAN 4, and R12 → R13 → R10 → R20 → R28 → R27 → R26 → R24 → R25 → R21 → LAN 3. Total multicast cost is 27. (c) R12 → R13 → R10 → R30 → R37 → R36 → R34 → R33 → R32 → LAN 5. Total multicast cost is 31. 10. (a) See Figure 15.3. (b) The complexity of Boolean splitting multicast algorithm is less than the one for tree-based algorthm since it does not need two switch fabrics for routing and copying. But with the Boolean split- ting multicast algorithm if the number of packets increases, the congestion can happen because the copying and routing are func- tioned at the same time. Thus, the performance of the Boolean splitting algorithm is better than the tree-base algorithm only for making low number of packets. 11. (a) See Figure 15.4. (b) See Figure 15.5. s = k = logd n = log2 8 = 3 for 1 ≤ k 2 = 3 F = 5, fj = 1
- 137. 129 0 0 01 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 0 0 0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 1 0 0 0 0 1 1 1 0 1 1 1 0 0 j =1 j =2 j =3 j =4 0 0 0 1 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 0 1 1 0 0 1 1 1 1 Figure 15.3: Multicasting with F = 7 copies in a D16,2 copy network using Boolean splitting algorithm. j = 4 : F4 = F3 f3 = 5 1 = 5f4 = F4 dk−4 = 5 26−4 5 22 = 2 j = 5 : F5 = F4 f4 = 5 2 = 3f5 = F5 d6−4 = 3 2 = 2 j = 6 : F6 = F5 f5 = 3 2 = 2f6 = F6 d6−6 = 2 = 2 12. See Figure 15.6. To construct a 4 × 4 Crossbar switch, we need four 4 × 1 multiplexers. if in0 wants to send packets to Out1 and Out2, it just needs to set the control-bits C4-C6 to accept packets from In0.
- 138. 130 Chapter 15.Multicasting Techniques and Protocols 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 1 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 1 0 1 0 1 0 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 0 1 0 3-Stage Copying 3-Stage Routing (All Identical) Figure 15.4: Multicasting with F = 5 copies in a Ω8,2 copy network using Boolean splitting algorithm. 000 001 010 100 101 110 111 011 000 001 010 011 100 101 110 111 Figure 15.5: Cascading two Omega networks to form a copy network.
- 139. 131 Out 1 Out 2 Out3 Out 4 int0 int1 int2 int3 C 0 C 1 C 2 C 3 C 4 C 5 C 6 C 7 Figure 15.6: Internal srtructure of the a multicast switch.
- 140. 132 Chapter 15.Multicasting Techniques and Protocols
- 141. Chapter 16 VPNs, Tunneling,and Overlay Networks 1. (a) The advantage to having egress nodes estimate routing is that the routing computations are distributed throughout the MPLS net- work, and require fewer entries in the forwarding table for each egress node. This could provide faster assignment of labels (and therefore faster transmission) to packets entering the MPLS net- work, due to the smaller tables. (b) The advantage of a preassigned router is that convergence of the best routes would be faster and synchronized, as updates to the topography would not need to be sent between all routers in the MPLS network (such as when LSRs go up and down which initiate topography changes). 2. N/A 3. N/A 133
- 142. 134 Chapter 16.VPNs, Tunneling, and Overlay Networks 4. N/A 5. N/A 6. N/A
- 143. Chapter 17 Compression ofDigital Voice and Video 1. See Figure 17.1. g(t) = sin(200πt) s(t) = +∞ n=−∞ [t−nTs τ ] = +∞ n=−∞ [2000t − 2n] gs(t) = g(t) × s(t) = +2 n=−2 [2000t − 2n] sin(200πt) G(f) = j[δ(f − 100) + δ(f + 100)] S(f) = 1 2 +∞ n=−∞ sinc[5 × 10−4(f − 1000n)] Gs(f) = G(f) × S(f) = 1 2 +2 n=−2 sinc[5 × 10−4(f − 1100n)] + sinc[5 × 10−4(f − 900n)] 2. N/A 3. Mean: E[X] = 0 Rate Capacity: R = 4 b/sample Variance: V [X] = 2 (a) For this source, we know Db = V [X]2−2R. With R = 4, we obtain 135
- 144. 136 Chapter 17.Compression of Digital Voice and Video 0 1 2 -1 -2 -2 -1 0 1 2 1 -2 -1 0 1 2 -100 100 - 1000 1000 -1100 - 100 1100 t t f f f G(f) S(f) G(f) s g(t) s(t) g(t) s t Figure 17.1: Solution to exercise. Sampling process in time and frequency domains. Db=0.0078 (b) If tolerable distortion becomes Db = 0.05 by using the same for- mula we obtain R = 1 2 log2 1 Db = 2.66 Thus, the required transmission capacity=2.66 b/sample 4. Variance = σ2 = V [X] = 10 N = 12
- 145. 137 (a) Δ √ V [X] =0.4238 Δ = 0.4238( √ 10) = 1.34 (b) ai = −aN−i = −(N 2 − i)Δ a1 = −a11 = −(6 − 1)1.34 = −6.7 a2 = −a10 = −(6 − 2)1.34 = −5.36 a3 = −a9 = −(6 − 3)1.34 = −4.02 a4 = −a8 = −(6 − 4)1.34 = −2.68 a5 = −a7 = −(6 − 5)1.34 = −1.34 a6 = 0 (c) x̂i = −x̂N+1−i = −(N 2 − i + 1 2)Δ x̂1 = −x̂12 = −(6 − 1 + 1 2)1.34 = −7.37 x̂2 = −x̂10 = −(6 − 2 + 1 2)1.34 = −6.03 x̂3 = −x̂9 = −(6 − 3 + 1 2)1.34 = −4.69 x̂4 = −x̂8 = −(6 − 4 + 1 2)1.34 = −3.35 x̂5 = −x̂7 = −(6 − 5 + 1 2)1.34 = −2.01 x̂6 = −x̂6 = −(6 − 6 + 1 2)1.34 = −0.67 (d) D V [X] = 0.0.1885 D = 0.1885 (e) For the source, we have Db = V [X]2−2R. Here, we are asked to find Db given R. N = 12 R = log2 N ≈ 4 Db = V [X]2−2R = 0.039 5. Note to instructors: this problem needs to be corrected to “· · · repeat Problem 4, this time using a 16-level optimal uniform quantizer · · ·”
- 146. 138 Chapter 17.Compression of Digital Voice and Video Variance = σ2 = V [X] = 10 N = 16 (a) Δ √ V [X] = 0.4908 Δ = 0.4908( √ 10) = 1.06 (b) ai = −aN−i = −(N 2 − i)Δ a1 = −a15 = −(8 − 1)1.06 = −7.42 a2 = −a14 = −(8 − 1)1.06 = −6.36 a3 = −a13 = −(8 − 1)1.06 = −5.3 a4 = −a12 = −(8 − 1)1.06 = −4.24 a5 = −a11 = −(8 − 1)1.06 = −3.18 a6 = −a10 = −(8 − 1)1.06 = −2.12 a7 = −a9 = −(8 − 2)1.06 = −1.06 a8 = 0 (c) x̂i = −x̂N+1−i = −(N 2 − i + 1 2 )Δ x̂1 = −x̂16 = −(8 − 1 + 1 2)1.06 = −7.95 x̂2 = −x̂15 = −(8 − 2 + 1 2)1.06 = −6.69 x̂3 = −x̂14 = −(8 − 3 + 1 2)1.06 = −5.83 x̂4 = −x̂13 = −(8 − 4 + 1 2)1.06 = −4.77 x̂5 = −x̂12 = −(8 − 5 + 1 2)1.06 = −3.71 x̂6 = −x̂11 = −(8 − 6 + 1 2)1.06 = −2.65 x̂7 = −x̂10 = −(8 − 6 + 1 2)1.06 = −1.59 x̂8 = −x̂9 = −(8 − 6 + 1 2)1.06 = −0.53 (d) D V [X] = 0.01154 D = 0.1154 6. N/A
- 147. 139 7. The samplingrate is fs = 80, 000 meaning that we take 80,000 samples per second. Each sample is quantised using 16 bits so the total number of bits per second is 80,000 × 16. For a music piece of duration 60 min=3000 sec the resulting number of bits is 80, 000 × 16 × 3000 = 3.8 × 109 8. We define Λ(x) as follows: Λ(x) = ⎧ ⎪ ⎨ ⎪ ⎩ x + 1 −1 ≤ x ≤ 0 −x + 1 0 ≤ x ≤ 1 0 otherwise We define the PDF as follows: fX(x) = 1 2 Λ(1 2x) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 2(x 2 + 1) = x+2 4 −2 ≤ x ≤ 0 1 2(−x 2 + 1) = −x+2 4 0 ≤ x ≤ 2 0 otherwise Q(X) is in fact the quantiztion function denoted by X̃ in the book. Thus X̃ = Q(X). We define the quantization error by a new random variable Y = X − X̃: For − 2 x ≤ −1 ⇒ x̃1 = −1.5 fX(x1) = x1+2 4 = (y1−x̃1)+2 4 = (y1−1.5)+2 4 = y1+0.5 4 For − 1 x ≤ 0 ⇒ x̃2 = −0.5 fX(x2) = x2+2 4 = (y2−x̃2)+2 4 = (y2−0.5)+2 4 = y2+1.5 4 For 0 x ≤ −1 ⇒ x̃3 = 0.5 fX(x3) = −x3+2 4 = −(y3−x̃3)+2 4 = −(y3+0.5)+2 4 = −y3+1.5 4
- 148. 140 Chapter 17.Compression of Digital Voice and Video For 1 x ≤ 2 ⇒ x̃4 = 1.5 fX(x4) = −x4+2 4 = −(y4−x̃4)+2 4 = −(y4+1.5)+2 4 = −y4+0.5 4 To find fY (y), we use the important property of the PDF and a func- tion of a random variable Y = g(X), as: fY (y) = n i=1 fX (xi) dy dx We know that dy dx = 1. Thus: fY (y) = 4 i=1 fX(xi) = y1+0.5 4 + y2+1.5 4 + −y3+1.5 4 + −y4+0.5 4 =1 9. See Table 17.1 and Figure 17.2. Table 17.1: Encoded words. Input: ABC Output: XYZ +7 = 111 110 +5 = 110 111 +3 = 101 101 +1 = 100 100 -1 = 011 000 -3 = 010 001 -5 = 001 011 -7 = 000 010 10. (a) See Figure 17.3 (b) 513 Cc − 46, 2 Cc0, 3 -2 Cc0, 2 -2 0 1 0 1 0 -1 Cc0, 4 -1 Cc0, 43 11. HX(x) = − 0 −1(x + 1) ln(x + 1)dx − 1 0 (−x + 1) ln(−x + 1)dx
- 149. 141 A B C X Y Z 1 1 11 0 0 0 0 0 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 X=A Y = A + B = A B + A B BC A BC A BC A Z = B + C = B C + B C Figure 17.2: PCM encoder design: Karnaugh maps and the logic circuit. = 1 2 12. Sample space (alphabet) = {a1, a2, a3, a4, a5} Corresponding probabilities = {0.23, 0.30, 0.07, 0.28, 0.12}. (a) Entropy H(x) = − 5 i=1 Pi log Pi = −(0.23 log2 0.23 + 0.3 log2 0.3 + 0.07 log2 0.07 + 0.28 log2 0.28 + 0.12 log2 0.12) = 2.157 b/sample (b) {a1, a2, a3, a4, a5} {1 5, 1 5, 1 5 , 1 5, 1 5} H2(x) = − 5 i=1 Pi log Pi = −5 × 1 5 log2 1 5 = 2.32 b/sample
- 150. 142 Chapter 17.Compression of Digital Voice and Video 513 -46 0 -2 0 -1 0 0 -46 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 -2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Figure 17.3: Quantization of a still image to produce Matrix Q[i][j], and the order of matrix elements for transmission. Entropy of a uniformly distributed source is more compared to the above entropy of a non-uniformly distributed suorce. 13. (a) PX (1) = PXY (1, 1) + PXY (1, 2) + PXY (1, 3) = 0.1+ 0.2+ 0.4 = 0.7 PX (2) = PXY (2, 1) + PXY (2, 2) + PXY (2, 3) = 0.1 + 0 + 0.2 = 0.3 PX (3) = PXY (3, 1) + PXY (3, 2) + PXY (3, 3) = 0 H(X) = −(0.7 log 0.7 + 0.3 log 0.3) = 0.881 PY (1) = 3 x=1 PXY (x, 1) = 0.1 + 0.1 = 0.2 PY (2) = 3 x=1 PXY (x, 2) = 0.2 PY (3) = 3 x=1 PXY (x, 3) = 0.4 + 0.2 = 0.6 H(Y ) = −(0.2 log 0.2 + 0.2 log 0.2 + 0.6 log 0.6) = 1.371 (b) The marginal entropy shows the average information that we re- cieve from one source if we know the other one. (c) H(X, Y ) = −(0.1 log 0.1 + 0.2 log 0.2 + 0.2 log 0.2 + 0.1 log 0.1 +
- 151. 143 0.4 log 0.4)= 2.122 (d) Joint entropy shows the average information thet we recieve from combination of two source. 14. H(X|Y ) = H(X, Y ) − H(Y ) H(Y |X) = H(X, Y ) − H(X) 15. (a) Sample output a4 (b) The information content of samples a1 and a5 is: I = I(P1) + I(P5) = − log2(0.1) − log2(0.2) = 5.79 b (c) Least prob. seq. = {a4, a4, a4, a4, a4, a4, a4, a4, a4, a4} And, its probability = (0.05)10 = 9.76 × 10−14 No, it is not. (d) H(X) = − 7 i=0 Pi log2 Pi = 2.501 b/symbol F = 50 HZ × 2 = 100 symbol/sec H(X) = 100 symbol/sec × 2.501 b/symbol = 250 b/s (e) number of typical seq. = 2nH(X) = 225 = 3.4 × 107 number of non-typical seq. = number of seq.-number of typical seq. = Nn − 2nH(X) = 28.7 × 107 − 3.4 × 107 16. Sample Space (Alphabet) = {a1, a2, a3, a4} P ∈ {0.15, 0.20, 0.30, 0.35}
- 152. 144 Chapter 17.Compression of Digital Voice and Video (a) H(x) = − N k=1 Pk log2 Pk = −(0.15 log2 0.15 + 0.2 log2 0.2 + 0.3 log2 0.3 + 0.35 log2 0.35) = 1.926 Number of typical sequences = 2nH(x) = 2100(1.926) = 9.6 × 1057 (b) Total number of sequences = Nn = 4100 = 1.607 × 1060 Total number of non-typical sequences = 1.607 × 1060 − 9.9 × 1057 ≈ 1.597 × 1060 Number of typical sequences Number of non-typical sequences = 9.6×1057 1.597×1060 = 0.006 (c) P[X̄] = 2−nH(x) = 2−100(1.926) = 1.04 × 10−58 (d) Number of bits to represent typical sequence= nH(x) = 100(1.926) = 193 b (e) Most probable sequence is: {a4, a4, · · · , a4} with probability Pn 4 = (0.35)100 = 2.55 × 10−46 17. (a) See Figure 17.4. The compressed codes are: a0 = 0, a1 = 110, a2 = 10100, a3 = 100, a4 = 1011, a5 = 111, a6 = 10101. (b) The entropy of the source is computed as H(X) = − 6 k=0 Pk log2 Pk = 2.07 The average code length is: R̄ = 6 i=0 Pii = 1(0.55) + 3(0.1) + 5(0.05) + 3(0.14) + 4(0.06) + 3(0.08) + 5(0.02) = 2.1 As it is expected: H(X) ≤ R̄ ≤ H(X) + 1
- 153. 145 0 1 0 1 0 0 1 a0 a6 a2 a4 a5 a1 a3 1 0 1 0 1 0.13 0.07 0.45 0.13 0.07 0.18 0.18 0.55 0.02 0.05 0.06 0.08 0.10 0.14 0.08 0.10 0.06 0.55 0.14 0.14 0.13 0.55 0.18 0.27 0.27 0.55 Figure 17.4:Huffman encoder. Code efficiency: η = H(X) R̄ = 2.07 2.1 = 0.98 18. {−3, −2, −1, 0, 2, 3, 5} P ∈ {0.05, 0.1, 0.1, 0.15, 0.05, 0.25, 0.3} (a) H(x) = − 6 k=0 Pk log2 Pk = −(0.05 log2 0.05+0.1 log2 0.1+0.1 log2 0.1+0.15 log2 0.15+0.05 log2 0.05+ 0.25 log2 0.25 + 0.3 log2 0.3) = 2.528 bits/sample (b) fs = 4000 guard = 200 Entropy rate = (2fs + guard)H(x) = (4000 × 2 + 200)H(x) = 20, 731 bits/sec
- 154. 146 Chapter 17.Compression of Digital Voice and Video (c) See Figure 17.5. The generated codes are: 5(00), 3(01), 0(100), −1(101), −2(110), 2(1110), −3(1111). -3 2 -2 -1 0 3 5 0.05 0.05 0.1 0.1 0.15 0.25 0.3 0 0.1 0 0.2 0.1 0.15 0 0.25 0.25 0.2 0 0.45 0.45 0.3 0.25 0 0.45 1 0 1 1 1 1 1 0.55 0.55 Figure 17.5: Huffman encoder. (d) R̄ = 6 i=0 Pii = (0.3 + 0.25)(2) + (0.1 + 0.1 + 0.15)(3) + (0.05 + 0.05)4 = 2.55 Cr = R̄ log2 N = 2.55 3 = 0.85 η = H(X) R̄ = 2.528 2.55 = 0.99 19. Source sequence is 0, 1, 01, 00, 001, 000, 11, 111, 0010, 10, 101, 1111, 010, 0101, 0101 Find the smallest phrases that have not appeared. Phrases are encoded Table 17.2. 20. See Table 17.3.
- 155. 147 Table 17.2: Lempel-Zivcoding process Parser Output Location Encoded Output 0 0001 00000 1 0010 00001 01 0011 00011 00 0100 00010 001 0101 01001 000 0110 01000 11 0111 00101 001 1000 01111 010 1001 01010 10 1010 00100 101 0110 10101 111 1011 10001 010 1100 00110 101 1110 11011 1010 1111 11100
- 156. 148 Chapter 17.Compression of Digital Voice and Video Table 17.3: Lempel-Ziv coding process Parser Output Location Encoded Output 1 00001 000001 11 00010 000011 110 00011 000100 0 00100 000000 01 00101 001001 010 00110 001010 10 00111 000010 101 01000 001111 1101 01001 000111 0111 01010 000101 100 01011 001110 0101 01100 001101 01010 01101 011000 00 01110 001000 1111 01111 010101 010100 10000 011010 001 10001 011101
- 157. Chapter 18 VoIP andMultimedia Networking 1. (a) Voice B.W. for telecommunication = 4 KHz Number of samples per second = 2 × 4= 8 Ksamples/s 8 b/sample = 8 × 8 =64 kb/s (b) RTP header=12 bytes (c) RTP encapsulation: RTP(12),UDP(8),IP(≥ 20) Header=12+8+20=40 byte Data=0.5 sec ⇒ 0.5 × 64 kb/s = 32 kb = 4 kbyte Packet = 4 kbyte + 40 byte ≈ 4040 byte 2. (a) Assume packets (segments) of 1,500 byte long including IP, UDP, and RTP headers. The RTP header consists of: • 12 B Common RTP header: 1 B (V+P+X+CSC), 1 B (M+Payload Type), 2 B (Seq. No.), 4 B (Time Stamp), and 4 B (Sync Source ID). • 8 B Contributing Source RTP header: 4 B (Contributing Source1 ID), and 4 B (Contributing Source2 ID). Thus, the total packet payload size = 1,500 B - [20B (IP header) 149
- 158. 150 Chapter 18.VoIP and Multimedia Networking + 12B (UDP header) + 20B (total RTP headers)]= 1,448 B (b) Total combined two source data rates = 2×31 Kb/s ×(1/8) B/b= 7.75 KB/s Time required for one packet = 1,448 B 7.75 KB/s = 186.8 ms Number of packets generated in 5 minutes = 5×60 186.8 ms = 1,605 packets. 3. N/A 4. 1,280×1,024 pixel blocks 1 packet = 0.1 row 1 chunk = 1 pixel cblock (a) Sample sapce ⇒ 77 samples ⇒ each sample (pixel) = 7 b A chunk = 1 pixel block = (8×8 pixels) × 7 b = 448 b A chunk + header = 448 b + (4 B) × 8 (b/B) = 480 b (b) 1 packet = 0.1 row = (0.1)(1,280) = 128 pixel blocks = 128 chunks = 128 chunks × 480 b/chunk = 61,440 b A packet + header = 61,440 b + 12 × 8 (b/B) = 61,536 b (c) Video clip = 4 minutes 1 sec = 30 images Total number of images/min = (4 min) × 60 (sec/min) × 30 = 7,200 images Total number of packets/image = (10 packets/row) × 1,024 (rows) = 10,240 packets/image Total number of packets/4 min = (7,200) × (10,240) = 73,728,000
- 159. 151 Total number ofbits/4 min = (73,728,000) × (61,440 b/packets) = 4.53 Tb Bandwidth = 4.53Tb 4 min = 18.8 Gb/s 5. Data of Chunk: 1 pixel block × 10 phrases × 5 bits/phrase = 50 b Header of Chunk: = 4 B Chunk = 50 + 4 × 8 = 82 bits (a) Row data = 1280 pixel block × 82 b/pixel block SCTP header = 12 bytes Packet = 1280×82 8 + 12 = 13, 132 byte ≈ 12.8 kbyte (b) One frame = 12.8 kbyte/row × 1024 rows = 12.8 Mbyte Required B.W. = 12.8 Mbyte/frame × 30 frames/s × 8 b/B = 3 Gb/s (c) Total data = 3 Gb/s × 2 × 60 × 60 = 2, 700 GB/h 6. (a) FY (t)(y) = P[Y (t) ≤ y] = P[X(t) + 2t ≤ y] = P[X(t) − 2t] = FX(t)(y − 2t) ⇒ fY (t)(y) = FX(y − 2t) = 1 √ 2παt e−(y−2t)2/2αt (b) FY (t)Y (t+1)(y1, y2) = P[X(t) + 2t ≤ y1, X(t + 1) + 2(t + 1) ≤ y2] = FX(t),X(t+1)(y1 − 2t, y2 − 2(t + 1)) ⇒ fY (t)Y (t+s)(y1, y2) = fX(t),X(t+1)(y1 − 2t, y2 − 2(t + 1)) = fX(t)(y1 − 2t)fX(t)(y2 − y1 − 2) = 3−(y1−2t))2/2αt √ 2παt × e−(y2−y1−2)2/2α √ 2πα 7. 70 cycles/min 6 pulses/cycle 1 pulse/chunk
- 160. 152 Chapter 18.VoIP and Multimedia Networking Q,R,S → 4 samples P,T,U → 1 sample (a) We use variable size chunk. Assume each sample is encoded by 8 bits. A packet consists of: 12B SCTP header + 6 chunks each with 16B chunk header. Total bits of the 3 chunks made by P,T,U pulses = 3(16B (chunk header) × 8 b/B + 1 sample/pulse × 8 b/sample) = 408 b Total bits of the 3 chunks made by Q,R,S pulses = 3(16B (chunk header) × 8 b/B+ 4 samples/pulse × 8 b/sample) = 480 b Total SCTP packet size = (408b + 480b) + [20B (IP header) + 12B (SCTP header)] × 8 b/B = 1144 b/(SCTP packet) Bandwidth = 70 cycles/min = 70×1144 b/packet 60s = 1335 b/s (b) H = Max. Number of Human’s Heartbeat Cycles G = Max. number of Patients L = Max. Link Bandwidth We can choose the above parameters as long as we must can sat- isfy: L = H×G×1144 60s (c) The second option reduces the transmission overhead (packet over- head) but requires a larger bandwidth. 8. N/A
- 161.
- 162. 154 Chapter 18.VoIP and Multimedia Networking
- 163. Chapter 19 Mobile Ad-HocNetworks 1. N/A 2. N/A 3. N/A 4. N/A 5. N/A 155
- 164. 156 Chapter 19.Mobile Ad-Hoc Networks
- 165. Chapter 20 Wireless SensorNetworks 1. N/A 2. N/A 3. N/A 4. N/A 157