IV. RESULTADOS
4.1. Análisis descriptivo de los datos
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 [email protected]. 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
About the Author
Nader F. 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
vi CONTENTS Mir has published 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.
Part I
Fundamental Concepts
1
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
2.3×108m/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
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
Dc = dconn-req + dconn-accep + dconn-release
(a) Here, the problem askes that trbe defined as the processing time for each packet. Therefore, tr= 20, 000× 0.2 = 4,000 s
5 Dc = dconn-req = dconn-accep = dconn-release
= [np+ (nh− 2)]tf + nhtr+ (nh− 1)tp
Data forms two packets:
(9960 + 40) bytes for packet1 (2040 + 40) bytes for packet2
tf1−packet1 = tf2= transfer times for control packets = 500b
109b/s = 5× 10−7s tp = c = 500 miles×1.61×103m
2.3×108m/s = 3.5× 10−3s (a) request + accept time:
t1+ t2 = 2 ([np+ (nh− 1)]tf2+ (nh− 1)tp+ nhtr]) = 2.06 s (b) t3 = 12(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−
6 Chapter 1. Packet-Switched Networks
7
Figure 1.1: Signaling delay in connection-oriented packet-switched environ-ment.
9. See Figure 1.1.
10. D = s[1−ρd+h
d/d(d+h)]
8 Chapter 1. Packet-Switched Networks (a) ∂D∂h = 0
Thus: hopt = d(1−2ρ2ρdd)
(b) Queueing delay
Chapter 2
Foundation of Networking 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
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
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
12 Chapter 2. Foundation of Networking Protocols (b) The longest match is 180.19.16.0/20.
7. (a) N1 L11: 1100 0011 . 0001 1001 . 0000 0000 . 0000 0000 (b) Packet: 1100 0011 . 0001 1001 . 0001 0001 . 0000 0011
L11: 1100 0011 . 0001 1001 . 0000 0000 . 0000 0000
13
8. (a) IPv4 address field has 32 bits
Total number of IP addersses available = 232
Number of IP addresses per person for 620 million people
= 620×10232 6 = 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
= 620×102128 6 = 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
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
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
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−1i=n
17
= 1+8/3+28/9+56/27+70/8170/81 ≈ 9%
(d) E[C] = 4j=1jpj = 1(0.275) + 2(0.32) + 3(0.213) + 4(0.0889) ≈
18 Chapter 3. Networking Devices
7. ρ = tta
a+td = 0.9
for m = 11 and n = 10 : Pc = the clipping prabability
=i=nm−1
9. See Figure 3.1.
1 1 0 0 1 1 1 0 1
Figure 3.1: Line coding.
19 10. See Figure 3.2.
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+50r ⇒ ∂d∂r∂T2 = 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).
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
Chapter 4
Data Links and Transmission
1. Tprop = 5000×103m
3×108m/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× 108s≈ 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
22 Chapter 4. Data Links and Transmission
(a)
(b) 3 14 1 2
0
11 4 10
3 1 2
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
23 5. The hardware is shown in figure 4.2.
1-bit Shift
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,0G |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
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,CRCG |2 shows 0 0 0 0 indicating no errors:
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
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,0G |2
D,CRC=1010 1101 0101 111,10100 (b) D,CRC = 1010 1101 0101 111,10100
G=1110 10
25 Remainder = 0
The data is correct.
7. v = 3× 108 m/sec
26 Chapter 4. Data Links and Transmission
Figure 4.3: Answer to exercise. The efficiency trend.
(a) Stop-and-Wait protocol:
E = ttf = t tf
f+2tp = 4×104×10−4+2(.2)−4 = 0.0010≈ 0.1%
(b) Sliding window protocol, w = 6:
E = w
Required condition on R3-R4: Link R3-R4 must transfer slower
27
28 Chapter 4. Data Links and Transmission
Chapter 5
Local-Area Networks and 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
=
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
=
= 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
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 ×601)(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
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 + 0.375km
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= lc = 180 b
100 Mb/s = 1.8× 10−6 s
5. (a) α = tTp β = λT
32 Chapter 5. Local-Area Networks and Networks of LANs
(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}
33 pi = pc(1− pc)i = 0.387(1− 0.387)7 = 0.0116
(e) na= 4 E[c] = 1−pp c
c = 1−0.3870.387 = 1.52
Listen to Medium Transmit data
Wait for 512g
Listen to Medium Transmit with Prob P for Max
Figure 5.2: Answer to exercise.
8. See Figure 5.3.
34 Chapter 5. Local-Area Networks and Networks of LANs
Hub Network
Analyzer
Bridge To
Internet
R1
To Other Buildings
Hub Repeater
Figure 5.3: Answer to exercise.
Chapter 6
Wireless Networks and 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
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
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 dbis 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
37
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
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.
38 Chapter 6. Wireless Networks and Mobile IP
Average Speed of A Vehicle, mph
Probability of Requirement For Handoffs
α01 = 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
39 increases, the probability of requiring a handoff decreases.
7. N/A
40 Chapter 6. Wireless Networks and Mobile IP
Chapter 7
Routing and
Inter-Networking
1. (a) See Figure 7.1 (a).
min=2 max=2
H = min+max2 = 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 n2 + 2 H = min+max2 = 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
42 Chapter 7. Routing and Inter-Networking
(a)
(c) (d)
(b)
Figure 7.1: Answer to exercise. Four different network topologies to connect two users.
43
2. Using Dijkstra’s Algorithm
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
44 Chapter 7. Routing and Inter-Networking
Figure 7.2: Answer to exercise.
Figure 7.3: Answer to exercise.
45
Figure 7.4: Answer to exercise.
Figure 7.5: Answer to exercise.
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) {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.
47
Figure 7.6: Answer to exercise.
R1
Figure 7.7: Answer to exercise.
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
49
Figure 7.8: Answer to exercise.
PCE = (0.3)(0.3)(0.6) = 0.054
50 Chapter 7. Routing and Inter-Networking
Chapter 8
Transport and End-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
52 Chapter 8. Transport and End-to-End Protocols
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
1,940B ≈ 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
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 log2n− 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
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 = Δf = 2 MB
208ms = 76.9 Mb/s
(d) With the additive increase/multiplative decrease protocol, ρu= Br = 76.9Mb/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
55
Source Destination
0.5 sec
LOSS OF SEGMENT 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.
56 Chapter 8. Transport and End-to-End Protocols
Chapter 9
Applications and Network 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
2. (a) To obtain the file name in a remote machine, the DNS server