Computer Networks Test 1

Concept:

In subnet mask: Network Id are all ones(1’s) and host Id of IP address are all zeros(0’s).

Calculation:

IP address consist of host id and network id

Host Id + Network id = 32 bits

IP address: 172.26.17.1/25

Network ID =25 bits

Therefore, Host ID= 32- (25) = 7

172.26.17.1 ≡ 172.26.17.00000001

Subnet Mask (binary) = 11111111. 11111111. 11111111.10000000

Subnet Mask = 255.255.255.128.

Data:

L = size of frame

BW → bandwidth = 10 Mbps = 10 × 106 b/s

Tp → propagation time = 50 μs = 50 × 10-6 s

Tt → transmission time

Formula:

For CSMA/CD,

Tt ≥ 2 × Tp

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

Calculation:

\(\frac{{\rm{L}}}{{{\rm{BW}}}} \ge 2 \times 40 \times {10^{ - 6}}\)

L ≥ 2 × 50 × 10-6 × 10 × 106

L ≥ 1000 bits

L ≥ 125 bytes

Lmin = 125 bytes

Therefore, the minimum size of a frame in the network is 125 bytes. 

IP address is 210. 32. 64. 79/26

Comparing with a.b.c.d/n

where n is number of bits in network id.

0 ≤ a, b, c, d ≤ 255

Network ID + Host ID = 32

∴ Host ID = 32 – 26 = 6

First address:

First address is obtained by making right most 32 – n bits to 0. Here n is 26

32 – 26 = 6, make right most 6 bits to 0.

210. 32. 64. 79 = 11010010.00100000.01000000.01001111

First address = 11010010. 00100000. 01000000. 01000000

In decimal form = 210.32.64. 64

Last address

It is obtained by making rightmost 32 – n bits to 1. Make rightmost 6 bits to 1

It becomes: 11010010. 00100000. 01000000. 01111111

In decimal form = 210. 32. 64. 127

Total number of address

Total number of addresses possible are: 2^32-n

Data:

Packets needed to be sent = p = 500

Error = e = 20% = 0.2

Total number of packets transmitted by the sender = n

Concepts:

In stop and wait protocol, packet loss during transmission is the only packet retransmitted.

Formula:

n = p + e × p + e× (e × p) + e × (e× e × p) + e × (e × e× e × p)

n = p(1 + e + e² + e³ …)

\(n = \frac{p}{1 - e}\)

Calculation:

\(n = \frac{500}{1 - 0.2}\)

n = 625

Data:

Bandwidth = BW = 2 ×10⁹ bps

Distance = D = 5 × 105 m = 500 k m

Speed = V = 2.5 × 10⁵ km/s

Formula:

\({\rm{Propagation\;Delay}} = {{\rm{T}}_{\rm{p}}} = \frac{{\rm{D}}}{{\rm{V}}}{\rm{\;}}\)

Calculation:

\( {{\rm{T}}_{\rm{p}}} = \frac{{\rm{D}}}{{\rm{V}}} = \frac{{500}}{{2.5 × {{10}^5}}} = 2 × {10^{ - 3}} = 2\;ms\)

The propagation delay in this link is 2 ms.

Concept:

a.b.c.d/NID

Therefore 23 bits are in NID and only 9 bits for Host id.

Explanation:

Last two-byte of IP address is represented in binary

Ethernet: 172.16.2.1/23

Range of IP address: 172.16.0000 00010.0000000 to 172.16.0000011.11111111

Range of IP address: 172.16.2.0 to 172.16.3.255

First and last IP address is reserved

Valid Host IP address: 172.16.2.1 to 172.16.3.254

Statement 1: 172.16.1.100

It is not valid since it does not fall in the range

Statement 2: 172.16.1.198

It is not valid since it does not fall in the range

Statement 3: 172.16.2.255

It is valid since it does fall in the valid range

Statement 4:172.16.3.0

It is valid since it does fall in the valid range

Hence option 3 is the correct answer.

Important Point:

Number of P = 3

Number of Q = 2

IPv4 address size = 32 bit

Number of IP address possible = 232

x = 50

y = 12.5

DNS: (Domain Name System)

• The Domain Name System is a hierarchical and decentralized naming system for computers, services, or other resources connected to the Internet or a private network.
• DNS protocol is used for Domain names to IP Address mapping.

HTTP: (Hypertext Transfer Protocol)

• It is an application-layer protocol for transmitting hypermedia documents, such as HTML.
• It was designed for communication between web browsers and web servers, but it can also be used for other purposes

SMTP: (Simple Mail Transfer Protocol)

• SMTP is used when email is delivered from an email client, such as Outlook Express, to an email server or when email is delivered from one email server to another.
• SMTP uses port 25.

POP (Post Office Protocol)

• POP is a type of computer networking and Internet standard protocol that extracts and retrieves email from a remote mail server for access by the host machine.

FTP (File Transfer Protocol)

• FTP stands for file transfer protocol. It is provided by TCP/IP for copying a file from one host to another. FTP can transfer a file through one of the following three modes: stream mode, block mode and compressed mode. Stream mode is the default mode.
• FTP uses two TCP connections one for data and another for control.

Therefore option 3 is correct

Concepts:

Bytes are numbered in TCP protocol

Calculation:

4 Gb → 1 second

\(1 \;b \to \frac{1}{{4 \times {{10}^9}\;}}\;seconds\)

\(8 \;b \to \frac{8}{{4 \times {{10}^9}\;}}\;seconds\)

\(1 \;B \to \frac{8}{{4 \times {{10}^9}\;}}\;seconds\)

Sequence number = 32 bits

\(2^{32} \;B \to \frac{2^{32}\times8}{{4 \times {{10}^9}\;}}\;seconds\)

\({2^{32}}B \to 8.58\;seconds\)

The minimum time before this sequence number can be used again is 8.58 seconds.

Concept:

CRC stands for cyclic redundancy check.  It is an error detection technique which is used in digital networks such as WAN and LAN as well as in some storage devices.

Explanation:

CRC is used to detect data units, such as packets with burst errors and sometimes after error correction.

It is based on modulo-2 division, in that addition is performed by XOR operation.

It is said that a CRC can detect burst errors of fewer than r + 1 bits where r is the degree of the polynomial. A burst of length greater than r + 1 bits is detected with probability of 1- 2^r.

Both statements 1 and 3 are true.

The physical layer is concerned with the connection of the devices to the media. In a point to point configuration, two devices are connected through a dedicated link. In a multipoint configuration, the sender link is shared among devices.

The physical layer defines the direction of transmission between two devices: simplex, half-duplex or full-duplex. In simplex mode, only device can send; the other can only receive. In half-duplex, two devices can send and receive but not at the same time. In a full-duplex mode, two devices can send and receive at the same time.

The transport layer can be either connectionless or connection-oriented. A connectionless transport layer treats each segment as an independent packet and delivers it to the transport layer at the destination machine. A connection-oriented transport layer makes a connection with the transport layer at the destination machine first before delivering the packet. When the data is transferred, the connection is terminated.

Generate the cyclic patter

23¹ (mod 29) ≡ 23

23² (mod 29) ≡ 7

23³ (mod 29) ≡ 16

23⁴ (mod 29) ≡ 20

23⁵ (mod 29) ≡ 25

23⁶ (mod 29) ≡ 24

23⁷ (mod 29) ≡ 1

23⁸ (mod 29) ≡ 23

23⁹ (mod 29) ≡ 7

Cyclic pattern = (23, 7, 16, 20, 25, 24, 1, 23 …)

Data:

Maximum Segment Size (MSS) = 2 KB

Round trip time (RTT) = 20 ms

Congestion window Size (CWS) = 64 KB

Formula:

\({\rm{Threshold}} = \frac{{{\rm{CWS}}}}{{{\rm{MSS}}}}\)

Calculation

\({\rm{Threshold}} = \frac{{{\rm{CWS}}}}{{{\rm{MSS}}}} = \frac{{64}}{2} = 32\;KB\)

Number of round trips = 21 – 20 = 20

Total time taken = 20 × 20 ms = 400

The third mask (/20) is applied to the destination address 210.128.22.67

Prefix length = 20

210.128.22.67   ⇒     11010010     10000000     00010110      01000011

Mask                 ⇒     11111111       11111111       11110000       00000000

Network add    ⇒     11010010     10000000     00010000      00000000   ⇒   210.128.16

The result is 210.128.16.0, which matches the corresponding network address. Therefore, a packet will be forwarded on interface 2.

Let p and q be two prime number that generated the modulus parameter n in RSA cryptosystem

n = p × q = 15251

φ(n) = φ(p × q) = (p - 1) × (q - 1) = 15000

p × q – p – q + 1 = 15000

15251 – p – q + 1 =15000

p + q = 252

p + 15251/p = 252

p² – 252p + 15251 = 0

(p – 101) (p – 151) = 0

∴ p = 101 or p = 151

if p = 101 then q = 151 and vice versa

Smallest prime factor of n is 101

Alternate Method:

n = p × q = 15251

From question: one question is less than 200 and greater than 150, that is, 150 < p < 200

Prime number less than are: 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199

Now find which prime number will give remainder 0 on division with 15251

15251%151 = 0

∴ 151 is the prime number

Other prime number is 15251/151 = 101

Smallest prime factor = 101

Tips to generate prime number:

In leaky bucket algorithm, the packet arrival rate is varying the output rate is always constant. The bucket leaks at a constant rate, i.e. the network interface transmits packets at a constant rate.

The token bucket algorithm is less restrictive than the leaky bucket algorithm, as it allows bursty traffic. However, the limit of burst is restricted by the number of tokens available in the bucket at a particular instant of time.

The implementation of basic token bucket algorithm is simple; a variable is used just to count the tokens. This counter is incremented every t seconds and is decremented whenever a packet is sent. Whenever this counter reaches zero, no further packet is sent out.

Bandwidth = 30 Mb/s =3 ×107 b/s

In 1 second, 10 7 bits can be sent

∴ 1 s → 3 × 107 b

1 ms →  3 ×104 b

100 ms → 300 × 104 b

Frame size = 1000 B = 1000 × 8 bit

Number of frames = ⌈(300 × 104) ÷ (1000 × 8)⌉

∴ Number of frames = 375 

Data:

File size = 1000 bytes

Length of packet = L = 500 bytes = 500 × 8 bit

Number of packets = 2

Bandwidth = BW = 10⁷ b/s

Distance = d₁= 50 km

Distance = d₂ = 50 km

Propagation speed = v = 2 × 10⁸ m/s = 2 × 10⁵ km/s

Transmission delay = T_t

Propagation delay = T_r = 0.1 ms

Processing time at switch = 0.1 millisecond

Formula:

\({T_t} = \frac{L}{{BW}}\)

\({T_p} = \frac{d}{v}\)

Calculation:

Packet 1:

Link 1:

\({T_{t1}} = \frac{{500 \times 8}}{{{{10}^7}}} = 0.4\;ms\)

\({T_{p1}} = \frac{{50}}{{2 \times {{10}^5}}} = 0.25\;ms\)

T_r = 0.1 ms

Link 2:

\({T_{t2}} = \frac{{500 \times 8}}{{{{10}^7}}} = 0.4\;ms\)

\({T_{p1}} = \frac{{50}}{{2 \times {{10}^5}}} = 0.25\;ms\)

Delay for packet 1:

T1 = 0.4 + 0.25 + 0.1 + 0.4 + 0.25 = 1.4 ms

Packet 2:

Transmission is started at 0.4 ms (after the transmission of 1^st packet)

Link 1:

\({T_{t1}} = \frac{{500 \times 8}}{{{{10}^7}}} = 0.4\;ms\)

\({T_{p1}} = \frac{{50}}{{2 \times {{10}^5}}} = 0.25\;ms\)

T_r = 0.1 ms

Link 2:

\({T_{t2}} = \frac{{500 \times 8}}{{{{10}^7}}} = 0.4\;ms\)

\({T_{p1}} = \frac{{50}}{{2 \times {{10}^5}}} = 0.25\;ms\)

T₂ = 0.4 + 0.25 + 0.1 + 0.4 + 0.25 = 1.4 ms

Total delay to transmit data = 0.4 + T₂ = 0.4 + 1.4 = 1.8 ms

Statement 1: TRUE

The computational overhead in link state protocols is higher than in distance vector protocols because LSR is based upon global knowledge whereas DVR is based upon Local information.

Statement 2: FALSE

Persistent looping can be avoided with the help of split horizon in DVR but there is no concept of persistent looping in LSR, in LSR only temporary loop exist and can automatically solved by system or router.

Statement 3: TRUE

After a topology change, a link state protocol will converge faster than a distance vector protocol.

Therefore option 4 is correct.

Important Point:

LSR → Link State Routing

DVR → Distance Vector Routing

Data:

Length of packet = L = 16 × 10⁶ bits = 2 × 10⁶ bytes

Bandwidth (BW) = 40 × 10⁷ B/s

Distance = d= 400 km

Propagation speed = v = 2 × 10⁷ m/s

Efficiency = η

Formula:

\({\rm{\eta }} \le \frac{{\rm{N}}}{{1 + 2a}}\)

\({\rm{a}} = \frac{{{\rm{Tp}}}}{{{\rm{Tt}}}}\)

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

\({{\rm{T}}_{\rm{p}}} = \frac{{\rm{d}}}{{\rm{v}}}\)

Minimum bits = ⌈log₂ N⌉

Calculation:

\({{\rm{T}}_{\rm{t}}} = \frac{{2 \times {{10}^6}}}{{40 \times {{10}^7}}} = 5\;ms\)

\({{\rm{T}}_{\rm{p}}} = \frac{{400 \times {{10}^3}}}{{2 \times {{10}^7}}} = \;20\;ms\)

\({\rm{a}} = \frac{{20}}{5} = 4\)

Since efficiency is maximum

\(1 = \frac{{\rm{N}}}{{1 + 2a}}\)

N = 1 + 2(4) = 9

Subnet mask: 255.255.255.254→ 255.255.255.1111 1110

X’s IP address: 200.20.15.3 →      200.20.15. 0000 0011

Subnet Mask & X’s IP address →  200.20.15.2

∴ Subnet Id = 200.20.15.2

Y’s IP address: 200.20.15.1 →       200.20.15.0000 0001

Subnet Mask & Y’s IP address →  200.20.15. 0

∴ Subnet Id = 200.20.15. 0

Z’s IP address: 200.20.15.2 →      200.20.15.0000 0010

Subnet Mask & Z’s IP address → 200.20.15.2

∴ Subnet Id = 200.20.15.2

X and Z belongs to same subnet 200.20.15.2

Tips and Tricks:

& → bitwise AND

Data = 4400 Byte

Router MTU = 900 byte

Data + header = 900

Data = 900 – 20

∴ Data = 880

Number of fragments = \(\frac{{4400}}{{880}} = 5\)

Since initial fragment offset is 100.

Fragment offset of 2^nd fragment = \(100 + \frac{{880}}{8} = 210\)

Fragment offset of 3^rd fragment = \(210 + \frac{{880}}{8} = 320\)

Fragment offset of 4^th fragment = \(220 + \frac{{880}}{8} = 430\)

Fragment offset of 5^th fragment = \(430 + \frac{{880}}{8} = 540\)

penultimate fragment = 2^nd last fragment = 430

Shortcuts:

penultimate fragment = Initial fragment offset + 110 × (n - 2)

Since both 1 and 2 transmit and collide

After the collision, Using the Back-off algorithm:

Range (0, 2¹ -1) = Rang (0, 1) for both X and Y

Station whose time is less wins the or the person whose time is more back-offs

No one wins, after the first collision, means they have collided for the second time 

Range (0, 2² -1) = Rang (0, 3) for both X and Y

 The probability that neither X nor Y wins = \(\frac{4}{{16}} = \frac{1}{4}\)