Solution

For class C default mask is 255.255.255.0 => 255.255.11111111.00000000

We require 16 blocks. So, we have to take 4 bits (2^4 = 16) from the 3rd octet and change it to zero. Therefore, the supernet mask is 11111111 11111111 11110000 00000000 => 255.255.240.0

Solution

The given supernet 255.255.248.0 = 255.255.11100000.00000000 has 21 no. of 1’s.

The default mask of class C has 24 no. of 1’s.

Since the difference is 3 no. of 1’s, so there are 2^3 = 8 blocks in this supernet.

The blocks are 214.216.32.0 to 214.216.39.0.

The first address is 214.216.32.0 for NID, and the last address is 214.216.39.255 for DBA.

Solution

Given IP is class B address.

So, 16 bits are for the NID part, and the remaining 16 bits are the HID part.

The company needs 1000 subnets.

To make 1000 subnets, we have to choose at least 10 bits from the HID part.

Now NID + SID = 16 + 10 = 26 bits, therefore, remaining 32 - 26 = 6 bit would be for HID for each subnet.

Therefore, possible no. of addresses in each subnet is 2^6 = 64

Note: we subtract 2 ( 2^6 - 2 ), only when it will ask possible no. of Host.

The new Subnet mask = 26 no. of 1's followed by 6 0's

11111111 11111111 11111111 11000000 = 255.255.255.192.

Solution

When a quick response is needed, UDP is preferred.

DNS uses the UDP at the transport layer with port number 53.

Solution

The prefix length (NID bits) is 27 bits, which means that we must keep the first 27 bits as it is, and change the remaining last 5 bits to 0s to get a network address. The 5 bits affect only the last octet. The last octet is 82 = 01010010. Changing the last 5 bits to 0s,

we get 01000000 = 64. The network address is 156.188.161.64/27.

Solution

Length and checksum can be modified when IP fragmentation happens. Time To Live is reduced by every router on the route to the destination. Only Source Address can not change.

Solution

Given-

• Bandwidth = 2 Mbps
• RTT = 40 msec
• Packet size = 1000 byte

Transmission delay(Tt) = L / B = 1000 x 8 bits / 2 x 10^6 bps = 4 msec

Propagation Delay(Tp) = RTT/ 2 = 20 msec

So, a = Tp / Tt = 20 / 4 = 5

Link Utilization or Efficiency (η) = 1 / 1+2a {For stop and wait protocol}

=> 1 / 1 + 2 x 5 = > 0.0909 = 9.09%

Solution

The HLEN value is 5, which means the total number of bytes in the header is 5 × 4 = 20 bytes. The total length is 0028₁₆ = 40 bytes,

Which means the packet is carrying 20 bytes of data (40 −20).

Solution

In class C address from the first octet, the first three bits (110) are reserved.

Therefore, total no. of address is (32 - 3 = 29) = 2^29

Solution

1-(iii), 2-(i), 3-(iv), 4-(ii)

Solution

We require four subnets, which means we need 2 bits. And add to subnet bits (26 + 2 = 28)

Now, the subnet bit is /28.

Therefore, the remaining 32 - 28 = 4 bits would be for HID for each subnet.

Possible no. of address in each subnet is 2^4 = 16

Note: we subtract 2 ( 2^4 - 2 ), only when it will ask possible no. of Host.

Now, with 2 bits the possible subnet bits are 00,01,10,11

Subnet S1 - 00: 141.211.210.64/28 to 141.211.210.79/28.

Subnet S2 - 01: 141.211.210.80/28 to 141.211.210.95/28.

Subnet S3 - 10: 141.211.210.96/28 to 141.211.210.111/28.

Subnet S4 - 11: 141.211.210.112/28 to 141.211.210.127/28.

Note - first, and last address used for NID and DBA respectively, remaining are used for the Host address.

Solution

Method I Total time takes:

L = 1000 GB = 1000 x 10^9 bytes,

B.W = 1 MBps =10^6 bytes/s

Tt = L / B = (1000 x 10^9 bytes) / (10^6 bytes/s) = 10^6 sec = 11.5 days(approx)

Method II Total time takes:

L = 1000 GB = 1000 x 10^9 bytes, B.W. = 100 MBps = 100 x 10^6 bytes/s

Tt = (1000 x 10^9 bytes) / (100 x 10^6 bytes/s) + 5 days = 5.1 days

Method II is 6.4 days faster than Method I.

Solution

Initial threshold is 32.

Slow start operating up to threshold Then congestion avoidance begins.

Slow start operating interval is (1,6)

First congestion avoidance interval (6,16)

Second threshold is 21

Second congestion avoidance interval (17, 22)

Solution

Solution

Congestion window size = 32 KB

Slow start threshold = 32 KB / 2 = 16 KB

Starting with window size = 1 segment = 2 KB

At RTT 1, cwnd = 2 KB

At RTT 2, cwnd = 4 KB

At RTT 3, cwnd = 8 KB

At RTT 4, cwnd = 16 KB (time-out) (As the threshold is reached, the Additive increase starts)

At RTT 5, cwnd = 18 KB

At RTT 6, cwnd = 20 KB

At RTT 7, cwnd = 22 KB

At RTT 8, cwnd = 24 KB

At RTT 9, cwnd = 26 KB

At RTT 10, cwnd = 28 KB

At RTT 11, cwnd = 30 KB

At RTT 12, cwnd = 32 KB

Time taken to reach 32 KB = Number of segments X RTT

600 msec= 12 X RTT

=> RTT = 50 msec

Solution

Both statements are correct.

Solution

a) False - Piggyback is only for efficiency. If there's no data packet to be piggybacked to, then B will just send the acknowledgement packet.

b) False - The sequence number of the subsequent segment depends on the number of 8-byte characters in the current segment.

C) False The acknowledgement number has nothing to do with the sequence number. The ack. number indicates the next sequence number A is expecting from B.

Solution

Group 1: each customer needs 256 addresses.

This means the suffix length(HID bit) is 8 (2^8 = 256)

The prefix length(NID bit) is then 32 - 8 = 24

The addresses are:

01: 168.215.0.0/24 to 168.215.0.255/24

02: 168.215.1.0/24 to 168.215.1.255/24

……………………………………………

31: 168.215.30.0/24 to 168.215.30.255/24

32: 168.215.31.0/24 to 168.215.31.255/24

Total = 32 × 256 = 8,192

Group 2: each customer needs 128 addresses.

This means the suffix length(HID bit) is 7 (2^7 = 128).

The prefix length(NID bit) is then 32 − 7 = 25.

The addresses are:

01: 168.215.32.0/25 to 168.215.32.127/25

02: 168.215.32.128/25 to 168.215.32.255/25

03: 168.215.33.0/25 to 168.215.33.127/25

04: 168.215.33.128/25 to 168.215.33.255/25

…………………………………………………

63: 168.215.63.0/25 to 168.215.63.127/25

64: 168.215.63.128/25 to 168.215.63.255/25

Total = 64 × 128 = 8,192

Group 3: each customer needs 64 addresses.

This means the suffix length(HID bit) is 6 (2^6 = 64).

The prefix length(NID bit) is then 32 − 6 = 26.

The addresses are:

001: 168.215.64.0/26 to 168.215.64.63/26

002: 168.215.64.64/26 to 168.215.64.127/26

003: 168.215.64.128/26 to 168.215.64.191/26

004: 168.215.64.192/26 to 168.215.64.255/26

…………………………

129: 168.215.159.192/26 to 168.215.159.255/26

128: 168.215.159.192/26 to 168.215.159.255/26

Total = 128 × 64 = 8,192

Number of granted addresses: 65,536

Number of allocated addresses: 24,576

Number of available addresses: 40,960

Solution

If a DF(do not fragment ) bit is set and a router needs to fragment the datagram to send it further then ICMP error message must be sent to the sender.

If TTL is exceeded then also it should report to the sender by sending ICMP error message.

Solution

IRTT = α (IRTT) + (1- α) ARTT

IRTT1 = 0.9×30+(1–0.9)×26 = 29.6

IRTT2 =0.9×29.6+(1–0.9)×32 = 29.84

IRTT3 =0.9×29.84+(1–0.9)×24 = 29.256

Solution

Given-

• Distance = 6000 km
• Bandwidth = 2.25 Mbps
• Packet size = 16 bytes
• Propagation speed =5 μsec / km

Solution

Total data = 60 x 8 Mb (convert byte to bit)

Time for computer to transmit data = 60 x 8 Mb / 5 Mbps = 96 sec

Maximum network router transmission rate = 4 Mbps

Actual data sent on network in 96 sec = 4 Mbps x 96 sec = 384 Mb = 48 MB (change into byte)

Bucket must hold = 60 MB - 48 MB = 12 MB

Solution

Given p = 5, q=17, d=13

Plain Text = IIT

N = p x q = 85

n = (p-1)(q-1) = (5-1)(17-1) = 64

Now, (e x d) mod 64 = 1 = > Therefore, e = 5

Sum of integer in cipher text message = 59 + 59 + 5 = 123

Solution

Perform the bitwise AND of the given IP address with the subnet mask from the table.

If it matches more than two destination IP addresses, then according to the concept, we will choose which has the highest no of 1's Matching.

Therefore, here I - 2 has the highest no of 1's Matching.

Solution

Solution

Perform bitwise AND (&) operation between IP address and subnet mask result should have the same Network ID to be in the same network.

By performing AND operation, all options are getting different network IDs. The only subnet mask 255.255.128.0 is giving the same network ID 11.110.0.0.

Solution

The Maximum Transfer Unit = 800 bytes = 780(Data) + 20(Header)

But data should be divisible by 8 so, it is 776 bytes

Each datagram size = 3625 bytes (Datagram means data with header)

3625 bytes = 3605 bytes data + 20 bytes header

Then the datagram divides into multiple frames depending on MTU.

First Fragment → 776 bytes data + 20 bytes header

Second Fragment → 776 bytes data + 20 bytes header

Third Fragment → 776 bytes data + 20 bytes header

Fourth Fragment → 776 bytes data + 20 bytes header

Fifth Fragment → 521 bytes data + 20 bytes header

⇒ Total number of fragments = 5

Offset is the number of data bytes ahead of a particular fragment and divide by 8.

⇒ The offset of the 3rd fragment is (776+776)/8 = 194 bytes.

Solution

Given-

• Sender window size = Receiver window size = 10
• Packet size (L) = 2000 bytes
• Transmission delay (T_t) = 100 sec
• Propagation delay (T_p) = 400 sec

We know,

Transmission delay = Packet size / Bandwidth

So, Bandwidth = Packet Size / Transmission delay (T_t)

= 2000 bytes / 100 s

= (2000 x 8 bits) / (100 sec)

= 160 bps

a = T_p / T_t

a = 400 sec / 100 sec

a = 4

Optimal window size = 1 + 2a = 1 + 2 x 4 = 9

Efficiency (η) = Sender window size / Optimal window size = 8 / 9 = 0.8888 = 88.88 %

Solution

RSA: An algorithm for digitally signing data with a private key such that anyone with possession of the corresponding public key can verify the signature.

MAC: A symmetric-key algorithm for ensuring that a message has not tampered.

Message integrity: The security goal of ensuring that a message arrives at the recipient is identical to what the sender transmitted.

AES: Standardized symmetric key encryption algorithm.

Solution

MTU 1= 200 => 180(D) + 20(H)

180 should be multiple of 8, so its 176

MTU 2= 110 => 90(D) + 20(H)

90 should be multiple of 8, so its 88

176 + 176 + 148 = 500

Solution

For GB- N- 1 2 3 4 5 6 7 8 9 6 7 8 9 10 11 8 9 10 11 12 13 10 11 12 13 12 13 = 27

No. of transmissions = 27

For SR

1 2 3 4 5 6 6 7 8 9 10 11 11 12 13 = 15

No. of transmissions = 15

Therefore, total no. of transmission for both = 15 + 27 = 42

Solution

Physical Layer ……….... Bits

Data Link Layer ……….. Frames

Network Layer ………... Packets

Transport Layer ……….. Segments

Solution

Here No congestion is mentioned, so no need for the threshold value.

Starting with window size = 1 segment = 2 KB

At RTT 1, cwnd = 2 KB

At RTT 2, cwnd = 4 KB

At RTT 3, cwnd = 8 KB

At RTT 4, cwnd = 16 KB

At RTT 5, cwnd = 32 KB

At RTT 6, cwnd = 64 KB (38 KB limited by receiver window)

However, it'll exceed the receiver's window. Therefore, the amount of time it takes BEFORE the 6th RTT (or full window, that is, 38KB) is 5 x 15 = 75 msec.