## GATE CSE Computer Network Previous Year Solutions

#### Show Answer With Best Explanation

Explanation:

Let’s understand it with exmaple:
Suppose, we have an Ethernet. Each machine on an Ethernet has unique physical (or MAC) Address Labelled E1 through E5 and each machine has unique Logical (or IP) Address Labelled I1 through I5 .

Let’s suppose, Host 1 want to communicate with Host 4 and Host1 knows the destination IP Address (i.e. I4 ).

Now, Upper Layer of Host 1 sent a packet to Host 4 using IP Address But Lower Layer(DLL) of Host 1 need MAC Address of Host 4 to sent a frame. For this purpose, ARP comes into picture.

An ARP request Packet sent from Host 1 to all other Host on an Ethernet. That’s why ARP Request is Broadcast Packet. The Broadcast packet will arrive at every machine on an Etherent & each one will check it’s IP Address. Host 4 alone will respond with it’s MAC Address (or E4). That’s why

• Source MAC Address (E4 in this example)
• Destination MAC Address (E1 in this example)

ARP Request packet has:
• Source MAC Address.(E1 in this example)
• Destination MAC Address.( E2, E3, E4, E5 in this example)

So, Option(IV) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
The maximum transfer unit (MTU) value of the link between P and R = 1500 bytes
The maximum transfer unit (MTU) value of the link between R and Q = 820 bytes.

Segment of size = 1400 bytes
IP identification value = 0x1234
IP header size = 20 bytes
Don’t Fragment (DF) flag value = 0

Logic
Since Segment size is 1400 B and MTU value of the link between P and R is 1500 bytes. So, Segment is successfully arrive at router R. MTU value of the link between R and Q is 820 bytes. Now here at router R, Fragmentation is required.

Segment size 1400 Byte is divided into two parts or two fragment
1400 = 800 + 600

1st fragment size = Data + Header = 800 + 20 = 820 Bytes
2nd fragment size = Data + Header = 600 + 20 = 620 Bytes

Option(I): If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234.(False)
If second or any fragment is lost, then whole TCP segment or entire fragment has to be sent again. If any fragment is lost, the sender need to retransmit, But sender does not know about any fragment. Because fragmentation is done at intermediate router. So, sender has to retransmit entire fragment or whole TCP segment and for the next time in fragmentation, IP identification value will be changed.

Option(II): TCP destination port can be determined by analysing only the second fragment.(False)
Each fragment of TCP Segement is send seperately.Each fragment contains TCP destination port. So, TCP destination port can be determined by analysing any fragment.

Option(III): Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.(True)
1st fragment size = Data + Header = 800 + 20 = 820 Bytes
2nd fragment size = Data + Header = 600 + 20 = 620 Bytes

Option(IV): If the second fragment is lost, P is required to resend the whole TCP segment. (True)
If second or any fragment is lost, then whole TCP segment or entire fragment has to be sent again.

So, Option(III) and Option(IV)are correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Processing delay of data frame = 0
Processing delay of Acknowledge frame = 0

Size of Data frame = 2000 bits
Size of Acknowledge frame = 10 bits
Data rate (or Bandwidth) = 106 bits per second
Propogation Delay (Tp) = 100 ms
Efficiency (or Link utilization) = 50% or 1/2

Sender window sizr (N) = ?

Formula,

where,
N= window size
Tt (data) = Transmission Delay for Data Frame
Tt (Ack) = Transmission Delay for Acknowledge Frame
Tp = Propogation Delay

Let’s Calculate,

Note: 1 second = 1000 millisecond

So, The minimum value of the sender’s window size is 51.

#### Show Answer With Best Explanation

Explanation:
Given,
Random Sequence number chosen by P is X.
Random Sequence number chosen by Q is Y.

P sends a TCP connection request to Q with TCP Segment having,
SYN bit = 1, SEQ (Sequence) number = X, ACK bit = 0

Let’s briefly understand some terms,
Sequence Number: The Sequence Number for each segment is the number of the 1st Byte carried in that segment. Host can choose any Random Sequence Number.

SYN bit: It is used during connection to synchronize the connection. First TCP segment send by any Host have SYN bit =1.
TCP is full Duplex protocol. So, 1st Segment sent by Host P to Host Q and 1st Segment sent by Host Q to Host P contain SYN bit = 1.

ACK bit:
The ACK bit =1, indicate the segment contains Acknowledge.
The ACK bit =0, indicate the segment does not contain Acknowledge.

ACK(Acknowledge) number: ACK number is the Byte Number expected next.
If the receiver of the segment has successfully received Byte Number x from other party, if defines X+1 as the Acknowledge Number.

TCP Segment Header that is sent by Q to P have :

SYN = 1 , Because it is 1st Segment sent from Q to P, it is used to Synchronize the connection.

Seq number = y , It is given that y be Sequence number chosen by Q.

Ack number = X+1 , SYN flag consume one Sequence number. Byte number X is consume by Q. So, Q expect X+1 next Byte number.

FIN bit = 0 , when FIN bit = 1, Host want to terminate connection. when FIN bit = 0, Host does not want to terminate connection.

So, Option(III) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Generator Polynomial = X3+X+1
The message m4m3m2m1m0=11000

Checkbit sequence c2c1c0 = ?

Concept,
Generator Polynomial = X3+X+1
Generator Polynomial bits = 1. X3 + 0.X2 + 1.X1 + 1.X0 = 1011

Generator Polynomial consists of 4 bits. So, 3 bits of zeros will be append to the message.

Now, Message bits = m4m3m2m1m0c2c1c0 = 11000000

Let’s Solve,

100 will be appended to message bits.
c2c1c0 = 100

So, Option(II) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
During one iteration, R measures its distance to its neighbours X, Y and Z as 3, 2 and 5, respectively.

The distance to router P from routers X, Y and Z are 7, 6 and 5, respectively.

The distance to router Q from routers X, Y and Z are 4, 6 and 8, respectively.

Combine all the above three diagram:

The miinimum Distance from R to P =?
The miinimum Distance from R to Q =?

Calculation,
The minimum Distance from R to P = min{ R->X->P, R->Y->P, R->Z->P }
= min{ 3+7, 2+6, 5+5 }
= min{ 10, 8, 10 }
= 8
The minimum Distance from R to P is 8 and the next hop router for a packet from R to P is Y.

The minimum Distance from R to Q = min{ R->X->Q, R->Y->Q, R->Z->Q }
= min{ 3+4, 2+6, 5+8 }
= min{ 7, 8, 13 }
= 7
The minimum Distance from R to Q is 6 and the next hop router for a packet from R to Q is X.

So, Option(III) and Option(IV) are correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Frame size = 1000 bits
Transmission rate (or Bandwidth) = 1Mbps = 106 bps
System produces (or Poission Process) = 1000 Frame per second

Throughput of the System = ?

Formula,
The Efficiency of the pure ALOHA = G * e-2G
The maximum Throughput = 0.184 (or 18.4%), when G =1.
where, G is the average number of Frame produced by the System during one Frame Transmission time.

Let’s Calculate,

Now, System produces 1000 frame per second. Since, Tt (Transmission Time) is 1ms. Then How many Frames are produced by the System in 1 ms.
1 second = 1000 Frame
1000 ms = 1000 Frame
1 ms = 1 Frame

Efficiency = G * e-2G
= 1 * e-2
= 0.13534

Throughput: Throughput is defined as the average number of frames successfully transmitted per second.

Throughput = 1000 * 0.13534 = 135.34 = 135(Approximate)

#### Show Answer With Best Explanation

Explanation:
Statement (I): A router does not modify the IP packets during forwarding.(False)
Many of the IP Packets are Modify by the router during forwaring. One of the field is TTL(Time to Leave). During the forwaring, each router on the way decrement TTL value by 1.

Statement (II): It is not necessary for a router to implement any routing protocol.(True)
In order to forward a packet, instead of routing protocol, Flooding Technique can be used. In Flooding, Each packet is forwarded to every direction without know about best Route.

Statement (III): A router should reassemble IP fragments if the MTU of the outgoing link is larger than the size of the incoming IP packet. (False)
Router does not assemble the fragments. Reassembling of IP Fragments is done at Receiver site.

So, Option(I) is correct.

#### Show Answer With Best Explanation

Explanation:
HTTP have 2 connection:
1) Persistent Connection: In a persistent connection, The server leaves the connection open for more requests after sending a response. The server can close the connection at the request of a client or if a time-out has been reached.

2) Non-Persistent Connection: In a non persistent connection, one TCP connection is made for each request/response. For N different pictures in different files, the connection must be opened and closed N times.

Now, it is gievn that, HTTP request is in non-persistent mode. The web page contains text and five small images.

Total HTTP Connection = one connection for text + 5 connection for 5 images = 6 connection

So, the minimum number of TCP connection required to display the webpage completely in your Browser is 6.

#### Show Answer With Best Explanation

Explanation:
Given,
Number of Host = 1500
Given the ISP uses CIDR

CIDR (Classless Inter Domain Routing): For Subneting in CIDR, a 32-bits IP address have:
• Network id (to represent network)
• Subnet id (to represent subnet within a network)
• Host id (to represent Host within a subnet)

Data,
Network id bits = 17bits
In order to represent 1500 computer we need ceil of(1500)= 11bits, Host id bits = 11bits
Subnet id bits = 32 – (17 + 11) = 4 bits

Logic to solve,
We can allot 16 (24) subnet to any of the organization. To represent 16 subnet:
• Network id(17) bits are fixed.
• Host id(11) bits should be all 0’s.
• Subnet id(4) bits can be any combination between 0 to 15.

A. 202.61.84.0/21: not possible to allotted.

B. 202.61.104.0/21 : possible to allotted.

Network id and Host id bits are all satisfied

C. 202.61.64.0/21
: possible to allotted.

Network id and Host id bits are all satisfied

D. 202.61.144.0/21
: not possible to allotted.

So, Option(I) is correct

#### Show Answer With Best Explanation

Explanation:
Given Specification,
RTT(Round Trip Time)= 6ms
Thresold = 32KB
MSS(Maximum Segment Size) = 2KB

Congestion window size at time t+60 ms after all acknowledgements are processed= ?
Here, t+60 ms is nothing but t +(60/6) = t+10th RTT {After 10thRTT or at 11th RTT}

Logic to solve,
2) Increase the sender window size in multiple of MSS(2KB) and stops when thresold(32KB) reached.
3) Once Thresold reached, increased sender window size by 1MSS(2KB) till timeout occurs.

Let’s Solve,
1st Transmission = 2KB
2nd Transmission = 4KB
3rd Transmission = 8KB
4th Transmission = 16KB
5th Transmission = 32KB {Thresold reached}
Now increase sender window size by 1MSS(2KB)

6th Transmission = 34KB
7th Transmission = 36KB
8th Transmission = 38KB
9th Transmission = 40KB
10th Transmission = 42KB
11th Transmission = 44KB

So, Then the size of the congestion window (in KB) at time t+60 ms after all acknowledgements are processed is 44KB.

#### Show Answer With Best Explanation

Explanation:
SMTP: SMTP stands for Simple Mail Transfer Protocol and is a TCP/IP protocol used in `sending e-mail` . It’s a set of communication rules that allow software to send email over the Internet.

POP3(Post Office Protocol 3) and IMAP(Intractive Mail Access Protocol): POP3 and IMAP are two different protocols used to `access email`. POP3 and IMAP function very differently and each has its own advantages. POP3 is useful in checking emails from a computer that is in a specific location. IMAP is the better option when we need to check our emails from multiple locations, such as at work, from home, or on the road, using different computers.

MIME(Multipurpose Internet Mail Extensions): MIME is an Internet Standard that extends the format of e-mail to support-
• Text in character sets other than ASCII
• Non-Text Attachments.
• Message bodies with multiple parts

So, Option(IV) is correct.

#### Show Answer With Best Explanation

Explanation:
Fermat’s Little Theorem:
If p is a prime and a is an integer then
ap-1 ≡ 1 (mod p) , if p does not divide a
else
ap ≡ a (mod p)

Caculation:
351 mod 5
here, p=5, a=3 and p does not divide
≡ 34*12 +3 mod 5
≡ ((34)12 mod 5) * (33 mod 5)
≡ (112 mod 5) * (33 mod 5) {* 35-1 ≡ 1 mod 5 *}
≡ 27 mod 5
≡ 2

#### Show Answer With Best Explanation

Explanation:
Given,
IP addresses of M = 100.10.5.2
IP addresses of N = 100.10.5.5
IP addresses of P = 100.10.5.6

Logic to solve,
We will find Network Id by X-ORing of IP Address of Machine with Subnet Mask. Same Network Id of Machine belong to same Subnet.

`Machine M`:
100. 10 . 5 .0000 0010
255.255.255.1111 1100
——————————-
(100.10.5.0)

`Machine N`:
100. 10 . 5 .0000 0101
255.255.255.1111 1100
———————————
(100.10.5.4)

`Machine P`:
100. 10 . 5 .0000 0110
255.255.255.1111 1100
———————————-
(100.10.5.4)

Machine N and P belong to same subnet.

So, Option(II) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
n = 3007
ϕ(n) = 2880

Calculation,
Let p and q be two prime number such that n=p*q = 3007
ϕ(n) = ϕ(p*q) = ϕ(p)* ϕ(q) = (p-1)*(q-1) = 2880
=> pq -p-q-1 =2880
=> 3007-p-q-1 =2880
=> p+q =128 ————(1)
=> pq =3007 ————(2)
put value of q in eq(1)
=> p+(3007/p) =128
=> p2 -128p + 3007 =0

The prime factor of n greater than 50 is 97.

Explanation:

#### Show Answer With Best Explanation

Explanation:
UDP Header’s port number = 16 bits
Ethernet MAC Address = 48 bits
IPV6 Next Header = 8 bits
TCP Header’s Sequence number = 32 bits

So, Option(III) is correct.

#### Show Answer With Best Explanation

Explanation:
Statement(I): The cwnd increase by 2 MSS on every successful acknowledgement. (False)
It is False because cwnd increase by 1 MSS on every successful acknowledgement.

Statement(II): The cwnd approximately doubles on every successful acknowledgement. (False)
It is False as the same reason for above.

Statement(III): The cwnd increase by 1 MSS every round trip time. (False)

Statement(IV): The cwnd approximately doubles every round trip time. (True)

There is a confusion between statement(II) and statement(IV).
Let’s understand it example

It is cleared with the diagram that:
The congestion window is increase by 1MSS on every successful acknowledge and
The congestion window approximately doubles every round trip time.

So, Option(III) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Bandwidth = 1 Gbps (109 bits/second)

Calculation,
=> 1Gb = 1 sec
=> 109 bits = 1 sec
=> (109 / 8) Byte = 1 sec
=> (109 / 8) Sequence number = 1 sec {*TCP count each Byte*}
=> 1 Sequence number = (8 / 109 ) sec
=> 232 Sequence number = ( 232 * 8)/ 109 sec {*Total 232 sequence number possible* }
=> wrap arround time = 34.55 = 34(aproximate)

The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is 34.

#### Show Answer With Best Explanation

Explanation:
Given,
IP Packet Length = 4500 Bytes
MTU = 600 Bytes

Data,
IP Payload = 4500 – 20 = 4480 Bytes
MTU Payload = 600 – 20 = 580 Bytes
MTU Payload should be multiple of 8 But 580 is not multiple of 8. So, the nearest value of 580 which is multiple of 8 is 576.

Calculation,

The fragmentation offset value stored in the third fragment is 144.

#### Show Answer With Best Explanation

Explanation:
TCP connection state diagram (RFC 793)
* FIN-WAIT-1
Client-side TCP is waiting for an acknowledgment of the connection termination request or for a simultaneous connection termination request from Server-side TCP. This state is normally of short duration.

* FIN-WAIT-2
TCP client is sent FIN segment to the TCP server. Server-side TCP responds by sending an ACK, which is received by the client-side TCP.Client-side TCP connection wait for the FIN from the server-side TCP.

* TIME-WAIT
Server side TCP is waiting for enough time to pass to be sure the server side TCP received the acknowledgment of its connection termination request.

* LAST-ACK
Client-side TCP is waiting for an acknowledgment of the connection termination request previously sent to Server. This state is entered when client-side TCP received a termination request before it sent its termination request.

#### Show Answer With Best Explanation

Explanation:
Given,
Generator polynomial = x3 + x + 1
Message = 01011011

Calculation,
Generator polynomial = x3 + x + 1
Generator polynomial bits = 1.x3 + 0.x2 + 1.x + 1 = 1011

Generator polynomial consists 4 bits, so 3 zeo’s will be append to message.

The message bits will be: 0101 1011 101
So, Option(III) is correct.

#### Show Answer With Best Explanation

Explanation:
Formula,

Given Data,
Propagation Delay (Tp) = 0.75 ms
Bit rate = 1Mbps = 106 bits per sec
Number of bytes in the information frame = 1980
Number of overhead bytes in the information frame = 20
Number of bytes in the acknowledge frame = 20
Time to process a frame = 0.25 ms.

Calculation,
Useful Data = information frame + overhead = 1980+20 = 2000 B

Processing Delay( Tprocess ) = Tprocess (Data) + Tprocess (Acknowledge) = 0.25 + 0.25 = 0.5 ms

#### Show Answer With Best Explanation

Explanation:
RSA cryptosystem:
1) Select two prime number p and q. p=13, q=17
2) n= p*q
3) Calculate, ϕ(n)=ϕ(p*q)=ϕ(p)*ϕ(q) = (p-1)*(q-1)= 12*16= 192
4) public key (e) = 35
Calculate d,
=> e*d mod ϕ(n)==1
=> 35*d mod 192 ==1
=> 35*11 mod 192 == 1

Private key(d) = 11

#### Show Answer With Best Explanation

Explanation:
Given,
Data rate (or Bandwidth) = 106 bits/sec
Distance = 10,000 km
Speed = 2×108 m/sec
Data Length = 50,000 Bytes

Transmission Delay = ?
Propagation Delay = ?

Formula,

Calculation,

So,Option(I) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Bandwidth = 128 x 103 bps = 128 Kbps
Propagation Delay = 150 milliseconds
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1

The minimum number of bits required for the sequence number ﬁeld =?

Formula,
For selective repeat protocol,
Efficiency (η) =
where,
N= Sender window size = Receiver window size

Tp = Propogation Delay
Tt = Transmission Delay =

Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required for the sequence number = ceil of(2N)

Calculation,

Efficiency (η) =
=> 1 =
=> N = 1+2a
=> N = 1+2*2.4 = 5.8 = 6(approximate)

Sequence Number = sender window size + receiver window size = N+N = 6+6 = 12
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (12) = 4

#### Show Answer With Best Explanation

Explanation:
Statement(I): If the sequence number of a segment is m, then the sequence number of the subsequent segment is always m+1.(False)
1 Byte of Data consumes 1 Sequence number.If Sequence number of segment is m, then the Sequence number of subsequent segment is m+1,only if 1 Byte of Data is transferred. But it is already given that Host transfer karge file. So, the statement is False.

Statement(II): If the estimated round trip time at any given point of time is t sec, the value of the retransmission timeout is always set to greater than or equal to t sec.(True)
TCP sets a timeout when it sends Data and if Data is not acknowledged before timeout expires it retransmit data. Timeout is based on RTT. Timeout is always set to greater than or equal to RTT.

Statement(III): The size of the advertised window never changes during the course of the TCP connection.(False)
Receiver’s window buffer is never changed.But depending upon the available window size, Advertised window is keep changing.
Let’s undestand it with example,
Suppose receiver window size is 1000B, so the size of advertised window is 1000B. Again suppose sender have 400B of data to sent, it sends to receiver. Now receiver window have (1000-400) 600B of available space. Now receiver will sent acknowlege to sender with advertised window of 600B. So, The size of the advertised window keep change during the course of the TCP connection.

Statement(IV): The number of unacknowledged bytes at the sender is always less than or equal to the advertised window.(True)
Because sender will never sends number of Byte greater than the advertised window.

So, Option(II) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Bit rate (or Bandwidth) = 64 kilobits =64 x 103 bps
Processing Delay (or Tp) = 20ms
Link utilization (or Efficiency) = 50% = 1/2

Frame size = ?

Formula,
For stop-wait protocol,
Efficiency (η) =
where,

Tp = Propogation Delay
Tt = Transmission Delay =
Calculation,
η =
=>
=> 1+2a = 2
=> Tt =2*Tp
=>
=> L = 2 * Tp * Bw
=> L = 2* 20 ms * 64 x 103 bps
=> L = 2* 20* 10-3 s * 64 x 103 bps
=> L = 40 * 64 / 8 Byte = 320 Byte

#### Show Answer With Best Explanation

Explanation:
Given,
Bandwidth = 106 bps
Frame size = 1000 B = 1000* 8 bits
Efficiency = 25% = 1/4

Propogation Delay = ?

Formula,
For stop-wait protocol,
Efficiency (η) =
where,

Tp = Propogation Delay
Tt = Transmission Delay =
Calculation,

=> η =
=> 1/4 =
=> 1+2a = 4
=> Tp =3*Tt /2
=> Tp = 3*8ms/2
=> Tp = 12ms

#### Show Answer With Best Explanation

Explanation:
RTT: Round-trip delay (RTD) or round-trip time (RTT) is the amount of time it takes for a signal to be sent plus the amount of time it takes for acknowledgement of that signal having been received.

Bandwidth-Delay Product(BDP): The bandwidth-delay product is the product of a data link’s capacity (in bits per second) and its round-trip time (in seconds). BDP = RTT * bandwidth

Given,
Bandwidth(or data link capacity) = 1048560 bits/sec
TCP window scale option is needed. It is needed for efficient transfer of data when the Bandwidth-Delay Product (BDP) is greater than 64 KB
{* BDP = 64KB (BDP is maximum 216-1 =65535 B )*}

Calculation,
BDP = RTT * bandwidth
=> 65,535 B = α * 1048560 bits/sec
=>
=> α = 0.5 sec = 500 milliseconds

By using the window scale option, the receiver window size increased up to a maximum of 1GB.
Receiver window size from 64KB to 1GB i.e. 216B to 230B.
a 14bits shift count is used in TCP Header. So, the maximum possible window size gets increased from 216-1 to (216-1 * 214) or 65,535 to 65,535 * 214.

So, Option(III) is correct.

#### Show Answer With Best Explanation

Explanation:
Statement(I): TCP connections are full duplex.(True)
TCP offers full-duplex service, in which data can flow in both directions at the same time. Each TCP then has a sending and receiving buffer, and segments move in both directions.

Statement(II): TCP has no option for selective acknowledgment.(False)
TCP has option for selective acknowledgment. With selective acknowledgments, the data receiver can inform the sender about all segments that have arrived successfully, so the sender need retransmit only the segments that have actually been lost.

Statement(III): TCP connections are message streams.(False)
TCP connections are Byte stream protocol. While sending data, each Byte of segment is counted and First Byte number of segment is called Sequence number.Sequence number of TCP segment is stored in TCP Header.

So, Option(I) is correct.

#### Show Answer With Best Explanation

Explanation:
Given,
Distance = 8000 km = 8 * 106 meters
Bandwidth = 500 * 106 bps
Propagation Speed= 4 * 106 meters per sec
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1

The minimum number of bits required for the sequence number ﬁeld =?

Formula,
For selective repeat protocol,
Efficiency (η) =
where,
N= Sender window size = Receiver window size

Tp = Propogation Delay
Tt = Transmission Delay =

Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required for the sequence number = ceil of(2N)

Calculation,

Efficiency (η) =
=> 1 =
=> N = 1+2a
=> N = 1+2*2.4 = 5.8 = 6(approximate)

Sequence Number = sender window size + receiver window size = N+N = 6+6 = 12
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (12) = 4

#### Show Answer With Best Explanation

Explanation:
Given,
size of congestion window = 32KB
Round Trip Time = 100ms
Maximum segment size = 2KB

Logic to solve,
When Time-out occurs in TCP slow start phase, Thresold value is calculated by
Thresold = (Congestion window size)/2

So, start with 1MSS (or 2KB) as sender window size. Slow start phase will begin and window size will grow exponentially till Thresod reached. After that congestion avaidance phase will begin. Now window size will grow linearly till 16MSS (or 32KB).

Calculation,
Thresold = (Congestion window size)/2 = 32KB / 2 = 16KB

Slow start phase:
window size is 1 MSS (or 2KB)
After 1st Round Trip, window size is 2 MSS (or 4KB)
After 2nd Round Trip, window size is 4 MSS (or 8KB)
After 3rd Round Trip, window size is 8 MSS (or 16KB)
(Thresold reached)
Congestion avaodance phase:
After 4th Round Trip, window size is 9 MSS (or 18KB)
After 5th Round Trip, window size is 10 MSS (or 20KB)
After 6th Round Trip, window size is 11 MSS (or 22KB)
After 7th Round Trip, window size is 12 MSS (or 24KB)
After 8th Round Trip, window size is 13 MSS (or 26KB)
After 9th Round Trip, window size is 14 MSS (or 28KB)
After 10th Round Trip, window size is 15 MSS (or 30KB)
After 11th Round Trip, window size is 16 MSS (or 32KB)

Total Round Trip = 11
Total Round Trip Time = 11* 100 ms = 1100ms
The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is 1100.

#### Show Answer With Best Explanation

Explanation:
Given,
Frame size = 1 KB = 8 x 210 bits = 8*1024 bits
Bandwidth = 1.5 Mbps = 1.5 x 106 bps
Propagation Delay(Tp) = 50 milliseconds
Link utilization (or Efficiency)= 60% = 0.6

The minimum number of bits required to represent the sequence number ﬁeld =?

Formula,
For selective repeat protocol,
Efficiency (η) =
where,
N= Sender window size = Receiver window size

Tp = Propogation Delay
Tt = Transmission Delay =

Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required for the sequence number = ceil of(2N)

Calculation,

Efficiency (η) =
=> 0.6 =
=> N = 0.6(1+2a)
=> N = 0.6(1+2*9.15) = 11.58 = 12(approximate)

Sequence Number = sender window size + receiver window size = N+N = 12+12 = 24
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (24) = 5bits

#### Show Answer With Best Explanation

Explanation:
Given,
Data rate (or Bandwidth) = 500 Mbps = 500 x 106 bps
Frame size = 10,000 bits
Speed = 200,000 km/sec

Distance =?
Formula,
Length = 2 * Tp * Bandwidth

Calculation,
=> L = 2 * Tp * Bandwidth
=> L = 2 *(Distance/Speed) * Bandwidth

So, Option(II) is correct.