# GATE 2021 CS Computer Science and information technology Set-2

GATE 2021 CS Computer Science and information technology Set-2

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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.

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[Q11 – Q15 Multiple choice Question(MCQ),carry ONE mark each (no negative marks) ]

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[Q16 – Q25 Numerical Answer Type(NAT), carry ONE mark each (no negative marks) ]

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[Q26 – Q39 Multiple choice Question(MCQ),carry TWO marks each (for each wrong answer: – 2/3) ]

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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.

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[Q40 – Q47 Multiple select Question(MSQ),carry TWO marks each (no negative mark) ]

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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.

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[Q48 – Q55 Numerical Answer Type(NAT),carry TWO marks each (no negative mark) ]

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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)

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