GATE 2021 Computer Science and information technology Set-1

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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.

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

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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.

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

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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.

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