GATE 2018 CS Computer Science and information technology

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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.

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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.

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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.

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[Q26 – Q55, Carry TWO marks each ]

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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.