GATE 2015 SET-2 CS Computer Science and information technology

[Q1 – Q25 carry ONE mark each ]

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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

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[Q26 – Q55 carry TWO mark each ]

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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.

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