GATE 2014 SET-1 CS Computer Science and information technology

[Q1 – Q25 carry ONE mark each ]

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[Q26 – Q55 carry TWO mark each ]

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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.

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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