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# GATE | GATE-CS-2017 (Set 1) | Question 47

• Difficulty Level : Basic
• Last Updated : 22 Sep, 2017

The values of parameters for the Stop-and – Wait ARQ protocol are as given below.

```Bit rate of the transmission channel = 1Mbps
Propagation delay from sender to receiver = 0.75 ms
Time to process a frame = 0.25ms
Number of bytes in the information frame = 1980
Number of bytes in the acknowledge frame = 20
Number of overhead bytes in the information frame = 20

```

Assume that there are no transmission errors. Then the transmission efficiency ( expressed in percentage) of the Stop-and – Wait ARQ protocol for the above parameters is _________( correct to 2 decimal place) .

Note: This questions appeared as Numerical Answer Type.

(A) A number between 86.5 and 87.5
(B) A number between 82.4 and 82.5
(C) A number between 92.4 and 95.5
(D) A number between 96.4 and 97.5

Explanation: Given,

Number of bytes in the information frame = 1980
Number of bytes in the acknowledge frame = 20
Number of overhead bytes in the information frame = 20

Therefore, useful Data = Total Data – Overhead = 1980 – 20 = 1960

Tt(Data) = (1960*8) / 10^6 = 15.68 millisecond

Tt(ACK) = 20*8 / 10^6 = 0.16 millisecond

Two way round trip time = 2 * Tp = 2*0.75 = 1.5 millisecond

T(precess) = 0.25(for info) + 0.25(for ack) = 0.5 millisecond

Efficiency = Useful Time / (Tt(info) + + Tt(ACK) + 2* Tp + Tprocess )
= 15.68 / (15.84 + 0.16 + 2*0.75+ 0.5 )
= 0.8711111
= 87%

Option (A) is correct.

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