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The values of parameters for the Stop-and-Wait ARQ protocol are as given below :
Bit rate of the transmission channel = 1 Mbps.
Propagation delay from sender to receiver = 0.75 ms.
Time to process a frame = 0.25 ms.
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 places).
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- 66.33%
- 89.33%
- 39.33%
- 59.33%
Correct Option: B
Given data: B = 1 Mbps , Tprocess = 0.25 ms, TPt = 0.75 ms LIF = 1980 bytes , L(Overhead) = 20 bytes , LAF = 20 bytes
Efficiency (η) = ?
As we know that,
| Transmission time of data = | ||
| Bandwidth |
| So , TX = | = | = | |||
| B | B | 1 × 106( bits / sec ) |
(∵ 1 byte = 8 bit)
| = | ||
| 106( bits / sec ) |
TX = 16 m sec
Propagation time = 0.75 m sec.
| Transmission time of TACK = | = | ||
| B | 106 |
∴ TACK = 0.16 m sec.
Then transmission efficiency of stop and wait ARQ.
| ⇒ | = | ||
| (TX + TACK + 2TPt + TProcess) | (16 0.16 + 2 × 0.75 + 0.25) |
| = | = 0.8933 = 89.33% | |
| 17.91 |
Hence answer is 89.33%
