Question 3026 – Data-Interpretation
May 1, 2024Question 4920 – UGC NET CS 2014 June-paper-2
May 2, 2024Stop-and-Wait-ARQ
| Question 6 |
The value 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).
| 89.33% | |
| 89.34% | |
| 89.35% | |
| 89.36% |
Question 6 Explanation:
Given Data:
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
Tp=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
Tt = L/B = 2000Bytes/1Mbps = (2000×8bits)/(106 b/s) = 16msec
TAck = LAck/B = (20×8bits)/(106 bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = Tt/(Tt+TAck+2Tp+TProc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
Tp=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
Tt = L/B = 2000Bytes/1Mbps = (2000×8bits)/(106 b/s) = 16msec
TAck = LAck/B = (20×8bits)/(106 bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = Tt/(Tt+TAck+2Tp+TProc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933
Correct Answer: A
Question 6 Explanation:
Given Data:
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
Tp=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
Tt = L/B = 2000Bytes/1Mbps = (2000×8bits)/(106 b/s) = 16msec
TAck = LAck/B = (20×8bits)/(106 bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = Tt/(Tt+TAck+2Tp+TProc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
Tp=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
Tt = L/B = 2000Bytes/1Mbps = (2000×8bits)/(106 b/s) = 16msec
TAck = LAck/B = (20×8bits)/(106 bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = Tt/(Tt+TAck+2Tp+TProc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933
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