GATE CS 2017 Set 1 — Question 45
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Computer Networks → Data Link Layer → Efficiency Numericals
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Question
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).
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).
Correct answer
86.5 to 89.5
Solution
Given parameters:
Payload size = bytes.Step 2: Calculate Transmission Times
The total time for one Stop-and-Wait cycle includes transmission of the frame, propagation to receiver, processing, transmission of ACK, and propagation of ACK back to sender.Step 4: Calculate Efficiency
Efficiency () is the ratio of time spent transmitting useful data to the total cycle time.Rounding to two decimal places, we get 88.34%.(Note: If efficiency is calculated based on the total frame size including overhead, . Both values fall within the accepted range of 86.5 to 89.5.)
- Bandwidth () = 1 Mbps = bits/sec
- Propagation delay () = 0.75 ms
- Processing time () = 0.25 ms
- Information frame size () = 1980 bytes
- Acknowledge frame size () = 20 bytes
- Overhead bytes () = 20 bytes
Payload size = bytes.Step 2: Calculate Transmission Times
- Transmission time for Information frame ():
- Transmission time for Acknowledge frame ():
- Transmission time for Payload ():
The total time for one Stop-and-Wait cycle includes transmission of the frame, propagation to receiver, processing, transmission of ACK, and propagation of ACK back to sender.Step 4: Calculate Efficiency
Efficiency () is the ratio of time spent transmitting useful data to the total cycle time.Rounding to two decimal places, we get 88.34%.(Note: If efficiency is calculated based on the total frame size including overhead, . Both values fall within the accepted range of 86.5 to 89.5.)
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