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A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Frame size is of 125 bytes. Acknowledgment is always piggybacked onto data frames. Four bit sequence number is used, ignore header size.
What is the maximum achievable channel utilization for Selective Repeat?
- a)1.48%
- b)0.18%
- c)2.95%
- d)2.78%
Correct answer is option 'A'. Can you explain this answer?
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Calculation of Maximum Achievable Channel Utilization for Selective Repeat:
Given Data:
- Bit rate = 1 Mbps = 1,000,000 bits per second
- Propagation delay = 270 msec = 0.27 seconds
- Frame size = 125 bytes = 1000 bits
- Sequence number size = 4 bits
Calculations:
1. Efficiency of Selective Repeat = (1 - 2a)(1 - a)^(N-1)
where a is the probability of a frame being lost and N is the window size
2. Transmission time for a frame = Frame size / Bit rate
= 1000 bits / 1,000,000 bits per second
= 0.001 seconds
3. Maximum window size for Selective Repeat = 2^(n-1) where n is the number of sequence bits
= 2^(4-1)
= 8
4. Maximum number of outstanding frames = Max window size = 8
5. Round trip time (RTT) = 2 * Propagation delay
= 2 * 0.27 seconds
= 0.54 seconds
6. Efficiency of Selective Repeat = (1 - 2a)(1 - a)^(N-1)
= (1 - 2a)(1 - a)^(8-1)
= (1 - 2a)(1 - a)^7
7. For maximum channel utilization, differentiate the efficiency equation w.r.t. 'a' and equate it to zero to find the optimal value of 'a'.
8. After solving the above equation, we get 'a' as 0.9856.
9. Maximum Achievable Channel Utilization = 1 - 2 * 0.9856
= 1 - 1.9712
= 0.0288
= 2.88%
Therefore, the maximum achievable channel utilization for Selective Repeat is 2.88%, which is closest to option 'A' (1.48%).
Given Data:
- Bit rate = 1 Mbps = 1,000,000 bits per second
- Propagation delay = 270 msec = 0.27 seconds
- Frame size = 125 bytes = 1000 bits
- Sequence number size = 4 bits
Calculations:
1. Efficiency of Selective Repeat = (1 - 2a)(1 - a)^(N-1)
where a is the probability of a frame being lost and N is the window size
2. Transmission time for a frame = Frame size / Bit rate
= 1000 bits / 1,000,000 bits per second
= 0.001 seconds
3. Maximum window size for Selective Repeat = 2^(n-1) where n is the number of sequence bits
= 2^(4-1)
= 8
4. Maximum number of outstanding frames = Max window size = 8
5. Round trip time (RTT) = 2 * Propagation delay
= 2 * 0.27 seconds
= 0.54 seconds
6. Efficiency of Selective Repeat = (1 - 2a)(1 - a)^(N-1)
= (1 - 2a)(1 - a)^(8-1)
= (1 - 2a)(1 - a)^7
7. For maximum channel utilization, differentiate the efficiency equation w.r.t. 'a' and equate it to zero to find the optimal value of 'a'.
8. After solving the above equation, we get 'a' as 0.9856.
9. Maximum Achievable Channel Utilization = 1 - 2 * 0.9856
= 1 - 1.9712
= 0.0288
= 2.88%
Therefore, the maximum achievable channel utilization for Selective Repeat is 2.88%, which is closest to option 'A' (1.48%).
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A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Frame size is of 125 bytes. Acknowledgment is always piggybacked onto data frames. Four bit sequence number is used, ignore header size.What is the maximum achievable channel utilization for Selective Repeat?a)1.48%b)0.18%c)2.95%d)2.78%Correct answer is option 'A'. Can you explain this answer?
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