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A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 jasec/ km and trunk data rate is 1.544 Mbps. We ignore the time taken to receive the bits in the acknowledgment. Frame size is 64 Bytes.
If Go-Back-N protocol is used, in order to achieve an efficiency of 100%, what is the maximum window size at the sender's side?
- a)32
- b)63
- c)110
- d)219
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A 3000 km long trunk is used to transmit frames using a Go-Back-N prot...
For sliding window protocol with Go-Back-N
Data rate = 1.544 Mbps
Frame size = 64 bytes - 64 x 8 bits
= 512 bits
Propagation speed = 6x10-6 s/km Distance = 3000 km
Utilization = 100%
Data rate = 1.544 Mbps
Frame size = 64 bytes - 64 x 8 bits
= 512 bits
Propagation speed = 6x10-6 s/km Distance = 3000 km
Utilization = 100%
Most Upvoted Answer
A 3000 km long trunk is used to transmit frames using a Go-Back-N prot...
Propagation Time:
Propagation time can be calculated using the formula:
Propagation time = Distance / Propagation Speed
Given that the distance is 3000 km and the propagation speed is 6 jasec/km, we can calculate the propagation time as follows:
Propagation time = 3000 km / (6 jasec/km) = 500 jasec
Transmission Time:
Transmission time can be calculated using the formula:
Transmission time = Frame size / Trunk data rate
Given that the frame size is 64 Bytes and the trunk data rate is 1.544 Mbps, we need to convert the frame size to bits and the data rate to bps:
Frame size = 64 Bytes * 8 bits/Byte = 512 bits
Trunk data rate = 1.544 Mbps * 10^6 bps/Mbps = 1.544 * 10^6 bps
Now we can calculate the transmission time:
Transmission time = 512 bits / (1.544 * 10^6 bps) = 0.000331 s
Efficiency:
Efficiency can be calculated using the formula:
Efficiency = (Window Size * Frame Size) / (Propagation Time + Transmission Time)
We need to find the maximum window size that achieves an efficiency of 100%, so we can set the efficiency to 1:
1 = (Window Size * 512 bits) / (500 jasec + 0.000331 s)
Rearranging the equation:
Window Size = (500 jasec + 0.000331 s) / 512 bits
Calculating Window Size:
Substituting the values:
Window Size = (500 jasec + 0.000331 s) / 512 bits
We need to convert the propagation time to seconds:
Propagation Time = 500 jasec * (1 sec / 10^12 jasec)
Window Size = (500 * 10^12 sec + 0.000331 s) / 512 bits
Simplifying the equation:
Window Size = (500 * 10^12 sec + 0.000331 s) / 512 bits
Window Size = (5 * 10^14 + 0.000331 * 10^12) / 512 bits
Window Size = (5 * 10^14 + 331 * 10^6) / 512 bits
Window Size = (5 * 10^14 + 331 * 10^6) / 512 bits
Window Size ≈ 9.765 * 10^11 bits / 512 bits
Window Size ≈ 1.909 * 10^9
Rounding the Window Size:
Since the window size cannot be a fraction, we round down the window size to the nearest whole number:
Window Size ≈ 1,909,000,000
The maximum window size at the sender's side is approximately 1,909,000,000 bits.
Propagation time can be calculated using the formula:
Propagation time = Distance / Propagation Speed
Given that the distance is 3000 km and the propagation speed is 6 jasec/km, we can calculate the propagation time as follows:
Propagation time = 3000 km / (6 jasec/km) = 500 jasec
Transmission Time:
Transmission time can be calculated using the formula:
Transmission time = Frame size / Trunk data rate
Given that the frame size is 64 Bytes and the trunk data rate is 1.544 Mbps, we need to convert the frame size to bits and the data rate to bps:
Frame size = 64 Bytes * 8 bits/Byte = 512 bits
Trunk data rate = 1.544 Mbps * 10^6 bps/Mbps = 1.544 * 10^6 bps
Now we can calculate the transmission time:
Transmission time = 512 bits / (1.544 * 10^6 bps) = 0.000331 s
Efficiency:
Efficiency can be calculated using the formula:
Efficiency = (Window Size * Frame Size) / (Propagation Time + Transmission Time)
We need to find the maximum window size that achieves an efficiency of 100%, so we can set the efficiency to 1:
1 = (Window Size * 512 bits) / (500 jasec + 0.000331 s)
Rearranging the equation:
Window Size = (500 jasec + 0.000331 s) / 512 bits
Calculating Window Size:
Substituting the values:
Window Size = (500 jasec + 0.000331 s) / 512 bits
We need to convert the propagation time to seconds:
Propagation Time = 500 jasec * (1 sec / 10^12 jasec)
Window Size = (500 * 10^12 sec + 0.000331 s) / 512 bits
Simplifying the equation:
Window Size = (500 * 10^12 sec + 0.000331 s) / 512 bits
Window Size = (5 * 10^14 + 0.000331 * 10^12) / 512 bits
Window Size = (5 * 10^14 + 331 * 10^6) / 512 bits
Window Size = (5 * 10^14 + 331 * 10^6) / 512 bits
Window Size ≈ 9.765 * 10^11 bits / 512 bits
Window Size ≈ 1.909 * 10^9
Rounding the Window Size:
Since the window size cannot be a fraction, we round down the window size to the nearest whole number:
Window Size ≈ 1,909,000,000
The maximum window size at the sender's side is approximately 1,909,000,000 bits.
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A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 jasec/ km and trunk data rate is 1.544 Mbps. We ignore the time taken to receive the bits in the acknowledgment. Frame size is 64 Bytes.If Go-Back-N protocol is used, in order to achieve an efficiency of 100%, what is the maximum window size at the sender's side?a)32b)63c)110d)219Correct answer is option 'C'. Can you explain this answer?
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