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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.
What is the maximum number of bits of the sequence number?
- a)5
- b)6
- c)7
- d)8
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...
Bits for sequence no. = 2no of bits > Window size
=> W = 110
27 = 128 > 110
Hence 7 bits used for sequence number.
=> W = 110
27 = 128 > 110
Hence 7 bits used for sequence number.
Most Upvoted Answer
A 3000 km long trunk is used to transmit frames using a Go-Back-N prot...
Propagation Delay:
The propagation speed is given as 6 jasec/km, which means it takes 1/6 jasec to propagate 1 km. Therefore, the total propagation delay for a 3000 km long trunk is:
Propagation delay = (3000 km) * (1/6 jasec/km) = 500 jasec
Transmission Delay:
The trunk data rate is given as 1.544 Mbps, which means it can transmit 1.544 million bits per second. Since the frame size is 64 Bytes, the transmission delay for each frame can be calculated as:
Transmission delay = (64 Bytes) * (8 bits/Byte) / (1.544 Mbps) = 328.36 µsec
Round Trip Time (RTT):
The round trip time (RTT) is the total time it takes for a frame to be sent and the acknowledgment to be received. It includes the propagation delay and the transmission delay.
RTT = Propagation delay + Transmission delay = 500 jasec + 328.36 µsec
Maximum Sequence Number:
In a Go-Back-N protocol, the sender can have a maximum window size equal to 2^n - 1, where n is the number of bits in the sequence number.
The sender's window size determines the number of unacknowledged frames that can be sent before waiting for acknowledgments. It should be less than or equal to the maximum sequence number to avoid ambiguities.
To find the maximum sequence number, we need to calculate the maximum number of frames that can be in transit at any given time.
Maximum number of frames in transit = RTT / Transmission delay
Since the RTT is given in jasec and the transmission delay is given in µsec, we need to convert the RTT to µsec.
RTT (µsec) = RTT (jasec) * 10^6
Substituting the values, we get:
RTT (µsec) = 500 jasec * 10^6 = 500 * 10^6 µsec
Now we can calculate the maximum number of frames in transit:
Maximum number of frames in transit = RTT (µsec) / Transmission delay
Substituting the values, we get:
Maximum number of frames in transit = (500 * 10^6 µsec) / 328.36 µsec
Finally, we can calculate the maximum sequence number:
Maximum sequence number = 2^n - 1
Substituting the values, we get:
2^n - 1 = (500 * 10^6 µsec) / 328.36 µsec
Solving for n, we find n = 7
Therefore, the maximum number of bits in the sequence number is 7, which corresponds to option C.
The propagation speed is given as 6 jasec/km, which means it takes 1/6 jasec to propagate 1 km. Therefore, the total propagation delay for a 3000 km long trunk is:
Propagation delay = (3000 km) * (1/6 jasec/km) = 500 jasec
Transmission Delay:
The trunk data rate is given as 1.544 Mbps, which means it can transmit 1.544 million bits per second. Since the frame size is 64 Bytes, the transmission delay for each frame can be calculated as:
Transmission delay = (64 Bytes) * (8 bits/Byte) / (1.544 Mbps) = 328.36 µsec
Round Trip Time (RTT):
The round trip time (RTT) is the total time it takes for a frame to be sent and the acknowledgment to be received. It includes the propagation delay and the transmission delay.
RTT = Propagation delay + Transmission delay = 500 jasec + 328.36 µsec
Maximum Sequence Number:
In a Go-Back-N protocol, the sender can have a maximum window size equal to 2^n - 1, where n is the number of bits in the sequence number.
The sender's window size determines the number of unacknowledged frames that can be sent before waiting for acknowledgments. It should be less than or equal to the maximum sequence number to avoid ambiguities.
To find the maximum sequence number, we need to calculate the maximum number of frames that can be in transit at any given time.
Maximum number of frames in transit = RTT / Transmission delay
Since the RTT is given in jasec and the transmission delay is given in µsec, we need to convert the RTT to µsec.
RTT (µsec) = RTT (jasec) * 10^6
Substituting the values, we get:
RTT (µsec) = 500 jasec * 10^6 = 500 * 10^6 µsec
Now we can calculate the maximum number of frames in transit:
Maximum number of frames in transit = RTT (µsec) / Transmission delay
Substituting the values, we get:
Maximum number of frames in transit = (500 * 10^6 µsec) / 328.36 µsec
Finally, we can calculate the maximum sequence number:
Maximum sequence number = 2^n - 1
Substituting the values, we get:
2^n - 1 = (500 * 10^6 µsec) / 328.36 µsec
Solving for n, we find n = 7
Therefore, the maximum number of bits in the sequence number is 7, which corresponds to option C.
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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.What is the maximum number of bits of the sequence number?a)5b)6c)7d)8Correct answer is option 'C'. Can you explain this answer?
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