ECE/CS 356: Computer Network Architecture
Homework 5
Problem 1 (25 points; assign the proper grade according to the
correctness of the answer)
Under TCP Reno implementation, the sender has sent three segments to the receiver.
However, the first ACK is lost. Please, describe the rest of the communication process between
the sender and the receiver using the figure above. Plot all segments sent between the sender
and the receiver until the transmission of the three segments is completed.
Note:
1. Indicate the sequence number of each segment and the sequence number of the ACK
references. The sequence number can be calculated using TCP or Go-Back-N method.
The initial sequence number of your choice. Each segment has the same length 100
bytes. Not necessary to consider the header size in this exercise
2. Each mark in the timeline of the sender/receiver indicates one unit of time
3. The Timeout is fixed to 5 units of time (no consider Timeout updates)
4. The Round trip time is fixed to 2 units of time
5. No congestion window or receiver window considered (infinite buffers)
Sender
Receive
Problem 2 (25 points; assign the proper grade according to the
correctness of the answer)
Under TCP Reno implementation, the sender has sent 4 segments to the receiver. However,
the second segment is lost. Please, describe the rest of the communication process between
the sender and the receiver using the figure above. Plot all segments sent between the sender
and the receiver until the transmission of the 4 segments is completed.
Note:
1. Indicate the sequence number of each segment and the sequence number of the ACK
references. The sequence number can be calculated using TCP or Go-Back-N method.
The initial sequence number of your choice. Each segment has the same length 100
bytes. Not necessary to consider the header size in this exercise
2. Each mark in the timeline of the sender/receiver indicates one unit of time
3. The Timeout is fixed to 3 units of time (no consider Timeout updates)
4. The Round trip time is fixed to 2 units of time
5. No congestion control or flow control considered (infinite buffers)
Sender
Receiver
Problem 3 (25 points; assign the proper grade according to the
correctness of the answer)
Under TCP Reno implementation, the sender has sent 5 segments to the receiver. However,
the second segment is lost. Please, describe the rest of the communication process between
the sender and the receiver using the figure above. Plot all segments sent between the sender
and the receiver until the transmission of the 5 segments is completed.
Note:
1. Indicate the sequence number of each segment and the sequence number of the ACK
references. The sequence number can be calculated using TCP or Go-Back-N method.
The initial sequence number of your choice. Each segment has the same length 100
bytes. Not necessary to consider the header size in this exercise
2. Each mark in the timeline of the sender/receiver indicates one unit of time
3. The Timeout is limited 10 units of time
4. The Round trip time is fixed to 2 units of time (no consider Timeout updates)
5. No congestion window or receiver window considered (infinite buffers)
Sender
Receiver
Problem 4 (25 points; assign the proper grade according to the
correctness of the answer)
Under TCP Reno implementation, the sender has sent 10 segments to the receiver. However,
the second segment is lost. Please, describe the rest of the communication process between
the sender and the receiver using the figure above. Plot all segments sent between the sender
and the receiver until the transmission of the 10 segments is completed. Now, consider the
congestion control activated. The initial congestion window is 1 and the slow start threshold is
4 segments. The second segment sent will be lost.
Note:
1. Indicate the sequence number of each segment and the sequence number of the ACK
references. The sequence number can be calculated using TCP or Go-Back-N method.
The initial sequence number of your choice. Each segment has the same length 100
bytes. Not necessary to consider the header size in this exercise
2. Each mark in the timeline of the sender/receiver indicates one unit of time
3. The Timeout is fixed to 5 units of time (no consider Timeout updates)
4. The Round trip time is fixed to 2 units of time
5. No flow control considered (infinite receiver buffer)
Sender
Receive