1. Assume the following parameters for : (7 points) a) Round Trip Time (RTT) = 10 ms b) A segment is 1500 bytes. Suppose that Bob sends 4 segments to Alice using rdt 2.1. Suppose that the segment reaches Alice with errors when it is sent the first time, but reaches Alice correctly when it is sent the second time. All other segments reach Alice correctly in their first attempts. What is the average throughput obtained in sending all the segments (in bytes/sec)? Draw a diagram to show your work. 2. Resolve Problem 1 to find the average throughput (in bytes/sec) obtained in sending the 4 segments when Bob sends the segments using . Assume again that the RTT is 10 ms and that a segment is 1500 bytes. Draw a diagram to show your work.  (7 points) 3. Assume the following parameters for :  (11 points) a) For Go-Back-N, assume N = 3 b) Round Trip Time (RTT) = 10 ms c) A segment is 1500 bytes.

1. To calculate the average throughput obtained in sending all the segments using rdt 2.1, we need to consider the number of segments sent, the time taken for transmission, and the overhead due to error retransmissions.

Given that the RTT is 10 ms and a segment is 1500 bytes, we can first calculate the time taken for transmission.

Since Bob sends 4 segments to Alice, and one RTT is required for each segment, the total time taken for transmission is (4 * 10 ms) = 40 ms.

Now, we need to account for the error retransmission. One segment reached Alice with errors and had to be sent again. This means that a total of 5 segments were sent. Since each segment is 1500 bytes, the total number of bytes sent is (5 * 1500) = 7500 bytes.

To calculate the average throughput, we divide the total number of bytes sent by the time taken for transmission:

Average throughput = Total bytes sent / Time taken
= 7500 bytes / 40 ms
= 187.5 bytes/ms

Since the question asks for the throughput in bytes/sec, we need to convert milliseconds to seconds:

Average throughput = 187.5 * (1/1000) bytes/ms * 1000 ms/s
= 187.5 bytes/s

Therefore, the average throughput obtained in sending all the segments using rdt 2.1 is 187.5 bytes/sec.

To illustrate this calculation, we can draw a diagram as follows:

—-
| | Segment 1 | Segment 2 | Segment 3 | Segment 4 | Segment 5 | (Retransmission of Segment 1 due to errors)
| |———–|———–|———–|———–|———–|
| |_______________________|_______________________|
RTT = 10 ms
—

2. To resolve Problem 1 for the average throughput obtained in sending the 4 segments using, we follow a similar approach as in Problem 1 but consider the protocol.

Since the protocol is not mentioned, we would assume it as rdt (stop-and-wait). In stop-and-wait, only one segment is sent at a time, and the sender waits for an acknowledgment before sending the next segment.

Given that the RTT is still 10 ms and a segment is 1500 bytes, we can calculate the time taken for transmission. Since Bob sends 4 segments to Alice, and each segment requires one RTT, the total time taken for transmission is (4 * 10 ms) = 40 ms.

Since there are no error retransmissions mentioned, the total number of bytes sent is (4 * 1500) = 6000 bytes.

To calculate the average throughput, we divide the total number of bytes sent by the time taken for transmission:

Average throughput = Total bytes sent / Time taken
= 6000 bytes / 40 ms
= 150 bytes/ms

Converting milliseconds to seconds:

Average throughput = 150 * (1/1000) bytes/ms * 1000 ms/s
= 150 bytes/s

Therefore, the average throughput obtained in sending the 4 segments using is 150 bytes/sec.

A diagram for this calculation can be drawn as follows:

—-
| | Segment 1 | Segment 2 | Segment 3 | Segment 4 |
| |———–|———–|———–|———–|
| |_______________________|_______________________|
RTT = 10 ms
—