Consider A Source S And A Destination D Using The Selective
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Consider A Source S And A Destination D Using The Selective Repeat Pro
Consider a source S and a destination D using the Selective Repeat Protocol (SRP) with a window size of N=4. Each packet is bits. The delays are as follows: Each packet transmitted by S has a transmission time of 10 ms. The end-to-end delay from the time that S starts transmitting a packet until D completely receives that packet is 100 ms. Each ack transmitted by D has a transmission time of 1 ms. The end-to-end delay from the time that D starts transmitting an ack until S completely receives that ack is 100 ms. The timeout value is 300 ms. Here is what happens to each packet: Packet #0 is received by D, and the corresponding ACK is received by S. The first transmission of packet #1 is dropped by a router in between S and D. The second transmission of packet #1 is received by D, and the corresponding ACK is received by S. The first transmission of packet #2 is dropped by a router in between S and D. The second transmission of packet #2 is received by D, and the corresponding ACK is received by S. Packet #3 is received by D, and the corresponding ACK is received by S. Packet #4 is received by D, and the corresponding ACK is received by S. Packet #5 is received by D, and the corresponding ACK is received by S.
Paper For Above instruction
The scenario described involves the use of the Selective Repeat Protocol (SRP) to manage reliable data transmission between a source (S) and a destination (D). SRP allows multiple packets to be sent without acknowledgment of each immediately, leveraging a window size of N=4, which enables the sender to transmit up to four packets before requiring an acknowledgment. This mechanism, combined with timeout and retransmission strategies, ensures efficient and reliable data transfer, even amidst packet drops and delays.
Part (a): At time t = 150 ms, what is S's window?
The window at a specific time in SRP depends on the sequence numbers of the packets that have been sent but not yet acknowledged. Initial conditions: at t = 0 ms, S begins transmitting packets #0, #1, #2, and #3, as the window size is 4. Packet #0 is acknowledged quickly, as the first response arrives early, establishing a baseline for understanding the ongoing window. Packet #1's first transmission is dropped, but the second is received successfully at D, and ACKs are sent back along with delays. Packet #2 is also initially dropped but eventually received successfully during retransmission. Packets #3, #4, and #5 are received and acknowledged sequentially.
By t = 150 ms, considering the delays: the initial window of packets #0 to #3 has been transmitted. ACKs for #0, #1, #2, and #3 are in transit or received, considering the end-to-end delays. Since ACKs for packets #0 through #3 have been received or are in transit, the window has advanced to include packets #4 and #5, with #0 acknowledged and #4, #5 being the next to be sent or awaiting acknowledgment if within window limits. Given these facts, S's window comprises sequence numbers #4, #5, and any unacknowledged packets within the window. Furthermore, the window size remains limited to N=4, so the range of unacknowledged packets is from the smallest unacknowledged sequence number to smallest unacknowledged + 3.
Part (b): When does the retransmission of packet #1 start?
Packet #1 is first dropped during its initial transmission, which occurs at t = 0 ms. Given the transmission time of 10 ms, packet #1 is sent at t = 0 ms, but the first transmission is dropped. The second transmission of packet #1 occurs after the timeout period, which is 300 ms. Since the acknowledgment for packet #1 is not received within this period (due to the initial drop), the sender initiates a retransmission of packet #1 when the timeout expires, i.e., at t = 0 ms + 300 ms = 300 ms. This retransmission starts precisely at t = 300 ms, conforming to the timeout setting designated in the protocol configuration.
Part (c): When does the application layer at D receive packet #2?
Packet #2's initial transmission was dropped, but it was eventually received during retransmission. Considering the delays: each packet takes 100 ms end-to-end to reach from S to D, and ACKs take 100 ms to travel back from D to S, with each ACK transmission time being 1 ms. The initial transmission of packet #2 occurs at t ≈ 20 ms (assuming immediate sending after the preceding packets), but it is dropped, so D does not receive it. The second transmission of packet #2 occurs at approximately t = 20 ms + transmission time (10 ms) = 30 ms, but since delay is 100 ms end-to-end, D receives packet #2 at t ≈ 30 ms + 100 ms = 130 ms. By this time, the ACK for packet #2, which D sends after receipt, will be sent back and received by S after the delay of 100 ms plus 1 ms for ACK transmission, totaling about t ≈ 130 ms + 100 ms + 1 ms ≈ 231 ms.
It is important to note, however, that the application layer at D receives packet data once the entire packet is received, which is after the initial end-to-end delay of 100 ms following the actual transmission. Therefore, the application layer at D receives packet #2 at approximately t = 130 ms, when the complete packet is received.
Part (d): What is the throughput during the time from t=0 ms until t=500 ms?
The throughput is calculated as the total data successfully transmitted divided by the elapsed time. Assuming each packet is a fixed size of bits, and all packets #0 through #5 are eventually delivered successfully, the total data transmitted includes six packets. Since each packet is transmitted successfully, including retransmissions where necessary, the total data is 6 times the size of one packet.
Calculations involve summing the time taken for all successful transmissions, including retransmissions, delays, and acknowledgment cycles. From the given information, packets #0, #3, #4, and #5 are successfully received on time, with packet #1 and #2 being retransmitted due to initial drops. The total successful transmission time for these packets accounts for initial transmissions, retransmissions, delays, and acknowledgment travel time. Given the delay periods, the entire set of six packets is delivered within the 500 ms window.
Thus, total data transmitted is approximately 6 packets' worth of data. The throughput is estimated as total data divided by 500 ms (0.5 seconds). Since the exact packet size isn't specified in bits, assuming a standard packet size (e.g., 1000 bits), the throughput is roughly (6 * 1000 bits) / 0.5 seconds = 12,000 bits/sec. This reflects a high utilization of the connection, showcasing the efficiency of SRP despite retransmissions and delays.
Conclusion
The analysis demonstrates the effective operation of the Selective Repeat Protocol under specified network conditions, including packet drops, delays, and retransmissions. Understanding these interactions provides insights into optimal timeout settings, window sizes, and the importance of acknowledgment timing. The protocol's ability to selectively retransmit only lost or corrupted packets ensures high throughput even in adverse network environments, underscoring its importance in reliable data communication systems.
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