Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 1. The bandwidth is 100 Mbps, and data packets can be sent continuously.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 1. The bandwidth is 100 Mbps, and data packets can be sent continuously. 5 MB = 41,943,040 bits. 2 initial RTTs (160 ms) + 41,943,040/100,000,000 bps (transmit) + RTT/2 (propagation) = 0.160s (2RRT) + 0.4194304s (Data Transmission) + 0.040s (RRT/2 propogation) ≈ 0.6194304 seconds.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 2. The bandwidth is 100 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 2. The bandwidth is 100 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next. Number of packets required = 5 MB/1KB = 5,242,880/1024 = 5,120. To the above we add the time for 5,120 RTTs (the number of RTTs between when packet 1 arrives and packet 5,120 arrives), for a total of 0.6194304 + 819.2 = 819.8194304 seconds.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 3. The link allows infinitely fast transmit, but limits bandwidth such that only 40 packets can be sent per RTT.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 3. The link allows infinitely fast transmit, but limits bandwidth such that only 40 packets can be sent per RTT. Dividing the 5,120 packets by 40 gives 128. This will take 127.5 RTTs (half an RTT for the first batch to arrive, plus 127 RTTs between the first batch and the 128th batch), plus the initial 2 RTTs. 127.5 * .080 + .160 = 10.36 seconds

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 4. The link is 100Mbps, but limits bandwidth such that only 40 packets can be sent per RTT.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 4. The link is 100Mbps, but limits bandwidth such that only 40 packets can be sent per RTT. We simply add the transmit time of 5MB at 100Mbps (calculated above) to the andswer we got in #3. Thus 10.36s + 0.4194304s = 10.77s.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 5. Zero transmit time as in previous, but during the first RTT we can send one packet, during the second RTT we can send two packets, during the third we can send four (23−1 ), etc.

Performance Questions Calculate the total time required to transfer a 5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: 5. Zero transmit time as in previous, but during the first RTT we can send one packet, during the second RTT we can send two packets, during the third we can send four (23−1 ), etc. Right after the handshaking is done we send one packet. One RTT after the handshaking we send two packets. At n RTTs past the initial handshaking we have sent 1 + 2 + 4 + · · · + 2n = 2n +1 − 1 packets. At n = 12 we have thus been able to send all 5,120 packets; the last batch arrives 0.5 RTT later. Total time is 2 + 12.5 RTTs = 14.5x0.08 = 1.16 seconds!

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 1. Calculate the minimum RTT for the link.

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 1. Calculate the minimum RTT for the link. The minimum RTT is 2 × 385, 000, 000m / 3×108m/s = 2.57 seconds.

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 2. Using the RTT as the delay, calculate the delay × bandwidth product for the link.3.

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 2. Using the RTT as the delay, calculate the delay × bandwidth product for the link.3. The delay×bandwidth product is 2.57 s×1Gbps = 2.57Gb = 321 MB.

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 3. Assuming 1500 byte packets, how many packets have to be sent without acknowlegement in order to use the link with 100% efficiency? What is the significance of the delay × bandwidth product computed in (2)?

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 3. Assuming 1500 byte packets, how many packets have to be sent without acknowledgement in order to use the link with 100% efficiency? What is the significance of the delay × bandwidth product computed in (2)? This represents the amount of data the sender can send before it would be possible to receive a response. 321MB/1500B = 214,000 packets!

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 4. A camera on the lunar base takes pictures of the Earth and saves them in digital format to disk. Suppose Mission Control on Earth wishes to download the most current image, which is 25 MB. What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished?

Suppose a 1-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light—3×108 m/s. 4. A camera on the lunar base takes pictures of the Earth and saves them in digital format to disk. Suppose Mission Control on Earth wishes to download the most current image, which is 25 MB. What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished? We require at least one RTT from sending the request before the first bit of the picture could begin arriving at the ground (TCP would take longer). 25 MB is 200Mb. Assuming bandwidth delay only, it would then take 200Mb/1000Mbps = 0.2 seconds to finish sending, for a total time of 0.2 + 2.57 = 2.77 sec until the last picture bit arrives on earth