CNT-4007 Homework 1

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Florida Atlantic University *

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4104

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Electrical Engineering

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Feb 20, 2024

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Homework 1 Problem 4 (a) The bandwidth is 1.5 Mbps, and data packets can be sent continuously. Packet size = 1 KB = 1 * 1,024 * 8 bits = 8,192 bits 1.5 Mbps = 1.5 * 1,000 * 1,000 = 1,500,000 bits/s RTT = 100 ms = 0.1 s Transmit time per packet = Size of packet/Bandwith = = 8,192 bits/1,500,000 bits/s = 0.00546 s = 5.46 ms # of packets to be sent = Size of file / Size of packet 0 = 1,000 KB / 1 KB = 1,000 Transmit time for all packets = Transmit time per packet*# of packets = = 5.46 ms * 1000 = 5.46 s Total time = Initial 2 RTT + Transmit time for all packets + Propagation = 2 * RTT + Transmit time for all packets + RTT/2 = = 2 * 0.1 s + 5.46 s + 0.1 s/2= 0.2 s + 5.46 s + 0.05 s = 5.71 s (b) The bandwidth is 1.5 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next. We are sending 1000 packets, so we must wait for 999 RTTs, so we must add to the value found in a) the time for 999 RTTs: Total Time = 5.71 s + 999*RTT = = 5.71 s + 999 * 0.1 s = 5.71 s + 99.9 s = 105.61 s (c) We need to send 1000 packets, then we will need 1000 / 20 = 50 RTTs to transmit all the data, but note that for the last 20 packet only need (RTT/2), so the total RTTs required are 49.5. Total time = Initial 2 RTT + Required RTTs = = 2*RTT + 49.5 * RTT =

= 51.5 * RTT = 51.5 * 0.1 s = 5.15 s (d) The bandwidth is infinite, and during the first RTT we can send one packet (2^1-1), during the second RTT we can send two packets (2^2-1), during the third we can send four (2^3-1), and so on. Right after the handshaking of 200 ms (2 RTTs) we send one packet. One RTT after we send two packets, the next RTT we send four packets, etc. At n > RTTs past the initial handshaking we have sent: 1 + 2 + 4 + ... + 2^n = 2^(n+1)-1 packets. At n = 9 we have sent all 1000 packets. The last batch arrives 0.5 later. Then: Total time = Initial 2 RTT + 9 * RTT + 0.5 * RTT = = 2 * RTT + 9.5 * RTT = 11.5 * RTT = 11.5 * 0.1 s = 1.15 s Problem 14 (a) Calculate the minimum RTT for the link. The minimum RTT for the link is the time it takes for a bit to travel from the Earth to the Moon and back, which is: RTT = 2 * distance / speed_of_light RTT = 2 * 385,000 km * 1000 m/km / 3 x 10^8 m/s RTT = 2.567 seconds (b) Using the RTT as the delay, calculate the delay × bandwidth product for the link. The delay × bandwidth product is a measure of how much data can be transmitted over a link in one RTT. It is calculated as follows: delay × bandwidth product = RTT * bandwidth delay × bandwidth product = 2.567 seconds * 100 Mbps delay × bandwidth product = 256.7 Megabits (c) What is the significance of the delay × bandwidth product computed in (b)? The delay × bandwidth product is a measure of the link's capacity to transfer data. It is also a measure of the amount of data that can be transmitted over the link without causing congestion.

(d) 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? The minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished is: total_time = RTT + transfer_time transfer_time = file_size / bandwidth total_time = RTT + file_size / bandwidth total_time = 2.567 seconds + 25 MB 1024 kB/MB 1024 bytes/kB 8 bits/byte / 100 Mbps 1000 kb/Mb * 1000 bits/kb Therefore, the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished is 4.663 seconds. Problem 17 (a) To calculate the latency for a single store-and-forward switch in the path To consider the propagation delay and the time it takes for the switch to transmit the packet. Propagation delay: 10 μs Packet size: 5,000 bits Ethernet speed: 10 Mbps The time it takes to transmit the packet can be calculated using the formula: Transmission Time = Packet size/Ethernet speed Transmission Time = 5,000bits/10 Mbps In the case of a single store-and-forward switch in the path, the latency is determined by the propagation delay and the time it takes to transmit the packet. The total latency is the sum of the propagation delay and the transmission time. The total latency is the sum of the propagation delay and the transmission time: Total latency = Propagation delay + Transmission time =10 μs + 0.5 ms = 10.5 μs

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(b) To calculate the latency for three switches in the path, we need to consider the additional propagation delays introduced by each switch. Propagation delay: 10 μs Packet size: 5,000 bits Ethernet speed: 10 Mbps The time it takes to transmit the packet remains the same: Transmission time = 0.5 ms The total latency includes the propagation delays for each link and the transmission time: There are three switches in the path, the latency increases due to the additional propagation delays introduced by each switch. The total latency is calculated by multiplying the sum of the propagation delay and transmission time by the number of switches. Total latency = (Propagation delay + Transmission time) x Number of switches Total latency = (10 μs + 0.5ms) x 3 = 31.5 μs (c) With “cut-through” switching, the switch can begin transmitting the packet after receiving the first 200 bits This reduces the transmission time. Propagation delay: 10 μs Packet size: 5,000 bits Ethernet speed: 10 Mbps The remaining bits to be transmitted after the switch begins transmission: Remaining bits = Packet size - 200 bits = 4800 bits The time it takes to transmit the remaining bits: Transmission time = Remaining bits/Ethernet speed = 4800 bits/10 Mbps = 0.48 ms The total latency includes the propagation delay and the transmission time: With “cut-through” switching, where the switch can start transmitting after receiving the first 200 bits, the latency is reduced compared to store-and-forward switching. The transmission time is calculated for the remaining bits of the packet, resulting in lower total latency. Total latency = Propagation delay + Transmission time = 10 μs + 0.48 ms = 10.48 μs

Problem 20 (a) 10-Mbps Ethernet with a delay of 10μs. In this scenario: ● Bandwidth (B) = 10 Mbps = 10,000,000 bps ● Propagation Delay (D) = 10 μs = 10 × 10^(-6) seconds To calculate the Bandwidth-Delay Product (BDP): BDP = Bandwidth × Propagation Delay BDP = 10,000,000 bps × 10 × 10^(-6) seconds = 100 bits The BDP for this scenario is 100 bits. (b) 10-Mbps Ethernet with a single store-and-forward switch, packet size 5000 bits, and 10μs per link propagation delay. In this scenario, the packet size (P) is 5000 bits, and the propagation delay (D) is 10 μs. To calculate the BDP: BDP = Bandwidth × Propagation Delay BDP = 10,000,000 bps × 10 × 10^(-6) seconds = 100 bits The BDP for this scenario is 100 bits. (c) 1.5-Mbps T1 link, with a transcontinental one-way delay of 50 ms. In this scenario: ● Bandwidth (B) = 1.5 Mbps = 1,500,000 bps ● Propagation Delay (D) = 50 ms = 50 × 10^(-3) seconds To calculate the BDP: BDP = Bandwidth × Propagation Delay BDP = 1,500,000 bps × 50 × 10^(-3) seconds = 75,000 bits The BDP for this scenario is 75,000 bits. (d) 1.5-Mbps T1 link through a satellite in geosynchronous orbit, 35,900 km high. The only delay is speed-of-light propagation delay. In this scenario: ● Bandwidth (B) = 1.5 Mbps = 1,500,000 bps ● Propagation Delay (D) = Distance / Speed of Light D =3×108 meters / second 35,900,000 meters D =119.67 seconds To calculate the BDP: BDP = Bandwidth × Propagation Delay BDP = 1,500,000 bps × 119.67 seconds = 179,505,000 bits The BDP for this scenario is 179,505,000 bits.

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