P3. a) A circuit-switched network would be more efficient for this job because the line will stay open while there is a break between the time of the N-bit units of data being sent. Plus with the circuit-switched network you are guaranteed full bandwidth while sending the N-bits.
P3. b) There should be some form of congestion control because with packet-switched networks there is a chance of packet loss if the stream becomes too congested which could also lead to a long delay.
P5. a) If the caravan is traveling at 150km/hr and the toll booth can send through 5 cars per minute then the total delay between the 3 toll booths is (10 cars through 3 toll booths at 5 cars/min) = 6 minutes. Then the total end-to-end delay of the cars traveling through the toll booths not including the time waiting is 200km/(150km/hr) = 80 minutes plus the 6 minutes totaling 86 minutes.
P5. b) If the caravan has eight cars as opposed to the previous ten, the time traveled between the toll booths will still remain at 80 minutes because the caravan is still traveling 200km at 150km/hr. The only difference her is that the delay from the toll booths will be slightly less. Because there are two less cars in the caravan this time, the total delay from the toll booths will be (8 cars through 3 toll booths at 5 cars/min) = 4 minutes 48 seconds (1 min 36 sec/booth). Thus resulting in the total end-to-end delay of 84 minutes 48 seconds.
P6. a) d(prop) is equal to the distance traveled over the speed. In this case the distance is represented by m and the speed is denoted by s. So the answer would simply be m/s.
P6. b) d(trans) is equal to the length of the packet L divided by the speed of the bits/sec R. So the answer would be L/R
P6. c) Ignoring the processing and queuing delay, the expression to represent the total end-to-end delay would be d(nodal) = d(prop) + d(trans)
P6. d) At time t = 0 the last bit of the packet most likely has not been sent yet because at time t = 0 the first bit of the packet is just sending. Once the entire packet has been transmitted, that is when it will be sent.
P6. e) The first bit of the packet is on the propagation delay because for the first bit to begin on the prorogation delay, the whole packet has to be received at the the end of the transmission delay.
P6. f) Same as P6. e)
P6. g) d(prop) is represented by d/s or (d/2.5 * 10^8 m/s) and d(trans) is represented by L/R or (120 bits/56 kbps). In order to find d, the d(trans) and d(prop) must be equal so set (d/2.5 * 10^8 m/s) = (0.01465 kbps). Then multiply that by 2.5 * 10^8 m/s to get d = 3662500 meters.
P9. a) 1024 people can be supported on the link but because of the probability being 0.1, the link can only support at max, 921 users.
P9. b) N = M - (M * p)
P10. a) (L/R1 + d1/s1) + (L/R2 + d2/s2) + (L/R3 +d3/s3)
P10. b) (3(1500 bytes/2 mbps = 0.0007s) = 0.002s) + (5000000 m/2.5 * 10^8 m/s = 0.02s) + (4000000 m/2.5 * 10^8 m/s = 0.016s) + (1000000 m/2.5 * 10^8 m/s = 0.004s) + (3(0.003) = 0.009s) = 0.051s
P11. a) 0.051s - (d(proc)) 0.009s = 0.042s
P12. a) Because once a packet arrives it only has the wait the last half of the time for the packet in front to finish being transmitted, the answer would be the speed of the transmission divided by 2 and wait for the other 4 packets to be sent. So 1500 bytes/2 mbps = 0.0007s. But since it is already half done being sent the queuing delay will only be 0.0007s/2 + (4(0.0007)) = 0.00315s
P13. a) Due to the fact that the packets arrive simulatenously, one packet must wait the full time for one packet to finish transmitting. So the queuing delay will be the entire time of L/R.
P13. b) N(L/R)
P14. a) d(nodal) = (IL/R(1- I)) + (L/R)
P15. a) (a/u(1 - a)) + (a/u)
P19. a) 4 links are the same between the 2 traceroutes. The transatlantic link is the same.
P19. b) None of the links are the same except for the return traceroutes back to the easternct IP.
P19. c) None of the links are the same and they did not diverge before reaching China.
P22. a) (1 - p)
P23. a) LRs bits/sec
P23. b)
P24. a) 40 Terabytes would be 41943040 mbps which would take 419430.4 seconds to send completely. That 419430.4 secs is 6990.5 minutes which is 116 hours (roughly 5 days). So I would rather send the 40 terabytes on Fed-Ex over night delivery because it would be much quicker than to transfer the data over a 100 mbps link. If you wanted to send it overnight by data transfer, a good course of action would be to invest in Google Fiber to get a 1 gb/sec upload speed. Then the transfer would be significantly quicker (10 times quicker to be exact). So that 5 day transfer would become a 12 hour transfer (roughly).
P25. a) (20000000 m/(2.5 * 10^8 m/s)) / 2 = 0.04s
P25. b) Considering the probability of 0.1 the maximum number bits sending at any given time would be 1843.2 kbps or 1.8432 mbps
P25. d) 50m, no
P25. e) m/R
P27. a) (20000000 m /(2.5 * 10^8 m/s)) / 1000 = 0.00008s
P27. b) 900 mbps
P27. c) 25000m
P31. a) 8000000/ (209715200 * 8) = 0.0047s for 1 link, (0.0047s * 3) = 0.0143s for total time
P31. b) 10000/ (209715200 * 8) = 0.000005s for 1st packet, 0.000011s for 2nd packet
P31. c) (0.000005s * 800) = 0.0047s (No difference)
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