2IC10 Computer Networks Assignments week 4

Transcription

1 2IC10 Computer Networks Assignments week 4 Mark van Eijk, s Bas Kloet, s Physical Layer 1 Question What is the most used physical medium nowadays? Why? Does that specific medium have the best performance? Can it be used everywhere? If not, give an example. The most used medium is twisted pair, because it s cheap and easy to use. In performance however, it s easily surpassed by other options, like optical fiber. Of course, there are places, like in a boat on the open sea, where communication via twisted pair, or any other wire for that matter, is simply impossible. For those places communication via satellite would be a much more viable option. Other places that are unsuitable for twisted pair, are those with very high magnetic radiation. In those places optical fiber would be a much better option. 2.8 Question It is desired to send a sequence of computer screen images over an optical fiber. The screen is pixels, each pixel being 24 bits. There are 60 screen images per second. How much bandwidth is needed, and how many microns of wavelength are needed for this band at 1.30 microns. The amount of bandwidth needed is bits per second. I don t understand the second question Question At the low end, the telephone system is star shaped, with all the local loops in a neighbourhood converging on an end office. In contrast, cable television consists of a single long cable snaking its way past all the houses in the same neighbourhood. Suppose that a future TV cable were 10 Gbps fiber instead of copper. Could it be used to simulate the telephone model of everybody having their own private line to the end office? If so, how many one-telephone houses could be hooked up to a single fiber? 1

2 Since the latency of a fiber channel is extremely lower then that of twisted pair, the fact the fiber goes past multiple houses is no problem in simulation the star shaped network. Neither is the bandwidth a problem, since speech needs 64kbps 1 of bandwidth. This means that one fiber can transport / calls by using multiplexing Question A cable company decides to provide Internet access over cable in a neighbourhood consisting of 5000 houses. The company uses a coaxial cable and spectrum allocation allowing 100 Mbps downstream bandwidth to each house at any time. To attract customers, the company decides to guarantee at least 2 Mbps downstream bandwidth to each house at any time. Describe what the cable company needs to do to provide this guarantee. Guaranteeing 2Mbps downstream, means that at most 100/2 50 houses can be connected to any coaxial cable at any time. This means that there will need to be 5000/ head ends, all upgraded to be able to guarantee the desired Quality of Service. Question Calculate the utilization of the channel when communicating parties are using stop-and-wait protocol if the following parameters are known: packet size: 8000 bits transmission rate: 1 Gbps distance between the two parties: 3000 km the signal speed: km/s ACK packets: 16 bits What is the effective throughput? How to improve the utilization? The propagation delay of the channel: 3000 km km/s 8000 bits s. The transmission delay of the packet: bits/s s. 16 bits The transmission delay of the ACK-packets: bits/s. The ACK-packet corresponding to a sent packet is received after: 2 propagation delay + transmission delay packet + transmission delay ACK-packet 8000 bits s bits/s + 16 bits bits/s 8016 bits bits/s s 1 american telephone system 2

3 This results in a utilization of: And an effective throughput of: transmission delay total delay s packet size 8000 bits delay s s % bits/s. Hence the effective throughput is almost totally depending on the propagation delay, instead of the transmission rate. Improving utilization can be reached by sending more packets instead of first waiting for the corresponding ACK-packet. This can be done by using one of the versions of the Sliding Window protocol with a sender of at least 2. Data Link Layer 3.1 Question An upper-layer packet is split into 10 frames, each of which has an 80 percent chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through? On average, 8 out of 10 frames arrives undamaged. This means that to get 10 frames through, you need to send the message 10/ times. 3.9 Question Sixteen-bit messages are transmitted using a Hamming code. How many check bits are needed to ensure that the receiver can detect and correct single bit errors? Show the bit pattern transmitted for the message Assume that even parity is used in the Hamming code. The formula needed is m + r r with m message length in number of bits and r the number of parity bits. So in this case this means that r 5, because , but With parity bits at positions 1, 2, 4, 8 and 16, even parity and counting the positions from left to right 2 you get Question Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in one direction, with very short acknowledgements coming back the other way. What is the maximum throughput for s of 1, 7, 15 and 127? The earth-satellite propagation time is 270 ms. throughput utilization bitrate 2 like in Tanenbaum, but unlike in the sheets. 3

4 If: transmission delay + 2 propagation delay < transmission delay Then we have the following formula for the utilization: Otherwise utilization is 100 %. bitrate propagation delay framelength When we fill in these formula, we get the following results: utilization throughput % 6798 bits/s % bits/s % bits/s % bits/s Question Which checksum is obtained for the message when the following CRC polynomial is used? C(x) x 7 + x 5 + x Using the polynomial C(x) x 7 +x 5 +x 3 +1, we get the divisor This divisor is 8 bits long, so the CRC will be 7 bits long. We now add 7 0 s to the message: Then we divide this by , which results in CRC The message on which the checksum will now be calculated is: To make division into k sections, each of the same amount of bits, possible, we add a 0 to the end of the message , which we now divide in 3 sections, each 8 bits: , and We add these numbers using one s complement: , where the left-most 1 is the carry-bit. Because the carry-bit is set, it has to be added to the result without the carry-bit: This result has to be complemented, which results in the following checksum: Question Name at least 3 services implemented by the sliding window protocol. The sliding window protocol implements: preservation of data unit sequence: data units are delivered to the receiver s higher layer in the same sequence that they were given from the sender s higher layer 4

5 no duplication: no data unit should be delivered to the receiver s higher layer more than once error-free delivery: data units are delivered to the receiver s higher layer without errors Question Assume that there are never transmission errors over a particular physical channel with a large propagation delay. Which of the two Sliding Window protocols would you expect to have better performance across the channel? Explain Assume that each protocol uses a 3-bit sequence number. The two Sliding Window protocol variants: Go-back-N: sender has K and receiver has 1 Selective Repeat: sender and receiver both have the same Go-Back-N has a smaller than and Selective Repeat has a of at most If the channel would also deliver packets in the order they were sent, the Go-back-N protocol would be more efficient, since it can have a larger windows size and the larger the the better performance. If the channel on the other hand would not exhibit such behaviour, the Selective Repeat will be more efficient if large numbers of packets arrive out of order, since it has a larger at the receiving side. This ensures that less packets have to be resent. 5