Solution of Questions from Assignment No.1 Question 1: Name the seven layers of the OSI Reference Model. Why did the creators of the OSI Reference Model decide to have seven layers? Can you propose a better reference model for computer networks? Name four other models that have been proposed for the same purpose other than the OSI reference model.Answer:The seven layers of the OSI Reference Model are:1. Application Layer2. Presentation Layer3. Session Layer4. Transport Layer5. Network Layer6. Datalink Layer7. Physical LayerThe reasons for having seven layers include the following:The designers of the OSI Reference Model felt that the seven layers represent the seven different categories of tasks that need to be done for successful data communication. In choosing to have seven layers the designers may also have been influenced by the IBM SNA model which also had seven layers.Four other models designed for the same purpose include:1. TCP/IP Model2. AppleTalk3. IBM Systems Network Architecture - SNA4. Digital Equipment Corporation's DECNET5. ISO Protocols 6. Cisco Protocols7. Novel NetwareBetter models can be proposed that have fewer or more layers than the OSI Model. In practice sometimes the datalink layer is implemented as two layers making it eight layers altogether. One could implement a five layer model also where the upper three layers are all merged into the application layer. This is done in the original TCP/IP model.Question 2: What are the origins of the DOD model for computer communications and networks. Is it better than the OSI Reference Model?Answer:The DOD Model evolved from the development and implementation of the ARPANET and subsequently the Internet. The DOD model is simpler than the OSI model but it lacks the elegance of the OSI model. The DOD model has become far more popular than the OSI Model. The main advantage of the DOD model is its wide implementations and general availability. In recent times the DOD model has evolved to come closer to the OSI model. Hence the DOD model has accepted the elegance and superiority of the OSI model.Question 3: Is the network layer implemented in hardware or software and why is it done that way?Answer:The network layer is either implemented in software or in case of routers it is implemented in firmware. In recent years Layer 3 switches have been introduced which implement the network layer entirely in hardware. This trend is due to advances in VLSI technology.Question 4: Compare and contrast between the transport layer and the datalink layer.Answer:The transport layer is an end to end layer whereas the link layer operates on links where devices are directly attached. Link layer provides reliable and efficient delivery across a physical link whereas the transport layer provides reliable and efficient delivery from end to end. There may be several link layer entities (one for each link) from one device to another but generally there are only two transport layer entities, one in each end system.Question 5: What is the difference between a router and a gateway?Answer:A router is a device that connects two systems at the level of the network layer. A gateway on the other hand is a device (software) that converts between two different applications layer entities.Question 6: Name the highest layer on which each one of the following entities operates:Answer:1. Four Port Token Ring Bridge - Datalink Layer2. Modem - Physical Layer3. FTP Server - Application Layer4. FTP Client - Application Layer5. Netmeeting Software - Application Layer6. Network Interface Card - Datalink Layer7. Router - Network Layer8. Internet Explorer - Application Layer9. Ethernet Switch - Datalink Layer10. Ethernet Hub - Physical LayerQuestion 7: What is a PDU? Give examples with figures of two real world PDUs.Answer:PDU stands for Protocol Data Unit. Examples would be an Ethernet frame, a PPP frame, an HDLC frame, an IP datagram, a TCP segment, etc.Question 8: A color image uses 16 bits to represent a pixel. What is the maximum number of different colors that can be represented?Answer: The number of different colors you can represent with 16 bits is 216, or about 65,000 colors.Question 9: Assume six devices are arranged in a mesh topology. How many cables are needed?Answer: "Mesh topology" is a rather unspecific term. I suppose that what you mean is that you want every device to have a direct connection to every other device.Then each device needs to be connected to 5 other devices. So, each device needs to have 5 ports. Six devices times five ports equals 30 total ports.Each cable has two ends, so it connects to two ports. Thus, the number of cables is half of the total number of ports. Therefore, 15 cables are needed.Question 10: How many ports are needed for each device?For each of the following four networks, discuss the consequences if a connection fails.a. Five devices arranged in a mesh topologyb. Five devices arranged in a star topology (not counting the hub)c. Five devices arranged in a bus topologyd. Five devices arranged in a ring topologyAnswer: a). Mesh topology: If one connection fails, the others still work.b). Star topology: The other devices will still be able to send data through the hub; there will be no access to the device which has the failed connection to the hub.c). Bus Topology: All transmission stops if the failure is in the backbone.d). Ring Topology: The failed connection may disable the whole network.Question 11: You have two computers connected by an Ethernet hub at home. Is this a LAN, a MAN, or a WAN? Explain your reason.Answer: This is a LAN (Local Area Network). A WAN is a Wide Area Network that typically connects machines that are geographically remote. A Man is a Metropolitan Area Network - that connects together all machines in an enterprise, campus or town.Question 12: Draw a hybrid topology with a star backbone and three ring networks.Answer: Question 13: Draw a hybrid topology with a ring backbone and two bus networks.Answer: Question 14: Performance is inversely related to delay. When you use the Internet, which of the following applications are more sensitive to delay?a. Sending an e-mailb. Copying a filec. Surfing the InternetAnswer: a. E-mail is not an interactive application. Even if it is delivered immediately, it may stay in the mail-box of the receiver for a while. It is not sensitive to delay.b. We normally do not expect a file to be copied immediately. It is not very sensitive to delay.c. Surfing the Internet is the an application very sensitive to delay. We except to get access to the site we are searching.Question 15: When a party makes a local telephone call to another party, is this a point-to-point or multipoint connection? Explain your answer.Answer: In this case, the communication is only between a caller and the callee. A dedicated line is established between them. The connection is point-to-point.Question 16: In Figure 2.22, computer A sends a message to computer D via LANl, router Rl, and LAN2. Show the contents of the packets and frames at the network and datalink layer for each hop interface.

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Question 17: In Figure 2.22, assume that the communication is between a process running at computer A with port address i and a process running at computer D with port address j. Show the contents of packets and frames at the network, data link, and transport layer for each hop.

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Part B (Chapter 3)

Question 1: If a binary signal is sent over a 4 kHz telephone channel whose signal to noise ratio is 30 dB, what is the maximum achievable data rate?

Answer:Using Shannon's formulation for channel capacity:C = W * log2 (1 + S/N)Where log2 represent logarithm base 230 dB S/N = 1000 S/NC = 3100 * log2 (1 + 1000)C= 30,894 bpsHence the channel capacity is 30,894 bits per second.

Question 2: What signal to noise ratio is needed to put an T1 carrier on a 100 kHz line? For the data rate of an T1 signal consult a reference.

Answer:Once again using the Shannon's formulation:C = W * log2 (1 + S/N)For a T1 system the data rate is 1.544 Mbps.1544000 = 100000 * log2 (1 + S/N)15.44 = log2 (1 + S/N)S/N = (2^15.44) - 1S/N = 44452.2S/N = 46.48 dBHence the signal to noise ratio required is 46.48 dB.

Question 3: It is desired to send a sequence of computer screen images over an optical fiber. The screen is 800 X 600 pixels using 256 colors. There are 30 screen images per second. How much bandwidth is needed to transmit this signal using binary signaling?

Answer:The data rate required is:Data rate = 800 X 600 pixels/screen X 30 screens/second X 8 bits/pixelData rate = 115200000 bits/second = 115.2 MbpsIf binary signalling is used then using Nyquist formulation:C = 2*W* log2 (M)115200000 = 2 * W * log2(2)W = 57.6 MHzHence the bandwidth required is 57.6 MHz.

Question 4: How long does it take to transmit an 8 inch by 10 inch image by facsimile over an ISDN B channel? The facsimile digitizes the image into 400 pixels per inch and assigns 2 bits per pixel. For the data rate of an ISDN consult a reference.

Answer:Total data to be transmitted by the facsimile machine:400 pixels per inch means 400 X 400 = 160,000 pixels per sq. inch160,000 pixels/ sq. inch X 8 inch X 10 inch = 12,800,000 pixelsWith 2 bits / pixel total number of bits = 12,800,000 X 2 = 25.6 MbitThe ISDN B channel has a data rate of 64 kbps.So time required to transmit the facsimile page using the ISDN B channel is:Time = 25,600,000 Mbits / 64,000 bits per secondTime = 400 secondsTime = 6.66 minutes

Question 5: Television channels are 6 MHz wide. How many bits/sec can be sent if 8 level digital signals are used? Assume a noiseless channel.

Answer: C = 2*W*log2(M) C = 2 * 6 X 10^6 * log2(8) C = 36 X 10^6 Hz C = 36 MHz

Question 6: Given a channel with an intended capacity of 25 Mbps. The bandwidth of the channel is 4 MHz What signal to noise ratio is required in order to achieve this capacity?

Answer: C = W * log2(1 + S/N) 25 X 10^6 = 4 X 10^6 * log2(1 + S/N) 6.25 = log2(1+S/N) S/N = (2^6.25) -1 S/N = 75S/N dB = 10 * log (75) S/N dB = 18.75 dB

Question 7: Suppose that a digitized television picture is to be transmitted from a source that uses a matrix of 1600 by 1200 picture elements (pixels), where each pixel represents one of 65,536 colors. Assume that 30 pictures (frames) are sent per second. (a) Find the data rate R of the source in bits per second. (b) Assume that the picture is to be transmitted over a 10 MHz channel with channel capacity equal to the data rate required for the signal. If M-ary signaling is to be used to accommodate the video signal on the channel what smallest value of M is required?

Answer: (a)The data rate required is: Data rate = 1600 X 1200 pixels/screen X 30 screens/second X 8 bits/pixel Data rate = 460800000 bits/second = 460.8 Mbps

(b)If binary signalling is used then using Nyquist formulation: C = 2*W*log2(M) 460800000 = 2 * W * log2(2) W = 23.04 MHz Hence the bandwidth required is 57.6 MHz.

Question 8: Is Nyquist Theorem true for optical fiber, or only for copper wire?

Answer: The Nyquist theorem is true regardless of the physical medium.

Question 9: [For part (a) and (b) consult text book](a) Imagine a Signal travels through a transmission medium and its power is reduced to half. Calculate the attenuation (loss of power) in dB?(b) Imagine a Signal travels through an amplifier and its power is increased 10 times. Calculate the amplification (gain of power) in dB?(c) We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000Hz (300Hz to 3300Hz). The Signal-to-Noise ratio is usually 35dB. For this channel calculate the highest bit rate? If we want to send data faster than this, what can we do?

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Question 10: We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 20V, the noise is 6mV. What is the maximum data rate supported by this telephone line?

Answer: Capacity=B*log2(1+SNR)=4000 * (log(1 + (20 / 0.006)) / log(2)) = 46812.7305 bps

Question 11: What is the total delay (latency) for a frame of 10 million bits that is being sent on a link with 15 routers each having a queuing time of 2ms and the processing time of 1ms? The length of the link is 3000Km. The speed of the light inside the link is 2x10m/s. The link has a bandwidth of 6 Mbps. Which component of the total delay is dominant? Which one is negligible?

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Givenmessage size = 10 million bits = 10 106 bits = 10000000 bitsqueuing time = 2 s = 2 10-6 sprocessing time =1 s = 1 10-6 slength of link, distance d = 3000 Km = 3000 103 mThe speed of light (Propagation speed) = 2 108 m/sBandwidth , B = 6 Mbps = 6 106 bpsFor delay(latency), use following formulaDelay (latency) = propagation time + transmission time + queuing time + processing time (1)

Propagation time = distance / propagation speed= 3000x10^3 / 2x10^3= 0.015s

Transmission Time = message size / bandwith= 1000000 / 6x10^6= 1.67s

Now, from eq (1), we get

Delay (latency) = 0.015 + 1.67 + 2 10-6 + 1 10-6 Delay (latency) = 1.685 s In this case, because the massage size (frame size) is very long and bandwidth is not very high, the dominant factor is transmission time, not the propagation time and hence transmission time = 1.67 s propagation time is negligible (ignored) because it is very less time than transmission time. propagation time = 0.015 s Both these components are the factors of delay (latency).

Question 12: A TV channel has a bandwidth of 6MHz? if we send a digital signal using one channel, what are the data rates if we once harmonic, three harmonics, and five harmonics?

Answer: Using the first harmonic, data rate = 2 x 6 MHz = 12 MbpsUsing three harmonics, data rate = (2 x 6 MHz) / 3 = 4 MbpsUsing five harmonics, data rate = (2 x 6 MHz) / 5 = 2.4 Mbps

Question 13: A non-periodic composite signal contains frequencies from 10 to 40 KHz. The peak amplitude is 12V for the lowest and the highest signals and is 30V for the 20 KHz signal. Assuming that the amplitudes change gradually from the minimum to the maximum, draw frequency spectrum.

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Question 14: What is the bandwidth of a signal that can be decomposed into five sine waves with frequencies at 0, 30, 40, 120, and 300Hz? All peak amplitudes are the same. Draw the bandwidth.

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Question 15: The light of the sun takes approximately eight minutes to reach the earth. What is the distance between the sun and the earth?

Answer:To answer this question, we also need the speed of light, which is about 300 million metres/second. It will also help to convert the time into seconds, since they are part of the units we are using for speed. The time comes out to 480 seconds. We know that speed = distance / time, so to solve for distance we multiply speed by time, which gives us (300,000,000 m/s) x (480 s) = 144,000,000,000 metres. We can then divide by 1,000 to get an answer in kilometres, and the final result is about 144 million kilometres, or about 90 million miles.