March 27, 2018 | Author: blindman88585695 | Category: Wavelength Division Multiplexing, Network Topology, Cable Television, Bit Rate, Computer Network



Solutions to Chapter 41. A television transmission channel occupies a bandwidth of 6 MHz. Solutions follow questions: a. How many two-way 30 kHz analog voice channels can be frequency-division multiplexed in a single television channel? 6 × 106 / 30 × 103 = 200 channels b. How many two-way 200 kHz GSM channels can be frequency-division multiplexed in a single television channel? 6 × 106 / 200 × 103 = 30 channels c. Discuss the tradeoffs involved in converting existing television channels to cellular telephony channels? The biggest advantage of using existing television channels to provide cellular telephony channels is the very large bandwidth that they occupy. In theory, one could divide each television channel into many cellular telephone channels, as shown in parts (a) and (b). However, a significant trade-off would soon become apparent. . Cellular telephone networks use low-power base transmitters to create relatively small cells so that frequency re-use is possible. This dramatically increases the number of users that a network can accommodate. Television stations transmit at a very high power over very long distances (as one would expect for a broadcast medium). Thus, in the region that currently receives a given television network broadcast (say, the larger part of a city), only 200 analog channels or 30 GSM channels could be accommodated. 2. A cable sheath has an inner diameter of 2.5 cm. Solutions follow questions: a. Estimate the number of wires that can be contained in the cable if the wire has a diameter of 5 mm. Ignoring empty space between the wires: π( 25 2 ) 2 = 25 wires 5 2 π( ) 2 b. Estimate the diameter of a cable that holds 2700 wire pairs. Ignoring empty space between the wires: d π ( )2 2 = 2700 wires à d ≅ 260 5 2 π( ) 2 3. Suppose that a frequency band W Hz wide is divided into M channels of equal bandwidth. Solutions follow questions: a. What bit rate is achievable in each channel? Assume all channels have the same SNR. Each user uses W/M bandwidth. Using Shannon’s Channel Capacity formula: Bit rate = b. M ) log 2 1 + SNR) bps What bit rate is available to each of M users if the entire frequency band is used as a single channel and TDM is applied? In this case, the total bit rate afforded by the W Hz is divided equally among all users: Bit rate = c. W log 2 (1 + SNR ) bps M How does the comparison of (a) and (b) change if we suppose that FDM requires a guard band between adjacent channels? Assume the guard band is 10% of the channel bandwidth. Because of the guard band we expect that the scheme in (b) will be better since the bit rate in (a) will be reduced. In (a), the bandwidth usable by each channel is 0.9W/M. Thus, we have: Bit rate = (0.9 W ) log 2 (1 + SNR) bps M 4. In a cable television system (see Section 3.7.2), the frequency band from 5 MHz to 42 MHz is allocated to upstream signals from the user to the network, and the band from 550 MHz to 750 MHz is allocated for downstream signals from the network to the users. Solutions follow questions: a. How many 2 MHz upstream channels can the system provide? What bit rate can each channel support if a 16-point QAM constellation modem is used? The system can provide: (42 − 5)MHz = 18 upstream channels. Mhz 2 Channel By Nyquist, we have: R = 2 ∗ ( 2 × 10 ) 6 pulses bits ∗4 = 16 Mbps sec ond pulse b. How many 6 MHz downstream channels can the system provide? What bit rates can each channel support if there is an option of 64-point or 256-point QAM modems? Similarly, the system can provide: (750 − 500)MHz = 33 upstream channels. Mhz 6 Channel R46-point = 2 ∗ (6 × 10 ) 6 pulses bits ∗6 = 72 Mbps sec ond pulse pulses bits ∗8 = 96 Mbps sec ond pulse R256-point = 2 ∗ (6 × 10 ) 6 5. Suppose a system has a large band of available bandwidth, say 1 GHz, which is to be used by a central office to transmit and receive from a large number of users. Compare the following two approaches to organizing the system: Solutions follow questions: a. A single TDM system. The channel will provide a maximum bit rate, say b bps. Each user will use b/n bps where n is the number of users. The number of users will be limited by the noise on the channel and the switching speed of the hardware. b. A hybrid TDM/FDM system in which the frequency band is divided into multiple channels and time division multiplexing is used within each channel. Although in theory, the total bit rate provided by the system is the same, the hybrid solution may be practically better because, if we assume again n users, each user timeslot is at a lower bit rate and for a longer duration. Therefore, TDM synchronization and switching speed problems are less significant than in the pure TDM version. By leveraging both the time and frequency domain, the technological requirements posed by either one are lessened. 6. Suppose an organization leases a T-1 line between two sites. Suppose that 32 kbps speech coding is used instead of PCM. Explain how the T-1 line can be used to carry twice the number of calls. Solution: If a 32 kbps speech coding (such as Adaptive Differential Pulse Code Modulation (ADPCM)) is used, the bit rate required for each user is reduced to half of that required for PCM (64 kbps). The format in which a T-1 system carries information needs to be carried. For example a frame could now carry 48 4-bit samples. Aside: Chapter 12 explains how advanced speech coding schemes reduce the bit rate while maintaining speech quality. 7. A basic rate ISDN transmission system uses TDM. Frames are transmitted at a rate of 4000 frames/ second. Sketch a possible frame structure. Recall that basic rate ISDN provides two 64 kbps channels and one 16 kbps channel. Assume that one-fourth of the frame consists of overhead bits. Solution: Define the random variable X as the number of times the bit position needs to be observed before the alternating pattern is violated.1 + P0.0.b and c are observed in succession. Now suppose that the frame synchronizer begins at a random bit position in the frame. Calculate the average number of bits that elapse until the frame synchronizer locks onto the framing bit.1. This is done by alternating the value of the framing bit at each frame. Calculate the average number of times this bit position needs to be observed before the alternating pattern is found to be violated.1.0 + P1. Consider an information bit position in the frame. the frame could look as follows: D channel B channel B channel Overhead bits t [ms] 0 0.1 = 1/4 + 1/4 = 1/2 P[X = 3] = P1.c as the probability that a.1. Define Pa. Assume that each information bit takes a value of 0 or 1 independently and with equal probability.104 0. Framing is done by examining each of 193 possible bit positions successively until an alternating pattern of sufficient duration is detected.0 + P0. Solutions follow questions: a. P[X ≤ 1] = 0 P[X = 2] = P0.Assuming that the 16 kbps “D” channel is followed by the two 64 kbps “B” channels and then the overhead (which is not necessarily the case). Similarly. assuming that no other bits can sustain an alternating pattern indefinitely.25 8. The T-1 carrier system uses a framing bit to identify the beginning of each frame. define Pa.021 0.0.0 = 1/16 + 1/16 = 1/8 P[X = n] = 1/2 + 1/2 = 1/2 n n (n -1) Define p = P[a one bit is observed] = P[a zero bit is observed] = ½ E[X] = ∑ iP[ X = i] = ∑ ip i =2 i =2 ∞ ∞ i −1 =∑ i=2 ∞ d i d ∞ i d ∞ p = ∑ p = dp [∑ p i − 1 − p] dp i = 2 dp i =0 = d 1 1 1 [ − 1 − p] = −1 = −1 2 dp 1 − p (1 − p) (1 − 1 2) 2 =3 b.b as the probability that ‘a’ is observed the first try and ‘b’ on the second. .1875 0.b. in the time domain.0. Suppose the synchronizer observes the given bit position until it observes a violation of the alternating pattern.1 = 1/8 + 1/8 = 1/4 P[X = 4] = P1. In addition to its correct position. so n must be chosen sufficiently large so that the synchronizer is reasonably confident that it has indeed found the frame bit. Let p be the probability that a specific byte pattern is observed. . 9. the average number of bits elapsed until the synchronizer observes the framing bit is: Average number of bits elapsed until frame bit found = 91 x [(2 x 193) + 1] = 35 217 bits At this point. Suppose that all frames begin with the same byte pattern. the an occurrence of the frame byte pattern can begin at any of 256 – 8 – 8 = 240 locations Thus. Thus. Thus. Consider an arbitrary information bit position in the frame. a non-alternating pattern will be discovered on the third observation. What is the probability that this pattern occurs elsewhere in the frame? Assume that each information bit takes a value of 0 or 1 independently and with equal probability. P[pattern occurs elsewhere in the frame] = P[pattern occurrence begins in bit 9] + + P[pattern occurrence begins in bit 10] +…+ P[pattern occurrence begins in bit 249] = 240(1/2) = 0. It can’t be sure that it has located the framing bit until an alternating pattern has been observed for a sufficient number of bits. For each such incorrect bit location. p = (½8 ) Also. it will have to examine 182/2 = 91 incorrect bit locations before locking onto the framing bit. let us assume that the framing byte is chosen in such a way that any other occurrence of the byte pattern cannot include any of the bits in the frame byte. at which time the synchronizer will move one bit forward to the next bit. The CEPT-1 carrier system uses a framing byte at the beginning of a frame. say n. the synchronizer will begin observing the framing bit. if the frame synchronizer starts at a random location.9375 b. Calculate the average number of times that the byte beginning in this bit position needs to be observed before it is found to not be the framing byte. A CEPT-1 frame is 32 bytes = 256 bits long. The synchronizer can never be sure that it has locked onto the frame bit. the total number of bits elapsed before the synchronizer has locked onto the framing bit is: Average number of bits elapsed until locked onto frame bit = 35 217 + 193(n – 1) bits. Solutions follow questions: a.On average. Thus. 8 Define the random variable X as the number of times the bit position needs to be observed to discover that it is not the location of the framing byte. on average. Suppose the synchronizer observes the byte beginning in the given bit position until it observes a violation of the alternating pattern.004 c.004 observations.8 Mbps.P[X = 1] = 1 – p P[X = 2] = p(1 – p) P[X = i] = pi . Each flag bit indicates whether the corresponding data bit carries user information or a stuff bit because user information was not available at the input. To accommodate deviations from the nominal rate. say n.5 incorrect bit locations before locking onto the framing bit. For each such incorrect bit location. Calculate the average number of bits that elapse until the frame synchronizer locks onto the framing byte. How frequently are the stuff bits used? In this case. on average. each at a nominal rate of 1 Mbps. Thus. It can’t be sure that it has located the framing byte until the framing byte pattern has been observed a sufficient number of times. an incorrect pattern will be discovered after 1. we have: ) E[X] = 1. On average. Suppose that the two input lines operate at exactly 1 Mbps. At this point. This will entail n – 1 full frames elapsing plus the 8 bits to check the framing byte the nth time. Now suppose that the frame synchronizer begins at a random bit position in the frame. . The remaining four positions consist of two flag bits and two data bits. the stuff bits are always used because the information bits alone only provide an aggregate bit rate of 1. it will have to examine 255/2 = 127. at which time the synchronizer will move one bit forward to the next bit. Each group of 22 bits in the output of the multiplexer contains 18 positions that always carry information bits.2 Mbps as follows. the multiplexer transmits at a rate of 2. E[X] = ∑ iP[ X = i] = ∑ ip i =1 i =1 ∞ ∞ i −1 (1 − p ) = (1 − p )∑ i =1 ∞ d i d ∞ p = (1 − p) ∑ p i dp dp i =1 = = (1 − p ) d d ∞ i 1− p 1 1 [∑ p − 1] = (1 − p ) [ − 1] = = 2 dp 1 − p dp i =0 (1 − p ) 1− p Since p = (½8.06 + 256(n –1) 10. the synchronizer will begin observing the framing byte.1(1 – p) We see that this sequence rapidly decreases (geometrically). so we expect the average of X to be very close to 1. Solutions follow questions: a. so n must be chosen sufficiently large so that the synchronizer is reasonably confident that it has indeed found the framing byte.5 + 256(n – 1) + 8 = 258. the total number of bits elapsed before the synchronizer has locked onto the framing byte is: Average number of bits elapsed until locked onto frame bit = [(E[X] – 1) x 256 + 1] x 127. nine from each input. Suppose a multiplexer has two input streams. The synchronizer can never be sure that it has locked onto the framing byte. if the frame synchronizer starts at a random location. The high-tier ring must carry all inter-ring traffic form the lower tiers.84 810 155.720 2488. At the higher tier a bidirectional SONET ring connects the four lower-level SONET rings. Solutions follow questions: a.b.430 622.0 Mbps. Assume that each site generates traffic that requires an STS-3. How much does this multiplexer allow the input lines to deviate from their nominal rate? This multiplexer provides either 9 or 10 bits for each stream per 22-bit frame. Thus. Solution: A conventional voice channel is 64 kbps. the results are summarized in the table below: Bit rate of signal (Mbps) Voice Channels 51. Consider a SONET ring with four stations. Consider a set of 16 sites organized into a two-tier hierarchy of rings. Calculate the number of MPEG2 video channels that can be carried by these systems. Unidirectional: STS-6 Bidirectional: STS-6 13.32 38. Calculate the number of voice channels that can be carried by an STS-1. STS-12. The traffic between each pair of stations is one STS-1.520 MPEG2 Video 10 31 124 498 1.52 2.9 Mbps and as high as 1.991 STS-1 STS-3 STS-12 STS-48 STS-192 12. Suppose that tributaries are established between each pair of stations to produce a fully connected logical topology. 11. Each station produces three STS-1’s worth of traffic to the furthest station along the ring and no traffic to other stations.28 155. At the lower tier a bidirectional SONET ring connects four sites. Unidirectional: STS-6 Bidirectional: STS-3 b.08 9. Assuming a bit rate of 5 Mbps. Each station produces three STS-1’s worth of traffic to the next station in the ring and no traffic to other stations. The capacity required on the high-tier ring is thus statistically dependent on the amount of inter-ring traffic generated from the lower rigs and the distribution of such traffic. An MPEG2 video signal can vary in bit rate. Unidirectional: STS-3 Bidirectional: STS-3 c.880 9953. Find the capacity required in each hop of the SONET ring in the following three cases. STS-48. STS-3. assuming first that the ring is unidirectional and then that the ring is bidirectional. and STS-192. it allows either of the two input streams to transmit as low as 0. . The higher-tier ring will now support 4 times more traffic. Thus. since each source still produces and receives STS-3 of traffic. each low-tier ring sends an equal amount of traffic to each of the other low-tier rings. The required bandwidth on the lower-tier rings will not change significantly. Consider a four-station SONET ring operated using the four-fiber bidirectional line-switched approach.33 hops.8 × STS-3 ≈ STS-3.33 × 0. and no limit is placed on the occupancy in each working fiber. then the required bandwidth on the hightier ring will be increased. One fiber pair carries working traffic and one fiber pair provides protection against faults. then the required bandwidth on the hightier ring will be reduced as most traffic is confined locally to the lower-tier rings. Since each ring produces 4×0. Solutions follow questions: a. If 80% of the traffic is going to stay within the same ring. . 14. let us assume that the traffic generated is distributed evenly throughout the high-tier network – that is. the bandwidth requirement is around STS-12. each message will travel an average of (1 + 1 + 2)/3 = 1. STS-3 1. If 20% of the traffic is going to stay within the same ring.For this question. Discuss the bandwidth requirements that are possible if 80% of the traffic generate by each site is destined to other sites in the same tier ring. Since each site will generate as much traffic as it will receive. The lower-tier ring must accommodate 100% of the traffic sent from each of its sites. Discuss the bandwidth requirements that are possible if 80% of the traffic generate by each site is destined to sites in other rings. Thus. each site will send STS-3 of bandwidth. 20% of the traffic from each sub-ring will be routed equally to each other sub-ring. Solutions follow questions: a. The number of links to be traversed by the average message is between one and two. On the high-tier ring.5×STS-3 b.2 × STS-3 amount of traffic. the bandwidth required on each link will be around 1. Assuming that the traffic is distributed evenly on all of the links in the low-tier ring. Traffic can be routed between any two stations in either direction. Sketch a break in the ring and explain how the system can recover from this fault. The bandwidth required on each link of the hightier ring is on the order of 1. Adjacent stations are connected by two pairs of optical fibers.5 × STS-3. that in both cases. In 2-fiber SONET rings. no explicit protection fibers exist.13. so the protection fibers must be “embedded” within the working fibers. half of the available bandwidth on each of the two fibers must be reserved to provide line protection. the working two fibers of the 4-fiber SONET ring can carry traffic up to capacity. however. Explain how the system can recover from this fault. Note. .a a d Protection loops b Working loops d b c a c Fiber not used for traffic between a and b d b Fiber carrying traffic between a and b c If a break in the ring occurs and all four fibers are affected. This protection is commonly referred to as line protection. the system can simply route the information between the two adjacent stations over the protection fiber pair. 15. Explain how this setup recovers from failures in the link between the primary gateways. How does this ring operation differ from the ring operation in Figure 4. This type of protection is commonly referred to as link protection.12 in terms of available bandwidth? In terms of available bandwidth. The primary gateway transmits the desired signal to the other ring and simultaneously transmits the signal to the secondary gateway which also routes the signal across the ring and then to the primary gateway. the system recovers by rerouting the traffic between the two affected nodes around the ring over the protection fibers. A service selector switch at the primary gateway selects between the primary and secondary signals. Consider the operation of the dual gateways for interconnecting two bidirectional SONET rings shown in Figure 4. Suppose a failure affects only one of the fiber pairs connecting two adjacent stations. c. half of the total system bandwidth on each link is reserved for protection because. In the case of such a failure. in the 4-fiber case. b. Thus. the entire bandwidth of the protection fibers is reserved for protection. Calculate the bandwidth for the 100 nm window in the 1550 band. Show how 28 T-1 signals can be carried in an STS-1. we have: . This redundancy provides a backup in case of a link failure.Solution: Two logical connections are maintained between the primary and secondary gateway. wavelength window ∆λ. The next 84 columns are the seven tributaries and the last two columns are unused. Solution: For a channel of bandwidth W.17. Explain how the pointers in the outgoing STS-1 signals are determined. and center wavelength λ: Frequency Bandwidth = c∆λ λ2 Thus. If the primary gateway link fails. 18. 16. Consider the synchronous multiplexing in Figure 4. Show that the 100 nm window in the 1300 nm optical band corresponds to 18 teraHertz of bandwidth. as shown below: 87 Columns T1 T2 T3 T4 T5 T6 T7 9 rows t The first column is path overhead. Solution: Tributaries are multiplexed in stream. Each tributary carries four T-1 channels and occupies twelve columns of the SPE. Draw a sketch to explain the relationship between a virtual tributary and the synchronous payload envelope. in the 1300 nm optical band. The second is connected indirect via the secondary gateway as described in the question. the service selector switch can select signals for the alternate indirect path until the primary gateway link is restored. The first is a direct physical connection via the primary gateway link. Solution: The outgoing pointers are all synchronized to a common clock such that all the outgoing STS-1 signals are synchronized with respect to each other. 17. 21. Calculate the spacing between the WDM component signals in Figure 4. Explain whether these schemes can be used with WDM rings. we have: Frequency Bandwidth = (3 ∗ 10 8 )(100 ∗ 10 −9 ) = 12.552 × 106 MPEG2 Video Signals 49 766 99 532 199 065 Although. 200. What is the spacing in hertz and how does it compare to the bandwidth of each component signal? .49 THz. How many telephone calls can be carried by each of these systems? How many MPEG2 television signals? The number of telephone calls and MPEG2 video signals that can be carried each system is shown below: System 100 wavelengths 200 wavelengths 400 wavelengths Telephone Calls 3. Solutions follow questions: a.328 Gbps The available bandwidth of this window is 12.49THz (1550 ∗ 10 −9 ) 2 19. Thus. none of these systems even comes close to using all of the available bandwidth.32 Mbps of traffic.Frequency Bandwidth = (3 ∗ 10 8 )(100 ∗ 10 −9 ) = 17. b. Thus. they can still carry huge amounts of traffic.832 Gbps 497.664 Gbps 995. and 400 wavelengths operating at the 1550 nm region and each carrying an STS-48 signal.19.776 × 106 15. To use these schemes in WDM. Consider the SONET fault protection schemes described in problem 14 and also earlier in the chapter.888 × 106 7. 20. in the 1550 band. Solution: The fault protection schemes described in problem 14 are applicable to any ring network structure. How close do these systems come to using the available bandwidth in the 1550 nm range? One STS-48 signal carries 2488.75THz (1300 ∗ 10 −9 ) 2 Similarly. the exact same procedures can be carried out except that instead of allocating half of the physical fibers for protection. Consider WDM systems with 100. This shows how WDM provides the potential for huge amounts of bandwidth and radically expands the bandwidth of existing optical links. half of the wavelengths in the WDM link would be allocated for protection and remain unused unless a fault occurred. none the systems comes close to using all of the available bandwidth. the aggregate bit rate of each of these systems is: System 100 wavelengths 200 wavelengths 400 wavelengths Bit rate 248. In SONET. since tributaries are electrically defined. in turn. there are 16 wavelengths distributed equally over the spectrum from 1545 nm to 1557 nm. 23. 24.5 Gbps of traffic. the maximum achievable bit rate that each wavelength can achieve using on-off modulation is 2 × 93.65 = 187 Gbps. in the context of the hierarchical SONET ring topology. a switch.19. WDM without wavelength conversion has an inherent restriction imposed on the logical topologies that it can create. Discuss how WDM and SONET differ and explain what impact these differences have in the way logical topologies can be defined.65GHz (1550 ∗ 10 −9 ) 2 Thus. 22. what are the consequences of a single fiber providing large number of high-bandwidth channels? Solution: WDM allows for much more aggregate bandwidth on each fiber. so we see that because of the relatively large guard bands needed. How does WDM technology affect the hierarchical SONET ring topology in Figure 4.75 ∗ 10 −9 ) = 93. WDM allows it to achieve this increased service ability without the need to lay down more physical fibers over its links. Compare the operation of a multiplexer. Solution: Both WDM and SONET define logical topologies by setting up semi-permanent paths between nodes in the network. It takes N separate signals on N different inputs and combines them into one higher rate signal on one output port. the number of lower-tier rings that it can service.13? In other words. Each signal carries 2. and a digital crossconnect. WDM creates these paths by assigning lightpaths between nodes. Solution: A multiplexer is a 1:N device. an add-drop multiplexer. the high-tier ring can use WDM in order to increase its bandwidth and the number of channels it can support and. WDM wavelengths have a very large spacing compared to their width. WDM and SONET can be used to create various logical topologies over a given physical topology. Thus the spacing between each wavelength is: 1557 − 1545 = 0. .Solution: In Figure 4. the only restriction imposed on the topologies is the total bandwidth utilized on each link.75nm 16 The spacing in Hertz is: Frequency Bandwidth = (3 ∗ 10 8 )(0. Thus. because no to lightpaths that share the same link can use the same wavelength. The number of crosspoints needed in the three-stage switch described is: . usually configured by network operators rather than signaling processes. 25. If the traffic load on the inputs is relatively low. Traffic on the output lines is higher than that of the input lines by a factor of n/k on average. This switch is appropriate when the outgoing lines carry less traffic than the incoming lines. Under what conditions is this arrangement appropriate? 2 The number of crosspoints in an N x N crossbar switch is N . this configuration may save on hardware costs. The switch has inherent demultiplexing functionality. Explain why the switch is called a concentrator when n > k? Under what traffic conditions is this switch appropriate? This switch is a concentrator because traffic comes in on n lines and is concentrated onto k lines.An add-drop multiplexer takes in N signals on one input port and replaces one of them with a new signal from a separate input port. The switch has inherent multiplexing functionality. Solutions follow questions: a. Suppose an N x N switch consists of three stages: an N x k concentration stage. A switch can provide full nonblocking interconnections between the input and output ports. Digital cross-connects provide a basic network topology on which routing can be applied. The new aggregate signal is routed to an output port and the signal that was replaced is “dropped” to a separate output port. A switch takes in N inputs and routes them to N different outputs. Explain why the switch is called an expander when n > k? Under what traffic conditions is this switch appropriate? This switch is an expander because the number of lines is expanded from n to k. c. Consider a crossbar switch with n inputs and k outputs. A digital cross connect is similar to a switch except that it is semi-permanent. Generally. switches are configured using signaling that establishes paths across the network. b. and a k x N expansion stage. a k x k crossbar stage. but this is always the case. Their configuration is also done on a larger time scale (days or weeks). One such example is at the egress of a backbone network to an access network. Estimate the percent of time p that a line between the first and second stage is busy.23 with N = 16. and blocking can occur in the first stage. n = 4 and k = 2. Consider the multistage switch in Figure 4. we have the following switch architecture: 4x2 4x4 4x2 2x4 2x4 4x2 4x4 4x2 2x4 2x4 Thus. so 16 connections can be supported at a time. if more than k inputs require a path to one of the third stage’s n outputs at the same time. since there are only 16 inputs and 16 outputs. 26. then no blocking will occur at any stage in the switch. When does the three-stage switch in part (c) fail to provide a connection between an idle input and an idle output line? Assuming unicast traffic. eight connections can be supported at a time. it is clear that each input-output pair can be connected through any one of the k second-stage switches. 27. Solutions follow questions: a. the second stage is the bottleneck. Thus. b. In the multistage switch in Figure 4. If k = 10. especially if n is much greater than k. For a given set of input-output pairs. an input line is busy 10% of the time. clearly only up to 16 connections can be supported at one time. Thus. is there more than one way to arrange the connections over the multistage switch? As shown in the picture in part (a). since it only has k outputs. 2k2 + kn may be less than n2. d.23. Solutions follow questions: a. k = 2. then there are ten 4 x 4 switches in the second stage. n = 4. . there are numerous ways to arrange the connections over a multi-stage switch. However. blocking will occur in the first stage. If k = 4. For N = 16.2k2 + kn Depending on the values of n and k. What is the maximum number of connections that can be supported at any given time? Repeat for k = 4 and k = 10. Assuming independence of all lines. since there are k possible paths from each input to each output (one different path possible through each second-stage switch). P[either one of two jumps busy] ≈ 2p For none of the paths to be available. the greater the value of p. and n = 2. Since we assumed that all are independent of each other. the greater the chance that such blocking will occur. The total number of crosspoints is 2Nk + k(N/n)2. This is the case because the second stage switches are kxk components. p′ = p = (n/k) 10% For a given input and output line. N 32 32 32 32 n 16 8 4 2 k 31 15 7 3 Number of Crosspoints needed 2108 1200 896 960 . on average. Although the control and routing algorithm of the switch can minimize the chance of this occurring. all k must be busy. we require knb = 2n – 1.23 with N = 32. for any one of the k paths between a given input and the output to be unavailable. Thus. what is the proportion of time p′ that a line between the second and third stage is busy? Assuming that the blocking probability of the second stage switches is small. Thus p = (n/k) 10% b.Although. n = 8. Compare the number of crosspoints required by a nonblocking switch with n = 16. How does this p affect the blocking performance of the intermediate crossbar switch? Blocking occurs at the second switches when two of its inputs are to be routed to the same output. the busyness of each output line of the first stage depends on the switch control. d. How is p affected by n and k? p is directly proportional to the ratio of n to k. c. we can estimate that each of the output lines will have n/k the amount of traffic as the input lines. the percentage of time that that the output lines on a switch in the second stage are busy is equal to the percentage of time that its input lines are busy. Solution: For any switch to be non-blocking. Consider the multistage switch in Figure 4. Supposing that the blocking probability of the intermediate crossbar is small. n = 4. P[none of the paths available] = (2p)k 28. The resulting number of crosspoints necessary for different values of n is shown below. what is the probability that none of the N/n paths between the input and output lines are available? In this question N/n should be k. which is not the case generally. one of the 2 jumps must be unavailable. For a one-stage N x N switch with n = 32. if multicasting is implemented in a switch. for k much less than n. A multicast connection involves transmitting information from one source user to several destination users. Solutions follow questions: a. c. just as was noted in question 25c. Thus the probability of blocking is increased because the traffic inside the switch is higher. Explain how a multicast connection may be implemented in a crossbar switch. the number of crosspoints in a three-stage non-blocking Clos switch can be calculated as: 2N(2n – 1) + (2n – 1)(N/n)2 This number can be minimized with the correct choice on n by taking the derivative of the above expression with respect to n and setting the result to zero: d 2N 2 N 2 [4nN − 2 N + − 2 ]=0 dn n n 2N N2 4N − 2 + 2 3 = 0 n n . branching should be done as late as possible. multistage switches can provide good hardware economy while remaining non-blocking. flow in multicast switches is not conserved. 29. Unicast traffic is also affected by this increased flow because fewer resources in the switch are available. Generally. Thus. we would require 1032 crosspoints. How does the presence of multicast connections affect blocking performance of a crossbar switch? Are unicast calls adversely affected? Unlike unicast switches. to keep traffic load in the switch to a minimum. the total traffic flow out of the switch is more than the flow into the switch. each branching procedure represents an increase in traffic flow beyond the branch point. Thus we see that. b. Show that the minimum number of crosspoints in a three-stage nonblocking Clos switch is given by n equal to approximately (N/2)1/2 Solution: Given the number of inputs N. multicasting can be implemented by copying the multicast input to multiple outputs. In a crossbar switch. Explain how multicast connections should be implemented in a multistage switch. Should the multicast branching be done as soon as possible or as late as possible? In a multistage switch with a given multicast traffic profile. 30. 544 Mbps 24 slots 24 slots 24 slots 24 slots A B C D A A A A 6.176 Mbps 1. What is the delay incurred in traversing a TSI switch? Solution: The arriving frame is written onto the switch register and after the frame is completely written in. one half as many slots per frame can be supported as compared to when each slot requires one PCM sample. it needs to be read out in permuted order. the maximum number of channels that can be supported is: Max = 125µs 4 × mem In other words.N (1 − n) = 2 n3 Assuming n>>1.544 Mbps D A 33.544 Mbps B A 1. Each frame that is output is of length 4n = 96 slots. Explain how the TSI method can be used to build a time-division multiplexer that takes four T-1 lines and combines them into a single time-division multiplexed signal. Suppose that the frame structure is changed so that each frame carries two PCM samples. Thus. The input frame size is n = 24 slots. Solution: Each input line carries T-1 traffic. Does this affect the maximum number of channels that can be supported using TSI switching? Solution: Yes. Be specific in terms of the number of registers required and the speeds of the lines involved.544 Mbps C A 1. . this reduces to: n2 = N/2 n= N 2 31. the slots are read out in the outgoing line. Now each slot requires two memory reads and two memory writes. 32. After a full set of 96 slots is written into the 4 registers. A A 1. for a total switching time of T.34. The implicit tradeoff. which in turn places a limitation on N. Consider n digital telephones interconnected by a unidirectional ring. smaller physical size in the resulting switching system. Suppose each telephone has a designated slot number into which it inserts its PCM sample in the outgoing direction and from which it extracts its received PCM sample from the incoming direction. Examine the time-space-time circuit-switch architecture and explain the elements that lead to greater compactness. each of the users inserts outgoing PCM samples on the slot and extracts incoming PCM samples from the same slot. however. Explain how a TSI system can be used to provide the required connections in this system. If user A requests a connection to user B. To communicate. say M. 36. the switching matrix needed at any given time is reduced in size accordingly and greater hardware compactness is achieved. How would connections be established in parts (a) and (b)? In the system described in part (a) the TSI switch monitors all of the slots continuously for connection request made by the users. that is. n inputs are switched simultaneously in a switching time of T seconds. The central switching mechanism would assign each pair of users a timeslot to . Solution: In a space switch. The maximum number of slots in a TSI switch. By switching the inputs one at a time instead of simultaneously. This limitation means that N/n ≤ M. Solution: A space-time-space implementation would have the middle stage replaced by a TSI switch stage. Consider the three-stage switch in problem 25c. TSI is used to swap the information in the time slots of all pairs of connected users b. Solutions follow questions: a. The space-time circuit switch switches each of the n inputs in succession with a switching time of T/n each. The limitation on the number of slots M suggests that the approach is suitable when concentration can be used in the input stage because of light loading in the input lines. 35. is limited by the length of the memory cycle. In the system described in part (b) a mechanism is needed to coordinate the sharing of slots between all the pairs of users. for any connected pair of users we can assign the same timeslot to both. The most efficient mechanism would have half as many timeslots as users. c. and thus reduce hardware costs. the central switch controller makes the connection if the slot corresponding to user B is available. Identify the factors that limit the size of the switches that can be built using this approach. is that higher line and switch speeds are needed. At some point in the ring. Explain how the TSI system can be eliminated if the pairs of users are allowed to share a time slot? If all user pairs share a time slot. Explain why a space-time-space implementation of this switch makes sense. Suppose that transmissions in the ring are organized into frames with slots that can hold one PCM sample. the number of inputs to the switch (since n has a maximum that is determined by technological constraints). Consider the application of a crossbar structure for switching optical signals. Consider the multistage switch structure in Figure 4. λn. The optical transmission rate is R bps and the actual information bit rate is Reffective. The jth such signal is routed to jth crossbar switch in the middle stage. λ2. the frame sizes are 1000 times longer. Solutions follow questions: a. For T = τ = 1 millisecond. a call could be initiated by sending a sample on the intended receivers time slot (provided it is free). Assuming that the users have time slot tunability. so the incoming optical signals must have guardbands to allow for this. b. each with a different wavelength. the number of bits in each “frame”. we have: Reffective = (total bit rate of connection)(proportion of useful bits) = R ∗( T n )= T +τ T +τ The value of T that yields 50% efficiency is T = τ. Suppose an optical signal contains n wavelength-division multiplexed signals at wavelengths λ1. For T = τ = 1 microsecond. each single wavelength signal inputted to the second stage can be directed to any one of the third stage elements. …. Consider a 2 x 2 crossbar switch and suppose that the switch connection pattern is (1 → 1. Explain how the resulting switch structure can be used to provide a rearrangeable optical switch. and letting n be the number of bits per frame (including overhead). Which functions are the crosspoints required to implement? The crosspoints must be able to open. The drawback of this mechanism is that no more than half of the available bandwidth would ever be used. find values of T that yield 50% efficiency in the use of the transmission capacity. into a WDM signal. Communication would occur on the receiver’s time slot in this scenario. Each third stage element corresponds to a single output port. The operation of this switch is very . 37. Call setup would require all users to communicate with the central switch on a setup slot that would have to be accessed via a MAC protocol. An alternative simple distributed mechanism would involve each user monitoring their own timeslot. close. and hold closed. Thus. 2 → 1) for T seconds. we have n = RT which gives frame size of n = 1000 bits at 1 Gbps and n = 1000000 bits at 1 Tbps. and the values T and τ? For R in the range from 1 gigabit per second to 1 terabit per second and τ in the range of 1 microsecond to 1 millisecond. It is assumed that the third stage consists of elements that recombine the n incoming signals. and the hold time is T seconds. hold open. the second-stage inputs to any given switching element have the same wavelength.communicate over. 38.23 and suppose that the first stage switches consist of an element that splits the incoming optical signal into n separate optical signals each at one of the incoming wavelengths. The actual bit rate between each input/output port is Reffective/2. Suppose it takes τ seconds to change between connection patterns. The transition time is τ seconds. Solution: In the resulting architecture. 2 → 2) for T seconds and (1 → 2. Calculate the relationship between the bit rate R of the information in the optical signals. Suppose first that the local loop carries analog voice. Repeat part (a) in the case where the two telephone calls are in different LATAs. the telephone set is connected by a pair of wires to the telephone office.similar to that of the switch in figure 4. Sketch a diagram showing the various equipment and facilities involved. The digital signal travels along a fixed path from the telephone switch in the originating office to the telephone switch in the destination office. The operation of the line card would differ. c. 39. The wires are terminated in a line card that digitizes the voice signal and transfers the digital signal to a digital switching system. In parts (a) and (b) identify the points at which a call request during setup can be blocked because resources are unavailable. as the arriving signal would already be in digital form.23. A modem can then deliver information at 56 kbps to the user. Suppose that an Internet service provider has a pool of modems located in a telephone office and that a T-1 digital leased line is used to connect to the ISP’s office. Modem Twisted Pair User’s comp Telephone Network T-1 line ISP office Modem bank . b. Solutions follow questions: a. then suppose it carries digital voice. 40. If the two telephones are in different LATAs then the digital voice signal must traverse tandem switches that connect the two LATAs. Recall that the user cannot send at 56 kbps because of limitations on the available SNR. Solution: The ISP office is connected to the user’s local central office. These resources can include time slots and crosspoints in the switches and time slots in the digital lines. Explain how the 56K modem (that was discussed in Chapter 3) can be used to provide 56 kbps transfer rate from the ISP to the user. The two or more offices are interconnected by digital transmission lines. it differs in that there is only one possible route for each input-output-wavelength combination. Consider the equipment involved in providing a call between two telephone sets. If the telephone set were to generate digital voice then a digital signal would be transmitted along the local loop that connects the telephone to the central office. In an analog connection. Call requests can be blocked when there are insufficient resources to complete the call. Sketch a diagram showing the various equipment and facilities between the originating telephone through a single telephone switch and on to the destination telephone. The ISP can use the T-1 line to send digital information at a rate of 56 kbps into and across the telephone network to the line card that connects to the user. which is in turn connected to the user. However. No. Since the television network is inherently a broadcast. bandwidth-limiting elements. must be removed. c. 42. In this problem we compare the local loop topology of the telephone network with the coaxial cable topology of cable television networks (discussed in Chapter 3). a method such as the cable modem method described in chapter 3 would be needed to accommodate the traffic of predominantly point-to-point telephone service. An ADSL modem is required at each end of the user’s twisted pair connection. voice may be routed over the phone network and data over the Internet). Explain how ADSL is deployed in the local loop. shared medium. 43. how does the network know which inter-exchange carrier is to be used to route a long distance call? Solution: The inter-exchange carrier that handles the long-distance calls for a specific user is specified a priori by the user’s service contract. It would be very difficult to provide cable television on the local loop network due to the huge bandwidth currently provided by the cable broadcast network. ADSL and ISDN use overlapping parts of the signal spectrum so they cannot operate simultaneously. Solutions follow questions: a. The only effective means of providing full cable service over the local loop would likely involve sending a single channel to .41. What happens after the twisted pairs enter the telephone office? At the telephone office. This power is stored in the form of wet batteries that can alternately be charged by a backup battery in the event of a power failure at the central office. and it is the responsibility of the local service provider to route the call over the appropriate inter-exchange carriers. In Figure 4. 44. Solutions follow questions: a.32b. Why does the telephone still work when the electrical power is out? Solution: The telephone company supplies each of its telephone lines with power at the central office. such as loading coils. The user “signs-up” with one long distance plan over another. b. Since ADSL uses higher frequencies for data transmission. the voice and data are separated/filtered and the demultiplexed signals are routed accordingly (for example. Explain how cable television service may be provided using the local loop. b. The mechanism used would include a MAC protocol of some form to provide a fair method of partitioning the total available bandwidth among callers. Can ADSL and ISDN services be provided together? Explain why or why not. ADSL was designed to provide high-speed digital access using existing telephone facilities. Explain how telephone service may be provided using the cable television network. In a ring topology. this architecture should be used if the connections are very reliable. Thus. Currently ADSL typically provides bandwidth of a few Mbps. the total bandwidth of the network is divided evenly among all of the users connected to the final branching point of the network. c. Compare both topologies in terms of providing Internet access service. which would then replace the originally transmitted channel with the requested one. When a user is not transmitting any information in a star topology. Cable modems have the advantage of providing huge peak bandwidth via that large bandwidth (up to 30 Mbps) of the cable network. 45. . since the cable network is a shared broadcast medium. Changing channels would require an upstream request to the local switch. the bandwidth provided does not scale well to many users. In general. As a result. active users can make use of the extra available bandwidth provided by the inactive users. However. Compare the star-star topology with a star-ring topology and a ring-ring topology. as has been discovered by many subscribers to this service. Solutions follow questions: a. ring networks are more efficient.each user. The three topologies under examination are shown below. each user’s service access rate is independent of the number of users. since the bandwidth is shared. During peak hours. Explain how information flows in these topologies and consider issues such as efficiency in use of bandwidth and robustness with respect to faults. service for many users is lost. the bandwidth of the line to the user is wasted. Thus. Star topologies generally provide efficient use of bandwidth because their hierarchical routing structure provides near shortest path routing. in situations with low load. Both these networks currently provide Internet access. The local loop was described as having a star topology in the feeder plant and a star topology in the distribution plant. However. such architecture is only good if the lines are very reliable. This leads to noticeable and significant decrease in service. Access to the Internet via the phone network using DSL modems has the advantage of providing a dedicated access line to its users. If one of the lines on a high tier star fails. The tradeoff is the lower bandwidths available. pure star topologies provide no mechanism for fault line recovery because there is not backup path. The star-star topology is efficient when the load on each of the lines is high. In this case. the total bandwidth on the fiber must be at lest 500 × 8 Mbps = 4 Gbps.5 Gbps on a single wavelength fiber is common and 10 Gbps is available. but the traffic in the “star” links is high. Trace the flow of PCM samples to and from the user. 46. Solutions follow questions: a. Let’s consider an approach for providing fiber-to-the-home connectivity from the central office to the user. and each fiber handles 500 users. . Discuss how Internet access service might be provided using this approach.The star-ring topology is good when the traffic produced by each user is relatively sporadic. Thus.5 Gbps. N = 10 à 640 kbps minimum capacity N = 100 à 6. The ring-ring topology is very robust to faults because every path can be backed up by forming the backup path around the ring in the other direction.4 Mbps Currently bit rates of 2. What bit rate does the optical fiber have to carry if each pedestal handles 500 users? If ADSL is used. the bandwidth increases to 55 Mbps x 500 = 27. Each user receives the entire TDM signal and retrieves its own signal. What bit rates are required if N=10. 47. This situation could arise if the bandwidth provided in “star” links is limited in capacity. a bandwidth of 55 Mbps can be provided. the fiber must have enough bandwidth to support N connections. How can SONET equipment be used in this setting? SONET can be used at the physical layer in the optical fiber so that the multiplexing of the 500 lines is simplified using tributaries. 100? Compare these to the bit rate that is available? Because a passive splitter is used. Suppose that the local loop is upgraded so that optical fiber connects the central office to the pedestal and twisted pair of length at most 1000 feet connects the user to the pedestal. What role would SONET transmission systems play in the above topologies? SONET transmission systems could be used in the architectures described. b. Solutions follow questions: a. If VDSL is used. b. It still has the same vulnerability as the star-star network in term of large scale loss of service if one of the high-tier links fails. It would be particularly useful in the rings to provide OAM functions and fault protection. Since the twisted pair length is only 1000 feet. bandwidth is shared and therefore used efficiently. b. The telephone conversations of users are time-division multiplexed at the telephone office and broadcast over a “passive optical network” that operates as follows. The TDM signal is broadcast on a fiber up to a “passive optical splitter” which transfers the optical signal to N optical fibers that are connected to N users. This functionality could be extended to the pedestal to provide OAM control of the large aggregate bandwidth connecting it to the central office. VDSL may be used. What bandwidth can be provided to the user? Using ADSL a bandwidth of 8 Mbps can be provided. c. In every link in the network. A 4 kHz analog connection across the telephone network. Physical Layer: the actual 4 kHz analog signal exists only in the physical layer of the OSI reference model. Discuss how cable television service might be provided using this approach. flow-control. the hardware costs required would be offset by the benefits.1 in Chapter 1 and explain the signaling events that take place inside the network as a call is set up and released. it would also be on the fiber to the user. The additional wavelengths provide an additional degree of multiplexing flexibility. 49. If the huge amounts of available bandwidth were used to carry heterogeneous traffic. the aggregate bandwidth of the optical fiber would be greatly increased. Solution: • • • Caller picks up phone. but the 64 kbps signal that carries user information is analogous to the 4 kHz signal used over the twisted pair that runs to the user’s premises. What role could WDM transmission play in the connection between the central office and the optical splitter? in the connection all the way to the user? If WDM is used. telephone. The additional bandwidth might still be leveraged in a number of ways. For example. Explain where the following fit in the OSI reference model: Solutions follow questions: a. b. Data-Link Layer: a 33. The resulting network could carry television. Physical Layer: the digital link across the network is controlled by many higher layer functions. Because the architecture is a passive broadcast system. on the order of 1000 channels could be provided by this network to each user. c. a WDM receiver typically represents significantly higher hardware costs.Internet access could be very simply provided by replacing the PCM samples with Internet traffic. in effect. Since MPEG2 television signals are 2-5 Mbps. A 64 kbps digital connection across the telephone network.6 kbps modem uses framing. c. and Internet traffic and. Identify the equipment and facilities involved in each phase of the call. 48. replace all of the currently existing access networks. Although WDM provides increased bandwidth. d. triggering the flow of current to on the twisted copper wire to the central office Current is detected and dial tone transmitted to user indicating that line is ready User pushes the keys on the phone and sends the number to the central office (in form of tones or pulses) . Refer to Figure 1. The users could change channels by resynchronizing their receivers to the appropriate time slot. A 33. and error correction to connect a user to the switch. Each TDM slot could correspond to one channel. the users could tune into one wavelength at a time (with tuning done on a relatively large time scale) and each wavelength could carry a “cable package“ that includes a subset of the provided channels.6 kbps modem connection across the telephone network. if WDM were used between the central office and the passive splitter. Call answer – Takes voice mail message if you are on the phone or do not answer the phone. the analog voice signal is digitized into PCM pulses The samples are carried via a number of DCCs within the network between the central offices When either party hangs up. When the ring signal is sent to the destination phone. . when a user is on the phone and a call arrives (assuming no call waiting). if the call is not answered. In another scenario. The results of the query are sent to the SCP. via the signaling network to the receiver phone. or the call is denied. The appropriate recorded message is sent back to the user. While the phone is ringing. the call is in transfer phase and a dedicated connection is established between the two users At the wire card interface. Once the receiver picks up the phone. c. Solution: Before a connection is made. a. then number used for routing can also be forwarded. the database is referenced to find the forwarding number and the call is rerouted accordingly. The credit card number is verified and may be used to check for the authorization of funds. if it subscribes to caller ID. 51. which handles the caller accordingly. When the call request reaches the SCP and call forwarding is activated. The receiver phone must have the capability to interpret the signal. The user inputs the digits and the tones are converted to numbers and the message sent to a database in the appropriate SCP. Although the call can be incrementally rerouted from this point. Call forwarding – Allows you to have calls transferred from one phone to another where you can be reached. Identify the equipment involved in each phase. instead of generating a busy signal. Based on the query.• • • • • • • • User information is split at the SSP to the control plane and the user plane The tones/pulses are received at the central office where the switch converts them to a number that specifies the address of the destination of the call The call is routed via the control plane to the receiving user using SS7 MIP3 A ring signal is sent to both sides of the connection. Sketch the sequence of events that take place in the setting up of a credit-card call over the intelligent network. After a certain number of rings (specified by the user a priori and stored in an SCP database). the call is accepted and a connection made. identifies the services that have been subscribed). the number of rings must be counted somewhere in the signaling network. a recorded message requests the credit card number for the call. the switch of the hanging up party terminates the call and the resources are freed for another call. 50. b. bandwidth may be conserved by rerouting the connection again from the start to the forwarding number. Explain how the intelligent network can provide the following services: Solutions follow questions: Before a connection is made every call has its signaling information in the control plane routed through an SCP that stores the destination phone's profile (that is. Caller identification – A display on your phone gives the telephone number or name of the incoming caller. it is forwarded by a mechanism similar to call forwarding to a voice mailbox center. the call is forwarded automatically to the voice mailbox. both calls share the line that makes up the last hop to the user. however. In the signaling network. Thus. Assume that two switches are involved in the connection. similar to Morse code. Which of these components increase as the volume of connection requests increases? Solution: . but rather a sequence of entries in forwarding tables in the packet switched network. and a second connection is made from the dialing party to the user. You can place your current call on hold. the current connection is “held” one hop upstream from the user at the local switch. a ring is sent to the dialer and receiver. a distinctive tone informs that you have an incoming local or long-distance call. 52. Identify the components in the delay that transpires from when a user makes a request for a telephone connection to when the connection is set up. Only the physical layer in the user plane of the end systems and the two switches are involved in the 64 kbps connection. while the other number may be two short rings. a different ring signal can be sent to the phone. The circuit does not exist as a physical entity. The phone may have only one ringer. If a line is busy. Called-party identification – Each member of a family has a different phone number. If call waiting is activated. Explain how the control plane from ISDN can be used to set up a virtual connection in a packetswitching network. if a line is free. Assume that the two switches in part (a) are involved in the call setup. Thus. Solutions follow questions: a. and application layers of the control plane are involved in the setting up of the connection. 54. Sketch the layers of the protocol stack in the control plane that are involved in the setting up of the connection. a tone from the local switch is sent to the receiving party (which briefly interrupts the current conversation). At this point. multiple numbers can all be routed to the same phone. e. it sends a ring back to the dialing party and voicemail and call forwarding may still be activated). each packet in the call follows the same route through the network. so to differentiate the numbers each ring can have a different duration with different length link. the other is maintained in its incomplete state while the caller is on hold. a busy signal is sent back to the dialer. 53. Consider a 64 kbps connection in ISDN. Call waiting – If you are on the phone. one number may be one long ring. speak briefly to the incoming party. Is a separate signaling network required in this case? Solution: In order to setup a virtual connection in a packet switched network. the ISDN control plane must first reserve resources during a call setup phase. The control plane treats this tone as a ring (that is. the user can only talk to one party at a time.d. Generally. As such. b. and return to the original call. Incoming calls have a different ring for each phone number. When one of the two connections is being used. For each number. The physical. Sketch the layers of the protocol stack in the user plane that are involved in the connection. If the user accepts the new call. a path must be selected for the call using the signaling network. 55.a) 0. the connection can be made and the delay is equal to the processing time in the switch and the short propagation delay of the signal.820 0. Hint: Use the recursion in problem 56. the signaling network will have to juggle more requests simultaneously and the routing procedure will take longer.013 57.564 0.215 0.120 0. as call requests increase. The amount of time required for the user to answer the does not depend on the volume of connection requests in the network.084 0. The delay from this stage includes. that is. the receiving user has to pick up the phone. Although there is a fixed minimum propagation delay. As shown in figure 4.After the user requests a connection by dialing the phone number. if the two phones are attached directly to the same switch. assuming the switch is non-blocking. Discuss the fault tolerance properties of the STP interconnection structure in the signaling system in Figure 4. propagation delay of the signals through the network. Thus.036 0. a = 90 and c = 100. a) B (0.40. A set of trunks has an offered load of 10 Erlangs. . How many trunks are required to obtain a blocking probability of 2%? Use the following recursive expression for the Erlang B blocking probability: B (c . Compare the blocking probabilities of a set of trunks with offered load a = 9 and c = 10 trunks to a system that is obtained by scaling up by a factor of 10. the propagation delay may also increase because the shortest path route may not be available. 56.022 0. if there are many simultaneous requests for connections.732 0. a ) = aB(c − 1. Solution: The mesh structure means that any STP is connected to any other STP by more than one path.a) 1.163 0.000 0.02. However. Before a connection can be made.485 0.909 0. a) c + aB(c − 1. and any routing delay overhead.647 0. processing time in each of the signaling nodes involved. the signaling traffic can be automatically rerouted over another path. c 0 1 2 3 4 5 6 7 8 a 10 10 10 10 10 10 10 10 10 B(c. If the two phones are not connected to the same switch. This also contributes to the overall connection setup time.057 0. 16 trunks are required to a obtain PB = 2%. the call setup procedure involves finding a path from the source to the destination.338 c 9 10 11 12 13 14 15 16 17 a 10 10 10 10 10 10 10 10 10 B(c. if a fault occurs on one link. a) = 1 Solution: Given a = 10 Erlangs and PB = 2% = 0.32a.409 0.273 0. If an STP fails. paths are available to a backup STP. 100 a 90 90 90 90 90 90 90 90 . a lower probability of blocking is achieved by scaling up the system. We use a similar method for the case when a = 90 and c = 100.989 0. Calls have an average duration of 25 minutes. c 0 1 2 3 4 5 6 7 8 9 10 a 9 9 9 9 9 9 9 9 9 9 9 B(c. The results of this process are summarized in the table below. .168 c 0 1 2 3 4 5 6 7 .967 0.105 j =0 c b.516 calls/minute 1.945 0.Solution: To find the blocking probability of trunks with offered load a = 9 and c = 10.978 0. Consider dynamic nonhierarchical routing (DNHR).706 0.a) 1. the utilization a = 2/(1/25) = 50 and c = 50.614 0.000 0. Calls arrive to a pool of 50 modems according to a Poisson process.289 0.802 0. 90 B(c.027 As shown. the maximum handled arrival rates. 58. .525 0.a) 1.224 0. we recursively calculate B(10. .9). Solutions follow questions: .934 0. What is the maximum arrival rate that can be handled if the maximum acceptable blocking probability if 1%? 10%? If the maximum acceptable blocking probability is 10% and 1%. The probability of blocking is: Pb = ac c! aj ∑ j! = 0.9) starting from B(0.441 0.000 0.982 calls/minute 59.362 0.956 0. What is the probability an arriving calls finds all modems busy if the arrival rate is two calls per minute? For this system. 0. given a service rate of 1/25 calls per minute are: Pb 1% 10% Load [Erlangs] 1.900 0.923 . Solutions follow questions: a. and assuming that this probability is independent of all others. explain why it makes sense to give priority to call requests that are almost completed rather than to locally originating call requests. the business day begins and long-distance calls are made between cities on the East Coast. The west-coast business day has not yet begun. but all resources are used by partially completed calls. . The opposite can be done in the evening when most people are in the residential areas and the telephone traffic in the business area is low volume. Thus they should be given priority. including guard bands. P[every circuit free] = P[1st free]P[2nd free]…P[Nth free] = pN b. resources are already locked up. One can envision a situation where very few completed connections exist. 60. Solutions follow questions: a. b. 61. Suppose that each segment is available with probability p. Evaluate the number of channels in each cell for the AMPS system. so if the call is denied. What is the probability that a call request can be completed? For a call to be completed. so the links from east to west are relatively under-utilized. Find an expression for the number of channels available in each cell. ever segment in the circuit must be free. For calls that are almost completed. suppose locally originating calls are given a higher priority. b is the bandwidth required by each channel. Explain how DNHR can be used to exploit the time differences between different time zones in a continent. Thus. In the morning in the east. there is a large overhead penalty. The opposite can occur in the evening after the business day in the east ends and the under-utilized eastern trunks are used for western traffic.a. To illustrate. Explain how DNHR can be used to exploit different business and residential activity patterns during the day. traffic may be re-routed through relatively non-busy lines in residential area to free up resources in the business areas. assuming that each switch and link has a probability p of being available. Suppose that the setting up of a call requires reserving N switch and link segments. Consider a cellular telephone system with the following parameters: B is the total bandwidth available for the system for communications in both directions. and a is the fraction of channels used for set up. Solutions follow questions: a. B (1 − a) b Number of channels available in each cell = channels/cell R b. Eastern traffic can be re-routed through the western free lines to provide additional capacity for the traffic between eastern cities. In allocating switch and transmission resources. For local calls during business hours. R is the re-use factor. 059 Explain why requests for channels from handoffs should receive priority over requests for channels from new calls. the number of channels in each cell is: B (1 − a) b n R b. How many Erlangs of traffic can be supported by the channels in a cell with a 1% blocking probability? 5%? The probability of blocking is given by the equation: Pb = ac c! ∑ aj j = 0 j! c where in the AMPS system c = 56 and a is the number of Erlangs of traffic in a cell. thus. a higher Pb. The calculations in part (a) assumed equal priority for all calls. Assuming that there will be n times more control channels in this new scheme. Solutions follow questions: a. How does this change the Erlang load calculations? In telephone conversations. Solutions follow questions: a. the maximum number of Erlangs that can be handled is: Pb 1% 10% b. Consider the AMPS system and assume that it is converted to digital format using n = 3. Newly attempted calls in this scenario have a lower priority and. Thus the number of channels per cell is: (416 –21)/7 = 56 channels per cell 62. Suppose that an analog cellular telephone system is converted to digital format by taking each channel and converting it into n digital telephone channels. Consider the AMPS system in problem 61. there are 416 total channels.342 56. Load [Erlangs] 43. except that now the number of trunks is tripled. The reuse factor is 7. How many Erlangs of traffic and the new system support at 1% blocking probability? 5%? This calculation is exactly the same as that in problem 4.In AMPS. the Erlang load calculations are more complicated. If priority is given to current calls that are being handed off. service interruption is less acceptable than denial of service due to a busy network. If the blocking probability is 10% and 1%. Find an expression for the number of channels that can be provided in the digital system using the parameters introduced in problem 61. 63. . so c = 160. 21 of which are used for call setup.62. To support 1% blocking or 5% blocking. Suppose the per capita traffic generated in a city is 0. so we have: c = 395 a = Erlangs of traffic The probability of blocking in this system is given by: Pb = ac c! aj ∑ j! j =0 c Thus we have: Pb 1% 5% Load [Erlangs] 370.856 b.410Erlangs/cell) = 270 Cells For a probability of blocking of 5%.10 Erlangs/person)(1 000 000 people) / (401. For a probability of blocking of 1%. How many Erlangs of traffic can be supported in each cell by this system at 1% blocking probability? 5%? This CDMA system has 416 channels. but with a reuse factor of 1. Assuming that 21 channels are still used for call setup and the reuse factor is 1. Total number of cells in city = (0.410 401.167 157. The city has a population of 1 million residents and the traffic is generated uniformly throughout the city.10 Erlangs during the busiest hour of the day.046 64. Suppose that a CDMA system has the same number of channels as the digital system in problem 63. Estimate the number of cells required to meet the cities traffic demand using the system in part (a). Solutions follow questions: a.856 Erlangs/cell) = 249 Cells . Total number of cells in city = (0. each cell has 395 channels. the traffic can be: Pb 1% 5% Load [Erlangs] 141.10 Erlangs/person)(1 000 000 people) / (370. 65. Suppose first that the mobile carries analog voice. Sketch a diagram showing the various equipment and facilities between an originating mobile telephone to a wireline destination telephone. Consider the equipment involved in providing a call between mobile and wireline telephones. then suppose it carries digital voice. Solutions follow questions: a. Analog case: Mobile Station Base Station Mobile Switching Center Base Station Mobile Station Mobile Station Base Station Mobile Switching Center Telephone Switch To telephone network Digital case: Mobile Station Base Station Mobile Switching Center Base Station Mobile Station Mobile Station Base Station Mobile Switching Center Telephone Switch To telephone network a Home Location Register Authentication Center . . Analog case: Mobile Station Base Station Mobile Switching Center Base Station Mobile Station Digital case: Mobile Station Base Station Mobile Switching Center Base Station Mobile Station Home Location Register c.b. it is unlikely that the call can be blocked between the mobile switching center and the telephone switch (in part a). Explain the signaling events that take place when call is set up and released in a cellular telephone system. 66. For this same reason. The call cannot likely be blocked between the Base station and the mobile switching center since it is assumed that this link has a capacity at least equal to that of the aggregate traffic from the base station. Whether or not the destination phone is in the same cell as the source phone. Consider the following cases. Repeat part (a) in the case where the two telephone calls are mobile. Solutions follow questions: a. so it may be possible that a call can be blocked at this point if the entire network experiences a lot of traffic. Identify the equipment and facilities involved in each phase of the call. because the base station does not know a priori whether the destination phone is in its cell. The call can be blocked between the mobile user and the base station if no frequency channels are available. The cellular service provider leases a link to the ground-based telephone network. Authentication Center In parts (a) and (b) identify the points where a call request can be blocked during call setup because resources are unavailable. The source and destination mobile phones are in the same cell. the procedure for call setup may still be the same. 68. Explain the signaling events that take place when call is handed off from one cell to another cell. the data must be reformatted to accommodate the signaling systems of the downstream network. the MSC communicates to neighboring MSCs to locate the mobile station. The base station also signals the MSC. but in the same MSC. Bob Smiley. which signals all neighboring base stations to monitor the setup channel. the base station is informed and the new connection is made through the new MSC. Mobile station sends a request in reverse setup channel along with user information such as phone number. The process is centralized if the mobile station moves to a cell that is connected to the same MSC as the original cell. other information about the signal is recorded at each base station and sent to the MSC. He turns on his cell phone to contact his Los Angeles client. The MSC contacts Bob’s home location register (in Chicago). Suppose first that the two cells are under the same MSC. and then that they are under different MSCs. and copies Bob’s database . If the number is not stored in the MSC database. When it drops below a certain threshold. and authentication Base station verifies information with MSC MSC sends call request to all base stations Base Stations broadcast request to all forward setup channels Receiving phone replies on reverse setup channel All base stations that receive this reply forward it to the MSC along with other information such as the power level of the received signal. At the cellular-wire-line network interface. Otherwise it is possible that a more distributed control mechanism is used. possibly. c. Solution: The signal level must be periodically monitored when a call is in progress. star salesman. Solutions follow questions: a. has just arrived in Los Angeles from his home base in Chicago. serial number. See answer for part (a). which sends user information to the MSC via the nearest base station. Sketch the sequence of events that takes place to set up his call. The source and destination mobile phones are in different cellular networks. If. after receiving the information about the signal quality at each base station. Bob Smiley turns on his phone. The service to the mobile phone is briefly interrupted during this process. the base station instructs the mobile station to transmit on a specific setup channel. This information helps the MSC to assign the connection channels MSC identifies which base station is best suited to service the call for the receiving phone The base station assigns a forward and reverse channel and the receiving phone is rung The source and destination mobile phones are in different cells. If the mobile station is found in a cell of one of these neighboring MSCs.• • • • • • • • b. it must go through the wire-line phone network which in turn will interface with the receiving phone’s cellular network as described above via the MSC and the closest base station. which uses it to decide which cell is best suited to receive the handoff. 67. The selected base station is informed and the old connection from the MSC to mobile station via the original base station is broken and a new connection via the new base station is established. The signal strength and. the MSC decides that none of the cells are suitable to receive the handoff. Assume he subscribes to roaming service. Consider the following two cases: . Now Bob can use his phone in the new network area. b. When the call arrives at Bob’s cellular network (in the form of signaling traffic).information into its own visitor location register (in LA). In this case it could be possible to bill them both locally. Compare cellular wireless networks to the local loop and to the coaxial cable television system in terms of their suitability as an integrated access network. say. A. Should this call be billed as local or long distance? Kelly’s long-distance service provider sees Bob’s Chicago area code and proceeds with connecting her to Bob’s Chicago cellular network. Bob’s home MSC is not aware that his calls should be forwarded to LA. Sketch the sequence of events that take place to set up this call. MSC Bob’s Mobile Station Base Station Telephone Network Phone Switch Visitor’s location register Wife’s phone c. If the three networks involved (two cellular and one wireline) are coordinated enough. She will pay the same as if she were dialing a regular Chicago number (which she is). Bob’s college roommate. and Bob will pay as if he is receiving a call from a Chicago caller. 69. Further. his wife. By using his phone in L. it is more likely that the connection will actually be made from Kelly to Bob’s home network and then back to L.) L. high-speed Internet access. Sketch the sequence of events that take place to set up this call. both Bob and Kelly will pay long distance charges. who now works in Hollywood calls Bob’s cell phone. Bob’s home MSC checks its home location register and sees that Bob is in LA (recall that it found out about his whereabouts in part a)). However. Kelly. and digital television service. The call is then forwarded back to LA where the connection can be made via Bob’s temporary MSC and its visitors’ location register. By dialing long distance. Next he calls home in Chicago to inform his wife that he forgot to take out his son’s hockey gear from the trunk of his car and to give her the parking spot where he left the car in the airport (“somewhere on the third level of parking lot A”). Note that Bob’s cell phone begins with the Chicago area code. Whether Bob or Kelley should be billed long distance depends on the setup of Bob’s “roaming service”. (Don’t concern yourself with the specifics of the conversation. Kelly implicitly agrees that she will pay the long distance charges for a call to Chicago. Bob expects to pay long distance fees for all calls he receives (even local ones) because if someone from Chicago calls him. In particular comment on the ability to support telephone service. A. In the meantime. the call must be carried by a long distance carrier. In this case. Kelly’s final connection could be routed directly from Hollywood to Bob’s mobile station directly. A. we examine the ability of cellular wireless networks to provide these services. If the band of frequencies in the network span 300 MHz (about 10x the current band).8 kbps voice transmission and 2. The main disadvantage was cost. and the frequency reuse factor is 1. assuming each cell has on the order of 100 users. most of the problems mentioned above would melt away. Service can most efficiently be provided in a broadcast network because it lacks the interactivity of phone and Internet. IP traffic can also be adaptive. Although the signal can be compressed using MPEG encoding. but due to the bandwidth restriction mentioned. The band of frequencies available spans 300 MHz. Compare the capabilities of the Iridium system with conventional cellular telephony networks. Telephone service is less constrained by bandwidth. Of course cells can be made small. b. . it does not interrupt service completely. but unlike the other two (phone and TV). the service quality is moderate. It would difficult to provide digital television using the above-described cellular network. point-to-point networks are well suited for carrying phone traffic. Current cellular networks have only moderate bandwidth. The main advantage of the Iridium system was that it provided truly global coverage with the network of satellites that communicated directly with mobile stations. Here.4 kbps or 4. the connection can usually be slowed down. This is slightly lower than that of conventional cellular systems. 70. Since the aggregate bandwidth of this system is 100 times that of the current cellular networks. a. and CDMA can be used so that the reuse factor is 1. If the band of frequencies available spanned 2 Ghz.4 kbps data transmission. the compressed signal still requires a minimum of 2 Mbps. Clearly. New protocols are currently being defined that reduce bandwidth for Internet services that run of devices of these networks. Each type of traffic on an integrated access network has very different characteristics. and each user can be sent the same information. and would likely be sufficient to serve as an integrated access network. but a bottleneck still exists. in the sense that if need be. The Iridium receivers communicated directly with satellites and were therefore very expensive and relatively large in physical size. Because of the high bandwidth required. the bandwidth available to each user is inversely proportional to the number of users. a broadcast network could provide digital television service. the local loop and coaxial cable network were compared in terms of their ability to act as integrated access networks. Digital television requires high bandwidth and low loss. all of the above obstacles would be significantly reduced. we see that a bandwidth on the order of 10 Mbps can be provided to each user (or to each channel in a 100-station digital television network). The band of frequencies available spans 2 GHz. Current high speed Internet access has medium bandwidth requirements. Because the vast majority of calls are between two users. Cellular networks share the same scalability problem as the coaxial television network. This is comparable to current ADSL in the local loop network. the cellular network can support telephone traffic. by partitioning the network into a number of parallel channels.Solutions follow questions: In question 44. This affects the quality of service. With a rough calculation. since it currently already does. Solution: Iridium provides 2. Successful efforts have been made to provide Internet service on mobile cellular devices. Compare the capabilities of the Teledesic network with terrestrial fiber networks. Solution: The link bandwidth provided by the Teledesic network will be considerably less than that of the terrestrial fiber network. Iridium users could expect to experience poorer voice quality (as compared to cellular phone users) via cumbersome handsets. It provides instant access to the Internet backbone for anyone. As a result. As such. Most users make a large percentage of their calls in one city or region. it is estimated that Terabit per second aggregate speeds can soon be achieved in the fiber optic backbone network. anywhere. Thus. Using WDM. Teledesic’s main benefit over the existing terrestrial fiber network is its global availability. . Just as with Iridium.As a result. it is no surprise that the Iridium network is no more. the advantage of true global coverage did not make the service any more attractive to most users. its initial target user would likely be somebody in a relatively isolated area where Internet access would be difficult to achieve by another means. 71.
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