1. The formula is the standard formula for any M/M/1 Markov queueing given in Sec. 4.1.1, namely, T = 1/(µC − λ). Here, C = 108 and µ = 10−4, so T = 1/(10000 − λ) sec. For the three arrival rates, we get (a) 0.1 msec, (b) 0.11 msec, and (c) 1 msec. For case (c) we are operating a queueing system with ρ = λ/µC = 0.9, which gives the 10× delay.
2. With pure ALOHA, the usable bandwidth is 0.184 × 56 kbps = 10.3 kbps.
Each station requires 10 bps, so N = 10300/10 = 1030 stations.
3. With pure ALOHA, transmission can start instantly. At low load, no collis-ions are expected so the transmission is likely to be successful. With slotted ALOHA, it has to wait for the next slot. This introduces half a slot time of delay.
4. Each terminal makes one request every 200 sec, for a total load of 50 re-quests/sec. Hence, G = 50/8000 = 1/160.
5. (a) With G = 2 Poisson’s Law gives a probability of e−2. (b) (1 − e−G)ke−G = 0.135 × 0.865k.
(c) The expected number of transmissions is eG= 7.4.
6. (a) From Poisson’s Law again, P0= e−G, so G = −lnP0 = −ln 0.1 = 2.3.
(b) Using S = Ge−G with G = 2.3 and e−G = 0.1, S = 0.23.
(c) Whenever G > 1, the channel is overloaded, so it is overloaded.
7. The number of transmissions is E = eG. The E events are separated by E − 1 intervals of four slots each, so the delay is 4(eG− 1). The throughput is given by S = Ge−G. Thus, we have two parametric equations, one for delay and one for throughput, both in terms of G. For each G value, it is possible to find the corresponding delay and throughput, yielding one point on the curve.
8. Slotted ALOHA achieves a higher maximum throughput than pure ALOHA but incurs higher wait times for a packet to be transmitted, leading to longer delays. Persistent and nonpersistent CSMA require additional hardware capa-bilities to sense the carrier (idle or busy) before transmitting a packet. For p-persistent CSMA protocols, lower value of p achieves a higher maximum throughput, but increases the wait time for a packet to be transmitted, leading to longer delays. Finally, nonpersistent CSMA provides higher maximum throughput than p-persistent CSMA but incurs longer delays.
9. (a) Signal propagation speed in twin lead is 2.46 × 108 m/sec. Signal propa-gation time for 2 km is 8.13 µsec. So, the length of contention slot is 16.26 µsec. (b) Signal propagation speed in multimode fiber is 1.95 × 108 m/s. Sig-nal propagation time for 40 km is 205.13 µsec. So, the length of contention slot is 410.26 µsec.
10. The worst case is where all stations want to send and s is the lowest-num-bered station. Wait time N bit contention period + (N − 1) × d bit for trans-mission of frames. The total is N + (N − 1)d bit times.
11. If a higher-numbered station and a lower-numbered station have packets to send at the same time, the higher-numbered station will always win the bid.
Thus, a lower-numbered station will be starved from sending its packets if there is a continuous stream of higher-numbered stations ready to send their packets.
12. Stations 2, 3, 5, 7, 11, and 13 want to send. Eleven slots are needed, with the contents of each slot being as follows:
Slot 1: 2, 3, 5, 7, 11, 13 Slot 2: 2, 3, 5, 7 Slot 3: 2, 3 Slot 4: 2 Slot 5: 3 Slot 6: 5, 7 Slot 7: 5 Slot 8: 7 Slot 9: 11, 13 Slot 10: 11 Slot 11: 13
13. The number of slots required depends on how far back in the tree one must go to find a common ancestor of the two stations. If they have the same parent (i.e., one level back), which happens with probability 2−n, it takes 2n + 1 slots to walk the tree. If the stations have a common grandparent, which happens with probability 2−n + 1, the tree walk takes 2n − 1 slots, etc. The worst case is 2n + 1 (common parent), and the best case is three slots (stations in dif-ferent halves of the tree). The mean, m, is given by
m =
i =0
Σ
n −12−(n − i)(2n + 1 − 2i)
This expression can be simplified to
m = (1 − 2−n)(2n + 1) − 2−(n − 1)
i =0
Σ
n −1i2i
14. Radios cannot receive and transmit on the same frequency at the same time, so CSMA/CD cannot be used. If this problem could be solved (e.g., by equipping each station with two radios), there is still the problem of not all stations being within radio range of each other. Only if both of these prob-lems can be solved, is CSMA/CD a candidate.
15. (a) Since all stations will see A’s packet, it will interfere with receipt of any other packet by any other station. So, no other communication is possible in this case.
(b) B/s packet will be seen by E, A and C, by not by D. Thus, E can send to D, or A can send to D, or C can send to D at the same time.
(c) This scenario is same as (b).
16. Yes. Imagine that they are in a straight line and that each station can reach only its nearest neighbors. Then A can send to B while E is sending to F.
17. (a) Number the floors 1–7. In the star configuration, the router is in the mid-dle of floor 4. Cables are needed to each of the 7 × 15 − 1 = 104 sites. The total length of these cables is
4
i =1
Σ
7j =1
Σ
15√
(i − 4)2+ (j − 8)2or about 1832 meters.
(b) For classic 802.3, 7 horizontal cables 56 m long are needed, plus one vert-ical cable 24 m long, for a total of 416 m.
18. Classic Ethernet uses Manchester encoding, which means it has two signal periods per bit sent. The data rate is 10 Mbps, so the baud rate is twice that, or 20 megabaud.
19. The signal is a square wave with two values, high (H) and low (L). The pat-tern is LHLHLHHLHLHLLHHLLHHL.
20. The round-trip propagation time of the cable is 10 µsec. A complete trans-mission has six phases:
1. Transmitter seizes cable (10 µsec) 2. Transmit data (25.6 µsec)
3. Delay for last bit to get to the end (5.0 µsec) 4. Receiver seizes cable (10 µsec)
5. Acknowledgement sent (3.2 µsec)
6. Delay for last bit to get to the end (5.0 µsec)
The sum of these is 58.8 µsec. In this period, 224 data bits are sent, for a rate of about 3.8 Mbps.
21. Number the acquisition attempts starting at 1. Attempt i is distributed among 2i − 1 slots. Thus, the probability of a collision on attempt i is 2−(i − 1). The probability that the first k − 1 attempts will fail, followed by a success on round k is
Pk = (1 − 2−(k − 1))
i =1
Π
k −12−(i − 1)
which can be simplified to
Pk = (1 − 2−(k − 1)) 2−(k − 1)(k − 2)/2
The expected number of rounds is then just
Σ
kPk.22. For a 1-km cable, the one-way propagation time is 5 µsec, so 2τ = 10 µsec.
To make CSMA/CD work, it must be impossible to transmit an entire frame in this interval. At 1 Gbps, all frames shorter than 10,000 bits can be com-pletely transmitted in under 10 µsec, so the minimum frame size is 10,000 bits or 1250 bytes.
23. The minimum Ethernet frame is 64 bytes, including both addresses in the Ethernet frame header, the type/length field, and the checksum. Since the header fields occupy 18 bytes and the packet is 60 bytes, the total frame size is 78 bytes, which exceeds the 64-byte minimum. Therefore, no padding is used.
24. The maximum wire delay in fast Ethernet is 1/10 as long as in Ethernet.
25. The payload is 1500 bytes, but when the destination address, source address, type/length, and checksum fields are counted, plus the VLAN header, the total is indeed 1522. Prior to VLANs, the total was 1518.
26. The encoding is only 80% efficient. It takes 10 bits of transmitted data to represent 8 bits of actual data. In 1 sec, 1250 megabits are transmitted, which means 125 million codewords. Each codeword represents 8 data bits, so the true data rate is indeed 1000 megabits/sec.
27. The smallest Ethernet frame is 512 bits, so at 1 Gbps we get 1,953,125 or al-most 2 million frames/sec. However, this only works when frame bursting is operating. Without frame bursting, short frames are padded to 4096 bits, in which case the maximum number is 244,140. For the largest frame (12,144 bits), there can be as many as 82,345 frames/sec.
28. Gigabit Ethernet has it and so does 802.16. It is useful for bandwidth effi-ciency (one preamble, etc.) but also when there is a lower limit on frame size.
29. Station C is the closest to A since it heard the RTS and responded to it by asserting itsNAV signal. D did not respond, so it must be outside A’s radio range.
30. RTS/CTS in 802.11 does not help with the exposed terminals problem. So, given the scenario in Figure 4-11(b), MACA protocol will allow simultaneous communication, B to A and C to D, but 802.11 will allow only one of these communications to take place at a time.
31. (a) Each set of 10 frames will include one frame from each station. So, all stations will experience a data rate of 54/50 Mbps = 1.08 Mbps. (b) Each station gets the same time to transmit. So, the 6-Mbps stations will get 0.6 Mbps, 18-Mbps ones will get 1.8 Mbps, and 54-Mbps ones will get 5.4 Mbps.
32. A frame contains 512 bits. The bit error rate is p = 10−7. The probability of all 512 of them surviving correctly is (1 − p)512, which is about 0.9999488.
The fraction damaged is thus about 5 × 10−5. The number of frames/sec is 11 × 106/512 or about 21,484. Multiplying these two numbers together, we get about 1 damaged frame per second.
33. It depends how far away the subscriber is. If the subscriber is close, QAM-64 is used for 120 Mbps. For medium distances, QAM-16 is used for 80 Mbps.
For distant stations, QPSK is used for 40 Mbps.
34. Uncompressed video has a constant bit rate. Each frame has the same num-ber of pixels as the previous frame. Thus, it is possible to compute very accu-rately how much bandwidth will be needed, and when. Consequently, con-stant bit rate service is the best choice.
35. One reason is the need for real-time quality of service. If an error is discover-ed, there is no time for a retransmission. The show must go on. Forward error correction can be used here. Another reason is that on very low-quality lines (e.g., wireless channels), the error rate can be so high that practically all frames would have to be retransmitted, and the retransmissions would proba-bly damaged as well. To avoid this, forward error correction is used to in-crease the fraction of frames that arrive correctly.
36. Like 802.11, WiMAX wirelessly connects devices, including mobile devices to the Internet at Mbps speeds. Also, like 802.11, WiMAX is based on OFDM and MIMO technologies. However, unlike 802.11, WiMAX base sta-tions are much more powerful than 802.11 access points. Also, transmissions in WiMAX are carefully scheduled by the base station for each subscriber without any possibility of collisions unlike CSMA/CA used in 802.11.
37. It is impossible for a device to be master in two piconets at the same time.
Allowing this would create two problems. First, only 3 address bits are avail-able in the header, while as many as seven slaves could be in each piconet.
Thus, there would be no way to uniquely address each slave. Second, the ac-cess code at the start of the frame is derived from the master’s identity. This is how slaves tell which message belongs to which piconet. If two overlap-ping piconets used the same access code, there would be no way to tell which frame belonged to which piconet. In effect, the two piconets would be merged into one big piconet instead of two separate ones.
38. A Bluetooth frame has an overhead of 126 bits for access code and header, and a settling time of 250 to 260 µsec. At the basic data rate, 1 Mbps, a set-tling time of 250 to 260 µsec corresponds to 250 to 260 bits. A slot is 625 µsec long, which corresponds to 625 bits at 1 Mbps. So, a maximum of 1875 bits can be transmitted in a 3-slot frame. Out of this, 376 to 386 bits are over-head bits, leaving a maximum of 1499 to 1509 bits for the data field.
39. Bluetooth uses FHSS, just as 802.11 does. The biggest difference is that Bluetooth hops at a rate of 1600 hops/sec, far faster than 802.11.
40. An ACL channel is asynchronous, with frames arriving irregularly as data are produced. An SCO channel is synchronous, with frames arriving periodically at a well-defined rate.
41. In a 5-slot Bluetooth frame, a maximum of 3125 (625×5) bits can transmitted at basic rate. Out of this, a maximum of 2744 bits are for data. In case of repetition encoding, data is replicated thrice, so the actual data transmitted is about 914 bits. This results in about 29% efficiency.
42. They do not. The dwell time in 802.11 is not standardized, so it has to be announced to new stations that arrive. In Bluetooth, this is always 625 µsec.
There is no need to announce this. All Bluetooth devices have this hardwired into the chip. Bluetooth was designed to be cheap, and fixing the hop rate and dwell time leads to a simpler chip.
43. We want to maximize the probability that one (and only one) tag responds in a given slot. Consulting Sec. 4.2.4, the best tag probability for 10 tags is 1/10.
This occurs when the reader sets Q equal to 10 slots. Consulting Fig. 4-0, the probability that one tag responds is roughly 40%.
44. One key security concern is unauthorized tracking of RFID tags. An adver-sary with an appropriate RFID reader can track the locations of the items tagged using RFID tags. This becomes quite serious if the item is sensitive in nature, for example, a passport, and the tag can be used to retrieve further information, for example, the nationality and other personal information of the person holding the passport. Another security concern is the ability of a reader to change tag information. This can be used by an adversary to, for ex-ample, change the price of a tagged item he plans to buy.
45. The worst case is an endless stream of 64-byte (512-bit) frames. If the back-plane can handle 109 bps, the number of frames it can handle is 109/512. This is 1,953,125 frames/sec.
46. The port on bridge B1 that connects to the white machine would need to be relabeled as W, and also the port on bridge B2 that connects to bridge B1 would need to be relabeled GW.
47. A store-and-forward switch stores each incoming frame in its entirety, then examines it and forwards it. A cut-through switch starts to forward incoming frames before they have arrived completely. As soon as the destination ad-dress is in, the forwarding can begin.
48. (a) B1 will forward this packet on ports 2, 3, and 4. B2 will forward it on 1, 2 and 3.
(b) B2 will forward this packet on ports 1, 3, and 4. B1 will forward it on 1, 2 and 3.
(c) B2 will not forward this packet on any of its ports, and B1 will not see it.
(d) B2 will forward this packet on port 2. B1 will not see it.
(e) B2 will forward this packet on port 4 and B1 will forward it on port 1.
(f) B1 will forward this packet on ports 1, 3 and 4. B2 will forward it on port 2.
49. Store-and-forward switches store entire frames before forwarding them.
After a frame comes in, the checksum can be verified. If the frame is dam-aged, it is discarded immediately. With cut-through, damaged frames cannot be discarded by the switch because by the time the error is detected, the frame is already gone. Trying to deal with the problem is like locking the barn door after the horse has escaped.
50. A bridge that does not have any station directly connected to any of its ports and is part of a loop is a candidate for not being a part of the spanning tree bridges. This can happen if the shortest paths to the root for all bridges con-nected to this bridge does not include this bridge.
51. No. Hubs just connect all the incoming lines together electrically. There is nothing to configure. No routing is done in a hub. Every frame coming into the hub goes out on all the other lines.
52. It would work. Frames entering the core domain would all be legacy frames, so it would be up to the first core switch to tag them. It could do this by using MAC addresses or IP addresses. Similarly, on the way out, that switch would have to untag outgoing frames.