This preview shows page 1. Sign up to view the full content.
Unformatted text preview: GEORGIA TECH LORRAINE GEORGIA INSTITUTE OF TECHNOLOGY School of Electrical and Computer Engineering ECE 6543 Fiber Optic Network Spring Semester 2009
Problem 1 1. From the relation established in class, the number of wavelength W is 3. However we are considering a two -ﬁber SPRING network, so this number needs to be doubled. The ﬁnal number of wavelengths needed is 6. 2. A two-ﬁber Spring, the traﬃc can be split between the two counterpropagating rings as half of the capacity of each is used. We need to write two routing tables one for clockwise and the other one for counterclockwise. Let consider the set λ1 , λ2 , λ3 , the clockwise wavelengths and λ4 , λ5 , λ6 , the counterclockwise wavelengths. The clockwise routing table is: ABCDE λ1 1 2 X 2 X λ2 2 X 1 1 1 λ3 X 1 2 X 2 The counterclockwise routing table is: AEDCB λ4 1 2 X 2 X λ5 2 X 1 1 1 λ6 X 1 2 X 2 3. If the NAS are elementary, they are connected to the ONN through a single-ﬁber-pair access link. We need to apply the limiting cut bound. We ﬁnd that W ≥ 4 wavelengths are required for full connectivity but this number may be tight. 1 If alternate routing is allowed, the answer is much more complicated. Let α1 denote the A–B trafﬁc routed on the direct path A–B, and α2 the A–B trafﬁc routed through C, that is, on the path A–C–B in the lightpath topology. Similarly, deﬁne β1 , β2 , γ1 and γ2 . Note that the trafﬁc γ2 is dropped to the IP router at node B and reinserted by it, whereas the trafﬁc γ1 passes through node B without touching the IP router at node B. Then, the supported values of α , β , and γ are those for which the following inequalities has a feasible solution. α1 + β2 + γ2 ≤ x α2 + β1 + γ2 ≤ x α2 + β2 + γ1 ≤ y Problem 2 See solution below
8.2 (a) The routing and wavelength assignment is as follows: Trafﬁc stream AB AD BC CD BD BD Wavelength λ1 , λ2 , λ3 λ1 , λ2 , λ3 λ1 , λ2 λ1 , λ2 λ3 , λ4 λ4 Path AB AD BC CD BCD BAD 60 WDM Network Design 59 (b) The minimum total trafﬁc load due to all the connections can be computed by using the minimum number of hops required for each connection as follows: Trafﬁc stream AB AD BC BD CD Total Trafﬁc 3 3 2 3 2 Min. hops 1 1 1 2 1 Trafﬁc load 3 3 2 6 2 16 Since there are only 4 edges to carry this load, the average load per edge is 16/4 = 4, and the maximum load per edge is therefore at least 4. Thus, at least 4 wavelengths are required. (c) Node A needs 3 ADMs, node B 4, node C 2, and node D, 4 ADMs. (d) Each node would need 4 ADMs. 8.3 Consider any source node. The N − 1 trafﬁc streams from that node to the other nodes, when routed on their shortest paths take up a total number of hops of hodd = 2 1 + 2 + 3 + · · · + for odd N and heven = 2 1 + 2 + 3 + · · · + N2 N −1 = 2 4 N −1 2 = N2 − 1 4 for even N . The trafﬁc between each pair of nodes is t/(N − 1), and so the average load due to this trafﬁc on each edge is
t hodd N −1 N 2N for odd N and
t heven N −1 N = N +1 t 8 2N for even N . 8.4 = N +1+ 8 1 N −1 t 2 Since two adjacent nodes use different paths along the ring, only N/2 nodes use any given edge 2 3 3 wavelengths 1 2 wavelengths 8.23 In a network using full wavelength conversion, a lightpath request is blocked if there is no free wavelength on some link in the path. The probability that no wavelength is free on any given link is π W . So the probability that there is no blocking on any of the H hops, using the link independent property, is given by = (1 − π W )H . Therefore, Pb,fc = 1 − (1 − π W )H . Problem 3 See solution below
8.24 We have πnc = 1 − 1 − Pb,nc
1/W 1/W 1/H . For small Pb,nc (small Pb,nc and W not large), using (1 − x)n ≈ 1 − nx , for small x , πnc = Also, πfc = 1 − (1 − Pb,fc )1/H
1/W Pb,nc H 1/W . . Again, using (1 − x)n ≈ 1 − nx for small x , for small Pb,fc , we get πfc ≈ Pb,fc H
1/W . The exact expression and the approximation for πnc are plotted versus the number of wavelengths W , for various values of Pb and number of hops, in the plots below. The approximation consistently underestimates the utilization so that the lower curve in each plot corresponds to the approximation. It can be seen that the approximation is accurate only for W ≤ 5 or so, when 3 69
Pb = 10−3 . When Pb = 10−5 , the range of accuracy of the approximation increases to around W ≤ 10. 69
0.25 0.2 Pb = 10 0.15 −3 . When Pb = 10 hops = 5 W ≤0.1 . 10 0.05 0.25 0 0.2 0.15 0.25 0.1 0.2 0.05 0.15 0.1 0 0.05 0.25 0 0.2 5 −3 10 Pb = 10 W hops = 5 Pb = 10 hops = 5 5 10 W 5 10 −4 Pb = 10 W hops = 5
−4 −3 0.25 0.2 0.1 0.05 15 20 0.25 0 0.2 0.15 0.25 0.1 0.2 0.05 0.15 20 0.1 0 0.05 15 20 0.25 0 0.2 0.15 0.25 0.1 0.2 0.05 0.15 20 0.1 0 0.05 15 20 0.25 0 0.2 0.15 0.1 Pb = Pb = 10 −5 , 0.15 range of10 hops = accuracy 10 the −3 0.25 0.2 0.1 0.05 of Pb = 10 hops = 20 0.15 the approximation −3 increases to around 5 −3 10 Pb = 10 W hops = 10 Pb = 10 hops = 10 5 10 W 5 10 −4 Pb = 10 W hops = 10 Pb = 10 hops = 10 5 10 W
−5 −4 15 20 0.25 0 0.2 0.15 0.25 0.1 0.2 0.05 0.15 20 0.1 0 0.05 20 0.25 0 0.2 0.15 0.25 0.1 0.2 0.05 0.15 20 0.1 0 0.05 20 0.25 0 0.2 0.15 0.1 5 −3 10 Pb = 10 W hops = 20 Pb = 10 hops = 20 5 10 W 5 10 −4 Pb = 10 W hops = 20 Pb = 10 hops = 20 5 10 W
−5 −4 15 20 15 15 15 20 15 15 20 0.15 0.25 0.1 −5 0.2 Pb = 10 0.05 0.15 hops = 5 5 10 0.1 0 W 0.05 0.25 0 0.2 0.1 0.15 15 15 15 20 5 −5 10 W Pb = 10 hops = 5 5 −5 10 Pb = 10 W hops = 10 15 5 −5 10 Pb = 10 W hops = 20 15 20 The approximation for Pb,fc is so accurate for Pb ≤ 10−3 that the curves for the approximate 0.05 0.05 0.05 and exact expressions are indistinguishable. Hence these curves are not shown here.
W W The probability that a wavelength is free on link k , given that it is free onW links 1, 2, . . . , k − 1, is given by 1 − πn , by the deﬁnition of πn . (It only matters that it is free on k − 1.) Therefore, the probability that a wavelength λ is free on link for P − πn . −The probability that it is approximate The approximation for Pb,fc is so accurate 1 is 1 ≤ 10 3 that the curves for the free on links b 2 1 and 2 is expressionsThe indistinguishable. Hence on allcurves are not− πn )H . here. probability and exact (1 − πn ) . are probability that it is free these H links is (1 shown So the that wavelength λ is not free is given by 1 − (1 − πn )H . Thus, 8.25 The probability that a wavelength is free on link k , given that it is free on links 1, 2, . . . , k − 1, is W givenbby 1 −1 −, (by− πn )H . P ,nc = πn 1 the deﬁnition of πn . (It only matters that it is free on k − 1.) Therefore, the probability that a wavelength λ is free on link 1 is 1 − πn . The probability that it is free on links Problem 4 See solution below 1 and 2 is (1 − πn )2 . The probability that it is free on all H links is (1 − πn )H . So the probability Gb/s C D that wavelength λ is not free B given by 1 − (1 − πE)H . Thus, is n A 2 3 1 2 W 8.26 (a)nc = 1 − (1 − πn )H B 1 4 2 . Pb, C 2 3 D 1 Gb/s B C D E (b) If we route each lightpath along its shortest path, starting from the top left of the matrix A 2 3 1 2 8.26 (a) B 1 4 2 C 2 3 D 1 8.25 0 5 10 15 20 0 5 10 15 20 0 5 10 15 20 (b) If we route each lightpath along its shortest path, starting from the top left of the matrix 4 70 WDM Network Design above, and going down row by row, and assigning the lowest possible wavelength to each lightpath, we get the following assignment: Lightpath AB AB AC AC AC AD AE AE BC BD BD BD BD BE BE CD CD CE CE CE DE Wavelength 1 2 3 4 5 1 2 3 1 1 2 3 4 5 6 1 2 3 4 7 2 (c) The most heavily loaded link is DE, with a total load of 7, which is also equal to the number of wavelengths. 5 chapter
Network Survivability 74 Network Survivability Problem hub is only C , where C is the link speed. Therefore trafﬁc patterns for which 5 See solution below the >C cannot be connection CE in the ﬁgure below. If link BC fails, we have the following supported, 10.1 Consider We will show that in both the UPSR restored BLSR/2, all trafﬁc patterns such that N 1 ti ≤ C , (a) path protection: Connection is and the along CDE (2 hops). i= can (b) supported (assuming trafﬁc from a single along in a BLSR/2 can be is very inefﬁcient, combe line protection: Connection is restored node CDEABAE, which split across two routes, if necessary). First consider the UPSR. Trafﬁc from node i uses capacity ti on every link in the ring pared to path protection. (considering both B working and protection trafﬁc). Therefore this trafﬁc can be supported provided N i =1 ti ≤ C . Now consider the BLSR/2. Note that only a capacity of C/2 on every link is available for working trafﬁc. Consider a trafﬁc pattern such that N 1 ti ≤ C . From node i , we route ti /2 units i= clockwise and ti /2 units counterclockwise on the ring to the hub. With this routing the trafﬁc load C A on each link is N 1 ti /2 ≤ C/2. Therefore this trafﬁc pattern can be supported. i= Therefore the UPSR and BLSR/2 can support the same set of trafﬁc patterns in this case. Thus a UPSR is superior for this application because it has the same trafﬁc carrying capacity as D a BLSR/2, and in addition, E supports OC-12c connections, has faster a 1 hop and, Next considerprotection,connection DE. If link DE fails, both path and line protection use is a restore the less expensive system. DCBAE to simpler and connection. In this case, both need the same amount of bandwidth for restoration. In general, path protection is adjacent pairs of nodes. So the capacity is NC , where The trafﬁc distribution has all trafﬁc betweenbetter (more efﬁcient use of bandwidth) at restoring multihop connections ﬁber and N the number of nodes. C is the bit rate on thethan line protection. N i =1 ti 10.4 Problem 6 See solution below
10.6 10.2 Consider a link trafﬁc case, the average hop length capacity. If that link fails, then is the no way 10.5 For the uniform carrying trafﬁc equal to its working is approximately N/4, where Nthere isnumber to restore trafﬁc reuse protection capacity = working capacity. of nodes. So the unlessfactor is approximately 4. So the capacity is 4C , where C is the bit rate on 10.3 the ﬁber. that if both types of rings operate at, say, OC-12 speeds, the maximum concatenated Note ﬁrst connection streamreader. be carried in a UPSR is OC-12c, whereas in a BLSR/2, it is OC-6c (a) Left to the that can (because half the bandwidth on each ﬁber is reserved for protection). This is true regardless of the (b) Left to the reader. trafﬁc pattern. (c) For UPSR, both the routes around the ring need to be used for work and protect. Thus Considerdemand utilizes the bandwidth on every link in the denote theSince the total demand each rings with N nodes and an additional hub. Let ti network. trafﬁc between node i and theis 80 STS-1s, this bandwidth is used by UPSRrouted to link hub, the working capacity into hub. First note that since all trafﬁc must be on every the in the network. For BLSR, use the shortest path between nodes. This yields a load of 24 on the links A–B, B–C and C—D, a load of 8 on D–E and 22 on E–A. The average load arising from 73 shortest path routing is a lower bound on the maximum load (from (8.10) together with the solution of Problem 8.7). Thus, under any routing scheme, the maximum load cannot be lower than (24 × 3 + 8 + 22)/5 = 21. We can get a better lower bound by reasoning as follows. Club the nodes D and E into one node “DE” to get a 4-node ring with the following demand matrix (ignoring the demand between D and E). STS-1 A B C B 12 C 6 8 DE 16 16 14 The average link load (rounded up) due to shortest path routing on this 4-node ring is (12 + 6 × 2 + 16 + 8 + 16 × 2 + 14)/4 = 24. This is a lower bound on the maximum load for the original 5-node ring. (To prove this, observe that if this is not the case and there is a routing scheme for the 5-node ring which yields a better maximum load, then the same 6 75 scheme can be applied to the 4-node ring leading to a contradiction.) Thus the maximum load of 24 obtained using shortest path routing is optimal. (d) UPSR requires an OC-192 ring whereas BLSR only requires an OC-48 ring. (e) BLSR is better since OC-48 rings are cheaper than OC-192 rings. 10.7 With 2 cuts, the network is partitioned into two clusters of nodes without any connection between the two clusters. Nodes within each cluster can communicate. Note that this is the case with all rings in general. The UPSR can handle multiple cuts in one of the two rings because the other ring will be still fully functional. While it is quite likely that both ﬁbers on a link get cut at the same time, this capability still enables the UPSR to continue providing service when a transmitter or receiver fails. Unlike the UPSR, the BLSR/2 cannot handle multiple cuts because the protection capacity is shared. The BLSR/4 can handle multiple failures of transmitter/receivers (one per span). It can handle simultaneous cuts of 1 ﬁber pair per span. Note that once span protection is used, line protection cannot be used any more to recover from another failure. This scheme works ﬁne under normal operation but cannot protect individual connections in case of a failure. For example, in Figure 10.4, if AB is cut, then receiver D must receive connections from A on the counter-clockwise ring but connections from B and C on the clockwise ring. The three approaches are illustrated in the ﬁgure below. There is no difference between them as far as line protection is concerned. Also, span protection in the case of equipment failures works the same way in all the approaches. However span protection in the case of ﬁber cuts works differently. Option (1) allows span protection to be used in case of a single ﬁber cut, whereas options (2) and (3) do not allow span protection to be used for this case. Therefore, we will pick option (1).
W BLSR/4 ADM W P P W W BLSR/4 ADM P P W W BLSR/4 ADM P P P P P P W W BLSR/4 ADM (3) W W P P W W BLSR/4 ADM (2) BLSR/4 ADM 10.8 10.9 (1) 7 ...
View Full Document
This note was uploaded on 09/01/2010 for the course ECE 6543 taught by Professor Boussert during the Spring '09 term at Georgia Institute of Technology.
- Spring '09