This preview shows page 1. Sign up to view the full content.
Unformatted text preview: EE555 Broadband Network Architectures (Spring 2009)
Homework 2 Solutions: Networking Models
Question 1. a) The routing matrix is given by: 1 1 2 3 4 b) The flow matrix is given by: 2 2 2 3 4 3 3 2 3 5 2 4 2 4 4 3 3 5 5 5 2 4 5 6 6 0 3 2 30 0 0 0 3 0 3 5 0 0 2 3 0 0 4 0 0 5 0 0 1 3 0 0 4 1 0 2 0 0 0 3 2 0 Question is poorly worded so two possible answers are allowed. Version 1: Assuming all link have same capacity: Links (2,4) and (4,2) will saturate first. Supposing the capacity of all the links is C, we need that 5 30 6 . So Version 2: Links (2,3) and (4,5) have capacity C/2 and the rest have capacity C. The utilization matrix is given by: 0 3 2 0 0 0 3 0 6 5 0 0 2 6 0 0 4 0 30 0 5 0 0 2 3 0 0 4 2 0 2 0 0 0 3 2 0 We see that [2,3] (and [3,2]) saturates first, so: 6 30 5 So c) Version 1: Supposing all links have capacity C. Delay is given by: 1 4 3 2 5 2 6 4 2 30 4 30 3 30 2 30 5 2 30 Version 2: Supposing the capacity of the thick links is C and that of the two thin links is C/2, the delay is given by: 1 5 4 3 2 6 2 2 4 6 2 30 5 30 4 30 3 30 2 30 6 Question 2. a) & b) The probability of connectivity for 0, 1, 2, and 3 failed components are as follows: Pr Pr Pr Pr 0 1 1 0 0 0 0.6634 1 0.2793 8 1 With two failures there are 3 combinations that result in a disconnected network ((1,2) and (2,3); (4,6) and (5,6); and (2,4) and (3,5)) 8 Pr 2 3 1 0.0459 2 0.0055 Pr 2 3 1 Counting for 3 failures is a little more tedious Pr Pr 3 3 8 26 3 26 1 1 0.0029 0.0025 Therefore, the bounds of probability of being connected are: 8 1 1 0.6634 0.9915 0.2793 25 3 1 1 0.0459 30 1 26 1 0.0029 0.9946 Pr Pr 1 0.0029 0.0025 Question 3. a) Average path length: 1 2 1 1 2 1 1 2 The total network traffic is and since we have a regular structure, we can assume that this is evenly distributed over the links. The maximum capacity of any link is , so: 2 2 ge h dd: b) Averag path length when n is od 2 1 Maximum traffic: m 2 1 2 1 1 2 8 4 Average p length when n is even path w n: 2 1 Maximum traffic: m 4 2 1 2 1 2 2 1 1 2 1 2 2 8 1 1 1 1 1 1 4 2 2 2 1 2 2 1 2 4 1 c) A 1D h hypercube is a line segmen with 2 node at the two e points. nt es end A 2D hyp percube is a sq quare with 4 n nodes at the 4 corners. A 3D hyp percube is a re egular cube w 8 nodes a the 4 corner with at rs. A 4D hyp percube is a regular (small cube insi another (b r ler) ide bigger) cube with the corr responding co orners connected and with 16 nodes at the 16 corners, an so on... d nd An illustr ration of the 4D hypercub is shown b be below. Numb the 16 nodes at the 16 corners in b ber 6 binary from 0000 to 1111. 0 To calcul late the avera path leng consider node (0000) and calculate the number of hops to every age gth, r other node e. 1 0001 , 0010 , 010 , 1000 00 : 0000 0011 , 0101 , 0110 , 1001 , 1010 , 110 : 0000 0 00 2 0111 , 1011 , 1101 , 1110 : 0000 0 3 1111 : 0000 0 4 Observe t that nodes th are k hop away from (0000), for hat ps m r numbering. The avera path lengt of the 4D h age th hypercube is g given by: 4 1 6 2 15 4 3 1 4 1,2,3,4, have k bits different in their s n 32 15 For the ge eneral kdimensional case, following a s similar proced dure, the aver rage path leng is given by: gth 1 2 1 1 1 2 2 3 3 To find th link flow, need to assume that traf is split e he , ffic evenly over a paths of e all equal length t get to uniform p pattern... d) Consid k even (for symmetry) der r For k=2, a average path length is 4/3. For k=4, av verage path len ngth is 44/15. For larger k ???? k, Question 4. a) Let T be t total numb of hours i a year: T = 24 x 365 = 8760 (or 8784 for a leap year). Hence, the ber in , b) For the 4 node (switch bidirectiona ring, we assume that the is a single link in both direction, so if the h) al ere e link fails, communicat tion in both d directions is c off. If we require that all nodes be operational (as is cut e t e usually th case since there will be customers connected to the nodes); the probabili of the net he o ity twork being con nnected is give by: en Pr P 4 1 (Alternate you could use the grap theoretic n ely, d ph notion of conn nectivity, i.e. we still have a connected graph when a node is down provided all other nodes can still com mmunicate; bu for a back ut kbone network the k, preferred solution is th one given a he above). c) For 2, and d 24, we have 8760 4 48 8760 0.9945 Method 1. For a path to exist from 1to 3 then (node 1 is up)&( node 3 is up)&( ((node 2 is up)&([1,2] is up)&([2,3] is up)) or ((node 0 is up)&([1,0] is up)&([0,3] is up))) 2 3 2 2 Pr([1,3] path exists} = aS (aS aL aL + aS aL aL  aS aL aL aS aL aL ) = aS aL (2  aS aL ) = 0.98876 (to 5 dp) Method 2. Following are the cases in which there is an availability of a path from node 1 to node 3. 1. (4 links are up) AND (4 nodes are up): can happen in 1 way 2. (4 links are up) AND (3 nodes are up): can happen in 2 ways 3. (Any 3 links are up) AND (4 nodes are up): can happen in 4 ways 4. (Any 3 links are up) AND (3 nodes are up): can happen in 4 ways 5. (2 appropriate links are up) AND (4 nodes are up): can happen in 2 ways: (1,0), (0,3) are up and (1,2), (2,3) are down; OR (1,2), (2,3) are up and (1,0), (0,3) are down 6. (2 appropriate links are up) AND (3 nodes are up): can happen in 2 ways: same as case 5 The respective probabilities for the 6 above cases are as follows (to 5 dp): Pr Pr Pr Pr Pr Pr 1 2 3 4 5 6 2 4 4 2 2 0.95683 1 1 1 1 1 1 0.01058 0.02117 0.00012 0.00006 1 0.00000 And thus the probability of an availability of a path from node 1 to node 3 is the sum of all the above 6 cases = 0.98876 Mean time per year node 1 and node 3 cannot communicate is given by: 8760 1 0.98876 98.4 . Question 5. a) Link co onnectivity be etween A and B is determ d mined by the n number of lin disjoint pat between A and nk ths B, which in this case is 3. s b) Networ link conne rk ectivity is the minimum nu e umber of link that need t be disconn ks to nected to mak the ke network d disconnected, which in this case is 2 (li inks connecte to node 6 c be cut to make the net ed can twork disconnec cted). c) Total fl that can be supported is 14. Links {[4,6], [4,7], [2,5]} saturat and indeed this set of lin is flow b te d nks a cutset (i disconnects A and B). i.e. Question 6. a) Assumi there is a single link in both directio the netwo availability is given by ing n ons, ork y: 8 1 0.9973 b) Try one chord conne e ecting node 1 to node 5 (no odes are num mbered 1 throu 8). ugh Network a availability is s: 9 4 4 9 a 9 + a 8 (1  a) +   a 7 (  a ) 2 = a 7 {a 2 + 9a (1  a ) + 24(1  a ) 2 } = 0.9988 (1 8 1 2 2 2 Without w working the details, it is lik that addin an additional chord from 3 to 7 will do it. d kely ng m ...
View Full
Document
 Spring '08
 Silvester

Click to edit the document details