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TCOM 5143

TCOM 5143. Lecture 4 Introduction to Network Design: A Three-Location Data Network Design Problem. Traffic. Time. Analysis of link delays in data networks.

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TCOM 5143

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  1. TCOM 5143 Lecture 4 Introduction to Network Design: A Three-Location Data Network Design Problem

  2. Traffic Time Analysis of link delays in data networks • Burstiness of data traffic over a network link refers to the observation that the data is not continuously transmitted over the link; that is data is transmitted for some period of time and no data is transmitted for another period of time, and so on.

  3. Traffic Time • For example, if burstiness is 10, then data is being transmitted over the link on 10% of the time. • Another example, if burstiness is 100, then data is being transmitted over the link on 1% of the time.

  4. Traffic (percent) Peak/busy Hour Hours of the day • Traffic on a link usually varies by hours of the day. The link should have sufficient capacity to handle traffic at the peak/busy hour.

  5. Queueing at a link occurs because bits, packets or messages share the link for transmission. When a bit/packet arrives and the link is busy, the bit/packet has to wait until the link become free. • Queueing at a communication channel/link is usually modeled as an M/M/1 queueing system with the following characteristics: 1. The message/packet/bit arrivals are independent of one another, the inter-arrival times are exponentially distributed with mean ; 2. The message/packet/bit service times are also exponentially distributed with mean ; and 3. There is one single server/link with unlimited queue.

  6. 2.0 . 2.0: . . . 1.0:  0.5: * * * . . 1.0 .   .  0.5 * .  * .  * * * * * .     . . . . . x • A continuous random variable x is exponentially distributed with the mean a if the probability density function of x is • The following figure shows the density functions for exponential distributions with a = 0.5, 1.0, and 2.0.

  7. M/M/1 queue (continued) • : mean arrival rate • : mean service rate • =/: traffic intensity (called also utilization factor) • Ts=1/: average service time • Probability that the system is empty: P0=1-/ • Probability that there are n customers in the system: Pn = (1-r) rn = (1 - l/ m) . (l / m)n • Expected number of customers in the system: L = r/(1- r) = l/(m - l)

  8. Expected number of customers in the queue: Lq = r2/(1- r) = l2/ (m(m - l)) • Expected total time in the system: W = 1/ (m - l) • Expected delay time in queue: Wq= l/ (m(m - l))

  9. W  0 1 Note that: • The following relationships are also correct: W = Ts+Wq = Ts/(1- r) • As 1, then W

  10. An example • Consider a web server which has a large number of users generating a request with an exponential interarrival time - mean of arrivals is one request every 500 milli seconds. The server requires exponential service time with average response in 475 milliseconds. What can we say about service that users will receive ?

  11. tbar = mean interarrival time = 0.500 seconds • xbar = mean service time = 0.475 seconds • hence :  = 1/tbar = 1/0.500 = 2.0 arrivals/second  = 1/xbar = 1/0.475 = 2.105 responses/second  = /  = 0.95 • Probability of “n” customers in system: Pn = (1-) n = (1 - / ) . ( / )n = (1-0.95)(0.95)n P0 = 0.05; P1 = 0.0475; P2 = 0.0451 and so on. • L = 0.95/(1-0.95) = 19 is the number (average) of requests in the system • W = 1/(2.105 - 2.0) = 9.52 seconds is the total waiting time (average) in the system

  12. Designing the data network • Problem goal: Design a data network to interconnect three given sites in the cheapest way.

  13. Traffic requirements

  14. Assumption about traffic pattern: • 20% of the internal email, WWW, and database traffic occurs in the peak (busy) hour and the external traffic email arrives evenly throughout the day. • Costs of network components:

  15. A terminal router is a small internet protocol (IP) router that can support multiple local area network (LAN) connections but only a single wide area link. • A transit router is a full-fledged router capable of switching transit traffic destined for other routes. Any site with 2 or more wide area network (WAN) links needs to have a transit router, whether or not it switches any traffic between the links, simply to terminate more than 1 telecommunication links.

  16. 64 Kpbs 64 Kpbs 64 Kpbs City A 96 employees City B 128 employees Gateway B Gateway A 64 Kpbs 64 Kpbs City C 72 employees 64 Kpbs Gateway C • Design Option: Interconnect the three nodes (cities) with 64 Kbps links, and connect each node to the internet with a 64 Kbps link. • Total cost: $6633/month

  17. Design principles • Design Principle 2.3. Seek to make a network where all the links have a 50% utilization. • Design Principle 2.4. Seek to make a network where all the links have about 50% utilization and as few links as possible are underutilized. • The utilization threshold is a fuzzy constraint - it should be considered in conjunction with other factors in the design such as cost.

  18. Representing Traffic Using Traffic Tables • The average waiting time and average total time in the system depend on the utilization of the system, which depends on different sources of traffic flowing through the system • We construct a traffic table that shows amounts of traffic (bandwidths) between the sources and destinations of traffic in the network. TRAFFIC TABLE SOURCE DEST BANDWIDTH COMMENT • We construct traffic tables for the different types of traffic during the peak hours: internal/external email, WWW-access, and database activities.

  19. 1. Traffic Table for Internal Email • The total (outbound and inbound) internal email volume per person during peak hours is • Hence the total email during peak hours is: (96 + 128 + 72)  266.67 bps = 78,934.32 bps.

  20. Internal email (continued) • This total traffic includes traffic generated by emails within the sites and traffic generated by emails between sites. We assume that the total traffic is proportionally distributed between the two groups of emails. • If we let B represent the number of email bits transmitted between any two employees, then we have • H96  96 + H96  128 + H96  72 + H128  96 + H128  128 + H128  72 + H72  96 + H72  128 + H72  72 = 78,934.42, • thus, H = 0.900913.

  21. Note that the internal email volume within A, within B, and within C does not enter the internode links or internet links. This gives us the following traffic table for internal email.

  22. Traffic Table for External Email • The outbound external email per person during peak hours is: • The computation for in-bound external mail per person during peak hours is similar: • This gives us the following traffic table for external traffic.

  23. Traffic table for external traffic

  24. 3. Traffic Table for WWW-Traffic • The outbound WWW traffic per person during peak hours is: • The inbound WWW traffic per person during peak hours is: • These give us the following traffic table for WWW-traffic.

  25. Traffic Table for WWW-Traffic (continued)

  26. 4.Traffic Table for Database Activities • We assume that information is randomly distributed over three servers in A, B, and C, hence a uniform probability of 1/3 for locating data in a particular server. • Also we assume that the database requests (queries and updates) in each site are served as follows: • 1/3 locally within the same site, 1/3 in the second site, and 1/3 in the third site.

  27. Traffic Table for Database Activities (continued) • The outbound database query traffic per person to each remote server during peak hours is: • The inbound database query traffic per person from each remote server during peak hours is:

  28. Traffic Table for Database Activities (continued) • The outbound database updates traffic per person to each remote server during peak hours is: • The inbound database updates traffic per person to each remote server during peak hours is: • These give us the following traffic table for database activities.

  29. Traffic Table for Database Activities (continued)

  30. Calculating the total link flows: The total flow on each link is computed as the sum of flows from all sources (internal email, external email, WWW, and database). • To comply with design principle 2.4 to improve the design, we replace lightly utilized (much below 50%) links, if possible.

  31. The Drop Algorithm • The Drop Algorithm consists of selecting and dropping/deleting the more expensive, lightly utilized links earlier in the process. • A candidate link is deleted only if the network without the link satisfies the traffic requirements. • When a link is deleted, no traffic can go through the link in any direction.

  32. A pseudocode description for the Drop Algorithm: 1. Initially mark all the links as Deletable. 2. Locate the most expensive Deletable link in the current network (break a tie by choosing the link with the lowest utilization). Such link is the candidate for possible deletion. The low utilization on the candidate link demands an easier rerouting/distribution of its flow to the remaining links.

  33. 3. Test for feasibility on the current network without the candidate link: (a) Reroute/redistribute the traffic flow (in both directions) on the candidate link to the remaining links. The rerouting may require heuristics, and the redistribution may require adjustment of capacities of the remaining links (and hence the network cost). (b) If the resulting network cannot satisfy the traffic requirements or its cost has increased, then the candidate link should not be deleted. Restore the network by adding the candidate link back, and mark the candidate link as Undeletable. Continue the deletion process back at step 2 until all links are marked as Undeletable.

  34. The rerouting of traffic flow on the candidate links may require heuristics such as Open Shortest Path First algorithm, network routing protocols, etc. • In this lecture (and textbook), the heuristic used in rerouting is to select paths with links of total low utilization. • Applying the Drop Algorithm to our current network yields the following iterations (rounds).

  35. B-GB; GB-B 0.12; 0.19 A-GA; GA-A 0.09; 0.14 A-B; B-A 0.24; 0.23 B GB GA A B-C; C-B 0.18; 0.19 A-C; C-A 0.14; 0.15 C C-GC; GC-C 0.07; 0.11 GC • Iteration 1: 1. All links A-B, B-A, A-C, C-A, B-C, C-B, A-Gateway A, Gateway A-A, B-Gateway B, Gateway B-B, C-Gateway C -C are marked Deletable. 2. Current candidate link is C-Gateway C because it is the most expensive Deletable link with utilization (0.07+0,11).

  36. Reroute the total flow on link (C-GC) through the route with links GA-A and A-C (in both directions). The network without the current candidate link has the following traffic flow at a lower cost Total cost = $[6633 – 1400 (internet link) – 111 (transit router) + 60 (terminal router)]/month = $5182/month

  37. B-GB; GB-B 0.12; 0.19 A-GA; GA-A 0.15; 0.25 A-B; B-A 0.24; 0.23 B GB GA A B-C; C-B 0.18; 0.19 A-C; C-A 0.25; 0.21 C Iteration 2: 2. Current candidate link is B-Gateway B because it is the most expensive Deletable link with utilization (0.12+0,19). 3. Reroute the total flow on link (B-GB) through the route with links GA-A and A-B (in both directions). The network without the current candidate link has the following traffic flow at a lower cost [-$1,400/month (64Kbps internet link)- $111/month (transit router)+$60/month (terminal router)]:

  38. Total cost = $[5182– 1400 (internet link) – 111 (transit router) + 60 (terminal router)]/month = $3731/month

  39. A-GA; GA-A 0.24; 0.44 A-B; B-A 0.43; 0.35 B GA A B-C; C-B 0.18; 0.19 A-C; C-A 0.25; 0.21 C Iteration 3: 2. Current candidate link is A-Gateway A because it is most expensive Deletable link with utilization (0.27+0,44). 3. The network without the current candidate link cannot satisfy the internet traffic. Restore the network by adding back the candidate link, and mark the link A-Gateway A as Undeletable.

  40. A-GA; GA-A 0.24; 0.44 A-B; B-A 0.43; 0.35 B GA A B-C; C-B 0.18; 0.19 A-C; C-A 0.25; 0.21 C Iteration 4: 2. Current candidate link is B-C because it is the most expensive Deletable link with utilization (0.18+0.19). 3. Reroute through links A-B, B-A, and A-C, C-A. The network without the current candidate link has the following traffic flow.

  41. The deletion of link B-C requires now that the capacity of link A-B be increased to the next level; that is, from 64 kps to 256 kps in order the total flow going through link A-B in both directions. • The deletion of the current candidate link B-C [-$700/month (64 Kbps internode link)] and the adjustment of capacity for the link A-B (+$700/month) do not justify the feasibility of the network without the candidate link. Furthermore, The deletion of link B-C makes the network less reliable because we will be left with only one route between all sites. So we restore the network by adding back the candidate link, and mark the link as Undeletable.

  42. Terminal router Transit router A-GA; GA-A 0.24; 0.44 A-B; B-A 0.43; 0.35 B GA A B-C; C-B 0.18; 0.19 A-C; C-A 0.25; 0.21 C Terminal router • Similar iterations for possible deletions of A-C and A-B result in marking these two links as Undeletable. • The final design using the Drop algorithm is the following design with total cost of: • $[3 * 700 (internode links) + 1400 (internet link) + 2 * 60 (terminal routers) + 111 (transit router)]/month = $3731/month.

  43. Terminal router Transit router B A GB 64 Kbps 64 Kbps 64 Kbps C Terminal router The Failure of the Greedy Drop Algorithm • Intuitively, since the most important node (with largest population and most traffic volume) is B, then the Gateway B should be selected as the (only) one connecting to the internet traffic. Consequently, the deletion of the link A-C will succeed. • The cost of this optimal design (see [Cah98] Exercise 2.7) is $[2 * 700 (internode links) + 1400 (internet link) + 2 * 60 (terminal routers) + 111 (transit router)]/month = $3031/month. • In this three-location problem example, the Drop algorithm has failed to produce the optimal solution.

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