Network Design - Chapter 3: Evaluating Technical Goals Project Costing - University of Pittsburgh

Nội dung text: Network Design - Chapter 3: Evaluating Technical Goals Project Costing - University of Pittsburgh

1. Evaluating Technical Goals Project Costing David Tipper Associate Professor Department of Information Science and Telecommunications University of Pittsburgh Slides 3 Technical Requirements & Constraints • Technical Goals – Scalability – Availability/reliability – Network Performance • Utilization, Throughput, Delay, Delay Jitter, packet loss rate, call/connection blocking rate – Security – Manageability/Interoperability – Affordability $$ • Need to determine reasonable goal for each category and the relative importance of each • For Availability and Performance simple models are useful to set goals and evaluate designs TELCOM 2110 Spring 06 2 1
2. Availability • Availability is the amount of time a network is available to users • Can be expressed as percent uptime – 165 hours in 168 hours/week = 98.21% • Availability Goals depend on application and user requirements – may vary with location – Highly available voice service at customer support call center – Lower available voice over IP service in engineering dept. – 99.999% => downtime = .00001 x 60 x 24 x 365 = 5.25 minutes per year! Availability level Downtime per year 99.999% 5.25 min 99.97% 157.68 min 99.9% 8 hours 46 min 99% 87 hours 4 min TELCOM 2110 Spring 06 3 Availability • Availability (A) – Ability of an item to perform stated function at over time – Fraction of the time that an item can be used when needed – Value in the 0.0 to 1.0 range – Mean Time To Repair (MTTR) • An average time to restore a full functionality to an item – This may include time to diagnose, isolate, remove and replace the failed part – MTTF: Mean-Time To Failure – MTBF: Mean-Time Between Failures – more a reliability measure MTTR MTTF MTTR MTTF ⎧⎫Uptime A = A = lim ⎨⎬ Tobs →∞ MTBF MTTF+ MTTR ⎩⎭Tobs TELCOM 2110 Spring 06 Failure Repair Failure Repair 4 2
3. MTBF – Physical Cable • Physical cables – MTBF can be specified using the Cable Cut (CC) metric • Average cable length that results in a single cable cut per year • CC = 450 km means that per 450 km cable, there will be on average on cable cut each year (CC × 365× 24) MTBF() hours = length of the cable (km) – Example, given CC = 450km and cable length = 260 km, 450km× 365× 24 h MTBF==15161.5 h cable 260km TELCOM 2110 Spring 06 5 Unavailability • Unavailability (U) – The fraction of the time that an item cannot be used when needed • U = 1 – A • Other expressions for unavailability – Downtime per year • Downtime in units of minutes per year • Obtained by multiplying U by minutes in a year – 0.99999 availability, – 0.00001 unavailability, – 5.256 downtime per year (in minutes), and TELCOM 2110 Spring 06 6 3
4. System Availability • System availability calculated from component availability Ai • If devices in series n Aseries= ∏ A i i=1 n 1 2 n UUsi≈ ∑ i =1 • If devices in parallel n 1 UUparallel= ∏ i i=1 2 n Aparallel=−1(1)∏ −A i n i=1 TELCOM 2110 Spring 06 7 System Availability - Example A single bidirectional line in WDM optical network Line System Line WDM OA OA WDM 80km 100km 80km Line System Equipment MTBF (hrs) MTTR (hrs) • The availability of the bidirectional line Bidirectional OA 5*105 24 system = ? Bidirectional 5*105 6 WDM Line System Equipment CC (km) MTTR (hrs) Terrestrial Fiber 450 24 Optic Cable TELCOM 2110 Spring 06 8 4
5. System Availability - Example • Devices in series • Availability of a bidirectional line (Aline) 22 AAline=×× cable AA OA line− system MTTR MTTR MTTR =− (1cable ) ×− (1 OA )22 ×− (1line− system ) MTBFcable MTBF OA MTBF lline− system 24hhh 24 6 =− (1 ) ×− (1 )22 ×− (1 ) 15161.5hh 5×× 1055 5 10 h = 0.998297 450km× 365× 24 h Note. MTBF== 15161.5 h cable 260km TELCOM 2110 Spring 06 9 Series-Parallel Reduction For complex systems need to apply series parallel reduction to determine overall availablity + series || parallel TELCOM 2110 Spring 06 10 5
6. Availability Analysis • General Methodology: 1) Get unavailability values of all components and sub-systems. 2) Draw parallel and series availability relationships 3) Reduce the system availability model by repeated applications of the parallel/series availability simplifications. 4) If not completely reduced, do quick unavailability lower bound estimation – Contributions of parallel elements to the unavailability are not taken into account B D F A C G H E Lower bound of Us: UA+UH TELCOM 2110 Spring 06 11 Network Performance • Several Performance measures – Utilization – Throughput – Accuracy (BER, Packet Loss) – Efficiency – Delay and Delay Jitter – Call Blocking for circuit switched networks • Typically look at measures during the busy period of the day set threshold values • Need to know how to estimate values • Approaches when designing network limited to – Queueing Analysis – analytical models – Simulation – measurements on computer model of the network design TELCOM 2110 Spring 06 12 6
7. Queueing Theory • Queueing theory : Mathematical analysis of waiting lines • Queueing Theory is the primary analytical framework for evaluating performance in the initial stage of system design. • Analytical Model of the system – based on stochastic processes • Approximates real system by focusing on contention at shared resources. • Examples: shared medium router, window flow controlled session, time shared computer system TELCOM 2110 Spring 06 13 Model of Router TELCOM 2110 Spring 06 14 7
8. Nomenclature of a Queueing System • The input process – how customers arrive • The system structure – waiting space – number of servers, etc. • The service process • Kendall’s Notation 1/2/3/4/5/6 – A Shorthand notation to describe a queueing system containing a queueing system. – 1 : Customer arriving pattern (Interarrival times distribution). – 2 : Service pattern (Service-times distribution). – 3 : Number of parallel servers. – 4 : System capacity. – 5 : Queueing discipline. – 6: Customer Population TELCOM 2110 Spring 06 15 Characteristics of the Input Process 1. The size of the arriving population – Infinite : the number of potential customers from external sources is very large as compared to those in the system. – Finite : the arrival process (rate) is affected by the number of customers already in the system. 2. Arrival pattern – Customers may arrive at a queueing system either in some regular pattern or totally random fashion • When customers arrive regularly at a fixed interval, the arriving pattern can be easily described by a single number − the rate of arrival • When customers arrive according to some random fashion, the arriving pattern is described by statistical distribution. TELCOM 2110 Spring 06 16 8
9. Characteristics of the Input Process • Probability distribution that are commonly used to describe the arrival process are: – M : Markovian (or memoryless), implies the Poisson process for arrivals – means the number of arrivals over a time interval has a Poisson distribution – this is equivalent to the time between customers arriving being exponentially distributed. – D : Deterministic, fixed interarrival times – Ek: Erlang distribution of order k – G : General probability distribution – GI: General and independent (inter-arrival time) distribution. TELCOM 2110 Spring 06 17 Characteristics of the Input Process 3. Behavior of the arriving customers ♦ Customer arriving at a queueing system may behave differently when the system is full (due to finite waiting space) or when all servers are busy. ♦ Blocking System : The arriving customers when the system is full are considered lost. ♦ Non-Blocking System: The arriving customers are placed in queues of infinite size. TELCOM 2110 Spring 06 18 9
10. Characteristics of the System Structure 1. Physical number and layout of servers ♦ Assume only parallel and identical servers ♦ A customer at the head of the queue can go to any server who is free, and leave the system after receiving this service from that server. 2. The system capacity ♦ The system capacity is the maximum number of customers that a queueing system can accommodate, inclusive of those customers at the service facility. TELCOM 2110 Spring 06 19 Characteristics of the Service Process 3. Characteristics of the Service process. ♦ Queueing discipline ♦ First-Come-First-Served (FCFS/FIFO) ♦ Last-Come-First-Served (LCFS) ♦ Priority ♦ Process sharing ♦ Random ♦ Service distribution • M : Markovian (or memoryless), implies the Poisson process. • D : Deterministic, constant service times • Ek: Erlang distribution of order k service time distribution • G : General service time distribution TELCOM 2110 Spring 06 20 10
11. Basic Queues • Four Basic Queueing models are used in network design • Data networks and database systems – M/M/1 – M/M/1/K • Telephony – M/M/C ⇒ Erlang C – M/M/C/C ⇒ Erlang B TELCOM 2110 Spring 06 21 Markovian Queues Analysis • Develop state transition diagram – System state is indicated by the number of customers in the system at time t ⇒ {n(t), t≥0} • Flow Balance Equations – Derive steady state probability πi = P{n(t) = i} ∑ flow in = ∑ flow out ∑ π i = 1 ∀i • Apply Little’s theorem to obtain mean performance metrics. L = λW TELCOM 2110 Spring 06 22 11
12. M/M/1 Queue • Single server system with infinite capacity. λ μ λ λ λ λ λ μ μ μ μ λπ 0 = μπ 1 j = 0 (λ + μ )π j = λπ j−1 + μπ j +1 j > 0 TELCOM 2110 Spring 06 23 M/M/1 Continued n π n = ρ (1− ρ ) where ⎛ λ ⎞ ⎜ ρ = ⎟ < 1 ⎝ μ ⎠ Mean Number in System L Mean Delay W ρ 1 L = ∑i×π i = W = i (1− ρ) (μ −λ) Variance of number in system σL Variance of Delay σW ρ 1 σ L = σ = (1− ρ)2 W μ2 (1− ρ)2 TELCOM 2110 Spring 06 24 12
13. M/M/1 Example • Consider a concentrator that receives messages from a group of terminals and transmits them over a single transmission line. • The packets arrive according to a Poisson process with one packet every 2.5 ms and the packet transmission times are exponentially distributed with a mean of 2 ms. That is the arrival rate = 1 packet/2.5 ms = 400 packets/sec • Service rate = 1packet/2ms = 500 packets/sec – Find the average delay through the system • Utilization = ρ = 400/500 = .8 – Delay W = 1/(500 – 400) = .01 secs = 10 msecs TELCOM 2110 Spring 06 25 M/M/1/K • The system has a finite capacity of size K. λe = λ(1− Pb) λ μ λ λPb • The state space will be truncated at state K. TELCOM 2110 Spring 06 26 13
14. M/M/1/K λ λ λ λ λ λ λ μ μ μ μ μ μ μ λπ 0 = μπ1 j = 0 (λ + μ)π j = λπ j−1 + μπ j+1 1≤ j < K μπ j = λπ j−1 j = K TELCOM 2110 Spring 06 27 M/M/1/K n n ≤ K π n = ρ π 0 where λ ρ = Normalized offered load μ 0 < ρ < ∞ Solving steady flow equations results in (1− ρ )ρ n π n = ρ ≠ 1 1− ρ K +1 1 π = ρ = 1 n K + 1 TELCOM 2110 Spring 06 28 14
15. M/M/1/K Probability of blocking (Pb) = Loss Rate K (1− ρ)ρ ρ ≠ 1 Pb = π k = 1− ρ K +1 1 P = π = ρ = 1 b k K +1 Portion of traffic dropped/rejected = λ⋅ Pb Effective throughput of the system λe= λ (1-Pb) ⇐ effective arrival rate Example M/M/1/10 Notice how it is nonlinear TELCOM 2110 Spring 06 29 M/M/1/K Effective server utilization : (actual utilization of the system) λ (1− P ) λ ρ = b = e e μ μ Average number in the system K ρ (K + 1)ρ K +1 L = iπ = − ρ ≠ 1 ∑ i K +1 i=0 1− ρ 1− ρ K k ⎛ 1 ⎞ L = ∑ iπ i = ∑ i⎜ ⎟ ρ = 1 i=0 i=0 ⎝ K + 1⎠ ⎛ 1 ⎞ k = ⎜ ⎟ ∑ i ⎝ K + 1⎠i=0 K = 2 TELCOM 2110 Spring 06 30 15
16. M/M/1/K Other performance measures L Mean Delay W= λe 1 Mean Queueing Delay W =W− q μ Mean Number in Queue Lq =L−ρe TELCOM 2110 Spring 06 31 M/M/1/K Example • Consider the queue at an output port of router. The transmission link is a T1 line (1.544Mbps), packets arrive according to a Poisson process with mean rate λ = 659.67 packets/sec, the packet lengths are exponentially distributed with a mean length of 2048 bits/packet. • If the system size is 16 packets what is the packet loss rate? model as M/M/1/16 queue with λ =659.67 , μ= 1.544 Mbps/2048 bits per packet = 753.9 packets/sec ρ = λ/μ= 0.875 Thus the packet loss rate = blocking probability (1− ρ)ρ K (1−.875).87516 P = = = 0.0165 b 1− ρ K +1 1−.87517 TELCOM 2110 Spring 06 32 16
17. M/M/C • C identical servers processes customers in parallel. • Infinite system capacity. μ λ λ TELCOM 2110 Spring 06 33 M/M/C λ λ λ λ λ λ μ 2μ 3μ (C −1)μ Cμ Cμ Flow Balance Equations λπ 0 = μπ1 j = 0 (λ + jμ)π j = λπ j−1 + ( j +1)μπ j+1 1≤ j < C (λ + Cμ)π j = λπ j−1 + Cμπ j+1 j ≥ C TELCOM 2110 Spring 06 34 17
18. M/M/C The server utilization (ρ) λ ρ = Cμ The traffic intensity (a) ⇐ offered load (Erlangs) λ a = μ The stability requirement a ρ = < 1 ⇒ a < C C With traffic intensity a Erlangs, C is the minimum number of servers requirement. TELCOM 2110 Spring 06 36 M/M/C Solving for the state probabilities yields 1 π 0 = C −1a n ac ∑ + n=0 n! (c −1)!(c − a) ai π = π i i! 0 ; 1≤ i < C ai π i = π 0 ; i≥C c!ci−c TELCOM 2110 Spring 06 37 18
19. M/M/C Probability of a customer being delayed C(c,a) ac ∞ (c − 1)!(c − a) C(c, a) = ∑ π j = c−1 a n ac j=c ∑ + n=0 n! (c − 1)!(c − a) C(c,a) ⇐ Erlang’s C formula Erlang’s delay formula Erlangs second formula In the telephone system, C(c,a) represents a blocked call delayed (BCD). TELCOM 2110 Spring 06 38 M/M/C Other performance measures ⎛ a ⎞ Lq = ⎜ ⎟ ⋅C(c,a) ⎝ c − a ⎠ L = Lq + a 1 C(c,a) Lq μ W = = q λ c − a 1 W = W + q μ TELCOM 2110 Spring 06 39 19
20. M/M/C Distribution of the waiting time in the queue −cμ (1− ρ )t P{}wq ≤ t = 1− C(c, a) ⋅ e The pth percentile of the time spent waiting in the queue tp ⎛ 1 − p ⎞ − ln⎜ ⎟ ⎜ C(c, a) ⎟ t = ⎝ ⎠ p cμ (1− ρ ) Note: p > 1 - C(c,a) TELCOM 2110 Spring 06 40 M/M/C Example • Consider a mail order retail company that receives phone orders according to a Poisson process at a mean rate of one call every 2 minutes. The calls have a mean duration of 3.5 minutes. How many operators are required to keep the percentage of customers placed on hold below 20%? • The load in Erlangs is a = .5 calls/minute x 3.5 minutes/ call = 1.75 calculating the probability a call is delayed C(c,a) we get C(2,1.75) = .8167 C(3,1.75) = .3337 C(4,1.75) = .1184 • Thus we need 4 operators to achieve the desired percentage of customers placed on hold. TELCOM 2110 Spring 06 41 20
21. M/M/C/C • C identical servers processes customers in parallel. • The system has a finite capacity of size C. μ λe = λ(1− Pb) λ λe λPb TELCOM 2110 Spring 06 42 M/M/C/C λ λ λ λ λ μ 2μ 3μ (C −1)μ Cμ λπ 0 = μπ1 j = 0 (λ + jμ)π j = λπ j−1 + ( j +1)μπ j+1 1≤ j < C (Cμ)π c = λπ c−1 j = C TELCOM 2110 Spring 06 43 21
22. M/M/C/C 1 π = 0 c n ∑ a n=0 n! ai ai π = π = i! ∀i = 1,2, c i i! 0 c n ∑ a n=0 n! TELCOM 2110 Spring 06 44 M/M/C/C Probability of a customer being blocked B(c,a) ac c! B(c,a) = π = c c n ⇐ Valid for M/G/c/c queue ∑ a n=0 n! B(c,a) ⇐ Erlang’s B formula Erlang’s blocking formula Erlangs first formula In the telephone system, B(c,a) represents a blocked call cleared (BCC). TELCOM 2110 Spring 06 45 22
23. M/M/C/C Erlang B formula can be computed from the recursive formula a ⋅ B(c −1, a) B(c, a) = c + a ⋅ B(c − 1, a) Erlang B formula can be used to computer Erlang C formula c ⋅ B(c, a) C(c, a) = c − a ⋅ (1− B(c, a)) Note : C(c,a) > B(c,a) TELCOM 2110 Spring 06 46 M/M/C/C The carried load λe = λ ⋅ (1− B(c, a)) ⇐ Effective throughput of the system a ρ = ⋅ (1− B(c, a)) Mean server utilization e c a L = ⋅ (1− B(c, a)) Mean number in the system μ 1 Average delay in the system W = μ TELCOM 2110 Spring 06 47 23
24. Traffic Engineering Erlang B table TELCOM 2110 Spring 06 48 M/M/C/C/ Example • Consider a corporate PBX connecting 250 phones to the local phone network. During the busy hour the company sees on average 1 outside call per phone with a average duration of 4 minutes per call. How many DS0s are need to connect the PBX to the local phone company and have a call blocking rate of 1%? • The load is per phone is aphone =1/60 calls per minute x 4 minutes/call = 0.0667 Erlangs • The load on the PBX is aPBX= 205 x 0.0667 = 16.667 Erlangs • From the Erlang B table we need 26 DSOs TELCOM 2110 Spring 06 49 24
25. Traffic Engineering Example TELCOM 2110 Spring 06 50 Project Costing • Review of Economics • Value of money changes with time – Inflation causes future dollars to be worth less than today’s dollars – Investment risk devalues future dollars proportionately to the risk •Elements – Future value (F) – Present value (P) – Rate (i) – Annuity (A) - A sequence of uniform payments – Net Present Value NPV sum of all cash flows moved to the present TELCOM 2110 Spring 06 51 25
26. Formulation • If an amount of money (P) were invested such that it grew at precisely the rate of inflation (i) for one time period, then F = P + Pi = P(1 + i) – That is, F is the equivalent future value of P • For 2 time periods, F = P(1+i) + P(1+i)i = P(1+i)(1+i) • Generalizing, for n time periods F = P(1+i)n – This is referred to as the future worth of a present amount TELCOM 2110 Spring 06 52 Cash Flow Diagrams i = % F dollars in future 12 3 N-1 N F = P (1 + i )N P dollars F = P ()i%, N deposited Example : $1000 today if invested in CD with 3% annual compound interest is worth ? in 5 years F= 1000(1+.03)5 = $1159 Can also find Present value of a Future Payment N ⎛ 1 ⎞ P = F ⎜ ⎟ ⎝ 1 + i ⎠ P = F ()i%, N TELCOM 2110 Spring 06 53 26
27. Cash Flow Series • Annuity - payments of A made at regular intervals • Compute future value i = % A A A A A A A 12 3 N-1 N 2 N F = A[1 + (1 + i) + (1 + i) + K + (1 + i) ] N ⎡ (1 + i) − 1⎤ Example: A company leases a PBX for $1000 a = A⎢ ⎥ quarter for 3 years. What is the value of the ⎣ i ⎦ contract at the end if inflation is 2% quarterly F = A()F / A; i%, N F = 1000( (1+.02)12 -1)/0.02) = $13,412 TELCOM 2110 Spring 06 54 Cash Flow of Series • Computing present value P of annuity A ⎡ (1 + i ) N − 1 ⎤ F = A ⎢ ⎥ ⎣ i ⎦ and F = P ()1 + i N N ⎡ (1 + i ) N − 1 ⎤ ⎛ 1 ⎞ ⎡ (1 + i ) N − 1 ⎤ ∴ P = A ⎢ ⎥ ⎜ ⎟ = A ⎢ N ⎥ ⎣ i ⎦ ⎝ 1 + i ⎠ ⎣ i ()1 + i ⎦ i = % A A A A A A 12 3 N-1 N F dollars P dollars Example Present value of PBX lease in future deposited P = 1000 [( (1+.02)12 -1)/(0.02(1+.02)12)] P = $10,575 TELCOM 2110 Spring 06 55 27
28. Cost Example • Move project cost either to net present value NPV or to Future Present Value to compare alternatives • Example buying PBX vs. leasing PBX for 10 year project (i = 5%) Buy PBX Lease PBX Purchase value = $27,000 Maintenance = $4000 yearly fee $1000 year Salvage value = $2000 NPV = 4000 [((1+.05)10 –1)/(.05(1+.05)10)] NPV = $27000 + = $30,887 1000[((1+.05)10 -1)/(0.05(1+.05)10)] - $2000 (1/(1+.05)10) Outcome may be different if NPV = $27000 + $7,722 - $1,228 include takes and if = $33,494 depreciation applicable to taxes! TELCOM 2110 Spring 06 56 Depreciation • Definitions of Depreciation – A System of Accounting which Aims to Distribute Cost or Other Basic Value of Tangible Capital Assets, Less Salvage Value, Over the Estimated Useful Life of a Unit in a Systematic and Rational Manner for the Purpose of Allocation (Paraphrased from ACPA) – Loss in Service Value Not Restored by Maintenance – Due to Normal Wear and Tear, Exposure and Decay, Technological Obsolescence, etc. • Depreciation Does Not Involve Actual Cash Outlays TELCOM 2110 Spring 06 57 28
29. Computing Depreciation Expense • Original Cost of Equipment • Estimated Service Life of Equipment – Data equipment 3-5 years lifetime – Telecom equipment 5-20 years lifetime • Estimated Net Salvage Value of the Equipment – Remaining Value at the End of the Service Life – Can Include the Cost of Removal • Depreciation Method – Retirement/Replacement • Not widely used because carry cost until retirement – Age-Life TELCOM 2110 Spring 06 58 Age-Life Methods • Designed to Provide More Consistent Expense Accounts from Year to Year • Straight-Line Depreciation – Depreciation Charge is Computed for Each Retirement Period – Draw a Straight Line Between Original Cost and estimated Salvage Value, and Allocate the Difference over Service Life Original Cost - Salvage Value Depreciation Charge = Service Life TELCOM 2110 Spring 06 60 29
30. Deprecation Concepts Cost ($) Original Cost Accrued Depreciation Annual Depreciation Charge Current Value Service Salvage Life Value 5 10 15 Time (Years) TELCOM 2110 Spring 06 61 Age-Life Methods • Accelerated depreciation allows higher depreciation early in the equipment life than straight line method • Sum-of-the-Year’s-digits # Years Remaining at Beginning of Year Depreciation Expense = x (Original Cost - Salvage Value) Total of the Digits of the Year’s Life – Subtract from current value (undepreciated value) – Repeat next year • Double declining balance – Double the depreciation rate of straight line – Subtract from current value – Depreciate remaining balance by straight line TELCOM 2110 Spring 06 62 30
31. Comparison of Depreciation Approaches Cost ($) Original Cost = $2M Straight-Line $2M -$100K = $1.9M $1.8M Sum of the Year’s Digits Double Declining Balance Service Life Salvage Value = $500,000 5 10 15 Time (Years) TELCOM 2110 Spring 06 63 Comments • Size of depreciation charge depends on – Service Life – Salvage value • Estimating both parameters in advance is difficult – Service life must take technological and usage factors into account – Actual salvage value depends on costs and prices at the time of decommissioning • Summarizing – Using GAAP the value of an asset is • Current Value = Purchase Value – Depreciation • Can be factored in to network design cost depends on organization TELCOM 2110 Spring 06 64 31
32. Network Design Cost •Network Cost •Project cost usually determined as Net Present Value (NPV) of all cash flows in the project •In WANs and metro access networks in addition to equipment cost link BANDWIDTH is a significant cost • Bandwidth is a reoccurring cost (treat like an annuity) Item Example Cost Terminal router $2000 purchase price Transit router $3700 purchase price WAN adapter $500 purchase price T1 1.544Mbps link $1000 to hook up + $1400/month TELCOM 2110 Spring 06 65 Bandwidth Cost •Tariffs • Tariff is a published rate for telecommunications services and facilities (in U.S. carriers file tariffs with FCC) • Types of Links • Usage-sensitive (fixed cost + variable cost charged per minute or per X bytes ) • Usage insensitive (leased line) (fixed cost + monthly fee) TELCOM 2110 Spring 06 66 32
33. Usage Sensitive Tariffs • A simple usage-sensitive voice tariff: • Fixed cost for the use of line. • Can place outgoing calls for an additional fee that depends on the time of day • Unlimited incoming calls (the calling party pays). • The tariff is specified in 1-minute increments. TELCOM 2110 Spring 06 67 Usage Sensitive Tariffs • Typical Usage Sensitive Tariff factors 1. Access fees ( the cost of maintaining a physical network connection). 2. Setup fees. 3. Teardown fees. 4. Usage fees, which depend on - channel capacity - usage (# phone calls, mean bit rate, peak bit rate, etc.) - distance (local, long distance, international) - time of day - national and administrative borders TELCOM 2110 Spring 06 68 33
34. Usage Sensitive Tariffs • Some data services are based on usage sensitive pricing – For example Frame Relay Service – Get peak rate and committed information rate (CIR) , charged usage fee per mean kbps above CIR – London – Manchester, UK, – Cable and Wireless 256K link , CIR = 64Kbps • Connection $16,393, • Rental (yearly) $17,591, • Bandwidth charge $ 974 x 1kbps/month • Many wireless data services have similar pricing TELCOM 2110 Spring 06 69 Usage Insensitive Cost • Leased Line Cost of Link Bandwidth depends on variety of factors • Tariffs • Service Provider • Capacity of link • fractional T1, T1, multiple T1, OC1, OC3, etc • Length • Technology (WATS, ISDN etc.) • Location • NYC-DC cheaper than Asheville, NC – Memphis, TN • QoS/Availability/survivability requirements TELCOM 2110 Spring 06 70 34
35. Leased Line Rates • U.S. leased line rates • ISDN 56 Kbps; fractional T1, e.g. 128 Kbps, 256 Kbps, 512 Kbps, full T1. and rest of synchronous digital hierarchy (SDH) in Table • Europe leased line rates • multiples of 64 Kbps, including half E1 = 1024 Kbps, multiple E1, E3 etc. • Note with leased line – get symmetrical bandwidth allocations • If go with data service (ATM, Frame Relay, MPLS) can get asymmetrical bandwidth allocations as part of a Service Level Agreement (SLA) TELCOM 2110 Spring 06 71 Usage Insensitive Tariffs • Suppose we have a call center located 70 miles from a large city, where we have several hundred customers. • Options – Leased line – Out-of-district line gives a line that appears to be in the city • Reduces the per-minute charge for outgoing calls from $0.095 per minute to $0.03 per minute. – Banded WATS (wide area telephone service) (4 hours/day) – allows up to 4 hours of calls per day to locations that are no more than a certain specified distance, for instance 250 miles. – Nationwide WATS - national coverage with unlimited usage. TELCOM 2110 Spring 06 72 35
36. Leased Line Tariffs • U.S. leased line rates • Tariff rate very in U.S. with amount of competition – rough approximation fixed cost + linear distance cost, in reality more complicated • In Europe tariffs are largely regulated and consistent within a country – Usage-insensitive data tariff for British Telcom in U.K. in table below • Software tools exist that incorporate detailed tariff data (e.g., Pricer at for analysis TELCOM 2110 Spring 06 73 Linear Distance Based Tariffs • As noted bandwidth costs are function of a variety of factors. • In practice use simple linear distance based model to represent cost of service • Cost = Fixed cost + distance cost x distance – For example, for a 128 Kbps link in the U.K. we can use the approximation of $757.09 + $2.40/km • If detailed tariff data available can develop linear model by using regression analysis on the tariff table data • Such an approach often results in a piecewise linear model TELCOM 2110 Spring 06 74 36
37. Distance Coordinate Systems ƒ Need to determine distance between sites to estimate cost – two coordinate system approaches Vertical and Horizontal (V&H) - a grid of lines defined by AT&T in 50’s for North America - allows for a simplified computation of distances - Widely used in Telco industry ƒ Latitude and Longitude (L&L) - defined for all locations on the surface of the earth. - The distance calculation is essentially an exercise in spherical geometry. C code and formula in book TELCOM 2110 Spring 06 75 V&H Coordinate System Given two cities coordinates (v1, h1), (v2,h2) find distance d apart 2 2 d = ceil ( (( v 1 − v 2 ) + 9 ) / 10 + (( h 1 − h 2 ) + 9 ) / 10 ) Can approximate by City name V H coordinate coordinate 2 2 New York 4997 1406 d = ceil( (v1 − v2) /10 + (h1 − h2) /10) Los 9213 7878 Angeles For example for simple network Chicago 5986 3426 discussed earlier , Dallas 8436 4034 DC (5622, 1583) – Denver (7501,5899) Pittsburgh 5621 2185 => d = ceil(sqrt(353064.1 + 1862785)) DC 5622 1583 d = 1489 miles Seattle 6336 8896 Miami 8351 0527 Denver – Dallas => d = 660 miles Atlanta 7260 2083 Boston 4422 1249 Dallas – DC => d = 1180 miles Denver 7501 5899 TELCOM 2110 Spring 06 76 37
38. Latitude and Longitude Coordinate • Distance D in degrees between two points X and Y on a sphere with latitude and longitude values (LatX, LongX), (LatY, LongY) found from • Cos(D) = Sin(LatX) Sin(LatY) + Cos(LongX)Cos(LongY)Cos(|LongY-LongX|) • Find D in degrees by D = cos-1(cos(D)) • Convert to kilometers multiply by 111.23 km/degree • Example: Paris, France (48.87oN, 2.33oE), Austin, Tx (30.27oN, 97.74oW) • cos D = [sin(48.87) * sin(30.27)] + [cos(48.87) * cos(30.27) * cos(|- 97.74 - 2.33|)] = 0.281 •Distance = 111.23 x cos-1(0.281) = 8,195.44 km TELCOM 2110 Spring 06 77 Simple Network Design Example • Example- company with offices in Dallas and Vienna, VA, • Factory in Denver • Appl A: Sales/inventory control •ApplB: CAM •ApplC: CAD • Appl D: video conference • Appl E: Intranet Voice TELCOM 2110 Spring 06 78 38
39. Applications Map • From Applications Map – get rough idea of traffic flows between network nodes • Get the beginnings of a traffic demand matrix across the Wide Area Network • If use Applications Monitoring Approach –gather data on each application • A: Mean rate = .1 Mbps, Peak = .15 Mbps • C: Mean rate = .5 Mbps, Peak = .75 Mbps • D: Mean rate = 2 Mbps, Peak = 2.5 Mbps TELCOM 2110 Spring 06 79 Traffic Demand Matrices • From the application map and associated matrix and the application monitoring data we have the mean traffic demand matrix and peak traffic demand matrix as below • Note, if the network monitoring approach is used get traffic demand directly. Mean data Dallas Denver Vienna Peak data Dallas Denver Vienna rate demands rate demands Dallas .1Mb 2.1 Mb Dallas .15Mb 2.65 Mb Denver .3 Mb .8Mb Denver .45 Mb 1.2Mb - Vienna 2.5 Mb 1.5 Mb Vienna 3.25 Mb 2.25 Mb TELCOM 2110 Spring 06 80 39
40. Example Network Design • Consider simple network design based on mean data rates • Objective: average link utilization 50% or less at each link • Link capacity is purchased in T1 or multiple T1 sizes. • A logical layer network design solution is a minimum spanning tree (discussed later) 1.6 Mb • The demands for each direction per link are 3T1 Denver given next to the directional arrow Vienna • In order to size link 2.8Mb – pick max demand in either direction 1.1Mb – double max demand to meet 50% utilization objective – Modularize into T1 multiples 4T1 • For example Dallas –Vienna Link 2.2Mb – Max = 2.8 Mb, double to 5.6 Mb => 4 T1 lines each 1.54 Mbps Dallas – Similarly Denver –Vienna link is 3 T1 lines – Need 7 Total T1 lines – Check shows peak demands can be carried TELCOM 2110 Spring 06 81 Example Network Design • Note many alternate network designs possible • A solution is a minimum spanning tree • If we root tree and Denver • Again to size link – pick max demand in either direction 6T1 4 Mb – double max demand to meet 50% Denver utilization objective Vienna – Modularize into T1 multiples 2.9Mb • For example Denver –Vienna Link – Max = 4 Mb, double to 8 Mb => 6 T1 2.8Mb 2.2Mb lines each 1.54 Mbps – Similarly Denver – Dallas link is 4 T1 lines 4T1 – Need 10 total T1 lines Dallas – Checking shows peak demands can be carried TELCOM 2110 Spring 06 82 40
41. Example Network Design • If spanning tree is rooted at Dallas • The demands for each direction per link are given next to the directional arrow • Again to size link – pick max demand in either direction Denver – double max demand to meet 50% Vienna utilization objective 4 Mb – Modularize into T1 multiples 1.1Mb • For example Dallas –Vienna Link 1.6 Mb 6T1 – Max = 4Mb, double to 8 Mb => 6T1 lines each 1.54 Mbps 3T1 2.9 Mb – Similarly Dallas - Denver link is 3 T1 lines Dallas – Need 9 T1 lines Total – Note peak demands can be carried TELCOM 2110 Spring 06 83 Example Network Design • Consider simple network design again – three options • Assume cost of T1 = $2406.00 +$0.49/mile per month • Dallas –Vienna Link – 4 T1 lines = 4 x(2406 +.49 x1180) = $11,937 1.6 Mb 3T1 Denver • Similarly Denver –Vienna link is 3 Vienna T1 lines 2.8Mb 1.1Mb – Cost = 3*(2406 +.49 x1489) = $9407 • Total Bandwidth Cost = $21,344 per month 4T1 2.2Mb • Similarly Cost of Other Designs Dallas • Denver Root Cost = $29,731 • Dallas Root Cost = $26,093 TELCOM 2110 Spring 06 84 41