1. 1 Lecture 06 EEE 441: Wireless And Mobile Communications BRAC University

2. Quiz2 today 2

3. Return HW2 3

4. 4 Queueing (or Trunking) Each user placed in queue for free channel upon trying to initiate a call Each call then assigned a channel by base station or MSC  (depending upon if channel assignment is fixed or dynamic) Potential problems:  Probability of call blocking  Probability of remaining in queue for more than certain amount of time Solutions:  Formulas proposed by Erlang based upon call arrival rate and distribution, serving rate, buffer size

5. 5 Queueing Theory Usually described by  A / B / S  A = Distribution of inter-arrival times  B = Distribution of serving times  S = Number of servers Common queue types studied in literature  M/M/m, M/G/m, M/M/1, M/G/1  M = memoryless (i.e. independent)  G = general  m = number

6. 6 Grade/Quality of Service (GoS/QoS) How can a large number of users be accommodated by a limited number of channels? Bottleneck occurs during busiest times day, month, year… What is the Probability of a call being blocked?  First kind of system: no queue maintained What is the Probability of being in a queue > t seconds?  Second kind of system: queue maintained

7. 7 The Erlang  Unit of traffic intensity  1 Erlang = a completely occupied channel  Let:  - Average number of call requests per unit time  H - duration of a call  Then traffic intensity of each user (in Erlangs) is:  A  = ?

8. 8 The Erlang and Single User Intensity  Erlang: unit of traffic intensity  1 Erlang = a completely occupied channel  Let:  - Average number of call requests per unit time  H - duration of a call  Then traffic intensity of each user (in Erlangs) is:  A u = H

9. 9 Total Traffic Intensity  If system has U total users,  then total traffic intensity is:  A = ?  If this total traffic intensity is equally divided among C channels in the system,  then average intensity per channel is:  A c = ?

10. 10 Total Traffic Intensity  If system has U total users,  then total traffic intensity is:  A = U A u = U H (in Erlangs)  If this total traffic intensity is equally divided among C channels in the system,  then average intensity per channel is:  A c = A / C = U A u / C

11. 11 Type 1: Blocked Call Cleared System No queuing for call requests If no channels are available, then requesting user is blocked and must try again later

12. 12 GoS or QoS: Defined by the Blocking Probability  GoS = Probability (call is blocked)  Given by Erlang B formula

13. 13 Type 2: Blocked Calls Delayed System Queue is provided to hold calls If a channel is not available immediately, the call request may be delayed System works as long as queueing delay is small

14. 14 GoS or QoS: Defined by Probability of Being Placed in Queue  Given by Erlang C formula

15. 15 Probability of Delay Exceeding a Threshold Avg delay D for all calls in a queued system is: D = H/(C-A). Pr[Delay>0]

16. 16 Example A hexagonal cell in a 4-cell system has a radius of 1.387km, and a total of 60 channels are used within the entire system. If the load / user is 0.029 Erlangs, = 1 call per hour, compute the following for an Erlang C system that has a 5% probability of a delayed call.  a. How many users per square km will the system support?  b. What is the Prob [ Delay > 10s ]?

17. 17 Solution Cell radius, R = 1.387 km Area covered per cell = 2.598 (1.387) 2 = 5 sq km Number of cells per cluster, N = 4 Total number of channels per cell = 60 / 4 = 15

18. 18 Solution (cont’d) a. From Erlang C chart, GOS = 0.05, C = 15,  => Traffic intensity per cell, A = 9.0 Erlangs  Hence Number of users in each cell = total traffic intensity / traffic per user = 9.0 / 0.029 = 310 users  Hence Number of users per sq. km = 310 / 5 = 62 users per sq. km.

19. 19 Solution (cont’d) b. Prob [Delay > 10s] = Pr [Delay > 0 ] e –(C-A) t / H = 0.05 x e –(15-9) 10 / H whereH = A  / = 0.029/1 hour = 0.029 x 60 x 60 seconds = 104.4 seconds Thus Prob [Delay > 10s] = 0.05 e –(15-9) 10 / 104.4 = 0.0281 = 2.81%

20. 20 Shannon’s Theorem The fundamental result in communication / information theory C = B log 2 (1+S/N) C is capacity in bits/s, B is bandwidth in Hertz, SNR is Signal Power (in Watts) over Noise Power (in Watts)

21. 21 Shannon’s Theorem: Interpretations Means that:  There exists an upper bound on the maximum amount of data (information) that can be transmitted within a specified bandwidth in the presence of the noise, with arbitrarily small error  It is possible to design a coding scheme at Tx rate R < C, such that Tx occurs with arbitrarily small P e  Contrarily if R > C is sought, then the error will increase without bound, and no useful information will be transmitted

22. 22 Notices Read:  Rappaport, Ch. 3.6  Special Attn: Examples 3.4 - 3.7 HW  HW3 assigned