1. Thought of the day Thought of the day Difference between science and spirituality is same as the difference between word and silence. Sameer Trapasiya.

3. What we have learned so far  Evolution of wireless communication  Call flow diagram  Cell Topology  Handoff  Umbrella cell  C0- channel cell Distance.  Interference

4. What we will learn today ?  Power control  Channel assignment Examples.  Trunking Theory  Erlang B formula  Examples

5. Power control for reducing interference  Practically all MS are transmitted same power to BS.  Power control means different user transmitting different power level.  User closer to BS transmit less power.  User far away from BS transmit more power.  It has following advantages  Save battery life at mobile terminal.  Reduce interference and increase S/I ratio.  It also reduce the effect of near-far problem.

6. Example Example There are two company A and B in the market. 832 channels are available to them. How channels are alloted to them ? 7 cell reuse scheme is implemented. Solution :- System A No. of voice channels 312(1-312) Extended block channels 83(667-716, 991-1023) No. of control Channels -21(313-333) System B No. of voice channels 312(355-666) Extended block channels 83(717-799) No. of control Channels -21(334-354)

8. Some basic Terms

9. Traffic Theory Trunking  Allows a large number of users to share a small number of channels  Channel allocated per call basis from a pool of available channels  Relies on statistical behavior of users so that a fixed number of channels (circuits) may accommodate a large random user community  Trunking theory is used to determine number of channels for particular area (users).  Tradeoff between the number of available channels and likelihood of call blocking during peak calling hours

10. Trunking Theory Trunking Theory  Developed by Erlang, Danish Mathematician, how a large population can be accommodated by a limited number of servers, in late 19 th century.  Today, used to measure traffic intensity  1 Erlang represents the amount of traffic intensity carried by a completely occupied channel  i.e. one call-hour per hour or one call-minute per minute  0.5 Erlang: Radio channel occupied 30 minutes during 1 hour

11. Grade of Service  GOS is a benchmark used to define performance of a particular trunked system  Measure of the ability of a user to access trunked system during the busiest hour.  Busy hour is based on the demands in an hour during a week, month or year.  Typically occur during rush hours between 4 pm to 6 pm.  GOS is typically given as likelihood of call blocking or delay experienced greater than certain queue time

12. Traffic Intensity  Traffic intensity is measured as call request rate multiplied by call holding time User traffic intensity of A u Erlang is A u = λ H (1) Where H is average call duration or holding time and λ is average number of call requests.  For system of U users and unspecified channels, the total offered traffic intensity A is A = UA u (2)  In a C channel trunked system, traffic equally distributed, traffic intensity per channel Ac is A c = UA u /C (3)

13. Block Calls Cleared  User is given immediate request if a channel is available.  If no channel available, the requesting user is blocked and free to try later  Assume call arrivals as Poisson Distribution  the Erlang B formula determines the probability that call is blocked with no queuing, is a measure of GOS for trunked system.

14. Erlang B Trunking GOS Number of Channel s C Capacity (Erlangs) For GOS Call Blocking Prob. = 0.01= 0.005= 0.002= 0.001 20.1530.1050.0650.046 40.8690.7010.5350.439 51.361.130.9000.762 104.463.963.433.09 2012.011.110.19.41 2415.314.213.012.2 4029.027.325.724.5 7056.153.751.049.2 10084.180.977.475.2

15. Erlang B chart Erlang B chart

16. Examples  How many users can be supported for 0.5% blocking probability for the following no. of trunked channel in a blocked call cleared system ? (a) 1 (b) 5 ( c) 10. Assume traffic intensity per user is 0.1 Solution :- (a) Given C=1, Au=0.1, GoS= 0.005 From the Fig. we obtain A= 0.005 Total No. of users = U= A/Au= 0.005/0.1=0.05 But one user can be supported one channel U=1; (b) Given C=5, Au=0.1, GoS= 0.005 From the Fig. we obtain A= 1.13 Total No. of users = U= A/Au= 1.13/0.1=11 ( C) Given C=10, Au=0.1, GoS= 0.005 From the Fig. we obtain A= 3.96 Total No. of users = U= A/Au= 3.96/0.1=40