- 2077 Chaitra · 2+2+2+3 marks
Numerical: trunked channels and blocking (Erlang B)
Short answer
a) Traffic per user is 0.25 Erlang. b) With one channel at 1% blocking, offered traffic is 0.010 E, supporting 0.04 user, effectively no user. c) With five channels, A=1.36 E, supporting 5.44, about 5 users. d) Doubling to 10 users gives A=2.5 E and about 7% blocking, which is unacceptable.
A mobile radio system where each user averages three calls per hour and each call lasting an average of 5 minutes. a) What is traffic intensity for each user? b) Find the number of users that could use the system with 1% blocking if only one channel is available. c) Find the number of users that could use the system with 1% blocking if five trunked channels are available. d) If the number of users you found in (c) is suddenly doubled, what is the new blocking probability of the five channel trunked mobile radio system? Justify whether the performance is acceptable or not.
Traffic (A) in erlangs for B%:
| No. of Trunks (N) | 0.1% | 0.2% | 0.5% | 1% | 1.2% | 3% | 5% | 7% |
|---|---|---|---|---|---|---|---|---|
| 1 | 0.001 | 0.002 | 0.005 | 0.010 | 0.012 | 0.031 | 0.053 | 0.075 |
| 2 | 0.046 | 0.065 | 0.105 | 0.153 | 0.168 | 0.282 | 0.381 | 0.470 |
| 3 | 0.194 | 0.249 | 0.349 | 0.455 | 0.489 | 0.715 | 0.899 | 1.06 |
| 4 | 0.439 | 0.535 | 0.701 | 0.869 | 0.922 | 1.26 | 1.52 | 1.75 |
| 5 | 0.762 | 0.900 | 1.13 | 1.36 | 1.43 | 1.88 | 2.22 | 2.50 |
Given: λ = 3 calls/hour, H = 5 min, blocked calls cleared (Erlang B).
a) Traffic intensity per user
Au = λH = 3 × 5/60 = 0.25 Erlang
b) One channel, 1 % blocking
From the table, N = 1, B = 1 %: A = 0.010 Erlang.
U = A / Au = 0.010 / 0.25 = 0.04 user
Fewer than one user can be supported: even a single user (0.25 E) would cause far more than 1 % blocking. So effectively no user (only 0.04 user) can be served at 1 % GOS with one channel.
c) Five trunked channels, 1 % blocking
From the table, N = 5, B = 1 %: A = 1.36 Erlangs.
U = 1.36 / 0.25 = 5.44 ≈ 5 users
d) Number of users doubled
U = 2 × 5 = 10 users
A = 10 × 0.25 = 2.5 Erlangs
From the table, N = 5 and A = 2.50 corresponds to B = 7 %. (Erlang B formula gives 6.97 %.)
If 5.44 users are doubled (10.88 users, A = 2.72 E), blocking is about 8.7 %.
Is it acceptable? No. The blocking probability rises from 1 % to about 7 %, i.e. 7 of every 100 call attempts fail in the busy hour. This is well above the usual design GOS of 1–2 %, so service quality is unacceptable. More channels are needed (from the Erlang B table, about 7 channels would be needed for 2.5 E at 1 %).
| Case | Channels | A (E) | Users | Blocking |
|---|---|---|---|---|
| b | 1 | 0.010 | 0.04 | 1 % |
| c | 5 | 1.36 | 5 | 1 % |
| d | 5 | 2.50 | 10 | ≈ 7 % |
Note: trunking 5 channels supports 5.44 users compared with 5 × 0.04 = 0.2 users on 5 separate single channels, showing the trunking gain.
✓ Answer checked by Claude Opus 5.5 (high effort)
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