---
title: "Numerical: users supported and market penetration"
subject: "Wireless Communication"
chapter: 2
key: "semester_7/wireless_communication:market-penetration-erlang"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/market-penetration-erlang
site: BCT Engineering
---

# Numerical: users supported and market penetration

Wireless Communication · Chapter 2: Cellular Mobile Communication Concepts

- Asked in: 2080 Baisakh, 5 marks; 2079 Chaitra, 5 marks (asked 2 times)
- Video: [Example#2.5 Wireless Communication by Theodore Rappaport Solved| Ibtisam Hasan |](https://www.youtube.com/watch?v=VGGoPBZcmmk) — Ibtisam Hasan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/market-penetration-erlang
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-2

## Question

An urban area has a population of two million residents. System A has 394 cells with 19 channels each. Find the number of users that can be supported at 2% blocking if each user averages two calls per hour at average call duration of three minutes. Assuming that trunk systems are operated at maximum capacity, compute the percentage market penetration of cellular provider.

Erlang B Traffic Table (Maximum Offered Load Versus B and N; B is in %):

| N/B | 0.01 | 0.05 | 0.1 | 0.5 | 1.0 | 2 | 5 | 10 | 15 | 20 | 30 | 40 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 16 | 5.339 | 6.250 | 6.722 | 8.100 | 8.875 | 9.828 | 11.54 | 13.50 | 15.18 | 16.81 | 20.30 | 24.54 |
| 17 | 5.911 | 6.878 | 7.378 | 8.834 | 9.652 | 10.66 | 12.46 | 14.52 | 16.29 | 18.01 | 21.70 | 26.19 |
| 18 | 6.496 | 7.519 | 8.046 | 9.578 | 10.44 | 11.49 | 13.39 | 15.55 | 17.41 | 19.22 | 23.10 | 27.84 |
| 19 | 7.093 | 8.170 | 8.724 | 10.33 | 11.23 | 12.33 | 14.32 | 16.58 | 18.53 | 20.42 | 24.51 | 29.50 |
| 20 | 7.701 | 8.831 | 9.412 | 11.09 | 12.03 | 13.18 | 15.25 | 17.61 | 19.65 | 21.64 | 25.92 | 31.15 |

(The Erlang B table was attached to the 2079 Chaitra paper.)

## Short answer

Each user offers 0.1 Erlang. At 2% blocking, 19 channels carry 12.33 Erlangs, so 123 users per cell, giving about 48,462 total users and 2.42% market penetration.

## Answer

**Given:** population = 2 000 000; cells = 394; channels per cell C = 19; GOS = 2 % blocking; λ = 2 calls/hour; H = 3 min.

**Step 1: Traffic per user**
```text
Au = λ × H = 2 × (3/60) = 0.1 Erlang
```

**Step 2: Traffic per cell from Erlang B table**
For N = 19 channels and B = 2 %:
```text
A (per cell) = 12.33 Erlangs
```

**Step 3: Users per cell**
```text
U = A / Au = 12.33 / 0.1 = 123.3 ≈ 123 users per cell
```

**Step 4: Total users supported**
```text
Total = 123 × 394 = 48 462 users
(Without rounding: 123.3 × 394 = 48 580)
```

**Step 5: Market penetration**
```text
Penetration = 48 462 / 2 000 000 × 100 = 2.42 %
```

**Answer: About 48 462 users (≈ 48 500) can be supported at 2 % blocking, giving a market penetration of about 2.42 %.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Traffic per user for 2 calls/hour of 3 minutes is:
   A. 0.1 Erlang
   B. 0.2 Erlang
   C. 6 Erlang
   D. 0.03 Erlang
   Answer: A. Au = 2 x 3/60 = 0.1 Erlang. (Key point: Traffic per user)

2. For 19 channels at 2% blocking, the Erlang B table gives per-cell traffic of:
   A. 14.32 Erlang
   B. 12.33 Erlang
   C. 11.23 Erlang
   D. 16.58 Erlang
   Answer: B. The N = 19, B = 2 entry is 12.33 Erlangs. (Key point: Table lookup)

3. Users per cell is found by:
   A. A divided by Au = 12.33/0.1 = 123
   B. A times Au
   C. A / 19
   D. A x 394
   Answer: A. Divide cell traffic by traffic per user. (Key point: Users per cell)

4. Using 123 x 394 = 48 462 users, penetration of 2 million residents is about:
   A. 24.2 %
   B. 0.24 %
   C. 2.42 %
   D. 48.5 %
   Answer: C. 48 462/2 000 000 = 2.42 %. (Key point: Penetration)

## Related in Wireless Communication

- [Numerical: trunked channels and blocking (Erlang B)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atrunking-five-channel-blocking) — Wireless Communication, chapter 2
- [Numerical: 32-cell system area and channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-32-cells-capacity) — Wireless Communication, chapter 2
- [Numerical: 12-cell system capacity at 2% GOS](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-capacity-12-cell-system) — Wireless Communication, chapter 2
- [Numerical: cluster size and users in service area](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acluster-size-users-service-area) — Wireless Communication, chapter 2
- [Numerical: cellular vs non-cellular voice channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-vs-noncellular-channels) — Wireless Communication, chapter 2
- [Evolution and worldwide market penetration](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aevolution-market-penetration) — Wireless Communication, chapter 1

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