Give step-by-step solution with explanation and final answer:
Summary Table
\(\ce{\begin{array}{|c|c|c|c|} \hline \text{Part} & \text{Parameter} & \text{Value / Calculation} & \text{Answer} \\ \hline \text{a} & \text{Channels per cell} & \text{ \lfloor 69/7 \rfloor } & \text{9} \\ \hline \text{b} & \text{Traffic per cell} & \text{Erlang B C=9 B=.01 } & \text{5. Erlangs} \\ \hline \text{c} & \text{Users per cell} & \text{ 5. / .05 } & \text{100 users} \\ \hline \text{d} & \text{Subscribers (omni)} & \text{ 500 \times 100 } & \text{50000 users} \\ \hline \text{e} & \text{Subscribers ( 120^\circ )} & \text{ 500 \times 220 } & \text{110000 users} \\ \hline \text{Trunking efficiency change} & \text{ 220-100/100 \times 100\% } & \text{120% improvement (not degradation)} & \text{} \\ \hline \end{array}}\)\begin{array}{|c|c|c|c|} \hline \text{Part} & \text{Parameter} & \text{Value / Calculation} & \text{Answer} \\ \hline \text{a} & \text{Channels per cell} & \text{ \lfloor 69/7 \rfloor } & \text{9} \\ \hline \text{b} & \text{Traffic per cell} & \text{Erlang B C=9 B=.01 } & \text{5. Erlangs} \\ \hline \text{c} & \text{Users per cell} & \text{ 5. / .05 } & \text{100 users} \\ \hline \text{d} & \text{Subscribers (omni)} & \text{ 500 \times 100 } & \text{50000 users} \\ \hline \text{e} & \text{Subscribers ( 120^\circ )} & \text{ 500 \times 220 } & \text{110000 users} \\ \hline \text{Trunking efficiency change} & \text{ 220-100/100 \times 100\% } & \text{120% improvement (not degradation)} & \text{} \\ \hline \end{array}
Notes
If you use a different reuse factor for sectoring (e.g., \(\ce{N=3 }\)N=3 ), the number of channels and users per cell would be even higher.
If the sectoring implementation splits the total channels among sectors (instead of tripling the channels per site), adjust accordingly. This answer follows the typical sectoring improvement by reducing the reuse factor.
Question:
Give step-by-step solution with explanation and final answer:
Summary Table
\(\ce{\begin{array}{|c|c|c|c|} \hline \text{Part} & \text{Parameter} & \text{Value / Calculation} & \text{Answer} \\ \hline \text{a} & \text{Channels per cell} & \text{ \lfloor 69/7 \rfloor } & \text{9} \\ \hline \text{b} & \text{Traffic per cell} & \text{Erlang B C=9 B=.01 } & \text{5. Erlangs} \\ \hline \text{c} & \text{Users per cell} & \text{ 5. / .05 } & \text{100 users} \\ \hline \text{d} & \text{Subscribers (omni)} & \text{ 500 \times 100 } & \text{50000 users} \\ \hline \text{e} & \text{Subscribers ( 120^\circ )} & \text{ 500 \times 220 } & \text{110000 users} \\ \hline \text{Trunking efficiency change} & \text{ 220-100/100 \times 100\% } & \text{120% improvement (not degradation)} & \text{} \\ \hline \end{array}}\)\begin{array}{|c|c|c|c|} \hline \text{Part} & \text{Parameter} & \text{Value / Calculation} & \text{Answer} \\ \hline \text{a} & \text{Channels per cell} & \text{ \lfloor 69/7 \rfloor } & \text{9} \\ \hline \text{b} & \text{Traffic per cell} & \text{Erlang B C=9 B=.01 } & \text{5. Erlangs} \\ \hline \text{c} & \text{Users per cell} & \text{ 5. / .05 } & \text{100 users} \\ \hline \text{d} & \text{Subscribers (omni)} & \text{ 500 \times 100 } & \text{50000 users} \\ \hline \text{e} & \text{Subscribers ( 120^\circ )} & \text{ 500 \times 220 } & \text{110000 users} \\ \hline \text{Trunking efficiency change} & \text{ 220-100/100 \times 100\% } & \text{120% improvement (not degradation)} & \text{} \\ \hline \end{array}
Notes
If you use a different reuse factor for sectoring (e.g., \(\ce{N=3 }\)N=3 ), the number of channels and users per cell would be even higher.
If the sectoring implementation splits the total channels among sectors (instead of tripling the channels per site), adjust accordingly. This answer follows the typical sectoring improvement by reducing the reuse factor.
This is AI generated Response.
Asked by: Aman
Created at: 2025-09-30 07:47:33
Final Answer:
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# Step-by-Step Solution & Explanation
is a detailed, stepwise solution each part of the problem using the summary table you provided. We'll include calculations and reasoning.
---
## Part (a): Channels per Cell
**Given:**
- Total available channels = 69
- Cluster size \(N = 7\) (reuse factor)
**Calculation:**
\[
\text{Channels per cell} = \left\lfloor \frac{69}{7} \right\rfloor = \left\lfloor 9.857 \right\rfloor = 9
\]
**Explanation:**
Divide the total number of channels by the number of cells in a cluster, then round down (since you can't have a fraction of a channel).
**Final Answer:**
**9 channels per cell**
---
## Part (b): Traffic per Cell (Erlang B)
**Given:**
- Number of channels per cell, \(C = 9\) (from part a)
- Blocking probability, \(B = .01\) (1%)
**Calculation:**
Use the Erlang B table or calculator for \(C = 9\), \(B = .01\):
\[
A = 5 \text{ Erlangs}
\]
**Explanation:**
The maximum carried traffic per cell at 1% blocking with 9 channels is approximately 5 Erlangs (from Erlang B formula/tables).
**Final Answer:**
**5 Erlangs per cell**
---
## Part (c): Users per Cell
**Given:**
- Traffic per cell = 5 Erlangs (from part b)
- Each user generates .05 Erlangs (average call rate)
**Calculation:**
\[
\text{Users per cell} = \frac{5}{.05} = 100
\]
**Explanation:**
Divide total traffic per cell by the average traffic per user.
**Final Answer:**
**100 users per cell**
---
## Part (d): Total Subscribers (Omnidirectional Cells)
**Given:**
- Number of cells = 500
- Users per cell = 100 (from part c)
**Calculation:**
\[
\text{Total subscribers} = 500 \times 100 = 50,000
\]
**Explanation:**
Multiply the number of users per cell by the total number of cells.
**Final Answer:**
**50,000 subscribers (omni)**
---
## Part (e): Total Subscribers with 120° Sectoring
**Given:**
- Each cell now supports 220 users (with sectoring)
- Number of cells = 500
**Calculation:**
\[
\text{Total subscribers} = 500 \times 220 = 110,000
\]
**Explanation:**
Sectoring reduces the reuse factor, increasing channels and users per cell.
**Final Answer:**
**110,000 subscribers (with 120° sectoring)**
---
## Trunking Efficiency Change
**Given:**
- Original users per cell = 100 (omni)
- New users per cell = 220 (sectoring)
**Calculation:**
\[
\text{Improvement} = \frac{220 - 100}{100} \times 100\% = \frac{120}{100} \times 100\% = 120\%
\]
**Explanation:**
Percentage increase in users per cell due to sectoring.
**Final Answer:**
**120% improvement (not degradation)**
---
## Notes
- If a different reuse factor is used for sectoring (\(N=3\), for example), the number of channels and users per cell would be higher.
- If sectoring splits channels among sectors (instead of reducing the reuse factor), the calculation would change. The above assumes the typical improvement from reduced reuse factor.
---
## Summary Table
| Part | Parameter | Value / Calculation | Answer |
|------|--------------------------|----------------------------|---------------|
| a | Channels per cell | \(\left\lfloor 69/7 \right\rfloor\) | 9 |
| b | Traffic per cell | Erlang B: \(C=9, B=.01\) | 5 Erlangs |
| c | Users per cell | \(5 / .05\) | 100 users |
| d | Subscribers (omni) | \(500 \times 100\) | 50,000 users |
| e | Subscribers (120°) | \(500 \times 220\) | 110,000 users |
| | Trunking efficiency change | \(\frac{220-100}{100} \times 100\%\) | 120% improvement |
---
### Visual Summary
*Alt text: Diagram showing hexagonal cellular network structure with sectoring.*
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