---
title: "Crosspoints in three-stage switch (4 inputs, 5 arrays)"
subject: "Telecommunications"
chapter: 4
key: "semester_8/telecommunications:crosspoints-4-input-5-second"
source: https://www.bctengineer.com/q/semester_8/telecommunications/crosspoints-4-input-5-second
site: BCT Engineering
---

# Crosspoints in three-stage switch (4 inputs, 5 arrays)

Telecommunications · Chapter 4: Digital Switching

- Asked in: 2079 Chaitra, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/crosspoints-4-input-5-second
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-4

## Question

What are the advantages of multi-stage switching over single-stage switching? Calculate and draw, how many cross points are found in three stages switching system, whereas 3 stages array of 4 input lines and 5 second stages array.

## Short answer

Multi-stage switching offers far fewer crosspoints, better crosspoint utilisation, alternative paths, modular growth, and concentration/expansion. For a 3-stage network with 3 arrays of 4 inputs and 5 middle arrays, total crosspoints: 3×4×5 + 5×3×3 + 3×5×4 = 165.

## Answer

#### Advantages of multi-stage switching over single-stage
- **Far fewer crosspoints:** a single-stage $N \times N$ switch needs $N^2$ crosspoints (e.g. $10^6$ for 1000 lines); a three-stage switch needs only a small fraction, so cost and size fall.
- **Better crosspoint utilisation:** in a single stage at most $N$ of the $N^2$ crosspoints can be in use at once (each input and output takes part in only one connection), so utilisation is at most $1/N$; in multi-stage each crosspoint is shared by many possible connections.
- **Alternative paths:** a call can go through any of several middle-stage arrays, so failure of one crosspoint does not isolate a subscriber (better reliability).
- **Modular growth:** capacity can be expanded by adding arrays.
- **Concentration and expansion** can be built into the stages to suit traffic.
- Drawbacks: some blocking may occur, and path search (control) is more complex.

#### Crosspoint calculation
**Reading of the data (assumed):** the first stage has **3 arrays of 4 input lines** each, so $N = 3 \times 4 = 12$ lines and $n = 4$, and there are **$k = 5$ second-stage (middle) arrays**. The network is symmetric (third stage = mirror of first).

Array sizes:
- First stage: $N/n = 3$ arrays of $n \times k = 4 \times 5$.
- Second stage: $k = 5$ arrays of $\frac{N}{n} \times \frac{N}{n} = 3 \times 3$.
- Third stage: 3 arrays of $k \times n = 5 \times 4$.

```text
  stage 1 (3 x [4x5])  stage 2 (5 x [3x3])  stage 3
 4 in->[4x5]-+-----> [3x3] -----+->[5x4]-> 4 out
             |-----> [3x3] -----|
 4 in->[4x5]-+-----> [3x3] -----+->[5x4]-> 4 out
             |-----> [3x3] -----|
 4 in->[4x5]-+-----> [3x3] -----+->[5x4]-> 4 out
 each 4x5 array has 1 link to each of the 5 middle
 arrays; each middle array has 1 link to each 5x4
```

$$
\begin{aligned}
N_x &= 2Nk + k\left(\frac{N}{n}\right)^2 \\
\text{Stage 1} &= 3 \times (4 \times 5) = 60 \\
\text{Stage 2} &= 5 \times (3 \times 3) = 45 \\
\text{Stage 3} &= 3 \times (5 \times 4) = 60 \\
N_x &= 60 + 45 + 60 = 165
\end{aligned}
$$

**Answer: 165 crosspoints** (60 + 45 + 60).

**Remarks:**
- A single-stage $12 \times 12$ matrix needs $12^2 = 144$ crosspoints; for such a small $N$ the three-stage network gives no saving. The saving appears for large $N$, e.g. $N = 128$, $n = 8$, $k = 15$ needs 7680 crosspoints against 16 384.
- With $k = 5 < 2n - 1 = 7$, this network is **blocking**; a strictly non-blocking Clos version needs $k = 7$, i.e. $2(12)(7) + 7(3)^2 = 231$ crosspoints.

## Check yourself

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

1. For N = 12, n = 4, k = 5, what are the first-stage and second-stage array sizes?
   A. First: 5 x 4, second: 3 x 3
   B. First: 4 x 5 (3 arrays), second: 3 x 3 (5 arrays)
   C. First: 4 x 4, second: 5 x 5
   D. First: 3 x 3, second: 4 x 5
   Answer: B. First stage has N/n = 3 arrays of n x k = 4x5; second stage has k = 5 arrays of (N/n)x(N/n) = 3x3. (Key point: Array sizes)

2. What is the total number of crosspoints in this three-stage network?
   A. 144
   B. 231
   C. 165
   D. 120
   Answer: C. 60 + 45 + 60 = 165, equal to 2Nk + k(N/n)^2 = 120 + 45. (Key point: Total crosspoints)

3. Why is this network blocking?
   A. Because N is too large
   B. Because k = 5 is less than 2n - 1 = 7
   C. Because it has only three stages
   D. Because the middle arrays are 3 x 3
   Answer: B. Strict non-blocking needs k = 2n-1 = 7; with k = 5 some calls can block. (Key point: Blocking condition)

## Close variants set in other exams

- [Switching functions and three-stage crosspoints](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aswitching-functions-crosspoints-4x4) — Telecommunications, chapter 4, asked 2×

## Related in Telecommunications

- [PCM switching and three-stage crosspoints](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Apcm-switching-crosspoints-4x5) — Telecommunications, chapter 4
- [Crosspoints and three-stage switching](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Acrosspoint-switching-array) — Telecommunications, chapter 4
- [Digital exchange and three-stage crosspoints](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adigital-exchange-crosspoints) — Telecommunications, chapter 4
- [Design 200×200 three-stage switch](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Athree-stage-200x200-clos) — Telecommunications, chapter 4
- [TST vs STS and three-stage design](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atst-sts-design-crosspoints) — Telecommunications, chapter 4
- [Three-stage combination switch](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Acombination-switch-3-stage) — Telecommunications, chapter 4