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

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 $k \times n = 5 \times 4$.

  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.