Design 200×200 three-stage switch

Three-stage network with $N$ lines, groups of $n$, and $k$ middle arrays:

• Stage 1: $N/n$ arrays of $n \times k$
• Stage 2: $k$ arrays of $(N/n) \times (N/n)$
• Stage 3: $N/n$ arrays of $k \times n$

$$N_x = 2Nk + k\left(\frac{N}{n}\right)^2$$

Design 1: $N = 200$, $n = 20$, $k = 4$

Number of groups $N/n = 200/20 = 10$.

$$N_x = 2(200)(4) + 4(10)^2 = 1600 + 400 = 2000$$

 20 in  [20x4] \                   / [4x20] 20 out
 20 in  [20x4] -->  4 x [10x10] -->  [4x20] 20 out
  ...     ...  /     middle      \    ...
 20 in  [20x4]  (10 arrays)          [4x20] 20 out

Each first-stage array has one link to each middle array; each middle array has one link to each third-stage array.

This needs only 5% of the $200^2 = 40\,000$ crosspoints of a single-stage switch, but only 4 of the 20 inputs of a group can be connected at once, so it is a blocking (concentrating) design, since $k = 4 < 2n - 1 = 39$.

Design 2: Clos non-blocking with minimum crosspoints

For strict non-blocking, $k = 2n - 1$. Crosspoints are minimum when

$$n = \sqrt{\frac{N}{2}} = \sqrt{\frac{200}{2}} = 10$$ $$k = 2n - 1 = 19, \qquad \frac{N}{n} = \frac{200}{10} = 20$$

Check with the Clos minimum formula:

$$\begin{aligned} N_{x,min} &= 4N\left(\sqrt{2N} - 1\right) \\ &= 4(200)\left(\sqrt{400} - 1\right) = 800 \times 19 = 15\,200 \end{aligned}$$

Answer: Design 1 needs 2000 crosspoints (blocking). The Clos design ($n = 10$, $k = 19$, 20 outer arrays) needs 15 200 crosspoints and is strictly non-blocking, which is 38% of a 40 000-crosspoint single-stage switch.