TST vs STS and three-stage design

Difference between TST and STS switches

Three-stage design

Reading of the data: the first stage has 4 arrays with 5 input lines each, so $N = 4 \times 5 = 20$ lines, $n = 5$; the second stage has $k = 6$ arrays.

• Stage 1: 4 arrays of $5 \times 6$
• Stage 2: 6 arrays of $4 \times 4$ (each middle array has one link from each of the 4 first-stage arrays)
• Stage 3: 4 arrays of $6 \times 5$

  5 in [5x6] \           / [6x5] 5 out
  5 in [5x6] --> 6 x [4x4] --> [6x5] 5 out
  5 in [5x6] -->  middle  --> [6x5] 5 out
  5 in [5x6] /           \ [6x5] 5 out
  (20 inputs)               (20 outputs)

Crosspoints

$$\begin{aligned} N_1 &= 4 \times (5 \times 6) = 120 \\ N_2 &= 6 \times (4 \times 4) = 96 \\ N_3 &= 4 \times (6 \times 5) = 120 \\ N_x &= 120 + 96 + 120 = 336 \end{aligned}$$

Check: $N_x = 2Nk + k(N/n)^2 = 2(20)(6) + 6(4)^2 = 240 + 96 = 336$.

A single-stage $20 \times 20$ switch needs 400 crosspoints, so the saving is 64 crosspoints (16%). Since $k = 6 < 2n - 1 = 9$, this network is blocking; 9 middle arrays would be needed for strict non-blocking.

Answer: total crosspoints = 336.