Derive STS and TST blocking probabilities

Blocking probability of multistage switches is found with Lee graphs: each path is drawn as links in series, parallel paths are independent, and each link is busy with probability $p'$.

STS switch

$N$ incoming TDM links enter an $N \times k$ space stage, then $k$ centre-stage TSIs, then a $k \times N$ space stage.

 A o---(k parallel paths, each = 2 links)---o B
        input S -> TSI_i -> output S

• Each path uses two internal links in series: input S → TSI and TSI → output S.
• If $p$ is the occupancy of external links, the internal link occupancy is

$$p' = \frac{p}{\beta}, \quad \beta = \frac{k}{N} \text{ (space expansion)}$$

• A path is free only if both links are free: probability $(1-p')^2$. Path busy: $1-(1-p')^2$.
• There are $k$ parallel paths (one per centre TSI); the call is blocked only if all are busy:

$$P_{B,STS} = \left[1 - (1 - p')^2\right]^{k}$$

TST switch

$N$ incoming links, each with $c$ slots, go to input TSIs; a time-multiplexed space stage with $l$ internal time slots; then output TSIs.

• The alternative paths are the $l$ internal time slots of the space stage. Each path uses two links: input TSI → S (in slot $x$) and S → output TSI (in slot $x$).
• Internal link occupancy:

$$p' = \frac{p}{\alpha}, \quad \alpha = \frac{l}{c} \text{ (time expansion)}$$

• Same Lee graph with $l$ parallel paths:

$$P_{B,TST} = \left[1 - (1 - p')^2\right]^{l}$$

Why TST has lower blocking

The expressions have the same form; the blocking depends on the number of parallel paths (the exponent) and $p'$.

1. In STS the exponent is $k$, the number of centre TSIs. Each extra path needs a whole TSI and adds crosspoints to both space stages ($2Nk$), so $k$ stays small.
2. In TST the exponent is $l$, the number of internal time slots. It is equal to or greater than the frame size $c$ (32 to 1024 in practice) and is raised just by running the space stage and memories faster, which is cheap.
3. So for a practical size, $l \gg k$ and the TST result is far smaller.

Example: $p = 0.7$ E, no expansion ($p' = 0.7$), $1-(1-0.7)^2 = 0.91$:

• STS with $N = k = 16$ links: $P_B = 0.91^{16} = 0.221$
• TST with $c = l = 32$ slots: $P_B = 0.91^{32} = 0.0489$
• TST with time expansion $l = 64$ ($p' = 0.35$): $P_B = 0.5775^{64} \approx 5.5 \times 10^{-16}$

Hence TST gives lower blocking at lower cost, which is why large digital exchanges use TST (or TSST/TSSST) structures.