Skip to main content

ST and TS drawbacks solved by STS

Short answer

Two-stage TS and ST networks suffer high blocking because the space stage must use a fixed slot. STS solves this by adding k centre TSIs: input space sends the sample in its original slot to any free TSI, TSI changes the slot, output space connects. Blocking becomes [1-(1-p')^2]^k, and if k≥2N-1 it is non-blocking.

What are the drawbacks of ST and TS switch and how are they solved by STS switch? Explain.

Drawbacks of ST and TS switches

A TS switch has a time stage followed by a space stage; an ST switch has a space stage followed by a time stage. Both are two-stage and give only one possible path for each connection, so they block heavily.

  • TS drawback: the input TSI moves the sample from slot ii to the wanted output slot jj on its own highway, and then the space stage must connect the highway to the output highway in slot jj. If another input highway already uses slot jj to reach the same output highway, the call is blocked, even though both A and B are free.
  • ST drawback: the space stage must connect input highway aa to output highway bb in the input slot ii. If slot ii is already used on output highway bb by a call from another input, the call is blocked. The output TSI could change the slot afterwards, but it never receives the sample.
 TS: [T] -> [S]   S must use slot j  (fixed)
 ST: [S] -> [T]   S must use slot i  (fixed)
 => one path only, blocking = P(that slot busy)

How STS solves this

An STS switch has an input space stage, a middle stage of kk TSIs and an output space stage.

 HW1 -->+------+  [TSI 1]  +------+--> HW1
 HW2 -->|N x k |->[TSI 2]->|k x N |--> HW2
 HWN -->+------+  [TSI k]  +------+--> HWN
  1. In input slot ii, the input S stage can send the sample to any free centre TSI mm (one with a free link in slot ii).
  2. TSI mm changes the slot from ii to jj.
  3. In slot jj, the output S stage connects TSI mm to output highway bb.
  4. The call is blocked only if all kk centre TSIs are busy in slot ii or slot jj.

So there are now kk alternative paths instead of one. Using a Lee graph with internal link occupancy p′=pN/kp' = pN/k:

PB=[1−(1−p′)2]kP_B = \left[1 - (1 - p')^2\right]^{k}

For p′=0.3p' = 0.3 and k=8k = 8: PB=0.518≈4.6×10−3P_B = 0.51^8 \approx 4.6 \times 10^{-3}, compared with about 0.30.3 for a single path. With k≥2N−1k \ge 2N - 1 the STS switch becomes strictly non-blocking.

(TST solves the same problem using ll internal time slots as alternative paths, and is usually cheaper.)

✓ Answer checked by Claude Opus 5.5 (high effort)

All Telecommunications chapter 4 questions →