Skip to main content

TST time slots for non-blocking operation

Short answer

Non‑blocking requires l ≥ 2c−1 = 239 slots. For c=120, P_B at p=0.6E: l=120 → 8.2×10^−10, l=150 → 2.9×10^−21, l=200 → 1.7×10^−46. At p=0.9E: 0.299, 4.8×10^−6, 2.2×10^−21. Even a small time expansion drastically reduces blocking.

A TST network is used in a digital switch and the secondary multiplex contains 120 time slots. How many time slots would be included in the time multiplexed space stage for non-blocking operation? What would be the blocking probability if the time multiplexed space stage contained 120 time slots, 150 time slots, and 200 time slots? Assume channel occupancies of 0.6E and 0.9E.

Given: time slots per frame on the external (secondary multiplex) link, c=120c = 120; occupancy p=0.6p = 0.6 E and 0.90.9 E.

Non-blocking condition

A TST switch is strictly non-blocking when the number of space-stage time slots satisfies l≥2c−1l \ge 2c - 1:

l=2(120)−1=239 time slotsl = 2(120) - 1 = 239 \text{ time slots}

Blocking probability

Lee-graph formula for TST:

PB=[1−(1−p′)2]l,p′=p clP_B = \left[1 - (1 - p')^2\right]^{l}, \qquad p' = \frac{p\,c}{l}

Step 1: internal link occupancy p′p'

llp′p' (p = 0.6)p′p' (p = 0.9)
1200.6000.900
1500.4800.720
2000.3600.540

Step 2: path-busy probability q=1−(1−p′)2q = 1 - (1-p')^2

llqq (p = 0.6)qq (p = 0.9)
1200.84000.9900
1500.72960.9216
2000.59040.7884

Step 3: PB=q lP_B = q^{\,l}

For example, l=120l = 120, p=0.9p = 0.9:

PB=0.99120=0.299P_B = 0.99^{120} = 0.299

and l=150l = 150, p=0.6p = 0.6:

PB=0.7296150=2.90×10−21P_B = 0.7296^{150} = 2.90 \times 10^{-21}
llPBP_B at 0.6 EPBP_B at 0.9 E
1208.19×10−108.19 \times 10^{-10}0.2990.299
1502.90×10−212.90 \times 10^{-21}4.80×10−64.80 \times 10^{-6}
2001.70×10−461.70 \times 10^{-46}2.24×10−212.24 \times 10^{-21}

Comment: even without time expansion (l=120l = 120) the blocking is negligible at 0.6 E, but at 0.9 E it is about 30%. A modest time expansion (150 or 200 slots) brings it down to a negligible value, using far fewer slots than the 239 needed for strictly non-blocking operation.

Answer: l=239l = 239 slots for non-blocking; PBP_B = 8.19×10−108.19 \times 10^{-10}, 2.90×10−212.90 \times 10^{-21}, 1.70×10−461.70 \times 10^{-46} (0.6 E) and 0.2990.299, 4.80×10−64.80 \times 10^{-6}, 2.24×10−212.24 \times 10^{-21} (0.9 E) for ll = 120, 150, 200.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Telecommunications chapter 4 questions →