--- title: "Design 200×200 three-stage switch" subject: "Telecommunications" chapter: 4 key: "semester_8/telecommunications:three-stage-200x200-clos" source: https://www.bctengineer.com/q/semester_8/telecommunications/three-stage-200x200-clos site: BCT Engineering --- # Design 200×200 three-stage switch Telecommunications · Chapter 4: Digital Switching - Asked in: 2078 Bhadra, 8 marks - Video: [Switching Engineering - Cross-points and Number of Switches - Calculations](https://www.youtube.com/watch?v=Bbm3lKI7O5Y) — KINEMARC - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/telecommunications/three-stage-200x200-clos - Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-4 ## Question Design a three-stage, 200×200 switch with division of input lines (N = 200) into groups with each group of n = 20 lines. Use k number of crossbars in the middle stage where k = 4. Also, redesign this three stage 200×200 switch using Clos criteria with a minimum number of cross points. ## Short answer With N=200, n=20, k=4, crosspoints = 2(200)(4) + 4(10)^2 = 2000 (blocking). For a non‑blocking Clos network with minimum crosspoints, choose n=10, k=2n−1=19, giving 20 outer arrays and total crosspoints = 2(200)(19) + 19(20)^2 = 15200, which is 38% of the 40000 required for a single stage. ## Answer 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 $N/n = 200/20 = 10$. | Stage | Arrays | Size | Crosspoints | |---|---|---|---| | 1 | 10 | $20 \times 4$ | $10 \times 80 = 800$ | | 2 | 4 | $10 \times 10$ | $4 \times 100 = 400$ | | 3 | 10 | $4 \times 20$ | $10 \times 80 = 800$ | | Total | | | **2000** | $$ N_x = 2(200)(4) + 4(10)^2 = 1600 + 400 = 2000 $$ ```text 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 $$ n = \sqrt{\frac{N}{2}} = \sqrt{\frac{200}{2}} = 10 $$ $$ k = 2n - 1 = 19, \qquad \frac{N}{n} = \frac{200}{10} = 20 $$ | Stage | Arrays | Size | Crosspoints | |---|---|---|---| | 1 | 20 | $10 \times 19$ | $20 \times 190 = 3800$ | | 2 | 19 | $20 \times 20$ | $19 \times 400 = 7600$ | | 3 | 20 | $19 \times 10$ | $20 \times 190 = 3800$ | | Total | | | **15 200** | 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For N = 200, n = 20, k = 4, the total crosspoints are A. 2000 B. 40000 C. 15200 D. 800 Answer: A. 2(200)(4) + 4(10)^2 = 1600 + 400 = 2000. (Key point: Design 1 crosspoints) 2. Why is the k = 4 design blocking? A. k = 4 < 2n - 1 = 39 B. N/n is too small C. The middle arrays are 10 x 10 D. It has three stages Answer: A. Strictly non-blocking needs k >= 2n - 1 = 39; only 4 of 20 inputs of a group can be connected at once. (Key point: Why Design 1 blocks) 3. For minimum crosspoints in a Clos network the group size is A. n = N/2 B. n = sqrt(N/2) = 10 C. n = sqrt(N) = 14 D. n = 2N - 1 Answer: B. Differentiating Nx gives n = sqrt(N/2), which is 10 for N = 200, so k = 19. (Key point: Optimum n) 4. The minimum Clos crosspoints for N = 200 is A. 4N(sqrt(2N) - 1) = 15200 B. 2000 C. 40000 D. 7600 Answer: A. 4(200)(20 - 1) = 15200, matching 3800 + 7600 + 3800. 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