3 Consider the combinational circuit shown below. The propagation delay for each unit is indicated in ns. Assume all other delays are zero. 8 ns C(x,y) 10 ns 10 ns y - 7 ns 5 ns 5 ns a) What is the latency and throughput of the circuit as shown? b) Indicate where storage registers should be placed to create a two-stage pipeline with maximum throughput. c) What is the latency and throughput of the two-stage pipeline in part b?

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**Combinatorial Circuit Analysis**

**Circuit Overview:**
The diagram illustrates a combinational circuit with specific propagation delays assigned to each component. The notations are as follows:
- Input x passes through a sequence of components with delays of 10 ns, 10 ns, and 8 ns, cumulatively contributing to the final output C(x,y).
- Input y passes through components with delays of 7 ns, 5 ns, and 5 ns, contributing to the same output.

**Questions for Analysis:**

a) **Latency and Throughput of the Circuit as Shown:**

   - *Latency*: This is the total time taken for an input to reach the output. For input x, it is calculated as 10 ns + 10 ns + 8 ns = 28 ns. For input y, it is 7 ns + 5 ns + 5 ns = 17 ns. Therefore, the longest path, determining the circuit latency, is 28 ns.
   
   - *Throughput*: This can be expressed as one completion every 28 ns given that throughput is limited by the longest latency path.

b) **Storage Registers Placement for Two-Stage Pipeline:**

   - To create a two-stage pipeline with maximum throughput, registers should be placed in locations that divide the longest path into stages as equally as possible. Potential placements include:
     - After the first 10 ns delay for input x.
     - After the 7 ns delay for input y. 

c) **Latency and Throughput of the Two-Stage Pipeline:**

   - With a two-stage pipeline, each stage should ideally balance latency for maximum throughput. Assuming perfect division, latency remains the total of 28 ns, but the throughput improves with reduced cycle time due to overlapping execution, ideally halving the cycle time compared to the non-pipelined version.
   
d) **Best Throughput via Pipelining and Pipeline Stages:**

   - Achieving optimal throughput involves creating as many stages as possible, each stage approximating the shortest delay. The most efficient scenario pipelines each component:
     - Three stages for input x: (10 ns), (10 ns), (8 ns).
     - Three stages for input y: (7 ns), (5 ns), (5 ns).
   
   - Ideal throughput improves to one completion every 8 ns, the duration of the longest stage.

This analysis helps understand the trade-off between latency and throughput and guides optimal pip
Transcribed Image Text:**Combinatorial Circuit Analysis** **Circuit Overview:** The diagram illustrates a combinational circuit with specific propagation delays assigned to each component. The notations are as follows: - Input x passes through a sequence of components with delays of 10 ns, 10 ns, and 8 ns, cumulatively contributing to the final output C(x,y). - Input y passes through components with delays of 7 ns, 5 ns, and 5 ns, contributing to the same output. **Questions for Analysis:** a) **Latency and Throughput of the Circuit as Shown:** - *Latency*: This is the total time taken for an input to reach the output. For input x, it is calculated as 10 ns + 10 ns + 8 ns = 28 ns. For input y, it is 7 ns + 5 ns + 5 ns = 17 ns. Therefore, the longest path, determining the circuit latency, is 28 ns. - *Throughput*: This can be expressed as one completion every 28 ns given that throughput is limited by the longest latency path. b) **Storage Registers Placement for Two-Stage Pipeline:** - To create a two-stage pipeline with maximum throughput, registers should be placed in locations that divide the longest path into stages as equally as possible. Potential placements include: - After the first 10 ns delay for input x. - After the 7 ns delay for input y. c) **Latency and Throughput of the Two-Stage Pipeline:** - With a two-stage pipeline, each stage should ideally balance latency for maximum throughput. Assuming perfect division, latency remains the total of 28 ns, but the throughput improves with reduced cycle time due to overlapping execution, ideally halving the cycle time compared to the non-pipelined version. d) **Best Throughput via Pipelining and Pipeline Stages:** - Achieving optimal throughput involves creating as many stages as possible, each stage approximating the shortest delay. The most efficient scenario pipelines each component: - Three stages for input x: (10 ns), (10 ns), (8 ns). - Three stages for input y: (7 ns), (5 ns), (5 ns). - Ideal throughput improves to one completion every 8 ns, the duration of the longest stage. This analysis helps understand the trade-off between latency and throughput and guides optimal pip
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