CA-2019Fall: Quiz 4

A

We can detect a load hazard after we have fetched the dependent instruction, and so this is the earliest point at which we can stall. We’d like to add a MUX between
our ID and IF stages. This MUX should current instruction to a NOP if a load hazard exists. Assume we have a new control bit LoadHazard which is 1 when a load hazard is present and 0 otherwise. Where should we connect tunnels A, B, and C? Select one option for each letter.

A:

1. PC
2. NO instruction
3. IMEM output
4. ID input (RegFile parser input)

B:

1. PC
2. NO instruction
3. IMEM output
4. ID input (RegFile parser input)

C:

1. PC
2. NO instruction
3. IMEM output
4. ID input (RegFile parser input)

B

For each of the questions below, please create a valid K-stage pipeline of the given circuit. Each component in the circuit is annotated with its propagation delay. Show your pipelining contours and place large black circles (●) on the signal arrows to indicate the placement of pipeline registers. Give the latency and throughput of each design, assuming ideal registers (t_PD=0, t_SETUP=0). Remember that our convention is to place a pipeline register on each output.

(1) Show the maximum-throughput 1-stage pipeline.
Latency (ns): __________
Throughput (ns^-1): __________

Diagram for B-1

(2) Show the maximum-throughput 2-stage pipeline using a minimal number of registers.
Latency (ns): __________
Throughput (ns^-1): __________

Diagram for B-2

(3) Show the maximum-throughput pipeline using a minimal number of registers.
Latency (ns): __________
Throughput (ns^-1): __________

Diagram for B-3

(4) Reimplement the slowest combinational component in the previous circuit (the one with a propagation delay of 4 ns) using two components with propagation delays of 2ns, as shown below. Show the maximum-throughput pipeline using a minimal number of registers.
Latency (ns): __________
Throughput (ns^-1): __________

Diagram for B-4

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Solutions

A

• A: 3
• B: 2
• C: 4

B

	Latency (ns)	Throughput (ns-1)
(1)	11	1/11
(2)	14	1/7
(3)	12	1/4
(4)	12	1/3