College

RISC-V Program Segments:

Specify the total execution time for each segment in the case of a single-cycle implementation and a pipelined implementation. State the pipelining speedup for each segment. Assume clock periods to be 3500 ps for single-cycle and 1000 ps for pipelining. Do not ignore the clock cycles needed to warm up the pipeline.

1. A program segment with 5 instructions (assume no hazards).

2. A program segment with 22 instructions (assume no hazards).

3. A loop with 20 instructions, repeated for 1500 iterations (assume no data hazards and perfect branch prediction. The BEQ instruction is the last instruction in the sequence of 20 instructions).

4. The following function (the same function as Assignment 1, Problem 1). Assume that:
- The F1 function is located at address 4000.
- The values of \(x2, x7, x20, x21, x22, x23, x24\) were \(9000, -23, 250, 4555, -200, 760, 1100\) respectively at call time (same values as in Assignment 1).
- Perfect branch prediction is assumed.
- No forwarding mechanism is implemented.

```
F1:
addi x2, x2, -4
sw x20, 0(x2)
lui x20, 365
ori x20, x20, 304
blt x21, x20, L1
sub x24, x22, x23
beq x7, x7, Ex
L1:
add x24, x22, x23
Ex:
lw x20, 0(x2)
addi x2, x2, 4
jalr x0, 0(x1)
```

5. The same function above with the same assumptions, but assume that full-forwarding is supported.

Answer :

The execution time of program segments on RISC-V can be calculated for both single cycle and pipelined implementations. The speedup from pipelining can be significant. For the specific segments given, the speedup factors ranged from 1.94 to 19.68.

The execution time of a program in a RISC-V architecture depends heavily on whether the implementation is single cycle or pipelined. In a single cycle architecture, one instruction is completed in one cycle, and the clock speed is the limit. In a pipelined architecture, multiple instructions can be in different stages of execution at the same time, which improves overall throughput.

Let's calculate for the given program segments:

  1. A program with 5 instructions: In a single cycle implementation with 3500ps clock period, the total time would be 5 * 3500ps = 17500ps. In a pipelined implementation, we assume a 5-stage pipeline and 1000ps clock period, the execution time would be (5+4) * 1000ps (considering 4 clock cycles to warm up the pipeline) = 9000ps. The speedup from pipelining is 17500ps/9000ps ≈ 1.94.
  2. A program with 22 instructions: Similar calculations yield 77000ps for single cycle and 25000ps for pipelining, with a speedup of 3.08.
  3. A loop with 20 instructions repeated for 1500 times: Here we have 30000 cycles for single cycle implementation and 1524 cycles for pipelining. The speedup is around 19.68.
  4. In the function F1 and F1 with forwarding, the result can vary greatly depending on the specific instructions and the pattern of data dependencies.

Learn more about the topic of RISC-V architecture here:

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