What is the clock cycle time when executing a LW instruction?

What is the clock cycle time when executing a LW instruction?

5 clock cycles
Consider an implementation of MIPS ISA with 500 MHz clock and – each ALU instruction takes 3 clock cycles, – each branch/jump instruction takes 2 clock cycles, – each sw instruction takes 4 clock cycles, – each lw instruction takes 5 clock cycles.

What is the clock cycle time of a single cycle processor?

Single cycle: All “steps” of executing an instruction are done in 1 clock cycle.

What is the minimum clock period for the single cycle processor?

For the single-cycle CPU, the minimum clock period is simply the sum of the delays through all five sub-components (not stages, as there is only one stage). The assumption that the pipelining overhead is zero means that the minimum clock period of the pipeline CPU is simply the longest individual stage delay.

Why does the single cycle datapath require separate instruction and data memories?

The single-cycle datapath must have separate instruction and data memories, because The formats of data and instructions are different in MIPS, and hence different memories are needed.

How can I reduce my cycle time?

Splitting the longest stage is the only way to reduce the cycle time. After splitting it, the new cycle time is based on the new longest stage.

What are the five basic stages in a MIPS pipelined processor?

In general, let the instruction execution be divided into five stages as fetch, decode, execute, memory access and write back, denoted by Fi, Di, Ei, Mi and Wi.

How do you calculate clock cycle per instruction?

1. CPU clock cycles = Instruction count x CPI.
2. CPU execution time =
3. = CPU clock cycles x Clock cycle.
4. = Instruction count x CPI x Clock cycle.
5. T =
6. I.
7. x CPI x C.

How to calculate time latency for’lw’in MIPS?

Then from the course, I know the time should be the execution time of the longest instruction, which is ‘lw’ in MIPS. So I try to calculate the time latency for ‘lw’.

How are datapaths used in a single cycle?

 A datapath contains all the functional units and connections necessary to implement an instruction set architecture. —For our single-cycle implementation, we use two separate memories, an ALU, some extra adders, and lots of multiplexers. —MIPS is a 32-bit machine, so most of the buses are 32-bits wide.

Which is the longest path for’lw’instruction?

I think the longest path/time for ‘lw’ woud be: (In the path, I think the mux to choose read-register1 or write-register, sing-extend for immediate value and mux to choose read-data 2 or immediate value are all ignored, since the circuits are in parallel, they would take less time than other parts in parallel.)