Pipeline Processor


Q41.

Consider a 4 stage pipeline processor. The number of cycles needed by the four instructions I1, I2, I3, I4 in stages S1, S2, S3, S4 is shown below: What is the number of cycles needed to execute the following loop? For (i=1 to 2) {I1; I2; I3; I4;}
GateOverflow

Q42.

Consider a non-pipelined processor with a clock rate of 2.5 gigahertz and average cycles per instruction of four. The same processor is upgraded to a pipelined processor with five stages; but due to the internal pipeline delay, the clock speed is reduced to 2 gigahertz. Assume that there are no stalls in the pipeline. The speed up achieved in this pipelined processor is_________.
GateOverflow

Q43.

Consider the following processors (ns stands for nanoseconds). Assume that the pipeline registers have zero latency. P1: Four-stage pipeline with stage latencies 1 ns, 2 ns, 2 ns, 1 ns. P2: Four-stage pipeline with stage latencies 1 ns, 1.5 ns, 1.5 ns, 1.5 ns. P3: Five-stage pipeline with stage latencies 0.5 ns, 1 ns, 1 ns, 0.6 ns, 1 ns. P4: Five-stage pipeline with stage latencies 0.5 ns, 0.5 ns, 1 ns, 1 ns, 1.1 ns. Which processor has the highest peak clock frequency?
GateOverflow

Q44.

Consider a pipelined processor with the following four stages: IF: Instruction Fetch ID: Instruction Decode and Operand Fetch EX: Execute WB: Write Back The IF, ID and WB stages take one clock cycle each to complete the operation. The number of clock cycles for the EX stage depends on the instruction. The ADD and SUB instructions need 1 clock cycle and the MUL instruction needs 3 clock cycles in the EX stage. Operand forwarding is used in the pipelined processor. What is the number of clock cycles taken to complete the following sequence of instructions? ADD R2, R1, R0 (R2 \leftarrow R1 + R0) MUL R4, R3, R2 (R4 \leftarrow R3 * R2) SUB R6, R5, R4 (R6 \leftarrow R5 - R4)
GateOverflow

Q45.

If a microcomputer operates at 5 MHz with an 8-bit bus and a newer version operates at 20 MHz with a 32-bit bus, the maximum speed-up possible approximately will be
GateOverflow