The pipeline consists of 5 stages with delays of 180, 250, 150, 170, and 250 nanoseconds. The delay due to the inter-stage latch is 10 nanoseconds. The pipeline clock cycle time is determined by the maximum delay stage plus the inter-stage latch delay. Therefore, compute the clock cycle time as follows:
Max stage delay = max(180, 250, 150, 170, 250) = 250 ns.
Clock cycle time = 250 ns + 10 ns = 260 ns.
To process 1000 instructions in a pipeline with no hazards, use the regular completion time formula:
Time for first instruction = number of stages × cycle time = 5 × 260 ns = 1300 ns.
Time for remaining instructions = (Number of instructions - 1) × cycle time = 999 × 260 ns.
Total time for 1000 instructions = 1300 ns + 999 × 260 ns = 1300 ns + 259740 ns = 261040 ns.
Convert this total time into microseconds:
Total time in microseconds = 261040 ns / 1000 = 261.04 μs.
This result falls within the given range of 260.2 to 261.2 microseconds, confirming its accuracy. Therefore, the time taken to process 1000 instructions is 261.04 microseconds.
Important Questions from Instruction Pipelining
- The right sequence of suboperations that are performed in arithmetic pipeline is-
A. Align the mantissas
B. Add or subtract the mantissas
C. Normalize the result
D. Compare the exponents
Choose the correct answer from the options given below: - Which one of the following is a technique of decomposing a sequential process into sub-operations, with each sub-process being executed in a special dedicated segment that operates concurrently with all other segments?
- The EX stage of a pipelined processor performs the memory read operations for LOAD instructions, and the operations for the arithmetic and logic instructions. Let $t_{EX}$ denote the time taken by the EX stage to perform the operation for an instruction. For each instruction type, the values of $t_{EX}$ and $M$ (the number of instructions of that type in a sequence of 100 instructions for a program P), are given in the table below.
The duration of the pipeline clock cycle is 1 nanosecond. Assume that the latch time for the interstage buffers in the pipeline is negligible.Instruction $t_{EX}$ in nanoseconds $M$ LOAD 1.8 15 IMUL 1.5 10 IDIV 2.5 5 FADD 1.7 10 FSUB 1.7 5 FMUL 2.8 15 FDIV 3.2 5 All other instructions Less than 1.0 35
When program P is executed, the number of clock cycles for which the pipeline is stalled due to structural hazards in the EX stage is ______. (answer in integer) - A non-pipelined instruction execution unit that operates at 1.6 GHz clock takes an average of 5 clock cycles to complete the execution of an instruction. To improve the performance, the system was pipelined with a goal of achieving an average throughput of one instruction per clock cycle. However, it could operate only at 1.2 GHz due to pipeline overheads. While executing a program in the pipelined design, 30% of instructions encountered a stall of 2 cycles due to pipeline hazards. The speed-up obtained by the pipelined design over the non-pipelined one for this program is ___________. (rounded off to two decimal places)
Note: 1G=$10^9$ - Pipelining improves performance by :
