Asap Consider a 5-stage pipeline with the followingfeatures:• The pipeline clock rate is 1 GHz• For a non-pipelined implementation, ALUoperations and branches take 3 clockcycles while memory operation will take 5clock cycles.• Relative frequencies of ALU operations,branches, and memory operations are 50%,30%, 20%.• In the pipeline due to clock setup time, theclock cycle time increases by 0.65ns.Calculate the estimated speed up.

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htatioh, ALU e 3 clock on will take E tatlon, operations,

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Suppose we could take the system of (Figure 1) and divide it into an arbitrary number of pipeline stages k, each having a delay of 300/k, and with each pipeline register having a delay of 20 ps. What would be the latency of the system? Express your answer in terms of k. What would be the throughput of the system? Express your answer in terms of k. What would be the ultimate limit on the throughput? Express your answer as in integer.
A nonpipelined system takes 300ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle of 60ns. Determine the speedup ratio of the pipeline for 100 tasks. What is the maximum speedup that could be achieved with the pipeline unit over the nonpipelined unit?
Consider a pipeline having 4 phases with a duration of 10, 50, 80, and 20 ns. Calculate the following Non-pipeline execution time for 1 instruction Sequential time for 1000 tasks Pipeline time for 1000 tasks Speed up ratio
We have 2 designs D1 and D2 for a synchronous pipeline processor. D1 has 6 stage pipeline with execution times of 1 ns, 3 ns, 2 ns, 4 ns, 2 ns, and 3 ns. While Design D2 has 7 pipeline stages each with 2ns execution time. The delay of an inter-stage register in the pipeline is 1 ns for both D1 and D2. Calculate: 1. Pipeline cycle time for D1 2. Pipeline cycle time for D2 3. Non-pipeline cycle time for D1 4. Non-pipeline cycle time for D2 5. Pipeline time for 500 instructions for D1 6. Pipeline time for 500 instructions for D2 7. Non-pipeline time for 500 instructions for D1 8. Non-pipeline time for 500 instructions for D2 9. How much time can be saved using design D2 over design D1 for executing 500 instructions?
Consider a non-pipelined processor with a clock rate of 2.5 gigahertz and average cycles per instruction of 4. 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 there are no stalls in the pipeline. What is the speed up achieved in this pipelined processor? 5
A nonpipelined system takes 200ns to process a task. The same task can be processed in a five-segment pipeline with a clock cycle of 40ns. Determine the speedup ratio of the pipeline for 200 tasks. What is the maximum speedup that could be achieved with the pipeline unit over the nonpipelined unit?
A non-pipelined system takes 100ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle of 20ns. a. Determine the speedup ratio of the pipeline for 100 tasks.b. Determine the speedup ratio of the pipeline for 20 tasks.c. Determine the speedup ratio of the pipeline for 1 task.d. What is the maximum speedup that could be achieved with the pipeline unit over the non-pipelined unit?
Pipelining with 1-stage Bypass Circuitry: Show the execution of the following sequence of MIPS instructions on a pipeline chart. Also show how source operands of ALUS are gathered for 12, 13, and 14 (i.e. read from the Register File at the Decode stage or forwarded from the first stage of the bypass circuitry) 11: LOAD R1, R8, #100 12: LOAD R2, R8, #500 13: ADD R3, R1, R2 14: STORE R3, R8, #1000
We have a single stage, non-pipelined machine and a pipelined machine with 5 pipeline stages. The cycle time of the former is 5ns and the latter is 1ns. a. Assuming no stalls, what is the speedup of the pipelined machine over the single stage machine?
Consider a 5-stage pipeline with stage delays 8 ns, 2 ns, 12 ns, 15 ns, 9 ns. A pipelined implementation of the processor requires buffering between each pair of consecutive stages with a delay of 2.5 ns. The speedup factor achieved by pipeline implementation over non - pipeline implementation is 1.05. Average execution time for non-pipeline implementation is 1911 ns. Find the number of instructions.
MIPS instructions classically take five steps: IF, ID, EX, MEM, WB. A certain MIPS pipeline has the following properties: - IF: 1 cycle - ID: 1 cycle - EX: 1 cycle for A type instructions, 2 cycles for B type instructions, 3 cycles for C type instructions. МЕМ: 1 суcle WB: 1 сycle EX stage of the pipeline has three separate datapaths, to accommodate for the various cycle counts. A-type instructions B-type instructions B-type instructions C-type C-type C-type instructions instructions instructions Run the following instruction sequences through this pipeline with and without forwarding (Assume full data hazard, meaning each instruction depends on data that was produced by the previous instruction, and assume no load/store instructions): a) AAAAAAAABBBBBBBBCCCCCCCC b) АВСАВСАВСАВСАВСАВСАВСАВС с) АСВАСВАСВАСВАСВАСВАСВАСВ
The instruction pipeline of a RISC processor has the following stages. Instruction Fetch (IF), Instruction Decode (ID), Operand Fetch (OF), Perform Operation (PO) and Writeback (WB). The IF, ID, OF and WB stages take 1 clock cycle each for every instruction. Consider a sequence of 100 instructions. In the PO stage, 40 instructions take 3 clock eveles each, 35 instructions take 2 clock cycles each, and the remaining 25 instructions take 1 clock cycle each. Assume that there are no data hazards and no control hazards. The number of clock cycles required for completion of execution of the sequence of instructions is