7. Consider a non-pipelined processor with aclock rate of 25 gigahertz and average cyclesper instruction of four. The same processor isupgraded to a pipelined processor with fivestages: but due to the intemal pipelined delay.the clock speed is reduced so 2 gigaher.Aasume that there are no valls in the pipeline.The speedup achieved in this pipelinodprocessor is

Here, we are going to find out speedup because of pipelined processor. Speedup=…

7. Consider a non-ppelined processor with a clock rate of 25 gigahertz and average cycles per instructon of four. The same processor is upgraded to a pipelinel processor with five stages: but due to the intemal pipelined delay. the clock speed is reduced so 2 gigaher. Aaume that there are no valls in the pipeline. The speedup achieved in this pipelinod processor is

7. Consider a non-ppelined processor with a clock rate of 25 gigahertz and average cycles per instructon of four. The same processor is upgraded to a pipelinel processor with five stages: but due to the intemal pipelined delay. the clock speed is reduced so 2 gigaher. Aaume that there are no valls in the pipeline. The speedup achieved in this pipelinod processor is

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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Ques 1: Consider a three stage pipeline with logic delays of 10ns, 20ns, 40ns for the three stages, respectively. Latch delay = 10ns. I am allowed to split any one of these three stages into two equal sized substages. Doing so, will make a four-stage pipeline. What is the maximum throughput (in MIPS) of this new pipeline?
What is the physical address of this instruction Source operand of MOV BL, [SI + BP - 9H], If you know that (ES) = 1AOOH , %3D (DS) = 4200H , (SS) = DOOOH , (BX) = 0200H %3D , (BP) = 1000H, (SP) = 1020H , (SI) = 0400H , %3D (DI) = 0300 H PA = E2503 H O PA = 424F7H PA = D13F7H O PA = 424F6H PA = 1D503H O
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We have a 5-stage (IF ID EX MM WB) SPARC CPU (which has delay slots). This CPU has the following characteristics: • The branches are solved at MM stage. • The jumps are solved at EX stage • The base CPI is 1.2 • 25% of the branches are taken • 60% of the delay slots are filled • 10% of the instructions are branches, 20% are jumps What is the overall CPI when control hazards are considered?