Print Page | Close Window

Computer Organization Printed From: One Stop GATE Category: GATE AT A GLANCE Forum Name: Basic GATE Info Forum Discription: All the basic information about GATE for the starters. URL: http://forum.onestopgate.com/forum_posts.asp?TID=1818 Printed Date: 11Feb2025 at 4:19pm Topic: Computer Organization Posted By: aparna Subject: Computer Organization Date Posted: 26Nov2007 at 2:33am Computer Organization In a certain machine data references constitute 40% of the mix, and that the ideal CPI of the pipelined machine is 1. Assume that the machine with structural hazards has a clock rate that is 1.05 times higher than the clock rate of machine without hazard. What is effective speedup with pipelining without structural hazard? Options 1. 1.05 2. 1.3 3. 1 4. none of above Ans: 2 Explanation: We major the average instruction time on both the machines to access speedup/slowdown. Avg. instruction time = CPI * clock cycle time Since there are no stalls/hazards, the average instruction time for ideal machine is the clock cycle times. The average clock cycle time for the machine with 40% structural hazard is M = (1+0.4 x 1) x Clock cycle time ideal/1.05 = 1.3 clock cycle time ideal Machine without structural hazard is 1.3 times faster.

Print Page | Close Window