{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment6Solution

# Assignment6Solution - Department of Computer Science...

This preview shows pages 1–2. Sign up to view the full content.

Department of Computer Science University of Minnesota, Twin Cities CSci 5103 - Operating Systems -- Fall 2010 (Instructor: Tripathi) Assignment 6 Solution Posted on: November 16 12:30 pm Problem 1: (a) Memory access time when TLB hit = 20 + 100 = 120 Memory access time with TLB miss = 20 + 100 + 100 = 220 Effective memory access = 0.9 x 120 + 0.1 x 220 = 108 + 22 = 130 nano seconds (b) Let p denote the probability of page-fault on a memory access. Average memory access time = A = (1-p) 100 + p [ prob[modified page to be replaced]x10 7 + prob[unmodified page is replaced] x 5 x 10 6 ] It is required that A be less than equal to 200 nano seconds. Therefore 100 - 100 x p + p [ 0.8 x 10 7 + 0.2 x 5 x 10 6 ] < = 200 -100 x p + p [ 9 x 10 6 ] <= 100 Ignoring -100p term. P <= 100/ (9 x 10 6 ) = 1.11 x 10 -5 Problem 2: Answer: Assume that the page size is in terms of words, and an integer is stored in one word. I.e. a page can store 200 integers. Assume that the two dimensional array is stored in row major form. I.e. the array elements A[i][0], A[i][1]…,A[i][99] are stored in consecutive locations.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}