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Problems in Memory Management CS 1550 Recitation October 9 th , 2002 The questions in this slide are from Andrew S. Tanenbaum's textbook page 264.

Problems in Memory Management CS 1550 Recitation October 9 th, 2002 The questions in this slide are from Andrew S. Tanenbaum's textbook page 264

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Page 1: Problems in Memory Management CS 1550 Recitation October 9 th, 2002 The questions in this slide are from Andrew S. Tanenbaum's textbook page 264

Problems in Memory Management

CS 1550 RecitationOctober 9th, 2002

The questions in this slide are from Andrew S. Tanenbaum's textbookpage 264.

Page 2: Problems in Memory Management CS 1550 Recitation October 9 th, 2002 The questions in this slide are from Andrew S. Tanenbaum's textbook page 264

11. If an instruction takes 10 nsec and a page fault takes and additional n nsec, give a formula for the effective instruction time if page faults occur every k instructions.

Consider k instructions: k-1 instructions execution without page fault:

10 * (k - 1) nsec1 instuction with page fault:

(10 + n) * 1nsecAverage:

(10*(k-1) + (10+n)*1) / k = 10 + n/k nsec

Page 3: Problems in Memory Management CS 1550 Recitation October 9 th, 2002 The questions in this slide are from Andrew S. Tanenbaum's textbook page 264

12. A machine has a 32-bit address space and an 8-KB page. The page is entirely in harware, with one 32-bit word per entry. The page table is copied to the hardware from memory, at one word every 100 nsec. If each process runs for 100 msec (including the time to load the page table), what fraction of the CPU time is devoted to loading the page tables?

1 nsec = 1/1,000,000,000 sec1 msec = 1/1,000 sec

Page size : 8-KB # of offset bits8-KB = 2^13 Bytes

Offset bits : 13 bitsPage number: 32 – 13 = 19 bits

Page 4: Problems in Memory Management CS 1550 Recitation October 9 th, 2002 The questions in this slide are from Andrew S. Tanenbaum's textbook page 264

How many pages are there?2 ^ 19 = 524288

Load time:524288 * 100 nsec = 0.0524288 sec

CPU time:100 msec = 0.1 sec

Fraction:0.0524288 / 0.1 = 52.4%

Real World …

32-bit address space == 32-bit word per entryAlways?