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Material Type: Assignment; Class: Operating System Concepts; Subject: Computer General; University: University of Central Florida; Term: Spring 2004;
Typology: Assignments
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1) (5 pts) Segmentation is a memory management scheme that does not reflect the programmer’s view of memory? TRUE **FALSE
5) (10 pts) A process has associated with it the following table. For each logical address shown indicate if the address is legal. If it is, compute the physical address. Addresses are in the form <p, d>. Assume each page/frame is 1000 bytes in size. Page # Frame # 0 24 1 14 2 6 3 55 Logical Address < 0, 321 > Is it legal? ___ Y ___ Physical Address ___ 24321 ____ (24 * 1000 + 321) Logical Address < 1, 3100 > Is it legal? ___ N ___ Physical Address ____________ (Displacement of 6100 is greater than size of one page) Logical Address < 4, 754 > Is it legal? ___ N ___ Physical Address ____________ (Page 4 is invalid. Does not appear in the page table) Logical Address < 2, 208 > Is it legal? ___ Y ___ Physical Address ___ 6208 _____ (6 * 1000 + 208) Logical Address < 3, 0 > Is it legal? ___ Y ___ Physical Address ___ 55000 ___ (55 * 1000 + 0) 6) (10 pts) The same page table as the previous problem. What happens when the size of the page is increased to 4000 bytes per page? Logical Address < 0, 321 > Is it legal? ___ Y ___ Physical Address ___ 96,321 _____ (24 * 4000 + 321) Logical Address < 1, 3100 > Is it legal? ___ Y ___ Physical Address ___ 59,100 ______ (14 * 4000 + 321) Logical Address < 4, 754 > Is it legal? ___ N ___ Physical Address ____________ (Page 4 is invalid. Does not appear in the page table) Logical Address < 2, 208 > Is it legal? ___ Y ___ Physical Address ___ 24,208 _____ (6 * 4000 + 208) Logical Address < 3, 0 > Is it legal? ___ Y ___ Physical Address ___ 220,000 ___ (55 * 4000 + 0)
9) (15 pts) For this problem, assume we are working with a decimal (base 10) computer. Given a paged memory system where: each page contains 100 bytes (10^2 ), the physical memory contains a total of 1,000,000 bytes (10^6 ), and logical memory (a program’s maximum logical address space) consists of 10,000 bytes (10^4 ). What is the maximum number of pages that can be used by a program? 104 / 10^2 = 10^2 or 100 pages How many frames are in the system? 106 / 10^2 = 10^4 or 10,000 frames Number of frames is equal to the size of physical memory (1,000,000) divided by the size of a frame (100). Since frames and pages must be equal in size, a frame for this question is 100 bytes in size. If every process that executes uses the maximum number of pages possible, what is the degree of multiprogramming in the system? 104 / 10^2 = 10^2 or 100 concurrent processes Divide the number of frames in the system (10,000) by the maximum pages per process (100) to get the maximum number of concurrent programs possible. **10) (5 pts) Pointers to page tables are kept where? ___________in a process’ PCB____________