How many bits are required for each virtual address?
There are 4KB or 22 * 210 addresses, so we need 12 bits
for a virtual address.
b. How many bits are required for each physical address?
Main memory is 512 B or 29, so we need 9 bits for a physical
address.
c. What is the maximum number of entries in a page table?
There are 212/27 pages in virtual memory, so the page table can
have 25 entries.
d. To which physical address will the virtual address
A4216ate?
A42 means 2626 in decimal.
page 0 contains addresses 0 to 127, page 1 contains 128 - 256),
like this 2626 will be present in page 20 ( which ranges from 2560
to 2687) . So page 20 maps to frame 1and page is present at offset
66(2626-2560). Page 20 maps to frame 1 , so 2626 maps to physical
address 2753 (2687+66).
e. B2A in decimal is 2858. It is in page 22 ranging between 2816 to
2943. Offset will be 42 (2858-2816)... It maps to physical address
2985 i.e 2943+42.
f. Which virtual address will translate to physical address
27610?
Physical address 276 is at offset 20(276-256) in virtual page2.
SN 6 A system implements a paged virtual address space for each process using a one-level...
A system implements a paged virtual address space for each process using a one-level page table. The maximum size of virtual address space is 8MB. The page table for the running process includes the following valid entries (the -> notation indicates that a virtual page maps to the givenpage frame, that is, it is located in that frame): Virtual page 2 -> Page frame4 Virtual page 4 -> Page frame 9 Virtual page 1 -> Page frame2 Virtual page 3...
17. A computer system implements a paged virtual memory system. Assume a 16-bit virtual address space and a 24-bit physical address space. Assume that the first 6 bits of a virtual address index the page table and the rest of the bits are the page offset. A process has the following indexed page table. Index Page Table Entry (PTE) 0x3800 0x3600 0x3200 0x1000 2 3 Each page table entry qives a hexadecimal page frame addresses. Translate the following two hexadecimal...
11. In a paged virtual memory system, can the computer 's physical memory address space be larger than a process's virtual memory address space? Explain your answer. (Note: A computer 's "physical memory address space" is the number of bits for the frame number plus the number of bits foir the offset. A process's "virtual memory address space" is the number of bits for a page number plus the number of bits for the offset.)
In a system using paged segmentation, the logical address space of each process consists of a maximum of 16 segments, each of which can be up to 64 Kbytes in size. Physical pages are 512 bytes. Determine how many bits are needed to specify each of the quantities below, justifying each of your answers by demonstrating the calculations. a) Segment number (segment address) b) Number of a logical page within the segment c) Displacement within a page e) Complete logical...
Exercise l: Suppose that we have a virtual memory space of 28 bytes for a given process and physical memory of 4 page frames. There is no cache. Suppose that pages are 32 bytes in length. 1) How many bits the virtual address contain? How many bits the physical address contain? bs Suppose now that some pages from the process have been brought into main memory as shown in the following figure: Virtual memory Physical memory Page table Frame #...
Address Translation Question [8 points] Suppose a computing system uses paging with a logical address of 24 bits and a physical address of 32 bits. The page size is 4KB. Answer each of the following. If an answer is a power of 2, you can leave it in the form of a power of 2. ... 2. [20 points] Memory address translation and TLB performance [8 points] Suppose a computing system uses paging with a logical address of 24 bits...
18. You have a byte-addressable virtual memory system with a two-entry TLB, a 2-way set associative cache, and a page table for a process P. Assume cache blocks of 8 bytes and page size of 16 bytes. In the system below, main memory is divided into blocks, where each block is represented by a letter. Two blocks equal one frame. Given the system state as depicted above, answer the following questions: a) How many bits are in a virtual address...
A computer system has a 36-bit virtual address space with a page size of 8K, and 4 bytes per page table entry. How many pages are in the virtual address space? What is the maximum size of addressable physical memory in this system? If the average process size is 8GB, would you use a one-level, two-level, or three-level page table? Why? Compute the average size of a page table in part c above
Suppose you have a byte-addressable virtual address memory system with 8 virtual pages of 64 bytes each, and 4-page frames. Assuming the following page table, answer the questions below: Page #Frame #Valid Bit0111312-03014215-06-07-0a) How many bits are in a virtual address? b) How many bits are in a physical address? c) What physical address corresponds to the following virtual addresses (if the address causes a page fault, simply indicate this is the case)? 1) Ox00 2) 0x44 3) OxC2 4) 0x80
Consider a virtual memory system with the following properties: 36 bit virtual byte address, 8 KB pages size, and 32 bit physical byte address. Please explain how you determined your answer. a. What is the size of main memory for this system if all addressable frames are used? b. What is the total size of the page table for each process on this processor, assuming that the valid, protection, dirty, and use bits take a total of 4 bits and...