Question

A certain byte-addressable computer system has 32-bit words, a virtual address space of 4GB, and a...

A certain byte-addressable computer system has 32-bit words, a virtual address space of 4GB, and a physical address space of 1GB. The page size for this system is 4 KB. Assume each entry in the page table is rounded up to 4 bytes.

a) Compute the size of the page table in bytes.

b) Assume this virtual memory system is implemented with a 4-way set associative TLB (Translation Lookaside Buffer) with a total of 256 address translations.

Compute the size of the TLB tag field and the size of the TLB index field.

0 0
Add a comment Improve this question Transcribed image text
Request Professional Answer

Request Answer!

We need at least 10 more requests to produce the answer.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)


All students who have requested the answer will be notified once they are available.
Know the answer?
Add Answer to:
A certain byte-addressable computer system has 32-bit words, a virtual address space of 4GB, and a...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
  • 3. Virtual Memory (20 points) An ISA supports an 8 bit, byte-addressable virtual address space. The...

    3. Virtual Memory (20 points) An ISA supports an 8 bit, byte-addressable virtual address space. The corresponding physical memory has only 256 bytes. Each page contains 32 bytes. A simple, one-level translation scheme is used and the page table resides in physical memory. The initial contents of the frames of physical memory are shown below. VALUE address size 8 bit byte addressable each byte of addressing type memory has its own address 32 B page size physical memory size 256...

  • Question 31 supus Given a computer using a byte-addressable virtual memory system with a two-entry TLB,...

    Question 31 supus Given a computer using 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 size 16 bytes. Assume pages of size 32 bytes and a main memory of 4 frames. Assume the following TLB and page table for Process P: TLB 03 4 هما 0 1 2 3 4 5 6 7 Page Table f Vali d 1 1 0 2...

  • A computer uses a byte-addressable virtual memory system with a four-entry TLB and a page table...

    A computer uses a byte-addressable virtual memory system with a four-entry TLB and a page table for a process P. Pages are 16 bytes in size. Main memory contains 8 frames and the page table contains 16 entries. a. How many bits are required for a virtual address? b. How many bits are required for a physical address?

  • A machine has a 16-bit byte-addressable virtual address space. The page size is 8 KB. How...

    A machine has a 16-bit byte-addressable virtual address space. The page size is 8 KB. How many pages of virtual address space exist? (25 points)

  • 18. You have a byte-addressable virtual memory system with a two-entry TLB, a 2-way set associati...

    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...

    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

  • please answer $5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose...

    please answer $5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, the bits. size of the set field is bits, and the size of the tag field is 5 pts Question 19 Suppose we have a byte-addressable computer...

  • Consider a virtual memory system with the following properties: 36 bit virtual byte address, 8 KB...

    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...

  • 16 It is known that computer system programs use 32-bit virtual addresses to access storage units....

    16 It is known that computer system programs use 32-bit virtual addresses to access storage units. If the physical memory space of the computer system is 1GB, and the paging management mechanism is adopted, the page size is 4KB, and each page table entry is 4B. If only one level of page table is used to realize the mapping from virtual address to physical address, how much memory space does the page table occupy? A. 1MB B. 4KB C. 1KB...

  • 1) Given a virtual memory system with: virtual address 36 bits physical address 32 bits 32KB...

    1) Given a virtual memory system with: virtual address 36 bits physical address 32 bits 32KB pages (15 bit page offset) Each page table entry has bits for valid, execute, read and dirty (4 bits total) and bits for a physical page number. a) How many bits in the page table? (do not answer in bytes!) Three digit accuracy is good enough. The exponent may be either a power of 2 or a power of 10. b) The virtual address...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT