Homework Answers
Page size = 64 bytes
- Page offset =
= 6 bits
= 6 bitsNumber of pages = 8
- Page number bits =
= 3 bits
= 3 bitsPage Frames = 4
- Frame number bits =
= 2 bits
= 2 bitsVirtual address = <Page Number, Page offset> = <3, 6> (Total 9 bits)
Physical address = <Frame Number, Page offset> = <2, 6>
Virtual Address = 0x44 = 001000100
- Page Number = 001 = 1st page
- Page offset = 000100
At page number 1, page table has frame number as 3 (11 in binary)
So, physical address = <11, 000100> = 11000100 = 0xC4 (Option C)
Add Answer to:
Question 2 Suppose you have a byte-addressable virtual address memory with 8 virtual pages of 64...
Suppose you have a byte-addressable virtual address memory system with 8 virtual pages of 64 bytes each
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
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...
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...
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 #...
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...
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?
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...
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 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...
Consider a logical address space of 8 pages; each page is 2048 byte long, mapped onto a physical memory of 64 frames.
Consider a logical address space of 8 pages; each page is 2048 byte long, mapped onto a physical memory of 64 frames.(i) How many bits are there in the logical address and how many bits are there in the physical address?(ii) A 6284 bytes program is to be loaded in some of the available frames ={10,8,40,25,3, 15,56,18,12,35} . Show the contents of the program's page table.(iii) What is the size of the internal fragmentation?(iv) Convert the following logical addresses 2249...
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...
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...
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 #...
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...
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...
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...
Most questions answered within 3 hours.
-
. A permanent magnet is dropped south-end-down through a horizontal
circular coil with a radius of...
asked 1 minute from now -
Bernie's Beverages purchased some fixed assets classified as
5-year property for MACRS. The assets cost $28,000....
asked 12 minutes ago -
How many ATPs are produced from the catabolism of a 10-C
molecule of fatty acid under...
asked 17 minutes ago -
Before practicing a routine on the rings, a 64.8 kg gymnast
hangs motionless, with one hand...
asked 18 minutes ago -
If the K b of a weak base is 6.3 × 10 − 6 , what...
asked 25 minutes ago -
Which of the following is the minimum amount of moles of NaOH
that must be added...
asked 28 minutes ago -
Stories about organizational ________ provide important clues
about cultural values and norms.
a. myths
b. heroes...
asked 30 minutes ago -
Explain the criteria used in selecting a target market
BUS220 Retail Management, thank you!
asked 32 minutes ago -
Convert/Calculate the following:
Determine the identity of an elemental gas if 4.55 L weighing
35.4g, under...
asked 35 minutes ago -
Consider the equilibrium C(s)+ CO2(g) ⇌2 CO(g)
A 2.0 L flask contains a mixture of 0.10...
asked 34 minutes ago -
MATLAB
Part 1 – randFloatValue.m This function accepts two numbers,
lower and upper, and returns a...
asked 40 minutes ago -
You have been asked to hide prizes around your house for your
3-year old nephew. His...
asked 41 minutes ago