2. Using the page table, give the physical address corresponding to each of the following virtual addresses: (a) 20 (b) 4100 (c) 8300 (d) 20600 Virtual address space 60K-64K X 56K-60K X 52K-56K X 48K-52K X 44K-48K 7 40K-44K X 36K-40K 5 32K-36K X 28K-32K X 24K-28K X 20K-24K 3 16K-20K 4 12K-16K 0 8K-12K 6 1 OK-4K 2 4K-8K Virtual page Physical memory address 28K-32K 24K-28K 20K-24K 16K-20K 12K-16K 8K-12K 4K-8K OK-4K Page frame

Q: Using the page mapping scheme we used in the lectures (see the figure below), give the physical…

A: So here is all information on the physical address so we can use this address for further…

Q: Suppose a byte-addressable computer using set associative cache has 2^24 bytes of main memory and a…

A: In a set-associative cache,  it can be defined in such a way that the cache is divided into units,…

Q: Consider the main memory size of 128 kB, Cache size of 16 kB, Block size of 256 B with Byte…

A: Main memory size = 128KB = 17 bits Total number of cache block = 16KB/256 = 64 Block size = 256 B.…

Q: 5. Suppose a byte-addressable computer using set-associate cache has 2^21 bytes of main memory and a…

A: If the number of address bits are n then, the size of the cache is 2n words. Since, the size of the…

Q: In an architecture with 18 bits of "virtual address" width, "page size" is given as 1024 bytes and…

A: We are given TLB as 2 -way set associative with 16 data blocks.  # sets= data blocks/…

Q: Table Q3 (a) lists the content of a program memory that consists of four (4) lines of AVR machine…

A: Answer: I have given answer in the handwritten format

Q: In a page addressing system of 15 bits, where eight bits are used for the page number, what would be…

A: The above question is solved in step 2 :-

Q: Suppose a computer using fully associative cache has 4G bytes of byte-addressable main memory and a…

A: Cache mapping refers to the technique used for bringing content from the main memory into the cache…

Q: A cache memory has a line size of eight 64-bit words and a capacity of 4K words. The main memory…

A: A 64-bit word means 8 byte.Line size: 8 words in a line, means 8 x 8 bytes = 64 bytes in a line = 26…

Q: Virtual Addressing. This question ex a machine with a 40-bit virtual addres protection bits. Write…

A: Given the information of a machine with a 40-bit virtual address, 16KB page size and 4 bytes per…

Q: Taska2: Logical to Physical Address Calculation: Question: Calculate the physical memory address…

A: Below is the answer of all subparts. I hope this will meet your requirement...

Q: Determine the number of page table entries (PTEs) that areneeded for the following combinations of…

A: Number of page table entries = address size / page size

Q: Suppose a byte-addressable computer using set associative cache has 4Mbyes of main memory and a…

A: Set-associative caches are an essential component of modern computer systems, aiming to improve data…

Q: ve a two-level page table where the first level is indexed by 15-bits, then how many entries does…

A: Dear Student, The answer to your questions is given below -

Q: mputer has a 16 bit address space A[15:0]. If all addresses having bits A[15:13]=111 are reserved…

A: Given : A computer has a 16 bit address space A[15:0]. If all addresses having bits A[15:13]=111 are…

Q: Consider the following segment table: What are the physical address for the following logical…

A: THIS IS A MULTIPART-BASED PROBLEM. ONLY THE FIRST THREE PARTS ARE SOLVED. KINDLY SEND THE REMAINING…

Q: Now suppose the system from problem 17 has 8Kb of physical memory. Process P1 has the following page…

A: Here in this question we have given  Virtual address = 16 bit  (from previous question) Page size =…

Q: Suppose a computer using fully associative cache has 224224 bytes of byte-addressable main memory…

A: Introduction: Faster access to memory may be achieved by putting frequently used data in the CPU's…

Q: If logical addresses are represented using m bits as shown below, where m=4 and n=2. What is the…

A: Actually, memory is used to stores the data.

Q: Assuming that no page faults occur, the average time taken to access a virtual address is…

A: Hi there, Please find your solution below, I hope you would find my solution useful and helpful.…

Q: Consider a computer which uses virtual addressing with 32 bit addresses and a two level page table.…

A: The size of the pages in this virtual addressing scheme can be determined by looking at the number…

Q: 9. Consider a system that uses 32-bit addresses and page table structures as discussed in class. If…

A: 32 bits= 2^5 256 pages= 2^8

Q: Problem 2. The following table gives the parameters for a number of different caches. For each…

A: The complete answer is below:

Q: A computer has a 16 bit address space A[15:0].  If all addresses having bits A[15:14]=11 are…

A: Computer architecture describes how a computer system's hardware and software are designed and…

Q: Our system is using virtual memory and has 48-bit virtual address space and 32-bit physical address…

A: (a) To determine the number of entries needed for a single-level page table, it need to calculate…

Q: For a 512k x 32 memory system, how wide does the incoming address bus needed to be in order to…

A: Here is the explanation about the width of the address bus:

Q: determine the offset in bytes

A: Given virtual address = 27KB => 27*2^10 B Page size  = (27 X 2 ^ 10)/( 2 ^ 10)…

Q: Suppose that an MMU has page table entries like the ones shown at right (formatted as in the slides…

A: The solution of the above question is:

Q: 3- Suppose that ECx=12345678h , EBx=87654321h ,and DS=1100h. Determine the contents of each address…

A: Suppose that ECx =12345678h, EBx=87654321h,and DS =1100h .

Q: Using the table shown in the image , give the physical address corresponding to each of the…

A: We need to relation between virtual address and physical memory address corresponding to the page…

Q: Problem 0. The following table gives some of the parameters for a number of different hardware…

A: Hardware caches play a critical role in improving the performance and efficiency of modern computing…

Q: Consider the following portion of a page table from a system with 4 KiB (i.e., 4096 byte) pages.…

A: We have given portion of page table and page size is 4KiB. Using virtual memory address, we have to…

Q: Consider the following segment table (all expressed and addressed in bytes) Segment Base Length 0…

A: Find the required answer given as below :

Q: According to the following table if the destination address is 11001000 00010111 00011100 10101010…

A: By doing prefix matching of the given destination address with the address ranges given in the…

Q: Show how the following values would be stored by machines with 32-bit words, using little endian and…

A: We need to show how the given values would be stored by machines with 32-bit words, using little…

Q: Suppose a byte-addressable computer using set associative cache has 224 bytes of main memory and a…

A: In a set-associative cache, the memory address is divided into three main fields: tag, set, and…

Q: Given a 16 bit virtual address space and a 10 bit low-order offset how many bytes is each page.

A:

Q: Given process 1 page table consisting of 6-bit page number and 10-bit offset. Process 1 Page Table…

A: Here is explanation regarding the physical address of both.

Q: In a system, with 10 bit addresses of which 4 bits is for the page number, give following page…

A: Here number of bits for page number=4 Thus number of bits for offset = 10-4 = 6. These are least…

Q: (i) The contents of the following segment registers are as given. CS = 1111H, DS = 3333H, SS =…

A: Segmentation is the process in which the main memory of the computer is logically divided into…

Q: Suppose a computer using fully associative cache has 224 words of main memory and a cache of 512…

A: As per the policy and  guidelines of Bartleby we are supposed to answer only first question or the…