Exercise:A computer has 4 frames. Page size is 2KB (2048). The loaded time, the R and M bits for eachpage are shown belowPageFrameLoaded timeMR1111531345576.33Question: compute the physical address of the following virtual addressa) 1000 using FIFOb) 3100 using NRUc) 3100 using LRUd) 6200 using Second chanceNote: consider each case separately

so we have to calculate physical address of given virtual addresses Which is calculated Below

Answered: Exercise: A computer has 4 frames. Page…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Computer Science Given Page requests: Reference string: 4, 3, 4, 1, 2, 4, 1, 2, 3, 2, 1, 1 Show the memory allocation state after each page reference for the given scheduling algorithm (given the respective number of page frames). What is the number of page faults for each algorithm? NFU / LFU Assume there are 3 page frames of physical memory AGING Assume there are 3 page frames of physical memory Also assume there are 4 bits in the counter for each page
The following data segment starts at memory address 1000h (hexadecimal) .data printString BYTE "ASSEMBLY IS FUN",0 moreBytes BYTE 25(DUP)0 dateIssued DWORD ? dueDate DWORD ? elapsedTime Word ? What is the hexadecimal address of dueDate ? a. 1045h b. 1029h c.1010h d. 102Dh
Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. How many bits are used for the page number? How many bits are used for the offset? 8 bits each. With this system, what’s the maximum number of pages that a process can have? 256 Suppose that each entry in the page table comprises 4 bytes (including the frame number, the valid bit, and miscellaneous “bookkeeping bits”). An OS uses an array to store the page table. What is the size of the page table? 1024 Bytes Furthermore, suppose the first 6 pages of a process map to frames 222 to 227 (as decimal numbers), and the last 5 pages of the process map to frames 1 to 5 (also decimal numbers). All other pages are invalid. Draw the page table, including the valid bit and the frame number. DONE Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert…
Assume that a memory module contains three holes of 10MB each. A sequence of 14 requests for 1MB each will be processed (See the diagram below). For each of the memory allocation methods listed below, draw a diagram representing how memory is allocated and determine the sizes of the remaining holes after all 14 requests have been satisfied. First fit Next fit C. Best fit d. a. b. Worst fit Hint - For the Next fit, the following allocation starts with the hole following the previous allocation 10 Mb 10 Mb 10 Mb
Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. Could you hand draw the page table, if possible a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page table
Q1). Suppose $s0 stores the base address of word array A and $t0 is associated with m, convert the following instruction into MIPS. A[240] = A[240+m]Q2). Suppose $t0 stores the base address of word array A and $s0 is associated with m, convert the following instruction into MIPS. m= 0 while (m <= 10): A[m] = A[m+4]*6 m = m + 5
Student information is stored in a data area, and each student entry is formatted as follows:The first nine bytes are the student ID in ASCII format.The following byte is the course score.The following phrase is the section identifier. There are over 300 such student records in memory, which have been loaded consecutively beginning at address $10000. To indicate the conclusion of the records, the last record loaded is a dummy record with a section identifier of $FFFF. Create an assembler subroutine GetMax that searches the complete list for the highest mark. If there is a tie, the student with the highest score on the list should be found. The beginning address of the subroutine should be returned.
Giventhe following assignment of some program’s virtual pages to physical pages in a system with 4 KiB byte pages, what physical memory address corresponds to virtual address 20000? (All values are given in decimal.)
Q.2 Suppose a system has a logical address space of size 25 (m=5), a page size of 4 bytes, and a physical memory of 128 bytes. Given the following page table, find out the respective physical addresses for each of the logical address listed in Table 2. (Note: all addresses are expressed in decimal number system) Table 1: Page table 0 2 5 7 0 15 1 2 3 4 5 18 6 20 7 22 Table 2: Fill out this table with corresponding physical memory addresses Logical Address Physical Address 0 6 10 17 23
Question 9
Computer science
Address Instruction Ox242 CALL sort Ох262 sort: MOVLW Ох90 Fill in the following blanks based on the above code segment: First word of machine code stored in locations 0x243 and Ox242: Ox * Second word of machine code stored in locations Ox245 and Ox244: Ox

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