physcal addresses are 4s ng4 Ame dat in a cetain compe, teaddresses can be translaled without y TLBentries At most how many ditina vidthe address translation peh has 12 vldThe Translation Look aside Bulfer (TLB)isine is kB and the word size ibyThe memory is word addresible. The pevirtual addresses are 64 bea long d thsine ismiss?

cenin Ame da ce The Translation Look aside Ber TL addreses can be tranlated wito wy TLA the address anslation pah has 1 vd e isAB and the word sile wby ces. At most how many nd vial addreses are 64 b The menory is word addree The physical addresses are 4 ng physcal addrees d he wond sie is sine is the address tran nd vin ma

cenin Ame da ce The Translation Look aside Ber TL addreses can be tranlated wito wy TLA the address anslation pah has 1 vd e isAB and the word sile wby ces. At most how many nd vial addreses are 64 b The menory is word addree The physical addresses are 4 ng physcal addrees d he wond sie is sine is the address tran nd vin ma

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

instruction is in the first picture please give me only implementation of int L1lookup(u_int32_t address) and int L2lookup(u_int32_t address) cacheSim.h #include<stdlib.h>#include<stdio.h>#define DRAM_SIZE 1048576typedef struct cb_struct {unsigned char data[16]; // One cache block is 16 bytes.u_int32_t tag;u_int32_t timeStamp; /// This is used to determine what to evict. You can update the timestamp using cycles.}cacheBlock;typedef struct access {int readWrite; // 0 for read, 1 for writeu_int32_t address;u_int32_t data; // If this is a read access, value here is 0}cacheAccess;// This is our dummy DRAM. You can initialize this in anyway you want to test.unsigned char * DRAM;cacheBlock L1_cache[2][2]; // Our 2-way, 64 byte cachecacheBlock L2_cache[4][4]; // Our 4-way, 256 byte cache// Trace points to a series of cache accesses.FILE *trace;long cycles;void init_DRAM();// This function print the content of the cache in the following format for an N-way cache with M Sets// Set 0…
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
Write a program in assembly language that copies three string “S1, S2, S3” to “S4” using indexed addressing insuch a pattern that String S4 would become: [SSStttrrriiinnnggg123]. The data is given below..dataS1 byte "String1",0S2 byte "String2",0S3 byte "String3",0S4 byte sizeof S1*3-2 Dup(0) solve this with assembly language
28. _____ method is used to map logical addresses of variable length onto physical memory. a. Paging b. Overlays c. Segmentation d. Paging with segmentation
By assuming that X is the last digit of your student number and 3X is a two digitnumber, consider memory storage of a 64-bit word stored at memory word 3X ina byte-addressable memory(a) What is the byte address of memory word 3X?(b) What are the byte addresses that memory word 3X spans?(c) Draw the number 0xF1234567890ABCDE stored at word 3X in both big-endianand little-endian machines. Clearly label the byte address corresponding to eachdata byte value.
mcq question The Physical address of the String Destination Memory Reference is: DS:DI, DS:SI, and DS:BX SS:SP, and SS:BP ES:DI only ES:DI and ES:SI
Modiflow y the beprogram given to include response time program;FCFS CPU SCHEDULING ALGORITHM #include<stdio.h>#include<conio.h>main(){int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg;clrscr();printf("\nEnter the number of processes -- "); scanf("%d", &n); for(i=0;i<n;i++){printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]);}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0]; for(i=1;i<n;i++){wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");for(i=0;i<n;i++){printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]);}printf("\nAverage Waiting Time -- %f", wtavg/n);printf("\nAverage Turnaround Time -- %f", tatavg/n); getch();}
Modiflow y the beprogram given to include response time program;FCFS CPU SCHEDULING ALGORITHM #include<stdio.h>#include<conio.h>main(){int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg;clrscr();printf("\nEnter the number of processes -- "); scanf("%d", &n);for(i=0;i<n;i++){printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]);}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0];for(i=1;i<n;i++){wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");for(i=0;i<n;i++){printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]);}printf("\nAverage Waiting Time -- %f", wtavg/n);printf("\nAverage Turnaround Time -- %f", tatavg/n); getch();}
Rspected Sir, please chek my qustioin 1. Calculate the sum of series of even numbers from the list of numbers. The length of the list is in ML 2200H and series itself begins from ML 2201H. Assume the sum to be greater than FFH. Store the results in ML 2300H and 2301H. 2. Discuss the working of CALL instruction in 8085 microprocessor by taking the suitable example. Also, mention various conditional CALL instructions of 8085 microprocessor.
MIPS Assembly Complete gcd_cur function, which recursively calculates the GCD (Greatest Common Divisor) of two given positive integers input using the following Euclidean algorithm (Links to an external site.). def gcd_recur(a, b): if b = 0: return a; else: return gcd_recur( b, (a mod b) ); >> a0: the 1st input argument, a>> a1: the 2nd input argument, b My Code: ############################################################################################################################################################################################## PART 2 (gcd_recur)#a0: input number#a1: input number###############################################################gcd_recur:############################### Part 2: your code begins here #### return value is v0move $t0, $a0 move $t1, $a1 loop: beq $t1, $0, done div $t0, $t1 move $t0, $t1 mfhi $t1 j loop done: move $v0, $t0…
Need help describing these: #define NAMELEN 2048 #define BUFSIZE 1024 #include <sys/types.h> #include <sys/stat.h> #include <time.h> #include <stdio.h> #include <stdlib.h> char *strmode(mode_t mode); // nasty function below this one int main(int argc, char *argv[]) { if (argc != 2) { fprintf(stderr, "Usage: %s <pathname>\n", argv[0]); exit(EXIT_FAILURE); } if(access(argv[1], F_OK)!=0){ // check file exists perror("Couldn't access file"); exit(EXIT_FAILURE); } struct stat sb; if (stat(argv[1], &sb) == -1) { // call to stat(), fills sb, distinguishes symbolic links perror("stat"); exit(EXIT_FAILURE); } printf("Filename: %s\n",argv[1]); printf("File type: "); if(0){} // checks to determine file type else if S_ISBLK (sb.st_mode) {…
.MODEL SMALL .STACK 100H .DATA ; The string to be printed STRING DB 'This is a sample string', '$' .CODE MAIN PROC FAR MOV AX,@DATA MOV DS,AX ; call reverse function CALL REVERSE ; load address of the string LEA DX,STRING ; output the string ; loaded in dx MOV AH, 09H INT 21H ; interrupt to exit MOV AH, 4CH INT 21H MAIN ENDP REVERSE PROC ; load the offset of ; the string MOV SI, OFFSET STRING ; count of characters of the; ;string MOV CX, 0H LOOP1: ; compare if this is; ;the last character MOV AX, [SI] CMP AL, '$' JE LABEL1 ; else push it in the; ;stack PUSH [SI] ; increment the pointer; ;and count INC SI INC CX JMP LOOP1 LABEL1: ; again load the starting; ;address of the string MOV SI, OFFSET STRING LOOP2: ;if count not equal to zero CMP CX,0 JE EXIT ; pop the top of…