Consider computing the overall MIPS * - for a machine A for which the following performance measures were recorded when executing a set of benchmark programs. Assume that the clock rate of the CPU is 400 MHz Instruction category Percentage of No. of cycles per occurrence instruction 25 ALU LOAD from Memory STORE to Memory 1 20 3 30 Branch 15 7 Call subroutine 10 CT=2.5 n sec ,CPI=3.6,MIPS=111.111 CT=0.25 n sec, CPI=3.06 ,MIPS=120.01 CT=25 n sec ,CPI=3.16,MIPS=251.01 O CT=2.5 n sec ,CPI=3.16,MIPS=101.211

Consider computing the overall MIPS * - for a machine A for which the following performance measures were recorded when executing a set of benchmark programs. Assume that the clock rate of the CPU is 400 MHz Instruction category Percentage of No. of cycles per occurrence instruction 25 ALU LOAD from Memory STORE to Memory 1 20 3 30 Branch 15 7 Call subroutine 10 CT=2.5 n sec ,CPI=3.6,MIPS=111.111 CT=0.25 n sec, CPI=3.06 ,MIPS=120.01 CT=25 n sec ,CPI=3.16,MIPS=251.01 O CT=2.5 n sec ,CPI=3.16,MIPS=101.211

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

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Consider the following set of processes A, B, C, D with the following CPU burst time and I/O. Find the average waiting time with RR of quantum 10ms and context switch time=2ms. What is the average waiting time? Write the waiting time for processes A,B,C,D.
Consider the following four processes with the length of CPU burst time in milliseconds (P1: 20, P2:15, P3:10, P4:5 ms). Calculate the average waiting time by using following algorithms FCFS ALGORITHM. 1. RR ROUND ROBIN (TIME QUANTUM 10 MS)
Consider Context Switch time of 2 secs and modify below program accordingly. 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],…
Consider Context Switch time of 2 secs and modify below program accordingly 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();}
hi can u anwser this qustion plesc ? Consider a Computer which has a memory which is capable of storing 4096 K words and each word in memory can be of size 32 bits. The computer supports a total of “K” addressing modes, and it has “M” computer registers. The computer supports instructions, where each instruction consists of following fields: Mode Operation code Register Register Memory AddressGiven that each instruction will be stored in one memory word, discuss with a suitable diagram the format of instruction by specifying number of bits for each field of instruction. Discuss each field of instruction.Note: Choose your own values for K (number of addressing modes) and M (number of Registers) k=8 m=50
Assume that a personal computer must get FIVE pieces of data from itsmemory for every operation (one piece of data indicates Which instructionto perform: the other is the intormation which is operated on). If thecomputer is the size of a typical PC. how long does it take to get these pieceof data? From this, what is the maximum rate at which the PC can executeuctions?
Suppose that a computer has a processor with two L1 caches, one for instructions and one for data, and an L2 cache. Let τ be the access time for the two L1 caches. Themiss penalties are approximately 15τ for transferring a block from L2 to L1, and 100τ fortransferring a block from the main memory to L2. For the purpose of this problem, assumethat the hit rates are the same for instructions and data and that the hit rates in the L1 andL2 caches are 0.96 and 0.80, respectively.(a) What fraction of accesses miss in both the L1 and L2 caches, thus requiring accessto the main memory?(b) What is the average access time as seen by the processor?(c) Consider the following change to the memory hierarchy. The L2 cache is removedand the size of the L1 caches is increased so that their miss rate is cut in half. Whatis the average memory access time as seen by the processor in this case
Consider the following processes with the given CPU arrival time and burst time. Please use the round-robin with quantum 10 to compute for the waiting times for each process.
Assuming a Round-Robin Scheduling works with a quantum time of 5, draw the timeline for CPU- and I/Obursts for the following three processes; A, B, C, with their arrival times, execution and I/O times. k=5 m=10 p=3
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();}
Two algorithms A and B report time complexities expressed by the functions n2and 2n , respectively. They are to be executed on a machine M that consumes 10–6 s to execute an instruction. What is the time taken by the algorithms to complete their execution on machine A for an input size of 50? If another machine N that is 10 times faster than machine M is provided for the execution, what is the largest input size that can be handled by the two algorithms on machine N? What are yourobservations?