Answer the given question with a proper explanation and step-by-step solution.operating system The arrival time of the process P1, P2, and P3 and their CPU-Burst time (unit time) are shown inthe following table. Answer the following questions:04-beta....isoProcessP1P2akhiP3WLAN{$papi/pn/43WATArrival TimeWITH03When using FIFO scheduling algorithm,(1) The turnaround time of P1 is:(a)20(b)(2) The turnaround time of P2 is:(a)20(b)(3) The turnaround time of P3 is:PUT(b)(a) 20(4) The average turnaround time is:(a) 20(b)232323CPU-Burst Time23203(c)(c)(c)(c)24242424(d)(d)(d)(d)22.322.322.322.3

A common scheduling principle, first-in-first-out (FIFO) can be simply implemented in either…

The arrival time of the process P1, P2, and P3 and their CPU-Burst time (unit time) are shown in the following table. Answer the following questions: Process PI P2 P3 Arrival Time 0 3 When using FIFO scheduling algorithm, (1) The turnaround time of P1 is: (a) 20 (b) (2) The turnaround time of P2 is: 20 (b) (3) The turnaround time of P3 is: (a) 20 (b) (4) The average turnaround time is: (a) 20 (b) 23 23 23 23 CPU-Burst Time 20 4 3 (c) (c) (c) 24 24 24 24 ê (d) 22.3 (d) (d) (d) 22.3 22.3 223

The arrival time of the process P1, P2, and P3 and their CPU-Burst time (unit time) are shown in the following table. Answer the following questions: Process PI P2 P3 Arrival Time 0 3 When using FIFO scheduling algorithm, (1) The turnaround time of P1 is: (a) 20 (b) (2) The turnaround time of P2 is: 20 (b) (3) The turnaround time of P3 is: (a) 20 (b) (4) The average turnaround time is: (a) 20 (b) 23 23 23 23 CPU-Burst Time 20 4 3 (c) (c) (c) 24 24 24 24 ê (d) 22.3 (d) (d) (d) 22.3 22.3 223

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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3. Job Scheduling: Consider the problem of scheduling n jobs of known durations t1, t2, ..., tn, for execution by a single processor. The jobs can be executed in any order, one job at a time. You want to find a schedule that minimizes the total time spent by all the jobs in the system. (The time spent by one job in the system is the sum of the time spent by this job in waiting plus the time spent on its execution.) Design a greedy algorithm for this problem. Does the greedy algorithm always yield an optimal solution?
The CPU clock is 1 GHz. One OP requires 5 clock cycles (for example arithmetic instruction and conditional). Each memory access (read or write) requires 100 clock cycles and the problem size is 1 million or n = 1,000,000. There is no memory access cost of loop index variable: don't count any arithmetic instructions for initializing or incrementing the loop index value. Assume the loop index variable is free of charge. There is an arithmetic operation involved in updating the accumulator variable. Algorithm Sum = 0 For i in range (0, n); Sum += i; Print (“sum =, sum) How many memory accesses and clocks are required by this algorithm?
The CPU clock is 1 GHz. One OP requires 5 clock cycles (for example arithmetic instruction and conditional). Each memory access (read or write) requires 100 clock cycles and the problem size is 1 million or n = 1,000,000.There is no memory access cost of loop index variable: don't count any arithmetic instructions for initializing or incrementing the loop index value. Assume the loop index variable is free of charge. There is an arithmetic operation involved in updating the accumulator variable. AlgorithmSum = 0A = list (range(0,N)) //vectorFor i in range (0, n);Sum += A[i];Print (“sum =, sum) In this algorithm how many memory accesses and clocks are required?
In a timesharing OS we have the following cpu timeline for two tasks X and Y. The timeslice is 1s.Both tasks are available in the system at the same time t=9:00:00.000 and order of arrival is the obviousX followed by Y. (The decimals reflect milliseconds if they showup in an indicated time reference.) There are no other processes (tasks) in the system other than X,Y. 1234567890 XYXYX--YXY The time line 1 indicates that at t=9:00:00s task X starts its execution and when t=9:00:01s is reached task Y takes over. The time line 1 indicates the 'first second' and time line 0 indicates the 'tenth second' above.Task Y completes its execution at t=9:00:10s, the completion of the tenth second since X started its execution. Task X has completed its execution earlier. (a) What is the total number of context switches starting from prior to t=9:00:00s (e.g. t=8:59:59.999) through thecompletion of $Y$? answer is 16s (b) What is the turnaround time for Y? answer is 10s (c) What is…
Let the processes P1, P2, P3, and P4 be given. They arrive in the system at the same time in this order. The processes have the following service times (in time units): Process P1 P2 P3 P4 Service time (CPU burst) 53 17 68 24 For the scheduling method "Round Robin" (RR), specify the average execution time (average turnaround time) and the number of context switches. The time quantum q is set to 20 time units. You must show your calculations. You will not have to draw Gantt charts. Note: For RR, processes execute in order of their ID (i.e., P1-P2-P3-P4).
Consider what happens when the CPU performs a load operationsuch as movq A,%rax
Q1: Consider two computers, P1 and P2, of the same instruction set (ISA). We have a program of 7.5x10° instructions and we want to run this program on P1 and P2. The P1 computer is a 5 GHz machine with CPI of 0.8. The P2 computer is a 6 GHz machine with CPI of 1.2. Which computer is faster?
Let the processes P1, P2, P3, and P4 be given. They arrive in the system at the same time in this order. The processes have the following service times (in time units): Process Service time (CPU burst) exe P1 53 und P2 17 P3 68 P4 all 24 ur ca ulations. TOU arts. For the scheduling method "Round Robin" (RR), specify the average execution time (average turnaround time) and the number of context switches. The time quantum q is set to 20 time units. You must show your calculations. You will not have to draw Gantt charts. Note: For RR, processes execute in order of their ID (i.e., P1-P2-P3-P4).
In Assembly code: We have seen that multiplication by a power of two is easy with binary numerals: Tomultiply a number by 2^p corresponds to p left shifts. This works both for signed andunsigned numbers as long as we are careful with the range.Below is an attempt on a subroutine to multiply by a positive power of two. As noted inthe contract, the assumption is that p is a strictly positive integer (i.e., 1, 2, 3 etc.). What happens when this subroutine is called with R5 = 3 and R6 = 11? Give a shortbut precise answer at the register level of the MCU
Refer to the code below determine whether this code is correct, and if not, correct any mistakes. Find the speedups and efficiencies of the parallel odd-even transposition. Does the program obtain linear speedups? Is it scalable? Is it strongly scalable? Is it weakly scalable? After correcting the mistakes, execute the code in CUDA or OpenMP and measure any performance improvement in terms of speedups and scalability. Process Pi evenprocess =(i%2 ==2); evenphase = 1; for(step =0; step < n; step++,evenphase = !evenphase) { if((evenphase && evenprocess) || (!evenphase && !evenprocess) { send (&a, Pi+1); recv (&x, Pi+1); if (x<a) a =x; } else { recv (&x, Pi-1); send (&a, Pi-1); if (x > a) a = x; } }
Please refer to this textbook: “A. Silberschatz, P. B. Galvin and G. Gagne, “Operating System Principles,”7th Edition, John Wiley & Sons Inc., 2006.” And answer the following questions: Question7: A practical way to handle network partitioning is to: (a) Have each process start an election. (b) Elect a leader only if a majority of nodes can be reached. (c) Run a normal election and multicast the result to the subset of processes that can be reached. (d) Divide and conquer: elect multiple coordinators.
Write a code in sim8085 for the following problem: The pressure of two boilers is monitored and controlled by a microcomputer works based on microprocessor programming. A set of 6 readings of first boiler, recorded by six pressure sensors, which are stored in the memory location starting from 2050H. A corresponding set of 6 reading from the second boiler is stored at the memory location starting from 2060H. Each reading from the first set is expected to be higher than the corresponding position in the second set of readings. Write an 8085 sequence to check whether the first set of reading is higher than the second one or not. If all the readings of first set is higher than the second set, store 00 in the ‘D’ register. If any one of the readings is lower than the corresponding reading of second set, stop the process and store FF in the register ‘D’. Data (H): First set: 78, 89, 6A, 80, 90, 85 Second Set:71, 78, 65, 89, 56, 75