11:42HomeTake Test: FinalQuestion Completion Status:-If the base register is loaded with value 12345 and limitregister is loaded with value 1000, which of the followingmemory address access will not result in a trap to theoperating system?A) 12500B) 12200C) 13346D) 12344Click Save and Submit to save and submit. Click Save All Answers to save allanswers.Save All AnswersSave and Submitbbhosted.cuny.edu 11:41HomeTake Test: FinalQuestion Completion Status:Computing systems need cache because||--A) accessing main memory is slow and cachespeeds it up.B) register access is slow and cache speeds it up.C) main memory is expensive and cache offsets thecost.D) All of the about.Click Save and Submit to save and submit. Click Save All Answers to save allanswers.Save All AnswersSave and Submitbbhosted.cuny.edu

Ans-1Option a: This option is correct because This is the primary reason for using cache. It reduces…

Answered: 11:42 Home Take Test: Final Question…

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

Assignment description: When an operating system has been running for a while without reboot, and more it keeps allocating and de- allocating pages, the quicker the memory becomes fragmented. And the kernel may not always be able to defragment enough memory for a requested size on time. If that happens, applications may not be able to allocate larger contiguous chunks of memory even though there is enough free memory available. Being a system admin and considering the following scenario, discuss the current proposed solutions to solve the problem caused in this case and critically analyze the performance of the proposed solutions to solve the problem. Provide a report considering the following outlines. Scenario: In the following snapshot of running operating system, we can see that there is a free memory space equal to (55 KB). Assume that new process (process#07) needs to be run inside the memory and required 50 KB but the memory (fragment) is not contiguous. Assigned Space Assigned…
We want to build an address book that is capable of storing name, address & phone number ofa person. Address book provides functionality in the form of a menu. The feature list includes:• Add – to add a new person record• Delete – to delete an existing person record by name• Search – to search a person record by name· Exit – to exit from application
Memory Management: THE RETURN OF TERRY Terry is back. After failing miserably as a Tunnel designer, Terry went and found a nice easy office job, but all is not well. Terry's head has only 2048 kilobytes of memory. There is a notebook which Terry carries around that has 200 pages in it each capable of holding 256 Kilobytes of information. Your mission should you choose to accept it: HELP TERRY 1. Design a system for Terry using what you know of memory management which will allow them to use the notebook to organize information. Include in your description how you might implement addresses, and the mapping of those addresses (MMU operation) 2. Using this analogy, in a real memory management system, what would be the virtual memory size of the Terry/Notebook system.
Memory Address Memory Content Øx20000013 Øx79 Øx20000012 Øxcd Øx20000011 Øxa3 Øx20000010 Øxfd ØX2000000F Өx12 According to the memory view given below, LDMIB r0 if RO = 0x20000004! after executing the {R7, r3, r9} Command, which of the following are the R9 and ro register contents in order (data placement big endian)? Øx2000000E Øxod ØX2000000D Øxe0 Øx2000000c Øxeb ØX2000000B Øx2c ØX2000000A Оx1а Øx20000009 Øxdc Øx20000008 Өx11 Øx20000007 Øx34 Øx20000006 Øxce Øx20000005 Øxfd Øx20000004 Øxe4 Øx20000003 Оха2 Øx20000002 Øx3f Øx20000001 Øx06 Øx20000000 Øxf5 Lütfen birini seçin: O A. R9= 0X79CDA3FD r0 = Ox2000000C O B. R9= 0X11DC1A2C r0 =0x20000010 O C. R9= 0XEBE0OD12 r0 =0x2000000C O D. R9= 0X120DEOEB r0 =0x2000000C O E. R9= 0XFDA3CD79 r0 =0x20000010
Java Programming Language DSA Don't use built in libraries. We want to build an address book that is capable of storing name, address & phone number ofa person. Address book provides functionality in the form of a menu. The feature list includes:• Add – to add a new person record• Delete – to delete an existing person record by name• Search – to search a person record by name· Exit – to exit from application
TASK1: DESCRIPTION One of the simplest methods for memory allocation is to divide memory into several fixed-sized partitions. Each partition may contain exactly one process. In this multiple-partition method, when a partition is free, a process is selected from the input queue and is loaded into the free partition. When the process terminates, the partition becomes available for another process. The operating system keeps a table indicating which parts of memory are available and which are occupied. Finally, when a process arrives and needs memory, a memory section large enough for this process is provided. When it is time to load or swap a process into main memory, and if there is more than one free block of memory of sufficient size, then the operating system must decide which free block to allocate. Best-fit strategy chooses the block that is closest in size to the request. First-fit chooses the first available block that is large enough. Worst-fit chooses the largest available…
"Can you give an overview of CPU scheduling strategies?" reads the query.
LAB MAKEUP 1 Initial value of shared variable m is equal to 0. Four threads A, B, Y, Z use this variable as follows: Processes A and B increase the value of m by 1. Processes Y and Z decrease the value of m by 1. Since m is shared, each process protects the critical region using semaphore S initialized to 2. Run your program to show that it is possible that m's value can be equal 2 after all processes run once. S=2 Proses A W(S) m=m+1 S(S) ADD TEXT OF YOUR CODE HERE Proses B W(S) m=m+1 S(S) ADD PRINTSCREEN OF YOUR OUTPUT HERE Proses Y W(S) m=m-1 S(S) Proses Z W(S) m=m-1 S(S)
Safari File Edit View History Bookmarks Develop Window Help 31% O Mon 11:29 PM Not Secure - codecheck.io a 3.1 The. Error -. Error -. Error -.- Error -. Error -. Math 1. Math 1. C Your As. * Start P. O Joe Ro... Error -. Click on the buttons below to view all parts of the assignment. 1 (100%) 2 (0%) It is March 14, and the bakery is holding a pie sale. All pies with a regular price of more than $10 are $3.14 off. All others are $1.00 off. Write a program that reads the price of a pie and prints the discounted price. Complete the following file: piesale.cpp 1 2 #include #include 4 5 using namespace std; int main() { cout « "Original price: double price; cin >> price; double discounted_price; 7 8 « endl; %3D 10 11 12 13 14 { 15 } else 16 17 6,802 28
Github work cycle involves the following actions (Commit, Modify, Resolve, Pull, Push). Order these actions based on when they would be performed in normal work cycle. You may use an action more than once.
Real Time Scheduling: Select all statements below that are true EDF scheduling assigns the highest priority to a process with the smallest remaining time until its deadline. RM scheduling assigns a higher priority to processes with shorter periods. Rate Monotonic (RM) scheduling works by dividing the total amount of time available into an equal number of shares, and then each process must request a certain share of the total when it tries to start. If a process has period p, it is activated every p units of time. A real-time system is schedulable under Earliest Deadline First (EDF) when its overall CPU utilization is less than or equal to 1. A real-time system is schedulable under RM when its overall CPU utilization is less than 1.
Match the tasks specified with the types of conditional jump that would be invoked: Jump to label L2 if the [ Choose ] doubleword in memory pointed to by EDI is even Jump to label L1 if the memory [ Choose ] word pointed to by ESI equals Zero Jump to a label if unsigned EAX [ Choose ] is greater than EBX Jump to a label if signed EAX is [ Choose ] greater than EBX [Choose] Jumps Based on Signed Comparisons Jumps Based on Specific Flag Values Jumps Based on Equality Jumps Based on Unsigned Comparisons へ > >

SEE MORE QUESTIONS

Recommended textbooks for you

Database System Concepts

Computer Science

ISBN:

9780078022159

Author:

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

Publisher:

McGraw-Hill Education

Starting Out with Python (4th Edition)

Computer Science

ISBN:

9780134444321

Author:

Tony Gaddis

Publisher:

PEARSON

Digital Fundamentals (11th Edition)

Computer Science

ISBN:

9780132737968

Author:

Thomas L. Floyd

Publisher:

PEARSON

Database System Concepts

Computer Science

ISBN:

9780078022159

Author:

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

Publisher:

McGraw-Hill Education

Starting Out with Python (4th Edition)

Computer Science

ISBN:

9780134444321

Author:

Tony Gaddis

Publisher:

PEARSON

Digital Fundamentals (11th Edition)

Computer Science

ISBN:

9780132737968

Author:

Thomas L. Floyd

Publisher:

PEARSON

C How to Program (8th Edition)

Computer Science

ISBN:

9780133976892

Author:

Paul J. Deitel, Harvey Deitel

Publisher:

PEARSON

Database Systems: Design, Implementation, & Manag…

Computer Science

ISBN:

9781337627900

Author:

Carlos Coronel, Steven Morris

Publisher:

Cengage Learning

Programmable Logic Controllers

Computer Science

ISBN:

9780073373843

Author:

Frank D. Petruzella

Publisher:

McGraw-Hill Education

SEE MORE TEXTBOOKS