Final (Memory Management) Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. The following,____, describes the first memory allocation scheme. a.|Each program to be processed was loaded into secondary storage, then swapped into memory in parts| b.|Each program to be processed was partially loaded into memory, then granted more memory as needed| c.|Each program to be processed was allocated a portion of memory and could negotiate with other programs to access more memory| d.|Each program to be processed was loaded in its entirety into memory and allocated as much contiguous space in memory as it needed| ____ 2. In a single-user system, jobs are processed ____. …show more content…
Assume the Memory Manager receives a request for a block of 200. When the best-fit algorithm is used, ____ is the beginning address of the block granted by the Memory Manager. Beginning Address Memory Block Size 4075 105 5225 5 6785 600 7560 20 7600 205 10250 4050 15125 230 24500 1000 a.|6785|c.|10250| b.|7600|d.|15125| ____ 16. ____ is how memory is deallocated in a fixed partition scheme. a.|Memory Manager releases the block and combines it with another free block.| b.|Memory Manager immediately gives memory to another program.| c.|Memory Manager adds block to free list and removes it from busy list.| d.|Memory Manager resets the status of the memory block where the job was stored to “free.”| ____ 17. In a dynamic partition scheme, ____, is how the Memory Manager deallocates a block that is between
i) CPU :CPU is an imparted aset as most servers, for example, file servers do some
a. The CPU tells the RAM which address holds the data that the CPU wants to
assign process identification, allocate address space, initialize process control block, set appropriate linkage, create or expand data structure
Memory segmentation is the division of a computer's primary memory information into sections. Segments are applied in object records of compiled programs when linked together into a program image and when the image is loaded into the memory. Segmentation sights a logical address as a collection of segments. Each segment has a name and length. With the addresses specifying both the segment name and the offset within the segment. Therefore the user specifies each address by two quantities: a segment name and an offset. When compared to the paging scheme, the user specifies a single address, which is partitioned by the hardware into a page number and an offset, all invisible to the programmer. Memory segmentation is more visible
In assembly language, the programmer must take a microscopic view of a task, breaking it down into tiny subtasks at the level of what is going on in individual _
ii)Memory buffer register(MBR), which contains the data to be written into memory or which receives the data read from memory.
Question 4 (d): The special case of the problem when each resource is requested by at most 2 processes.
3. The __ Processing Unit is a hardware component capable of quickly drawing items to the screen. Graphical Processing Unit
5. (Page 34)Where are programs and data temporarily stored when there is not enough RAM to hold all the information it is processing? Swap file
3. If a necessary program is using too much of the system resources and bogging down other applications, what can you do to fix the problem?
In computer organization and architecture, memory is the most important part of a computer. Every computer must have its own memory. Memory represents two stable or semi-stable states that are representing 1 and 0. It is capable of being written to at least once and read multiple times. In this lab, we will learn briefly on two parts of memory only, which are RAM and ROM. As we know that a memory can either be non-volatile (as seen in a read-only memory) or volatile (used in random access memory).
data table which houses all the static variables necessary for the program to run is generated and stored in RAM [1]. A few benefits that exist when the variable data is stored this way include the fact that the variable data can be accessed easily by multiple programs or several times from the same program for reading [1].
On (Russinovich, 2009)when the system can’t handle page faults,the kernel and device drivers use non-paged pool to store data that might be accessed. The kernel enters such a state when it executes interrupt service routines (ISRs) and deferred procedure calls (DPCs), which are functions related to hardware interrupts. Page faults are also illegal when the kernel or a device driver acquires a spin lock, which, because they are the only type of lock that can be used within ISRs and DPCs, must be used to protect data structures that are accessed from within ISRs or DPCs and either other ISRs or DPCs or code executing
e) All the three assignments are to be completed by due dates (specified from time