Question: Internal fragmentation occurs when a
Answer:
some of the memory allocated to process remain unused,which happens if a process is allocated more memory than it needed
MCQ: Internal fragmentation occurs when a
Correct Answer:
A. memory area remain unused because it is too large to be allocated
Explanation:
some of the memory allocated to process remain unused,which happens if a process is allocated more memory than it needed
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Related Questions
- 1. memory fragmentation results in_______
- 2. In External fragmentation,merging of free memory areas using boundary tags.which of the following statement stands TRUE. <ol> <li>Boundary tags ,is a status descriptor for a memory area.</li> <li>It consists of an ordered pair giving allocation status of the area;whether it is free or allocated.</li> <li>Boundary tags are identical tags stored at the start and end of memory area.</li> <li>when an area of memory becomes free ,the kernel checks the boundary tags of its neighboring areas.</li> </ol>
- 3. A relation called the 50-percent rule holds
- 4. Which of following statements stands true of a relation called 50-percent rule. <ol> <li>the number of free area is half the number of allocated area ie. m=n/2</li> <li>it helps in estimating the size of the free list</li> <li>it also gives us method of estimating the free area in the memory at any time</li> <li>if sf is the average size of free area of memory,the total free memory is sf×n/2</li> </ol>
- 5. In________memory binding are changed in such manner that all free memory area can be merged to form a single free memory area.
- 6. Which of the following statements stands true for memory compaction method. <ol> <li>it involves movement of code and data in the memory.</li> <li>it is feasible only if computer system provides relocation register;the relocation can be achieved by simply changing the address in the relocation register</li> <li>it does not involves movement of code and data in the memory</li> <li>it does not involves use of relocation register</li> </ol>
- 7. Buddy system and power of 2 allocators leads to ___________
- 8. In the Binary buddy system the sizes of memory blocks are ___________ and separate free lists are maintained for blocks of ___________sizes.
- 9. Which of the following statement is true for buddy system allocators? <ol> <li>Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.</li> <li>No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.</li> <li> When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2<sup>i</sup> for the smallest value of i such that 2<sup>i</sup> ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.</li> <li> When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2<sup>i</sup>.if the list is empty, it checks the lists for block of size 2<sup>i+1</sup>.it takes one block off this list and splits it into two halves of size 2<sup>i</sup>.it put one of these blocks into the free list of size 2<sup>i</sup>,and uses the other block to satisfy the request.</li> </ol>
- 10. Which of the following statement is true for Power-of-2 allocators? <ol> <li>Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.</li> <li>No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.</li> <li> When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2<sup>i</sup> for the smallest value of i such that 2<sup>i</sup> ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.</li> <li> When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2<sup>i</sup>.if the list is empty, it checks the lists for block of size 2<sup>i+1</sup>.it takes one block off this list and splits it into two halves of size 2<sup>i</sup>.it put one of these blocks into the free list of size 2<sup>i</sup>,and uses the other block to satisfy the request.</li> </ol>
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