Question: 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>
Answer:
Boundary tags ,is a status descriptor for a memory area.
It consists of an ordered pair giving allocation status of the area;whether it is free or allocated.
Boundary tags are identical tags stored at the start and end of memory area.
when an area of memory becomes free ,the kernel checks the boundary tags of its neighboring areas.
MCQ: 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>
Correct Answer:
A. 1,2 only
Explanation:
Boundary tags ,is a status descriptor for a memory area.
It consists of an ordered pair giving allocation status of the area;whether it is free or allocated.
Boundary tags are identical tags stored at the start and end of memory area.
when an area of memory becomes free ,the kernel checks the boundary tags of its neighboring areas.
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Related Questions
- 1. A relation called the 50-percent rule holds
- 2. 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>
- 3. In________memory binding are changed in such manner that all free memory area can be merged to form a single free memory area.
- 4. 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>
- 5. Buddy system and power of 2 allocators leads to ___________
- 6. In the Binary buddy system the sizes of memory blocks are ___________ and separate free lists are maintained for blocks of ___________sizes.
- 7. 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>
- 8. 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>
- 9. The buddy and power of 2 allocators are faster than the first-fit ,best-fit, next-fit allocators
- 10. The power of 2 allocators is faster than the Buddy System allocators.
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