Consider the following heap :
JOB1-50 FREE-150 JOB2-300 FREE-350 JOB3-600
The sequence of requests for blocks of sizes 300, 25, 125, 50 can be satisfied if we use
A. | either first fit or best fit policy |
B. | first fit, but not best fit |
C. | best fit, but not first fit |
D. | none of these |
Option: B Explanation : Click on Discuss to view users comments. Adarsh said: (3:17pm on Monday 6th July 2015)
how is it possible? Can you explain?
jamal said: (4:52pm on Monday 8th August 2016)
yes adarsh the best fit can not be used here because if we used the best case than to bring the first job that 300 it will occupied in the free space 350 wile livening 50 free space similarly in the second the job 25 will occupied the free space 50 and livening 25 free space and similarly 125 job will go to 150 free space livening 25 free space ,than for the job 50 there will be no free space so best case is not possible...
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Cached and interleaved memories are ways of speeding up memory access between CPUs and slower RAM. Which memory models are best suited (i.e. improves performance the most) for which programs ?
(i) Cached memory is best suited for small loops.
(ii) Interleaved memory is best suited for small loops.
(iii) Interleaved memory is best suited for large sequential code.
(iv) Cached memory is best suited for large sequential code.
A. | (i) and (ii) are true |
B. | (i) and (iii) are true |
C. | (iv) and (ii) are true |
D. | (iu) and (iii) are true |
Option: B Explanation : Click on Discuss to view users comments. |
Consider following page trace :
4,3,2, 1,4,3,5,4,3,2, 1,5
Number of page faults that would occur if FIFO page replacement algorithm is used with Number of frames for the Job M=3, will be
A. | 8 |
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B. | 9 |
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C. | 10 |
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D. | 12 |
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Option: B Explanation : When M-3
Total Page Fault = 9 Click on Discuss to view users comments. |
Consider following page trace :
4,3,2, 1,4,3,5,4,3,2, 1,5
Percentage of page faults that would occur if FIFO page replacement algorithm is used with
Number offrames for the Job M = 4, will be
A. | 8 |
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B. | 9 |
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C. | 10 |
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D. | 12 |
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Option: C Explanation : When m=4
Total no of page fault = 10 Click on Discuss to view users comments. |
Consider a logical address space of 8 pages of 1024 words mapped into memory of 32 frames.
How many bits are there in the logical address?
A. | 9 bits |
B. | 11 bits |
C. | 13 bits |
D. | 15 bits |
Option: C Explanation : Logical address will have 3 bits to specify the page number (for 8 pages). 10 bits to specify the offset into each page (210 = 1024 words) = 13 bits. Click on Discuss to view users comments. |