| A. | 60 sec |
| B. | 30 sec |
| C. | 45 sec |
| D. | 10 sec |
|
Option: C Explanation :
T = N instr x 1µs + 15,000 x 2,000 µs = 60s
N instr x 1µs = 60,000,000 µs – 15,000,000 µs = 30,000,000 µs
N instr = 30,000,000
The number of instruction between two page faults is
N instr /N PageFaults = 30,000,000/15,000 = 2,000
If the mean interval between page faults is doubled, the number of instruction between
two page faults is also doubled and is 4,000. Now the number of page faults is
30,000,000/4,000 = 7,500
T’ = 30,000,000 µs + 7,500 x 2,000 µs
= 30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s
Doubling the memory, doesn’t mean that the program runs twice as fast as on the first system. Here, the performance increase is of 25%.
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Consider a disk with 16384 bytes per track having a rotation time of 16 msec and average seek time of 40 msec. What is the time in msec to read a block of 1024 bytes from this disk?
| A. | 57 msec |
| B. | 49 msec |
| C. | 48 msec |
| D. | 17 msec |
|
Option: B Explanation :
Time in msec to read a block of 1024 bytes (Access time or Disk Latency) = seek time +average rotational delay + transfer time
If there are 16384 bytes per track there are 1024/16384 tracks to be read for this block.
Seek time = 40 msec
Rotational delay = 16 msec
Transfer time = (sectors_read/sectors per rev.) x rotational delay
= (1024/16384) x 16 = 1
average rotational delay = rotational delay/2 = 16/2 = 8
access time = 40 + 8 + 1 = 49
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| A. | contrab |
| B. | init |
| C. | gettydefs |
| D. | inittab |
|
Option: C Explanation : Click on Discuss to view users comments. |
A data cube C, has n dimensions, and each dimension has exactly p distinct values in the base cuboid. Assume that there are no concept hierarchies associated with the dimensions. What is the maximum number of cells possible in the data cube, C?
| A. | pn |
| B. | p |
| C. | (2n -1)p+1 |
| D. | (p+1) n |
|
Option: D Explanation :
the maximum number of cells possible in the base cuboid - pn
This is the maximum number of distinct tuples that you can form with p distinct values per dimensions.
the minimum number of cells possible in the base cuboid -p
You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.
the minimum number of cells possible in the data cube, C(2n -1)×p+1.
The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.
the maximum number of cells possible (including both base cells and aggregate cells) in the data cube, C (p+1) n
The argument is similar to that of part (a), but now we have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.
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Suppose that from given statistics, it is known that meningitis causes stiff neck 50% of the time, that the proportion of persons having meningitis is 1/50000, and that the proportion of people having stiff neck is 1/20. Then the percentage of people who had meningitis and complain about stiff neck is:
| A. | 0.01% |
| B. | 0.02% |
| C. | 0.04% |
| D. | 0.05% |
|
Option: B Explanation :
The computation is based on the simplified Bayes’ formula.
P{B|A} = (P{A|B}·P{B) / P{A}.
P{M|S} = probability that a person had meningitis, conditioned by the existence of stiff neck.
P{S|M} = probability that a person complains about stiff neck, conditioned by the existence of meningitis. = 50%=1/2
P{S} = proportion of people who complain about stiff neck. = 1/20
P{M} = proportion of people who had meningitis. = 1/50,000
Then:
P{M|S} = (P{S|M}·P{M}) / P{S} =( 1 / 2 x 1 / 50,000 ) / 1 / 20 = 0.0002 = 0.02%
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