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Recent posts tagged preparation-schedule

5
Probability that buckets other than $1$ are selected $= \dfrac{n-1}{n}$ This should happen $k$ times and each of the $k$ events are independent so $\dfrac{(n-1)^k}{n^k}$ When $k=1,$ probability of no collision $= \dfrac{n}{n}$ (for only one insert, there can ... $k>n+1$ probability of first collision $= 0$ (as it should have definitely happened in one of the previous $(n+1)$ insertions.
posted Nov 23, 2018 in Operating System Manoja Rajalakshmi A 526 views
11
$\text{Size of each segment} = \frac{2^{16}}{8}=2^{13}$ Let the size of page be $2^k$ bytes We need a page table entry for each page. For a segment of size $2^{13}$, number of pages required will be $2^{13-k}$ and so we need $2^{13-k}$ page table entries. Now ... $\text{Size of each page table entry} = 2$ bytes $=16$ bits $\text{Number of bits left for aging} = 16-12 = 4$ bits
posted Sep 17, 2018 in Programming & Data Structures Manoja Rajalakshmi A 1,385 views
14
Ans: because we have given wavy edges form MST So, for verification of option A we have to check that with MST how many cost to reach at a->b then we will get a->e->d->b = -2+5+3 = 6 so in given option a with cost(a,b)>= 6 this is posible coz , cost ... making MST so cost must be >6 not equal to 6 so option A is Need Not HOLD. Like wise if you check for other option then enequality is holding...
posted Aug 8, 2018 in From GO Admins Arjun 12,889 views
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