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1
Formula for no.of calls in f(n) =2f(n)-1 when f(0)=1 only applied fibonacci series where f(0)=1 f(1)=1 f(2)=2 i,e (f(0)+f(1)) f(3)=3............f(7)=21 so no.of calls 2(21)-1=41 remember when f(0)=0,f(1)=1 then formula changes as 2f(n+1)-1
posted Mar 20, 2020 in Exam Application KUSHAGRA गुप्ता 1,139 views
2
When window scaling happens, a 14 bit shift count is used in TCP header. sir why 14 bit only ?? any reason or just an assumption ??
posted Mar 14, 2020 in Others Dee 350 views
5
Sir, can we do like this Let there be 4 process P,Q,R,S and 5 resources. Let they be allocated one resource each initially.Now after allocation need of P is 1 , need of Q is 1, need of R is 2 and need of S is 2. Now only 1 resource is available as 1 has been ... to R.....and hence R could complete.....Now available becomes 4. We can now allocate 2 resources to S and hence S can complete......
posted Jul 2, 2018 in Project Work Arjun 1,355 views
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Instead of finding a function 'f' for the first MUX (in terms of z,x and y) and the second MUX (in terms of f , x and y), we can logically find out the answer.We can put in 4 values of A and B and analyse the output C. We can get the values appropriate values of x and y by drawing and comparing with a 2x1 MUX when A and B are given as inputs respectively.
posted May 4, 2018 in Preparation Advice Shikha Mallick 1,127 views
7
Super Key is any set of attributes that uniquely determines a tuple in a relation. Since $E$ is the only key, $E$ should be present in any super key. Excluding $E$, there are three attributes in the relation, namely $F, G , H$. Hence, if we add $E$ to any subset of those three ... the answer is $8$. The following are Super Keys: $\left \{ \substack{E\\EF\\EG\\EH\\EFG\\EFH\\EGH\\EFGH} \right \}$
posted Dec 10, 2017 in Job Openings Chhotu 739 views
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