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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
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
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
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
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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