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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 in Exam Application Kushagra गुप्ता 920 views
Total cache size $= 256\ KB$ Cache block size $=32\text{ Bytes}$ So, number of cache entries $=\dfrac{ 256\ K}{32}=8\ K$ Number of sets in cache $=\dfrac{ 8\ K}{4}=2\ K$ as cache is $4\text{-way}$ associative. So, $\log(2048) = 11\ bits$ are ... , we need $16\text{ tag bits}$ along with each cache entry to identify which of the possible $2^{16}$ blocks is being mapped there. Correct Answer: $C$
posted Feb 17, 2018 in From GO Admins Arjun 10,742 views
2*1*2*1=4 break it into determinant of smaller pieces where you are doing it in such a way that you should have to do lesser number of operations
posted Apr 30, 2017 in From GO Admins gatecse 394 views
by taking common I mean, use only one inverter, not two...and the circuit you designed above gives the cost of 65...
posted Nov 4, 2016 in Announcements Arjun 1,121 views
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