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 One Stop GATE Forum : GATE Technical Discussions : GATE CS |
 Topic: From set10 |
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Vimal
Newbie 
Joined: 02Feb2007 Online Status: Offline Posts: 26 |
   Topic: From set10Posted: 02Feb2007 at 11:45am |
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q38 The sorting algorithm randomized quicksort for n times has worst case running time of O(nlogn).....I think it is true...suggest me plz.
q67 If a=Pkt trans time b=token trans time c=ring latency d =prop delay b/w 2 stations then the time taken by the token to make it available to the next station in early token strategy is a+b+c+d a+b+d a+c+d a+b+c which one is correct??? q11. Given that (1,2,3) ia an eigen vector for the matrix q33 50 sub-networks are to be created from a given class B network 150.193.0.0 and each subnet is expected to have 750 hosts of assorted IP addressable network devices.calculate the required subnet mask. 255.255.0.0 255.255.253.0 150.193.252.9 none the given ans C(why???) 252 means 2^6 but there are 750 hosts...!! q44 A computer system has a word-addressable main memory consisting of 256K 16 bit words.It also has a cache organized in a set associative manner with 4 block frames per set and 64 words per block.. The no of tag bits 8 18 6 none...// givn answer 8 my doubts is this.. there are 4 sets an each set consists of 4 bl..so the set field is 2^3 i.e;8 is here tag=set ...is it always true for any case???? 2 3 -1 3 2 1 2 2 3 Find the corresponding eigen value ..the given answer is 5...how it is possible?? my approach is.. 2 3 -1 1 5 3 2 1 * 2 = 10 2 2 3 3 15 Post Resume: Click here to Upload your Resume & Apply for Jobs |
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Ajay
Newbie
Joined: 02Feb2007 Online Status: Offline Posts: 1 |
Posted: 02Feb2007 at 2:59pm |
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Quicksort with random pivots takes O(n log n) expected running time, but its worst-case running time is in Theta(n^2).
yes mee too agree with your answer
x+ 6 - 3 =0 x = -3 but x = 2-eigenvalue so EigenValue = 3+2 = 5
2^6 means 64 subnets. so the subnet masx x.x.252.x remaining 10 bits for the hosts so max possible host = 2^10 -2 in each subnet i.e 1022
256 K = 2^18 Cache Block size = 64 = 2^6 Total Cache Size = 4K = 2^12 Tag bit in case of Direct Mapped Cache would be 12-6 = 6 but since we have 4 way set associatve add 2 more bits to Tag hence 8 [/quote] _________________ Regards Ajay Kumar Edited by Ajay - 02Feb2007 at 3:00pm |
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