From set10
Printed From: One Stop GATE
Category: GATE Technical Discussions
Forum Name: GATE CS
Forum Discription: General Technical Discussions, Queries, doubts etc. for GATE in CS.
URL: http://forum.onestopgate.com/forum_posts.asp?TID=124
Printed Date: 10Mar2025 at 2:15am
Topic: From set10
Posted By: Vimal
Subject: From set10
Date Posted: 02Feb2007 at 11:45am
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
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Replies:
Posted By: Ajay
Date Posted: 02Feb2007 at 2:59pm
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| Quote: |
| 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. |
Quicksort with random pivots takes O(n log n) expected running time, but its worst-case running time is in Theta(n^2).
| Quote: |
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
I think it should be a + b + d
which one is correct??? |
yes mee too agree with your answer
| Quote: |
| q11. Given that (1,2,3) ia an eigen vector for the matrix |
x+ 6 - 3 =0
x = -3
but x = 2-eigenvalue
so EigenValue = 3+2 = 5
| Quote: |
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...!! |
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
| Quote: |
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???? |
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
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_________________
Regards
Ajay Kumar
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