In the cache memory three types of misses are present. 

(1) compulsory misses ( cold-start misses/ first-reference misses)---- this type of miss every first reference in cache memory  itself miss and  cache is initially miss. therefore it is called first-reference miss / compulsory miss/cold-start miss.

(2) conflict misses ( collision misses/ Inference misses)---- this miss will occur when the to many blocks are mapping into the same cache line.( it is consider only in the direct mapping technique or set -associative mapping technique but in full associative mapping tech. it is completely eliminated.)

(3) capacity misses-this type of miss due to the small size of cache memory .it can't allocate all the possible memory block at a time .therefore capacity misses will occur.

 In direct mapping technique---

(a) conflict misses reduce hit rate.

(b) required less area( only one comparator and fewer tag bits required)

But In fully -associative mapping technique----

(a) no conflict misses, therefore generally higher hit rate.( Usually have fewer misses.)

(b) need one comparator for each line in the cache.

According to the question fully- associative mapping technique given therefore 

total misses = capacity misses + compulsory misses= 10% +30%=40%

then find avg. access time = hit rate*cache memo.access time + miss rate*(cache memo.access time + main memo.access time)

                                    = 60%*10 + 40%*( 10+100) ns

                                    = 50 ns is right answer.

 

 

21 Nov 2014 - 6:23pm

Allotment of resources would be successful to all cases but in option

(B) where n = 21 & k = 12.

Let use take i = 10

So for P10 only R10 can be allotted whereas for P11 odd case (else).

R21 −11 i.e R10 can only be allotted.

This is an conflict.

Hence (B) is correct option.

20 Nov 2014 - 11:30am

question-consider the languages L1 , L2 and L3 as given below

L1={ 0p1q| p,q € N }

L2={0p 1q|p,q € N and p=q }

L3={ 0p 1q 0r| p,q,r € N and p=q=r}. which of the following statements is not true?

(a) PDA or DPDA can be used to recognize to L1 and  L2

(b) L1 is a regular lang.

(c) all the three lang. are CFL

(d) Turing m/c can be used to recognize all languages

answer-----

L1 -  If in a language having no any comparison or linear or having zero memory ( having no stack accept by FA M/C or DPDA M/C) here p,q belongs to natural no. and having no any comparison. therefore language L1 is regular.

L2- If in a language having one comparison or one stack is required ( accept by PDA M/C OR DPDA M/C ).

language will be CFL . In given L2 lang p,q belongs to N and one comparison between p and q i.e p=q. therefore L2 is CFL.

L3 -If in a language having two or more comparison and  two or more stacks are required (accepted by NPDA M/C OR LBA M/C). In given L3 language comparison between p, q and r will be p=q,p=r, q=r, p=r  ( 4 comparisons) . Therefore L3 language is CSL.

SO, choice(C) is correct a/c to question.

 

 

 

19 Nov 2014 - 12:55pm

Brother NPTL( IIT Madras lectures BY Professor kamala krithivasan)

19 Nov 2014 - 11:14am

yes in given question B-tree not B+-TREE

19 Nov 2014 - 11:03am

if(fork()==0) ( child process)

let assume a= x and &a = y

else part ( parent process)

let assume a=u and &a = v

let a=10

child process

a= a+5

a=10+5=15

therefore x=15

x=u+10

parent process

a=a-5

a=10-5=5

therefore u=5

hence choice (c) x=u+10 and v=y is correct.

16 Nov 2014 - 11:15am

lot of thanks my brother Sameer !!!!!

 

16 Nov 2014 - 11:01am

yes but sameer brother plz explain 

16 Nov 2014 - 8:05am

nice explain pritam brother

15 Nov 2014 - 5:45pm

see book PATTERSON HENNESSY ( 4TH EDITION) PAGE NO. 472

15 Nov 2014 - 4:57pm

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