pankaj.chawla

Answer is B since more subroutines means more branches which means more pipiline stalling.

ya my mistake. aneways thats great news. thats something to be really proud of :)

as i said earlier there can be 0,1.... or n exchanges in first pass. so the number of possibilities is (n+1) from 0 to n. mutliplying every possibility by its probability(rhymes here, huh!) which is 1/(n+1) because every case is equally likely to happen, we get the formula >> 1*1/(n+1) + 2*1/(n+1)....... n*1/(n+1) = (n*(n+1)/2)*1/(n+1) = n/2

take a look at this page>> http://en.wikipedia.org/wiki/One's_complement#Excess-N in general n is added to every number. for example 3(011 in binary) is added to every number in excess-3 notation. so to get the original exponent u hv to subtract 3 from it. for example if the exponent given is 011 it will be equal to 0 giving u 2^0 as the answer in base 2. i read somewhere that it is done to make sure that every exponent is representated as a positive quantity.

nevermind. its good to share some light moments while doing something so intense :)

atlast someone replied :) here is my input >> 12) when calculating probability, u consider all the cases. i mean to say that in the first pass 0 or 1 or 2 or 3..... or n exchanges will take place and all cases are equally likely. so the average number of exchanges in first pass will be 1*1/(n+1) + 2*1/(n+1)....... n*1/(n+1) = (n*(n+1)/2)*1/(n+1) = n/2 so the probability of exchange on first pass will be (n/2)*1/n = 1/2 now i dont know how to generalize it for ith pass but from the answer choices u can eliminate a) and d). the answer cannot be c because probability will decrease with increasing i which is not so in c) so the answer has to be b). even if u dont know how to calculate parobability, u can always make a good guess. probability will always decrease with increasing i so eliminate a) and c). now probability of exchange being 1 on first pass is unlikely while being 1/2 is more probable so the answer has to be b). 10) pavelbuet is correct. u hv to make all the parse trees. 13)pavelbuet is correct. 11) if option d) is n(log n) / log n*log n = n/log n. answer is d otherwise answer is b. pavelbuet d) is not decreasing but its growth rate is less than n. 15)pavelbuet is right. there is something wrong. if question was to find which is correct then the answer would hv been a).

maybe becoz we dont consider the process to be in the middle of the list of processes. u cannot say that a process's execution time started when it was 3rd or 4th in the list of processes.

well i think that n*T came from the following logic>> the process needs T sec to complete in the absence of competition. now with n processors in total every process will get 1 instruction executed in n seconds. so total time taken by a process is n*T. in short>> execution time when no competition = T secs when competition, 1 instruction = n secs so total time when competition = n*T secs the above solution considers that the process entered in the queue and it was the last process. smthing like this>> p1 p2 p3 | p1 p2 p3 | p1 p2 p3| p3 entered the queue and it took a total of n*T time to get it executed. if u consider the process as the first one in the queue u will arrive at the answer n*(T-1) + 1. now i dont know which answer to consider as the correct one.

can anybody give me a link to it or upload the material somewhere else. i m unable to find the link to it on http://www.gter.net

hey pavelbuet plz upload it on rapidshare. megaupload says all download slots are in use. btw u cant see anything or u cant understand what is being displayed. i m asking this becoz i too hv some files which i cant understand.

whatever u do it will always involve guesswork and that exactly is what i dont want to do. therefore i want to find a proper method to it.

i agree that it has a quadratic convergence but it is not exactly n^2. i mean that quadratic convergence does not imply that in the first iteration accuracy is upto 1 decimal place, in the second upto 4 decimal places and so on. also it depends on the approximate zero that we choose. there has to be a method to it. although in this question i think no other option would fit.

Newton Raphson method is used to calculate the square root of 2 using the iteration x(i+1) = (x(i) + 2/x(i))/2 if the intial value x(0) is chosen as 1.5, then the number of iterations needed to get an accuracy of 10^(-14) are (a) 2 (b) 4 © 8 (d) 16 (e) 32 again i dont know the correct answer. plz tell me how to solve this one.

i guess u r right...... my mistake :)

we cannot consider a*(b+c)* since it will violate the above restriction so only one a can appear in the string. if u still dont undertsand take an example suppose if u consider the string "aabcbc" formed from a*(b+c)* it will be followed by an a so it will be invalid. ya i m still an undegrad :)