@Harsh,
What type of discussion you would like to have?
e.g.
Study Methodology?
Resources?
GATE questions discussion?
Other?

Also, lets try to make such discussions open on I n I instead of email so that it will be helpful for everybody who needs it.

What do you think?

@Kapil : First of all thanks of bringing this forum idea up.

@Harsh : Buddy I’d suggest you create a local group in your city/locality for GATE preparations.

Hey I just made my first post on this forum! This is Aditya, PM10 batch for CSIT at ACE Academy, Hyderabad

Welcome on board Aditya.

Soon I am posting an interview with Nikhil Krishnan AIR 4 GATE 2011 ECE. He gave good idea about GATE study group. Hope that will help you all.
Just pray that I dont become lazy 🙁

I did it finally 🙂
Do check Nikhils interview at https://www.inspirenignite.com/how-to-prepare-for-gate-nikhil-krishnan-air-4-gate-2011-part-1/

Hope this give you all a direction..
Good Luck you all…

Interested..
Me too having no coaching or any good guide.. so dependent on net

HI FRIENDS THOSE WHO ARE BELIEVING IN SELF STUDY PLZ CHECK THIS PAGE HERE U GET ALL KIND OF INFO ABOUT HOW TO PREPARE FOR GATE,WHAT ARE THE STRATEGY AND TRICKS , DIRECT FROM TOPPERS, CHECK EACH BLOG CAREFULLY
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http://www.facebook.com/pages/GATE-Computer-Science/287782614585176

free.the question is

a computer has 6 tape drives with N process competing for them.. each process need 2 drives… then the max value for N to b deadlock free is??????
plzz temme hw to find solution for this

I answered your question in the other thread anusha. 🙂

An operating system contains 3 user processes each requiring 2 units of resource R .The minimum number of units of R such that no deadlocks will ever arise is
ans is given as 4 . i want the solution to solve this??????

You can use the formulae told by Ajay or you can think as follows:
3 user processes reqiuring 2 units of resource each. Meaning 3*2=6 resources will always suffice. But is it the minimum?? Lets try with 5..
If we have R=5, then 2 of the 3 process can hold their total requiremnt of 2 units, and 1 unit will be still remaining, which can be acquired by the 3rd process, now 3rd process will wait for another unit to get free. But since the other two processes have already got their 2 units, they can complete and release their resources, as soon as they complete, 4 resources will be freed and 3rd process can take any 1 of the 4 newly freed resources, and this way 3rd can also complete. So R=5 is alaso, but again, is it the minimum??
Letstry with 4… If each of the 3 process are holding 1 resource, their will be 1 recource left which any of the 3 can acquire and complete its execution and release its 2 units of resources, then the other two also complete… So 4 is the minimum R till now.
Lets try with R=3, IF each process is holding 1 resource then all the 3 units will be helded by some process and their will be no more units left, now each process will wait for 1 more resource to complete, which they will never get because of the hold and wait condition.. Thus 3 units of resource wont suffice here.
Thus 4 is the right answer.
Lol, If u r unable to understand this logic u can use the formulae as well.. i knew m bad at explaining things 😀

if we apply the formulae in the other link

then 3*2<3+R
6<3+R
3<R
SO CHOICES ARE GIVEN AS FOLLOWS 4,5,6
the above three options are satisfies so how can we knw 4 is the right answer

Anusha : here they are asking minimum number of R so its obviously 4 only in ur choice in the list (4, 5,6)
Any way its my own formula. it is not in any text book . u can apply and check and reply back

@lorin ahmed
i am unable to understand the logic u are trying to tell… can u plzz tell me in imple way..plzz

@lorin ahmed

i am unable to understand the logic u are trying to tell… can u plzz tell me in imple way..plzz