Ajay Kumar, Author at All About Education

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Viewing 7 posts - 1 through 7 (of 7 total)

1. Ajay Kumar
  Member
  
  IISc uploaded the Defective gate applications details .check out soon
  
  Nov 30, 2011 at 1:46 pm
1. Ajay Kumar
  Member
  
  can u tell me how did u get this 3 as the correct answer ..
  OR in which book did u seen this answer? so that i will check there is one more formula
  related to this type of problem in GATE CS 1993 paper S<(m+n) where s = max requirement of all processes
  m = number of resources
  n = number of processes
  
  Nov 23, 2011 at 6:14 pm
1. Ajay Kumar
  Member
  
  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
  
  Nov 22, 2011 at 4:46 pm
1. Ajay Kumar
  Member
  
  @Anusha : here they are asking minimum number of R so its obviously 4 only in ur choice in the list (4, 5,6)
  
  Nov 22, 2011 at 4:38 pm
1. Ajay Kumar
  Member
  
  its simple formula :
  N*D<R+N where N= total number of processes , D= each process need
  R= resoures available
  so in this problem N= 'N' number of processes
  D=3 and R=6
  so N*3<6+N
  so 3N<6+N by solving this equation u will get
  2N<6 therefore N<3 so N should be less than 3 so N=2
  
  Nov 21, 2011 at 5:21 pm
1. Ajay Kumar
  Member
  
  u can use this formula for ur previous problem also…..
  
  Nov 21, 2011 at 5:12 pm
1. Ajay Kumar
  Member
  
  its simple formula :
  N*D<R+N where N= total number of processes , D= each process need
  R= resoures available
  so in this problem N= 'N' number of processes
  D=3 and R=6
  so N*3<6+N
  so 3N<6+N by solving this equation u will get
  2N N<3
  therefore N should be 2
  u can aply this formula for other GATE 1998 CS and GATE 2005 CS
  
  Nov 21, 2011 at 5:00 pm

Viewing 7 posts - 1 through 7 (of 7 total)

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Ajay Kumar

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