# Ajay Kumar

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Viewing 7 posts - 1 through 7 (of 7 total)
1. IISc uploaded the Defective gate applications details .check out soon

1. 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

1. 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

1. @Anusha : here they are asking minimum number of R so its obviously 4 only in ur choice in the list (4, 5,6)

1. 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

1. u can use this formula for ur previous problem also…..

1. 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

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