assgn 2 - (1 R AC-CB(1 R AD-DB(1 R AE-EF-FB R parallel...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
world B A C D p F E A B C D ASSIGNMENT 2 10) Given a system with 3 parallel links b/w 2 routers with each link having an independent reliability of p. To find reliability for a system with 3 parallel links calculate- R=1-(1-p) 3 We need to assign a value for p such that R=0.999 0.999 = 1- (1-p) 3 (1-p) 3 = 0.001 p =0.9
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
The Independence assumption is false in the case when all 3 links in the same cable duct as all 3 links would not function if the cable is disconnected and therefore the system would go down. 11) R private n/w to G =1-(1-p) 2 (parallel configuration) R G =q R G to world =r Total reliability R= R private n/w to G * R G * R G to world (series configuration) R series =(1-(1-p) 2 )*q*r 13)
Background image of page 2
q p p p p p p The system is up if at least one of the links are up. R AB =q R AC-CB =p 2 R AD-DB =p 2 R AE-EF-FB =p 3 Calculate the total Reliability R of the system(at least one link to be up) R parallel =1-(1- R AB
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: )(1- R AC-CB )(1- R AD-DB )(1- R AE-EF-FB ) R parallel =1-(1-q)(1-p 2 ) 2 (1-p 3 ) 14) 100 100 100 100 100 100 100 100 100 100 100 100 The problem states that there should be at least 200mbps flowing b/w A and BB through either the upper link or the lower link for the system to function. This means that at least 2 of the 3 links on either of the paths should be up For the upper path AC the reliability is given as R AC = ∑ 3 C i p i (1-p) 3-i for i=2 to 3 For CB reliability is given as R CB = ∑ 3 C i p i (1-p) 3-i for i=2 to 3 For the lower path AD the reliability is given as R AD = ∑ 3 C i p i (1-p) 3-i for i=2 to 3 For DB the relability is given as R DB = ∑ 3 C i p i (1-p) 3-i for i=2 to 3 Therefore the Total Reliability is R=1-[(1-R AC R CB )*(1-R AD R DB )]...
View Full Document

{[ snackBarMessage ]}

Page1 / 5

assgn 2 - (1 R AC-CB(1 R AD-DB(1 R AE-EF-FB R parallel...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online