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Get the correct site link cost - CBTNuggets vs Formula?
nms
Registered Users Posts: 1 ■□□□□□□□□□
Hey guys, I'd like to ask your help about what's the correct (I mean the one which is correct on the exam, not necessarily in real life 
) way to count the site link cost number?
Let's say you have three links:
A-B: 1.54Mbit
B-C: 512Kbit
A-C: 128kbit
By James in the CBTNugget 70-297 he says that you have to count the difference like
- take the fastest link as base, 1.54Mbit -> Cost=100 (as this is the default value)
1) the link (512Kbit) between A-B is approximately 3x slower -> Cost=300
2) the link (128Kbit) between A-C is approximately 12x slower -> Cost=1200
In the nugget the base cost link is lower (5), but the AD console allows the cost to set up to 99999, so theoretically this way can work.
On the other hand I know the formula:
Cost=1024/log(available bandwidth in Kbit)
Taking the above example it would give these numbers:
1) 1.54Mbit(1577KBit) = 1024/3.1978316933289028653690938277842 = 320.21697769028887990880037685061
2) 512Kbit = 1024/2.7092699609758307569236500525204 = 377.96159657385100491324523286641
3) 128Kbit = 1024/2.1072099696478683664961722630715 = 485.95062416637986345988672797112
The numbers look valid like T1 is the fastest, etc - but ultimately these aren't the same numbers like using James' approach.
+ the bonus question: whichever is the correct way, how do you have to handle the rounding? One digit is enough, or two, etc
Thanks and keep the good work,
nm
) way to count the site link cost number?Let's say you have three links:
A-B: 1.54Mbit
B-C: 512Kbit
A-C: 128kbit
By James in the CBTNugget 70-297 he says that you have to count the difference like
- take the fastest link as base, 1.54Mbit -> Cost=100 (as this is the default value)
1) the link (512Kbit) between A-B is approximately 3x slower -> Cost=300
2) the link (128Kbit) between A-C is approximately 12x slower -> Cost=1200
In the nugget the base cost link is lower (5), but the AD console allows the cost to set up to 99999, so theoretically this way can work.
On the other hand I know the formula:
Cost=1024/log(available bandwidth in Kbit)
Taking the above example it would give these numbers:
1) 1.54Mbit(1577KBit) = 1024/3.1978316933289028653690938277842 = 320.21697769028887990880037685061
2) 512Kbit = 1024/2.7092699609758307569236500525204 = 377.96159657385100491324523286641
3) 128Kbit = 1024/2.1072099696478683664961722630715 = 485.95062416637986345988672797112
The numbers look valid like T1 is the fastest, etc - but ultimately these aren't the same numbers like using James' approach.
+ the bonus question: whichever is the correct way, how do you have to handle the rounding? One digit is enough, or two, etc
Thanks and keep the good work,
nm
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DragonNOA1
Member Posts: 149 ■■■□□□□□□□
Lowest cost wins. If one link has a large cost but two links added together are smaller than the one big link than it will take that path. That is my understanding. So your first example with the 100 & 300 cost links would be the one I'd use.
Great link here explaining it in more detail.
How Active Directory Replication Topology Works: Active DirectoryThe command line, an elegant weapon for a more civilized age