Re: EIGRP Metrics

Looking at:

(some random number) divided by (the link's reliability value + some
different random number)

Just how is it that it has anything to do with MTU?

If you are entering the metric on a command line, yes, the FIFTH value
is the mtu. Those are seeds for the route. Yes, MTU is carried to
determine the lowest MTU along a path for tie breakers, but in the math
presented there, the "K5" value has nothing to do with MTU.

I'd actually argue that the docs have things wrong (listing K4 as
reliability and K5 as a modifier) since mathematically 0/(any number) =
0, so K5 would really be the on/off switch while K4 would just be an
additive value due to the + sign and not a * sign in the algorithm.

But that's just me looking for math to make sense. ;)

I'm open to other thoughts though as it's not a widely explored topic,
but math is generally straightforward. At least with the values
presented in the equation!

*Scott Morris*, CCIE/x4/ (R&S/ISP-Dial/Security/Service Provider) #4713,

JNCIE-M #153, JNCIS-ER, CISSP, et al.

JNCI-M, JNCI-ER

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Joe Astorino wrote:
> K5 is MTU. In the formula to calculate the metric it is written as metric =
> [K1*bandwidth + (K2*bandwidth)/(256 - load) + K3*delay] * [K5/(reliability +
> K4)] ... but K5 still represents the MTU
>
> On Tue, Aug 25, 2009 at 3:06 PM, Cisco Fanatic <ebay_products_at_hotmail.com>wrote:
>
>
>> All,
>>
>> It is simple, but somehow I get confused with different wording in
>> different
>> docs.
>>
>> Is K5 MTU or (K4 and K5 both reliability)?
>>
>>
>>
>>
>>
>> ToS - 0
>>
>> Bandwidth - K1
>>
>> Load - K2
>>
>> Delay - K3
>>
>> Reliability - K4
>>
>>
>> MTU - K5
>>
>> -Yuri
>>
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Received on Tue Aug 25 2009 - 16:09:00 ART