Thanks Yoav,

I think I missed it somehow from the RFC.

Regarding the usage of EAP-MD5, it is just that the client should
support all types of authentication,
So even the weakest ones also and rfc-3748 says that md5-challenge is a
must.
Moreover, it is the server that has to decide what is the best suited
auth mechanisms for a particular client.


Regards,
Prashant

-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf
Of Yoav Nir
Sent: Tuesday, October 25, 2011 5:59 PM
To: Prashant Batra (prbatra); Glen Zorn
Cc: [email protected]
Subject: Re: [IPsec] eap-md5 based authentication

No, you don't use the same password for calculating the AUTH payload.
>From section 2.15:


   There are two types of EAP authentication (described in
   Section 2.16), and each type uses different values in the AUTH
   computations shown above.  If the EAP method is key-generating,
   substitute master session key (MSK) for the shared secret in the
   computation.  For non-key-generating methods, substitute SK_pi and
   SK_pr, respectively, for the shared secret in the two AUTH
   Computations.


So the client in your case is going to use SK_pi.

If you don't mind the question, how did this come up?  If you're writing
your own client, why not use something better than EAP-MD5 such as
EAP-EKE
or EAP-pwd?  If you're using a third-party client (like Microsoft's Win7
client) I think they're using EAP-MSChapv2.  Where did you find a client
with EAP-MD5?

Yoav


On 10/25/11 2:10 PM, "Prashant Batra (prbatra)" <[email protected]>
wrote:

>Thanks Yaov and Glen,
>
>I could successfully calculate the challenge response.
>Now, after the challenge response is successful, the server will send
>EAP-SUCCESS, then the client has to send a AUTH payload.
>As eap-md5 doesn't result in any key like eap-aka/sim, the client will
>use the same password(used for calculating challenge response) to
>calculate AUTH payload.
>If so, why there is an explicit auth required here. EAP-SUCCESS, can
>itself indicate that the client is authenticated.
>
>Maybe, it is required for some extra authentication?
>
>Regards,
>Prashant
>
>-----Original Message-----
>From: [email protected] [mailto:[email protected]] On Behalf
>Of Glen Zorn
>Sent: Tuesday, October 25, 2011 3:46 PM
>To: Yoav Nir
>Cc: [email protected]; Prashant Batra (prbatra)
>Subject: Re: [IPsec] eap-md5 based authentication
>
>On 10/25/2011 3:35 PM, Yoav Nir wrote:
>
>> Hi Prashant.
>>  
>> I think in the challenge request, the first byte is the challenge
>length
>> (usually 16) followed by the challenge itself, and then followed by
>some
>> server name. I guess the reasoning is that this allows the client to
>> choose the correct password based on the server name.
>
>The format is defined in Section 4.1 of RFC 1994
>
>...
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