I have been selected as the General Area Review Team (Gen-ART) reviewer for this draft (for background on Gen-ART, please see http://www.alvestrand.no/ietf/gen/art/gen-art-FAQ.html ). Please wait for direction from your document shepherd or AD before posting a new version of the draft. Document: draft-ietf-manet-olsrv2-metrics-rationale-03.txt Reviewer: Suresh Krishnan Review Date: 2013/04/23 IESG Telechat date: 2013/04/25 Summary: This document is almost ready for publication as an Informational RFC but I do have some comments you may wish to address. Major ===== * Section 5.6 Link Metrics I had a hard time understanding this section as a whole and specifically the metric calculation. Intuitively, it was clear to me that you cannot squeeze 2^24 values into 12 bits :-). So the question immediately became what values get left out. If I understand correctly for a given value of b, each increment of a increases the metric value by 2^b. If my understanding is correct (my apologies if it is not), please consider modifying the text to clarify the following -> The compressed metric is lossy and is not capable of representing *all* metric values in the range 2^24-2^m. The larger the value of b more the values that cannot be represented. -> Two metric values that are different can result in the same compressed metric and hence cannot be really compared for equality as described in the section. e.g. metric value 821 and 822 would be mapped into the same compressed metric and a comparison will conclude that they are equal. -> Describe a short algorithm for converting a metric value into a compressed metric. i.e. Given a metric value x, how to obtain a and b from it. e.g. b=log2(x+256)-9 * To achieve interoperability and to avoid loops, the values of e and m need to be the same across the network. The following sentence seems to allow for other values of e and m ("An appropriate pair..."). Suggest OLD: The required 24 bit limit can be achieved if 2^e+m = 24. An appropriate pair of values to achieve this is e = 4, m = 8. NEW: The required 24 bit limit can be achieved if 2^e+m = 24. Since e+m=12, solving for e and m will lead to e = 4, m = 8. Minor ===== * It is unclear to me how the metrics can be configured to handle the privileged relay aerial router case described in Section 4. Assume there is a cost metric on the link to/from the aerial router, would this be set to low or high (I would assume high)? If the metric is set high, the ground routers may route through several low quality links which seem to have a lower path metric but may end up underusing the aerial link. Are there any guidelines in this regard? Thanks Suresh