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All right, let's jump into this challenge together.

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If you haven't already done so, be sure to grab a piece of paper and a pencil

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and do

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some of this math to make sure we can translate some of these values back to

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decimal.

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It's going to be very important that we can do so.

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But first of all, let's answer this question about COS tags.

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What are two situations where a COS tag ends up getting lost?

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Well, the COS tag is inside not just the Ethernet header, but also specifically

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the 802.1q part

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of the header, which is optional.

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So anytime we lose this, either one of these elements, we lose the COS tag.
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So anytime we lose this, either one of these elements, we lose the COS tag.

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The two primary situations we'd be talking about here would be non-trunk links,

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in which

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case we lose the 802.1q header.

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And so we lose the COS tag.

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The other would just be non-ethronet links, such as T1s.

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Anytime we send a packet across one of those types of connections, we're going

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to lose

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either the Ethernet header or the 802.1q header.

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In either case, the COS tag is lost.

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So with that out of the way, let's go ahead and do some math.
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So with that out of the way, let's go ahead and do some math.

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So we're going to start with CS6.

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Class selector six, this is only using, remember, our diagram there with the

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bits.

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It's only using these first three bits.

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And so a value of six here would be 110.

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And we could do our binary math and say, okay, well, here's a bunch of zeros

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here, and we

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try to figure that all out.

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Or we could simply realize that again, this is the eights.

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And we're essentially saying that we have six of these eights.

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So math is simply six of the eights, and that would be a value of 48.
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So math is simply six of the eights, and that would be a value of 48.

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So DSCP 48 here would be the answer for CS6.

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Now things get a little bit fun.

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We're going to go into the Assured Forwarding World, and we're going to start

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with AF13.

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Well in order to do this math calculation, we want to remember our formula 8x

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plus 2y.

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And so we're going to look at 1 and 3.

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And so the math formula simply says 8x1 plus 2x3.

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And so that would be 8 plus 6.
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And so that would be 8 plus 6.

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And the answer here is 14.

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Then we go to AF32.

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Same deal.

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We're going to take 8x3 this time plus 2x2.

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So this is 24, this is 4, and the answer is 28.

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And so we see how we can calculate pretty seamlessly to decimal simply using

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this formula.

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And as we explained earlier, it's really just playing off of the binary and
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this situation

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over here where we have an eights and we have a twos because this is always a

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zero with

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a shared forwarding.

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Now lastly here, or not lastly lastly, but we do have the next one which is

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asking us

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to work our way backwards, which is kind of interesting.

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We're saying we want to start with decimal 36.

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And we want to go back to an AF number.

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So how would we go about doing this?

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This is a pretty interesting exercise and there's lots of ways to do this.

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We could translate 36 into binary and then look at it in this format here using
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We could translate 36 into binary and then look at it in this format here using

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6 binary

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digits and extract the x and y values.

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Or we could actually use this formula in reverse.

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We could pull out a number of eights knowing that there's going to be a

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remainder probably.

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And then the rest of that would be our twos.

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So for example, 36, if I think about multiples of 8, 8, 16, 24, 32, I could

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pull 32 out of

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this and that's 4 of our eights.

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And so that means that x is going to be 4.
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And so that means that x is going to be 4.

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Now after I take 32 out of 36, there should be a remainder of 4 and so that's

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how many

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twos?

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Well that would be two separate twos.

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And so that means my y is a 2.

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That means that the answer to this should be AF4 for x2 for y.

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And then I can turn around and insert this back into the formula in order to

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make sure

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it works.

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4 times 8 plus 2 times 2.

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That would be 32.

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That would be 4 and the answer is 36 which is what we started with.
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That would be 4 and the answer is 36 which is what we started with.

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So AF42 is our answer.

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Now the last question here is simply asking us do we know the decimal value of

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EF expedited

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forwarding?

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This is always going to have a value of DSCP 46.

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It's simply something to memorize.

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It's essentially the highest even value before we get into CS6, CS7 area which

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we typically

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see reserved for network traffic.

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And so that would be the number that it is if that helps us to remember it.
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And so that would be the number that it is if that helps us to remember it.

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But ultimately it's simply a number that we're going to have to memorize at

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some point.

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So this brings us to the end of this skill.

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That's a great look at how we perform classification and marking.

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And in the end we do want to be able to come away from this being able to

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understand the

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structure of our COS and our DSCP tags and to be able to perform some math

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operations

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to make sure that we can calculate out the per hot behavior or decimal values

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while also

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understanding what those per hot behaviors really mean.
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understanding what those per hot behaviors really mean.

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I hope this has been informative for you and I'd like to thank you for viewing.