Okay, so now using the information that we got, for one hop before, let's try to see if we can use what we just learned, to move up to see what the cost would be for two hops. so again, we're going to consider one router at a time, but we're going to use the case of one hop to try to build up to two hops. So, let's start with router A, see what we're talking about here. 'Kay so we'll, we'll have A say, let's, let's let A go and look at each of its outgoing neighbors, okay. So again D has a cost of eight after one hop, because you know D, D, D's cost is eight to get to F in one hop. E's cost is ten to get to F in one hop and those are just again over the direct links right there. So now A is going to look and A will see D, A sees C and A sees B. Okay, let's start with B for a second okay, well A can get to B in four hops but B can't get to the destination F in one hop. So that cost would still be infinity, because four plus infinity is infinity. So for B if you tried to go to B the cost would still be infinity. if you tried to go to C again C can't get there because the cost is still infinity. So the cost would be two plus infinity which is still infinity. Now for D on the other hand, D can get to F in one hop, and that cost is eight. So, the cost then becomes, the cost from A to D, which is six, plus the cost from D to F, which is eight. So, this cost becomes 14. So now, let's move on to router B. So now. But just to backtrack for a second, now we know that A can get to F in two hops, and the cost to get to F in two hops is 14. So we say the cost of A is 14. Now, for B, B can look at his neighbors, B has two. B can get to C and B can get to E. And remember the, the links are not bidirectional. So you, B can't get to A. Just because A can get to B doesn't mean that B is going to be able to get to A, and here B cannot get to A. So now from B, if B tried to go to C, C doesn't have a direct path like to F. So that costed B infinite. We tried to go to E on the other hand. E can get to F in one hop with that cost of ten. So this cost becomes four which is from B to E plus ten, so four plus ten which is also 14. And, so, then we'd say the cost of B, in two hops, is gong to be 14. Now, for C, we move on to C, C has two neighbors it can go to. C can go to D, and C can go to E. for D, each of these can get to F, to, for D the cost is two, which is the cost from C to D, and then eight, which is the cost from D to get to F. So this becomes for, for C going to D, that cost becomes two plus eight, which is 10. And then for C going E, the direct cost, the direct path cost to E is three, so it gets three plus, and then 10 is that advertised cost, plus ten, which is 13. So now this is the first time we actually have to make a decision here. We have to choose the smaller of the two. Clearly 10 is smaller than 13, so we're going to go with the D. So we would say that the cost for C right now is ten, and we can also write just the path links here. So A is going to go and forward directly to D in this case. B is going to forward to E. And, the cost of c is 10. And C would be forwarding to, D, in this case. Alright. So now, let's look at, D and E. This, again, is, it's pretty trivial because, for D what was just, analyzed there for completeness. D can only choose either A or F. Right, so D would look and say okay well I can get to A in cost of seven and then plus infinity, so that's through A would just be infinity. So I'm going to stick with what I'm currently doing, and say the cost of D is still eight, right. So, because nothing's better than that cost, yet at least. And the same thing for E. E is going to say a little, something a little different now because even though this is going to be the first time that we're going to see this decision happening where we can actually change in the path links to change when consider multiple different types of hops. And so writing this down again, for D the path is from D D is going to go directly to F still. So now, for E, let's think about this and look at this for a second. Okay. E, has a cost of one to get to D. Okay, and then D can get to F in one hop with a cost of eight. So if it goes through D, the cost becomes 8 plus 1 or 1 plus 8, which is 9. And, if he goes directly to F, then the cost is going to be 10. But now 9 is less than 10, so he is going to go to choose to go to D. So that we say, the cost of E now is going to be nine, which is less. And this is indicating that E is going to go to D, rather than going directly to F. So, again, we're not choosing the shortest, the shortest path in terms of the the, nu-, mu-, minimum number of hops. It's the shortest path in terms of the cost. And this cost is less than this cost to go here, so that's going to be the choice that we make. [BLANK_AUDIO]