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.