 In this video, we provide the solution to question number eight for practice exam number one for math 1030, in which case we're given a weighted complete graph. We're asked to solve the associated traveling salesman problem, and we're told to do this using the cheapest link algorithm. So what we're going to do is look for the cheapest links that are on the graph avoiding any premature circuits when possible. So the cheapest link I can see here is going to be C to B, which is worth nine. So I'm just going to write this down as I go. So nine plus something after nine, the next cheapest link appears to be 18, which goes between B and E like so. So we're going to get that one. So we then write down 18. The next notice notice that this moment views be twice now so any circuit can't use be more than once. So we're going to scratch out those numbers. So we're not tempted to use them anymore. I also noticed that E to A was also 18. So I'm going to grab that one as well. That's cheap. Turns out it didn't matter which of the two you grabbed. We end up with the same answer here. We've now used E twice, so we're not going to use any other edge associated to E. So we scratch those ones out. Then looking on the board, what is still with the cheapest option that haven't scratched out already looks like 22 is going to be my next cheapest one, which I'm allowed to do that one. So I'm going to take, I'm going to go to A to D there. So add 22 onto that one for which I don't. I'm not going to use this one now that I want to because I've already visited A twice now. And so honestly, there should just be one option left. I should go from C to D. Now I'll finish up my circuit. And if you do that, that would add up to be 25. Oh, sorry, you throw another 25 onto the sum. Now as you add these all together, nine plus 18 plus 18 plus 22 plus 25, that gives you 92. And so the correct answer would then be choice C.