That's precisely why I decided to use the idea of maximum friction - it eliminates the need to know the lengths.

Ha! And I was working the opposite way, and suggesting using the lengths/angles, to eliminate the need to know the friction!

I did realise you were estimating the change due to pushing the krab up, but didn't have time to do the calculation for that using angles this morning, so I just posted my logic in the hope that it might be useful to you.

I've now done a calculation, just to get a different perspective. As you say, I had to make some assumptions. These were that the bolts are 1m apart, and at the same level, with the pitch rope dropping midway between them after the deviation, and 60º angles as you did. So the upper part of the rope and the dev cord are both 30º from horizontal, and the tension in all 3 sections is the weight of the caver, which I took as 1000N, might be less but easy to follow the numbers.

I then assumed that the krab is pushed 10cm higher. This puts the dev cord at 19º to horizontal. As it moves up on a radius, the krab moves slightly nearer the main belay, so the rope above the krab is at 22.5º to horizontal. Solving the equations for those angles gives 1390N in the main rope and 1420 in the dev cord. So the force in the main rope would change by 390N at the krab. Is that reasonable? It's 0.27 times the force in the dev cord, so a coefficient of friction of more than 0.27 would hold the krab there. The force required to push the krab up the rope would be more than 390N; if the krab were a frictionless pulley, it would start at zero and build up to 390N, so the actually friction force would have to be added to that, during the push.

Initially I assumed the krab was pushed 15cm. That gave me 13.9º, 17.5º, 1830N and 1870N, and a change of 830N at the krab would be too much for friction to hold or to push up that far.