Variation on Bellman-Ford Algorithms?
Since we are checking edges in different order each time, the time taken will depend on the order of the edges being checked in Bellman-Ford algorithm.

We will get the final result in 2 steps if our order is $v_1v_2, v_2v_3, v_3v_4, ... v_{99}v_{100}$.

However we  will get the final result in 100 steps if our order is $v_{99}v_{100}, v_{98}v_{99}, ..., v_3v_2, v_2v_1$. We can prove that this is worst possible by noting that in Bellman-Ford algorithm with positive weights, in every iteration at least one shortest distance is calculated correctly. So in first iteration we will only determine the shortest distance to $v_2$, others will remain $\infty$, in second iteration we will determine the shortest distance to $v_2$ and $v_3$ while others still remain $\infty$, and so on so forth.