Pollard's p - 1 - how do you set the bound & how do you set the base numbers

In Pollard's p-1 algorithm for factoring N, you try to find a L such that p - 1 divides L. Then you check $gcd(pow(a,L,N)- 1, N)$. If 1 < gcd < N, then you have found one of the factors.

I have seen 2 methods to do this.

1. For n from 1 to Bound, try $L = n!$ (i.e. factorial(n)) & try the $gcd(pow(a,L,N)- 1, N)$ for each one.
2. for n from 1 to Bound, try $L = LCM(range(1,n))$ & try the $gcd(pow(a,L,N)- 1, N)$ for each one.

In either method, once you hit the Bound unsuccessfully without finding a factor, you redo the loop with a new $a$

I have a few questions

1. How do you choose the Bound for each of the 2 methods? You are trying to check if the factor is Bound-Powersmooth, but how do you arrive at what Bound you want to check - i.e. what powersmoothness you expect?
2. In both the methods, how do choose the $a$'s?
3. In both the methods, how many such $a$'s do you try before giving up (because p - 1 probably doesn't have any small factors)?

Pollard's p-1 is useful only when for one of the primes p, p-1 is smooth. If you have some random integer you want to factor, you would use ECM and GNFS. Which means, if you are trying p-1, you have a reason to suspect that p-1 is reasonably smooth and then you should already have an idea of how smooth it can be (the smoothness bound L). In any case, the more you try - the more chances to break you have, so you should set as large bound as you can afford to wait, but only if you have reasons to suspect p-1 to be smooth.

I believe choosing $a$ does not matter much, and changing $a$ is not useful at all, until you get a non-trivial $gcd$. The idea is that for new $a$ you have to multiply by all those $1,2,3,...$ again, while you've already done this work for previous $a$. You might get a new $a$ such that some large factor $d$ of $p-1$ is already removed, and then you need a smaller bound $L$ to work, but the chance of that is $1/d$ and you rather keep raising your original $a$ to next powers and reach power $d$ naturally.

The only issue that can occur - is that you will arrive at 1 mod $p$ and 1 mod $q$ simultaneously (i.e. get $a^L\equiv 1 \mod{n}$), which does not leak a factor. Then you try another $a$, but at least you learn that Pollard's $p-1$ is likely to work well on this number.