Assume that initially there are no ready processes, when the
first one, A, arrives. It has priority 0 to begin with. Since
there are no other accepted processes, A is accepted
immediately.
After a while another process, B, arrives. As long as b / a < 1,
B’s priority will eventually catch up to A’s, so it is accepted;
now both A and B have the same priority.
All accepted processes share a common priority (which rises at
rate b ); that makes this policy easy to implement i.e any new
process’s priority is bound to get accepted at some point. So no
process has to experience starvation.
Even if b / a > 1, A will eventually finish, and then B can be
accepted.