That *feels* like throwing information away, but if so that would to recapture the simplicity, and the deadlines have obviously been abstracted into the start times.
My reasoning was to guess to try to work out what S would be from scratch, and reason that you must be able to consider A and B as two separate blocks of time, each of which can be slid about as a unit. But block B must start no later than SB, and you know examining the internals can't help, as all block A must be finished by then, and any block C is less urgent, so it's at least as good to do all B as fast as you can first. So the only keys for S are (a) must start A by SA and (b) must start B by SB, ie. finish A by SB.
And a little munging makes me think that's equivalent to what you wrote.
S = min(SA, EDA-LA) S = min(SA, min(DA,SB)-LA) S = min(SA, min(DA-LA, SB-LA) ) S = min(SA, DA-LA, SB-LA)
But SA<=DA-LA, so
S = min(SA, SB-lA)
I had a different (wrong) simplification earlier which involved two scalars and one "you shipment of fail has arrived" bit, with successive logical ANDs, which was funny but didn't work. (Of course, if your log(N) is large, you may be able to escape early if any S is less than now if you like. But I don't know if that's ever worth it.)
no subject
L = LA+LB
S = min(SA,SB-LA)
That *feels* like throwing information away, but if so that would to recapture the simplicity, and the deadlines have obviously been abstracted into the start times.
My reasoning was to guess to try to work out what S would be from scratch, and reason that you must be able to consider A and B as two separate blocks of time, each of which can be slid about as a unit. But block B must start no later than SB, and you know examining the internals can't help, as all block A must be finished by then, and any block C is less urgent, so it's at least as good to do all B as fast as you can first. So the only keys for S are (a) must start A by SA and (b) must start B by SB, ie. finish A by SB.
And a little munging makes me think that's equivalent to what you wrote.
S = min(SA, EDA-LA)
S = min(SA, min(DA,SB)-LA)
S = min(SA, min(DA-LA, SB-LA) )
S = min(SA, DA-LA, SB-LA)
But SA<=DA-LA, so
S = min(SA, SB-lA)
I had a different (wrong) simplification earlier which involved two scalars and one "you shipment of fail has arrived" bit, with successive logical ANDs, which was funny but didn't work. (Of course, if your log(N) is large, you may be able to escape early if any S is less than now if you like. But I don't know if that's ever worth it.)