SeaQueen
Posts: 1164
Joined: 4/14/2007 From: Washington D.C. Status: offline

Well... okay... in the case of WRA, there is an optimum, and it actually does come down to mathematics. What's the probability of annihilating a raid consisting of N aircraft or weapons? When the probability of killing a single weapon is P. That's easy! Pa = P^N Since P<1, Pa is a decreasing function of the raid size. So now, the question is, what's the probability of killing that weapon, P as a function of salvo size, M and single shot probability of kill Pk < 1. Assuming that it takes one or more hits to knock down the weapon, then P = 1  (1  Pk)^M All that says is that the probability of scoring at least one hit for a given salvo is one minus the probability of all the missiles missing. Putting both equations together means: Pa = (1  (1  Pk)^M)^N That's the threshold value for what's is considered to be an acceptable risk of not completely annihilating the raid (in which case there might be leakers), of course we're assuming a single shot opportunity. With very long ranged SAMs that might not necessarily be a valid assumption, but lets say you would prefer not to have to take it even if you have the second opportunity, so you're going to optimize for probability of annihilating on the first shot opportunity. Therefore: M = (ln(1  Pa^(1/N)) / ln(1 Pk) There's your salvo size necessary to get a 0.9 probability of raid annihilation on the first shot opportunity. This is an increasing function of salvo size. So for fixed Pk, and Pa, bigger raids demand more and more missiles to ensure a fixed risk. The problem with this is that there's still not a single "optimum" choice. The solution needs to be constrained. Warships have fixed magazine sizes, so obviously you can't shoot more and more missiles at larger and larger raids indefinitely. Additionally, they don't want to have to withdraw immediately following a battle because their magazines are empty. Warships are there for a reason, they protect important things and attack important things. Protecting themselves is important, but ultimately if they have to withdraw it constitutes only a tactical victory because their mission can't continue. So let's assume a warship has W weapons in its magazines. If you're shooting salvos of M weapons at raids of size N, then you're expending M*N weapons. To continue the fight you need J weapons remaining. Let's suppose that you want W  N*M >= J. All that says is that there's at least J weapons left in your magazine to continue the fight with. There's the constraint. Therefore (WJ)/N >= M If you plot both curves and look at the intersection, you can find that mathematical optimum. But look at all the assumptions you have to make! I need to assume a single shot Pk, an acceptable level of risk, and a desired number of weapons remaining to continue my mission. While it's probably reasonable to make some engineering based assumption about Pks versus a given threat. What's necessary to continue my mission or what an acceptable level of risk might be are not necessarily clear. Personally, if it's me taking the hit, I'd say make the risk as small as possible, which is to say, I want J=0 and that gets me back to more is better, basically, with me just exhausting my magazine on every raid that comes in. If you start putting in reasonable values, though, what you end up with is for fairly substantial raid sizes using an Arleigh Burke sized destroyer, somewhere between 2 and 3 shots to be pretty good. For very large raids sometimes it gets up to 4 missiles per salvo, but that's big raids (7+ incomming). For smaller raids between 1 or 2 shots leaves you with plenty left in the magazine.
< Message edited by SeaQueen  4/30/2019 8:33:49 PM >
