Differintal equation & optimalization & NukeZone que
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- SWPIGWANG
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Differintal equation & optimalization & NukeZone que
To quote Sun Tzu: Know thy self, Know thy enemy, one hundred battles, one hundred victories.
Ok, here is for math heads: (other ppl read below)
General Form:
Goal: Maxiumize dM/dt
M is the money earned
dM/dt is the rate where money is earned
(dM/dt)^2 is the rate of change of money making rate
(dM/dt)^2 is proptional to M by a constant
Instead of letting M, grow exponentially, I can also make M grow at dM/dt = ((dMi/dt) + 7500), but in this case (dM/dt)^2 would be zero.
I can switch between the two whenever I want.
So when would be the optimal time to switch?
There are the inital conditions for the specific problem:
(dM/dt)^2 = (dMi/dt)^2 + 12M/125000
The choice between freebies are given
Case 1.
M = 500,000
dMi/dt = 2500
(dMi/dt)^2 = 6
Case 2.
M = 120,000
dMi/dt = 3940
(dMi/dt)^2 = 12
So whats the better starting condition? And at what point should I switch from pumping out money linearly at (dMi/dt) + 7500 and when do I change to pumping though improving dM/dt? (give in dM/dt terms please, and show work)
--------------------non-math section
So you people play nukezone don't math crunch? Fools, as one who does not know oneself can not obtain victory.
By the way, AH-64 has 0.029 attack to cost ratio, and 0.029 defense to cost ratio, while X-35 has 0.049 attack to cost ratio and 0.040 defense to cost ratio. Also note that Heavy Tanks have 0.047 attack to cost ratio and 0.044 attack ratio.
Meaning: AH-64 sucks, though not as bad as rifles at 0.026 ratio for both. Specialized units like SAM, Artillery or SH-60 gets 0.074-0.076 attack to cost ratio.
Since you only has one attack for muti-role units, they suck.
Ok, here is for math heads: (other ppl read below)
General Form:
Goal: Maxiumize dM/dt
M is the money earned
dM/dt is the rate where money is earned
(dM/dt)^2 is the rate of change of money making rate
(dM/dt)^2 is proptional to M by a constant
Instead of letting M, grow exponentially, I can also make M grow at dM/dt = ((dMi/dt) + 7500), but in this case (dM/dt)^2 would be zero.
I can switch between the two whenever I want.
So when would be the optimal time to switch?
There are the inital conditions for the specific problem:
(dM/dt)^2 = (dMi/dt)^2 + 12M/125000
The choice between freebies are given
Case 1.
M = 500,000
dMi/dt = 2500
(dMi/dt)^2 = 6
Case 2.
M = 120,000
dMi/dt = 3940
(dMi/dt)^2 = 12
So whats the better starting condition? And at what point should I switch from pumping out money linearly at (dMi/dt) + 7500 and when do I change to pumping though improving dM/dt? (give in dM/dt terms please, and show work)
--------------------non-math section
So you people play nukezone don't math crunch? Fools, as one who does not know oneself can not obtain victory.
By the way, AH-64 has 0.029 attack to cost ratio, and 0.029 defense to cost ratio, while X-35 has 0.049 attack to cost ratio and 0.040 defense to cost ratio. Also note that Heavy Tanks have 0.047 attack to cost ratio and 0.044 attack ratio.
Meaning: AH-64 sucks, though not as bad as rifles at 0.026 ratio for both. Specialized units like SAM, Artillery or SH-60 gets 0.074-0.076 attack to cost ratio.
Since you only has one attack for muti-role units, they suck.
- Durandal
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So find the point at which the exponential growth is exceeded by the constant growth. Case 2 will give you a higher initial rate of growth with a higher rate of growth acceleration.SWPIGWANG wrote:The Growth of one is constant, and the other is exponential.
And when the switch occurs, the M carries over.
By the way, I haven't taken differential equations yet.
Damien Sorresso
"Ever see what them computa bitchez do to numbas? It ain't natural. Numbas ain't supposed to be code, they supposed to quantify shit."
- The Onion
"Ever see what them computa bitchez do to numbas? It ain't natural. Numbas ain't supposed to be code, they supposed to quantify shit."
- The Onion
- The Dark
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I look not only at the cost/efficiency but also at the versatility of a unit. If I have a crapload of B-2s and get attacked, I'm screwed. Apaches may have a low attack/cost ratio, but they can hit the most common units (infantry and ground) as well as damage buildings. If I start running low on space, I might switch from F-22s and Apaches to F-22, X-35, and B-2, with extra flamethrowers to cover for the infantry loss.
BattleTech for SilCoreStanley Hauerwas wrote:[W]hy is it that no one is angry at the inequality of income in this country? I mean, the inequality of income is unbelievable. Unbelievable. Why isn’t that ever an issue of politics? Because you don’t live in a democracy. You live in a plutocracy. Money rules.
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I've read forums of nukezone, and high tier players don't use infantry that much, because only rocket is cost effect enough to do anything, but it spilt attacks REGUARDLESS if the enemy have the said unit type, appearently. So if you attack with vehicle only army for example, the rocket would only do 1/3 damage.
Same for AH64
Also, the cost effectiveness is bloody 1/2 or less than specialized units, so it is better to have 2 specialized units.
Same for AH64
Also, the cost effectiveness is bloody 1/2 or less than specialized units, so it is better to have 2 specialized units.
- The Dark
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Yeah, I ran some figures just on offense, and came up with a new model army .
All I'm using now are SAM tanks, SH-60 Seahawks, X-35 JSFs, Humvees, and Heavy Artillery.
In order (worst to best) of dollars/point of attack:
Rifle Infantry
AH-64 Apache
F-22 Raptor
Destroyer
Hover Tank
Cruiser
Rocket Soldier
Attack Boat
Heavy Tank
X-35 JSF
Submarine
F/A-18 Super Hornet
B-2 Stealth Bomber
Flamethrower
Heavy Artillery
SAM Tank
Humvee
I haven't done defense yet.
All I'm using now are SAM tanks, SH-60 Seahawks, X-35 JSFs, Humvees, and Heavy Artillery.
In order (worst to best) of dollars/point of attack:
Rifle Infantry
AH-64 Apache
F-22 Raptor
Destroyer
Hover Tank
Cruiser
Rocket Soldier
Attack Boat
Heavy Tank
X-35 JSF
Submarine
F/A-18 Super Hornet
B-2 Stealth Bomber
Flamethrower
Heavy Artillery
SAM Tank
Humvee
I haven't done defense yet.
BattleTech for SilCoreStanley Hauerwas wrote:[W]hy is it that no one is angry at the inequality of income in this country? I mean, the inequality of income is unbelievable. Unbelievable. Why isn’t that ever an issue of politics? Because you don’t live in a democracy. You live in a plutocracy. Money rules.
Yes you have, you just don't know it Have you gone from Newtonian to Lagrangian mech yet? Just curio. I tried taking intro to Newtonian (which is after basic gravitational, electromagnetic, and optical stuff) but it was before I had vector analysis so I got kind of lost and I dropped it, then since it wasn't my major I never went back. But I had some in math, and later some Lagrangian crap, but not like a real formal class in it.Durandal wrote:So find the point at which the exponential growth is exceeded by the constant growth. Case 2 will give you a higher initial rate of growth with a higher rate of growth acceleration.SWPIGWANG wrote:The Growth of one is constant, and the other is exponential.
And when the switch occurs, the M carries over.
By the way, I haven't taken differential equations yet.
I have being given A's for depleting Dragon ball Z the way it should be.
- Durandal
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We did Lagrangian multipliers back in multivariable calculus, and actually, I did do a bit of differential equations in physics last semester, and we're doing a bit of them this semester in thermodynamics. Basically, the solution to every differential equation is to integrate it.Malecoda wrote:Yes you have, you just don't know it Have you gone from Newtonian to Lagrangian mech yet? Just curio. I tried taking intro to Newtonian (which is after basic gravitational, electromagnetic, and optical stuff) but it was before I had vector analysis so I got kind of lost and I dropped it, then since it wasn't my major I never went back. But I had some in math, and later some Lagrangian crap, but not like a real formal class in it.Durandal wrote:So find the point at which the exponential growth is exceeded by the constant growth. Case 2 will give you a higher initial rate of growth with a higher rate of growth acceleration.SWPIGWANG wrote:The Growth of one is constant, and the other is exponential.
And when the switch occurs, the M carries over.
By the way, I haven't taken differential equations yet.
Damien Sorresso
"Ever see what them computa bitchez do to numbas? It ain't natural. Numbas ain't supposed to be code, they supposed to quantify shit."
- The Onion
"Ever see what them computa bitchez do to numbas? It ain't natural. Numbas ain't supposed to be code, they supposed to quantify shit."
- The Onion
- SWPIGWANG
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- Location: Commence Primary Ignorance
GRRRRRRRRRRRRRRRR JUMPS UP AND DOWN.....
*Tries to solve the original equation...*
M = Ce^ct should work.... to solve the differential equation.
Intergrating both sides would be easy in single order equations, but this is second order.
However the optimalization is totally beyond me. I have no idea how to even start solving the equation.
*GRRRRR 1st year = we are bunch of idiots*
HELP HELP HELP
*Tries to solve the original equation...*
M = Ce^ct should work.... to solve the differential equation.
Intergrating both sides would be easy in single order equations, but this is second order.
However the optimalization is totally beyond me. I have no idea how to even start solving the equation.
*GRRRRR 1st year = we are bunch of idiots*
HELP HELP HELP
I may have been kind of abrupt with my subject changes there. I wasn't referring to Lagrangian multipliers, although it's somewhat serendipitous that they arise in a talk abt optimization. I mean Lagrangian mechanics, where your point of reference is accelerating, not observing a moving frame from the outside but from the inside. You make the transformation from an outside observer, describe the real forces in terms of that transformation, and eliminate fictional forces. Newtonian mechanics at MTU was basically vector analysis (so you learn the linear algebra and how to make linear transformations), and a pre-req for Lagrangian mechanics.Durandal wrote:
We did Lagrangian multipliers back in multivariable calculus, and actually, I did do a bit of differential equations in physics last semester, and we're doing a bit of them this semester in thermodynamics. Basically, the solution to every differential equation is to integrate it.
I have being given A's for depleting Dragon ball Z the way it should be.
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Do you need to integrate? You have the correct form of solution. Differentiating it once would produce dM/dt, wouldn't it? In your original post, I'm not sure whatSWPIGWANG wrote:GRRRRRRRRRRRRRRRR JUMPS UP AND DOWN.....
*Tries to solve the original equation...*
M = Ce^ct should work.... to solve the differential equation.
Intergrating both sides would be easy in single order equations, but this is second order.
HELP HELP HELP
is. Can you clarify?There are the inital conditions for the specific problem:
(dM/dt)^2 = (dMi/dt)^2 + 12M/125000
(3.13, 1.49, -1.01)
- Durandal
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Those are the only Lagrangians I have experience with.Malecoda wrote:I may have been kind of abrupt with my subject changes there. I wasn't referring to Lagrangian multipliers, although it's somewhat serendipitous that they arise in a talk abt optimization. I mean Lagrangian mechanics, where your point of reference is accelerating, not observing a moving frame from the outside but from the inside. You make the transformation from an outside observer, describe the real forces in terms of that transformation, and eliminate fictional forces. Newtonian mechanics at MTU was basically vector analysis (so you learn the linear algebra and how to make linear transformations), and a pre-req for Lagrangian mechanics.Durandal wrote:
We did Lagrangian multipliers back in multivariable calculus, and actually, I did do a bit of differential equations in physics last semester, and we're doing a bit of them this semester in thermodynamics. Basically, the solution to every differential equation is to integrate it.
I'm actually taking linear algebra this semester.
Damien Sorresso
"Ever see what them computa bitchez do to numbas? It ain't natural. Numbas ain't supposed to be code, they supposed to quantify shit."
- The Onion
"Ever see what them computa bitchez do to numbas? It ain't natural. Numbas ain't supposed to be code, they supposed to quantify shit."
- The Onion
-
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Ok, I think I worked out what you're on about.
M'' = M''(0) + 12M/125000
(M'' = d^2M/dt^2)
The general solution to this diff. equation will require two steps. First, you find the general solution to the "homogenous" diff. eqn that corresponds to your (inhomogenous) D.E. - this equation is got from the first by just removing any terms that do not operate on or contain M:
M'' - PM = 0
(P = 12/125000)
To solve this, you form the "auxilliary equation" by representing the differential operator d/dt with D,
(D - sqrt(P)).(D + sqrt(P)).M = 0
This particular example (two real roots, sqrt(P) and -sqrt(P)) has the corresponding general solution
M = A.exp(t*sqrt(P)) + B.exp(-t*sqrt(P))
A and B are the arbitrary constants.
To solve the original (inhomogenous) equation, there is a second step. You must find a particular integral, a function that is a solution of the inhomogenous equatiion (it will not be a unique solution)
M'' - PM = M''(0)
Often, these can be done by inspection, as in this case. Try M = K (constant) so M'' = 0,
and substitution produces
K = - M''(0) / P
The general solution for the inhomogenous equation is the sum of the particular integral you found and the complementary function,
M = A.exp(t*sqrt(P)) + B.exp(-t*sqrt(P)) - M''(0)/P
From here, all you need to do is to establish A and B. Differentiate the solution,
M' = A*sqrt(P)*exp(t*sqrt(P)) - B*sqrt(P)*exp(-t*sqrt(P))
M'' = A*P*exp(t*sqrt(P)) + B*P*exp(-t*sqrt(P))
Since your cases provide initial conditions, consider t = 0:
M(0) = A + B - M''(0)/P
M'(0) = (A - B)*sqrt(P)
M''(0) = (A + B)*P
Obviously, you have plenty of equations with which to solve for A and B. Finding these and rearranging gets me
M' = (M''(0)/sqrt(P)) * sinh[t*sqrt(P)] + M'(0)*cosh[t*sqrt(P)]
as the most compact representation. Since sinh and cosh increase monotonically with positive t, I expect that there is only one intersection with your other possibility, M' = M'(0) + 7500, and so sophisticated optimisation is unnecessary. Simply find the intersection and you can be sure that for t > t(intersection) the exponential growth option will win. The simplest way is to set M'(0) + 7500 = (M''(0)/sqrt(P))*sinh .... and solve with a simple numerical method, such as Newton-Raphson. For your first case, I find that after t = 188.3 you should switch to the exponential. For the second, after t = 149.06, you should switch. Before these times the linear option is better.
You should really get hold of a textbook on this sort of thing, as this is not a good "how to solve DEs" explanation. Boas is quite good.
M'' = M''(0) + 12M/125000
(M'' = d^2M/dt^2)
The general solution to this diff. equation will require two steps. First, you find the general solution to the "homogenous" diff. eqn that corresponds to your (inhomogenous) D.E. - this equation is got from the first by just removing any terms that do not operate on or contain M:
M'' - PM = 0
(P = 12/125000)
To solve this, you form the "auxilliary equation" by representing the differential operator d/dt with D,
(D - sqrt(P)).(D + sqrt(P)).M = 0
This particular example (two real roots, sqrt(P) and -sqrt(P)) has the corresponding general solution
M = A.exp(t*sqrt(P)) + B.exp(-t*sqrt(P))
A and B are the arbitrary constants.
To solve the original (inhomogenous) equation, there is a second step. You must find a particular integral, a function that is a solution of the inhomogenous equatiion (it will not be a unique solution)
M'' - PM = M''(0)
Often, these can be done by inspection, as in this case. Try M = K (constant) so M'' = 0,
and substitution produces
K = - M''(0) / P
The general solution for the inhomogenous equation is the sum of the particular integral you found and the complementary function,
M = A.exp(t*sqrt(P)) + B.exp(-t*sqrt(P)) - M''(0)/P
From here, all you need to do is to establish A and B. Differentiate the solution,
M' = A*sqrt(P)*exp(t*sqrt(P)) - B*sqrt(P)*exp(-t*sqrt(P))
M'' = A*P*exp(t*sqrt(P)) + B*P*exp(-t*sqrt(P))
Since your cases provide initial conditions, consider t = 0:
M(0) = A + B - M''(0)/P
M'(0) = (A - B)*sqrt(P)
M''(0) = (A + B)*P
Obviously, you have plenty of equations with which to solve for A and B. Finding these and rearranging gets me
M' = (M''(0)/sqrt(P)) * sinh[t*sqrt(P)] + M'(0)*cosh[t*sqrt(P)]
as the most compact representation. Since sinh and cosh increase monotonically with positive t, I expect that there is only one intersection with your other possibility, M' = M'(0) + 7500, and so sophisticated optimisation is unnecessary. Simply find the intersection and you can be sure that for t > t(intersection) the exponential growth option will win. The simplest way is to set M'(0) + 7500 = (M''(0)/sqrt(P))*sinh .... and solve with a simple numerical method, such as Newton-Raphson. For your first case, I find that after t = 188.3 you should switch to the exponential. For the second, after t = 149.06, you should switch. Before these times the linear option is better.
You should really get hold of a textbook on this sort of thing, as this is not a good "how to solve DEs" explanation. Boas is quite good.
(3.13, 1.49, -1.01)
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- Jedi Knight
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Ummm, if all you really wanted was the answer as to when to switch the earning scheme, you could have gotten an approximation by simply iterating (I would suggest using a spreadsheet so that you can just copy and paste the cells for each iteration).
But if you want to drive the calculus nail through your head, then by all means...
But if you want to drive the calculus nail through your head, then by all means...
Time makes more converts than reason. -- Thomas Paine, Common Sense, 1776