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INTRODUCTION-EXAMPLES
المؤلف: Lawrence C. Evans
المصدر: An Introduction to Mathematical Optimal Control Theory
الجزء والصفحة: 5-10
2-10-2016
287
EXAMPLE 1: CONTROL OF PRODUCTION AND CONSUMPTION.
Suppose we own, say, a factory whose output we can control. Let us begin to construct a mathematical model by setting x(t) = amount of output produced at time t ≥ 0.
We suppose that we consume some fraction of our output at each time, and likewise can reinvest the remaining fraction. Let us denote
α(t) = fraction of output reinvested at time t ≥ 0.
This will be our control, and is subject to the obvious constraint that
0 ≤ α(t) ≤ 1 for each time t ≥ 0.
Given such a control, the corresponding dynamics are provided by the ODE
the constant k > 0 modelling the growth rate of our reinvestment. Let us take as a payoff functional
The meaning is that we want to maximize our total consumption of the output, our consumption at a given time t being (1−α(t))x(t). This model fits into our general framework for n = m = 1, once we put
A = [0, 1], f(x, a) = kax, r(x, a) = (1 − a)x, g ≡ 0.
As we will see later , an optimal control α∗(.) is given by
for an appropriate switching time 0 ≤ t∗ ≤ T. In other words, we should reinvest all the output (and therefore consume nothing) up until time t∗, and afterwards, we should consume everything (and therefore reinvest nothing). The switchover time t∗ will have to be determined. We call α∗(.) a bang–bang control.
EXAMPLE 2: REPRODUCTIVE STATEGIES IN SOCIAL INSECTS
The next example is from (CONTROLLABILITY, BANG-BANG PRINCIPLE) of the book Caste and Ecology in Social Insects, by G. Oster and E. O. Wilson [O-W]. We attempt to model how social insects, say a population of bees, determine the makeup of their society.
Let us write T for the length of the season, and introduce the variables
w(t) = number of workers at time t
q(t) = number of queens
α(t) = fraction of colony effort devoted to increasing work force
The control α is constrained by our requiring that
0 ≤ α(t) ≤ 1.
We continue to model by introducing dynamics for the numbers of workers and the number of queens. The worker population evolves according to
Here μ is a given constant (a death rate), b is another constant, and s(t) is the known rate at which each worker contributes to the bee economy.
We suppose also that the population of queens changes according to
for constants ν and c.
Our goal, or rather the bees’, is to maximize the number of queens at time T:
P[α(.)] = q(T).
So in terms of our general notation, we have x(t) = (w(t), q(t))T and x0 = (w0, q0)T .
We are taking the running payoff to be r ≡ 0, and the terminal payoff g(w, q) = q.
The answer will again turn out to be a bang–bang control, as we will explain later.
EXAMPLE 3: A PENDULUM.
We look next at a hanging pendulum, for which
θ(t) = angle at time t.
If there is no external force, then we have the equation of motion
the solution of which is a damped oscillation, provided λ > 0.
Now let α(.) denote an applied torque, subject to the physical constraint that
|α| ≤ 1.
Our dynamics now become
Define
Then we can write the evolution as the system
We introduce as well
for
We want to maximize P[.], meaning that we want to minimize the time it takes to bring the pendulum to rest.
Observe that this problem does not quite fall within the general framework described in 1.1, since the terminal time is not fixed, but rather depends upon the
control. This is called a fixed endpoint, free time problem.
EXAMPLE 4: A MOON LANDER
This model asks us to bring a spacecraft to a soft landing on the lunar surface, using the least amount of fuel.
We introduce the notation
h(t) = height at time t
v(t) = velocity = h˙ (t)
m(t) = mass of spacecraft (changing as fuel is burned)
α(t) = thrust at time t
We assume that
0 ≤ α(t) ≤ 1,
and Newton’s law tells us that
the right hand side being the difference of the gravitational force and the thrust of the rocket. This system is modeled by the ODE
We summarize these equations in the form
We want to minimize the amount of fuel used up, that is, to maximize the amount remaining once we have landed. Thus
P[α(.)] = m(τ ),
where
τ denotes the first time that h(τ ) = v(τ ) = 0.
This is a variable endpoint problem, since the final time is not given in advance.
We have also the extra constraints
h(t) ≥ 0, m(t) ≥ 0.
EXAMPLE 5: ROCKET RAILROAD CAR.
Imagine a railroad car powered by rocket engines on each side. We introduce the variables
q(t) = position at time t
v(t) = q˙ (t) = velocity at time t
α(t) = thrust from rockets,
where
−1 ≤ α(t) ≤ 1,
the sign depending upon which engine is firing.
We want to figure out how to fire the rockets, so as to arrive at the origin 0 with zero velocity in a minimum amount of time. Assuming the car has mass m, the law of motion is
We rewrite by setting x(t) = (q(t), v(t))T . Then
Since our goal is to steer to the origin (0, 0) in minimum time, we take
for
τ = first time that q(τ ) = v(τ ) = 0.
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