How Can i transform condition into constraint equation?

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Maria 2023-1-10

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评论： Walter Roberson 2023-1-17

Hello everyone!

i'm asking if i can change a condition line command into equation in order to use it in my problem. The condition is related to trajectory of vehicle in segment line, so if it arrives at the last point of the segment , it will come back to the initial point and keep going back and forth until the time ends.

in my code i wrote the condition as follow :

qu(1) = 0;   % qu the initial position of vehicule 
speed = 50;     % 50 is the speed of vehicule ( constant) 
k = 1;
while k < length(qu)
    xk = qu(k) + speed;
    if xk > 10000       %10000 is the length of the segment
        speed = -50;
    elseif xk < 0
        speed = 50;
    else
        k = k+1;
        qu(k) = xk;
    end
end

How can i transform this code into simple equation?

any help is appreciated!

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Walter Roberson 2023-1-11

I'm pretty sure they are using fixed timestep, and that the speed variable is distance per time step.

Maria 2023-1-11

@Walter Roberson exactly!

@Torsten i want to add the condition of the length of segment, it means that : when new position = length of the segment , the new position should return at the previous position.

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回答（1 个）

Walter Roberson 2023-1-10

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length(qu) starts as 1, k starts at 1, 1 < 1 is false, so the loop body is never entered. Therefore the code can be reduced to just the first two lines.

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Maria 2023-1-17

编辑：Maria 2023-1-17

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@Walter Roberson @Torsten hello!

i'm trying to solve an optimization problem but the hard part is discussed above, i tell you that i want to optimize the first location of vehicule but i didn't find how to write the constraint. I found a method to write the different location points based on the first one qu(t=1) the point that i want to optimize,

here is the clarification ,

because in each time or iteration qu(t=1) will be known so, i noted a= (L-qu(1))/v (v is constant)

e.g for qu(1)=60, L=1000, v=20m/s so a=47 and a1= L/20=50

now for 0<t<a+1

qu(t)=qu(1)+(t-1)*v.teta.dt , dt=1 and teta =1

e.g qu(20)=60+(19*20)*1 =440 m

now for t>a+1

for e.g

qu(49)=qu(1)+(a*v*teta*dt)-(1*v*teta*dt)
=60+(47*20)-(20) = 980

here i substracted (49-(a+1))=48 so for the 47 i put it in the positive part and the 1 in the negative part

another eg

qu(100)=qu(1)+(47*v*teta*dt )-(50*v*teta*dt)+(2*v*teta*dt)

60+940-1000+40 =40

here i subtracted (100-(a+1))= 52 , grater than a1 ,so for the 47 i put it in the positive part and the 50 in the negative part and the 2 in the positive part.

so i'm asking if there is a way to write constraint based in a and a1 to successively determine the location.

thanks in advance

Torsten 2023-1-17

Thus my question remains.

Is qu(1) a multiple of 20 ?

If no: How to deal with the case that the opposite wall is not exactly hit at a time instant t ?

Walter Roberson 2023-1-17

Are you wanting the model to reject certain proposed values of qu(1), to say "Ah, this set of proposed parameters does not lead to a feasible model, try a different set of parameters" ?

If so, if qu(1) is one of the model parameters, then set a lower bound on it of 0, and set an upper bound of (L-NumberOfTimeSteps*v) . [If (L-NumberOfTimeSteps*v) would be negative, then that implies the equations are impossible to satisfy]

But I don't think that is what you want. You want the direction to reverse when you hit L or 0 and you want to keep going. That is not a constraint and cannot be handled by upper/lower bounds or by linear inequalities or by nonlinear equalities.

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