I am new to the community and to the software (IBM ILOG CPLEX Optimization Studio). I am solving a production problem modeled as an integer program. Each integer variable is a three-dimensional decision variable:
dvar int p[order][center][week] in 0..1;
Certain orders need to be canceled before others, e.g. the order #1 needs to be fully processed by the second week, while order #2 could potentially wait until the sixth week. How can I express that in the software? I.e. p[i=1][center][week=3...6] = 0?
In the upper bounds OPL allows conditions and you can use the ternary operator ?:
range order=1..6;
range center=1..4;
range week=1..6;
dvar int p[o in order][c in center][w in week] in 0..(((o==1)&&(w in 3..6))?0:1);
maximize sum(o in order,c in center,w in week) p[o][c][w];
subject to
{
}
works fine
Related
I want to find a variable in C++ that allows a given nonlinear formula to have a maximum value in a constraint.
It is to calculate the maximum value of the formula below in the given constraint in C++.
You can also use the library.(e.g. Nlopt)
formula : ln(1+ax+by+c*z)
a, b, c are numbers input by the user
x, y, z are variables to be derived
variable constraint is that x, y, z are positive and x+y+z<=1
This can actually be transformed into a linear optimization problem.
max ln(1+ax+by+cz) <--> max (ax+by+cz) s.t. ax+by+cz > -1
This means that it is a linear optimization problem (with one more constraint) that you can easily handle with whatever C++ methods together with your Convex Optimization knowledge.
Reminders to write a good code:
Check the validity of input value.
Since the input value can be negative, you need to consider this circumstance which can yield different results.
P.S.
This problem seems to be Off Topic on SO.
If it is your homework, it is for your own good to write the code yourself. Besides, we do not have enough time to write that for you.
This should have been a comment if I had more reputation.
I am new to Cplex. I'm solving an integer programming problem, but I have a problem with objective function.
Problem is that I have some project, that have a due date D, and if project is tardy, than I have a tardiness penalty b, so it looks like b*(cn-D). Where cn is a real complection time of a project, and it is decision variable.
It must look like this
if (cn-D)>=0 then b*(cn-D)==0
I tried to use "if-then" constraint, but seems that it isn't working with decision variable.
I looked on question similar to this, but but could not find solution. Please, help me define correct objective function.
The standard way to model this is:
min sum(i, penalty(i)*Tardy(i))
Tardy(i) >= CompletionTime(i) - DueDate(i)
Tardy(i) >= 0
Tardy is a non-negative variable and can never become negative. The other quantities are:
penalty: a constant indicating the cost associated with job i being tardy one unit of time.
CompletionTime: a variable that holds the completion time of job i
DueDate: a constant with the due date of job i.
The above measures the sum. Sometimes we also want to measure the count: the number of jobs that are tardy. This is to prevent many jobs being tardy. In the most general case one would have both the sum and the count in the objective with different weights or penalties.
There is a virtually unlimited number of papers showing MIP formulations on scheduling models that involve tardiness. Instead of reinventing the wheel, it may be useful to consult some of them and see what others have done to formulate this.
I've been reading this website: http://www.csimn.com/CSI_pages/PIDforDummies.html and I'm confused about the proportional integral part. Here's what it says.
Proportional control
Here’s a diagram of the controller when we have enabled only P control:
In Proportional Only mode, the controller simply multiplies the Error by the Proportional Gain (Kp) to get the controller output.
The Proportional Gain is the setting that we tune to get our desired performance from a “P only” controller.
A match made in heaven: The P + I Controller
If we put Proportional and Integral Action together, we get the humble PI controller. The Diagram below shows how the algorithm in a PI controller is calculated.
The tricky thing about Integral Action is that it will really screw up your process unless you know exactly how much Integral action to apply.
A good PID Tuning technique will calculate exactly how much Integral to apply for your specific process - but how is the Integral Action adjusted in the first place?
As you can see, the proportional part is easy to understand it says that you multiply error by tuning variable. The part that I don't get is where you get the P and I from on the second part, and what mathematical operation you do with them. I don't have a degree in mathematics or advanced calculus knowledge, so I would appreciate it if you would try to keep it algebra level.
There is a big part missing from the text, the actual physical system that turns the control into a process and the actual physical variable.
Think of the integral as some kind of averaging operation that filters out small oscillations in the PV input. It also represents some kind of memory of the immediate past of the process.
A moving exponential average, for instance, can be thought of being a mix of integral and proportional action.
Staying with the car driving example, if you come to a curb where you need the steering wheel in a certain position to go in a circle, you don't just yank the wheel to that position, you move it gradually (most of the time). Exactly such ramp-up and -down actions are effects of using the integral action part.
I integral part is just summation also multiplied by some constant.
Analogue integration is done by nonlinear gain and amplifier.
Digital integration of first order is just:
output += input*dt;
second order is:
temp += input*dt;
output += temp*dt;
dt is the duration time of iteration loop (timer or what ever)
do not forget that PI regulator can have more complicated response
i1 += input*dt;
i2 += i1*dt;
i3 += i2*dt;
output = a0*input + a1*i1 + a2*i2 +a3*i3 ...;
where a0 is the P part
Now the I regulator adds more and more amount of control value
until the controlled value is the same as the preset value
the longer it takes to match it the faster it controls
this creates fast oscillations around preset value
in comparison to P with the same gain
but in average the control time is smaller then in just P regulators
therefore the I gain is usually much much smaller which creates the memory and smooth effect LutzL mentioned. (while the regulation time is similar or smaller then just for P regulation)
The controlled device has its own response
this can be represented as differential function
there is a lot of theory in cybernetics about obtaining the right regulator response
to match your process needs as:
quality of control
reaction times
max oscillations amplitude
stability
but for all you need differential math like solving system of differential equations of any order
strongly recommend use of Laplace transform
but many people also use Z transform instead
So I-regulator add speed to regulation
but it also create bigger oscillations
and when not matching the regulated system properly also creates instability
Integration adds overflow risks to regulation (Analog integration is very sensitive to it)
Also take in mind you can also substracting the I part from control value
which will make the exact opposite
sometimes the combination of more I parts are used to match desired regulation response shape
I can't really find any answer in the Modelica specification so ill ask you guys. The specification states that
A tool is free to solve equations, reorder expressions and to not evaluate expressions if their values do not influence the result (e.g. short-circuit evaluation of Boolean expressions). If-statements and if-expressions guarantee that their clauses are only evaluated if the appropriate condition is true, but relational operators generating state or time events will during continuous integration have the value from the most recent event.
If a numeric operation overflows the result is undefined. For literals it is recommended to automatically convert the number to another type with greater precision.
Now, I wonder, can the tool choose to evaluate an expression several time in an integrator step? For example (probably not an valid example, just to give you guys an idea of what I was wondering :) )
Real x;
equation
der(x) = -t;
Modelica.Utilities.Streams.print(String(time));
This will print the same time for several times, so I figured that there is some kind of iteration going on. But I would really like to have it confirmed by some source.
That is normal.
Variable step size solvers (like dassl) can go back
and forth in time to find the direction of the curve.
Also, if you have events more values can be generated
at the same time.
If you want to print time or values just at exact time instants you need when equations:
when sample(0, 1) then
Modelica.Utilities.Streams.print(String(time));
end when;
Read more in the Modelica Spec about sample.
Is also possible to use fixed step size solvers like Euler or so.
I just implemented the numerical integration for a set of coupled ODEs
from a discretized PDE using the odeint C++ library. It works nicely and
is lightning fast, but there is one issue:
My system of ODEs has, so-called, absorbing boundary conditions: the time
derivatives of my state variable, n, which is a vector of N doubles
(a population density) gets calculated in the system function, but before that happens
(or after the time integration) I would like to set:
n[N]=n[N-2];
n[N-1]=n[N-2];
However, of course this doesn't work because the state variable in the system
function is declared as const, and it looks as if this could not be changed
other than through meddling with the library... is there any way around this?
I should mention that setting dndt[N] and dndt[N-1] to zero might look like a
solution, but it doesn't really help as it defies the concept of absorbing boundary
conditions (n[N] and n[N-1] would then always have the values they had at t=0, rather
then the value of n[N-2] at any point in time), and so I'd really prefer to change n.
Thanks for any help!
Regards,
Michael
Usually absorbing boundary condition manifests itself in the equations of motion. n[N] = n[N-1] = n[N-2], so can insert n[N]=n[N-2] and n[N-1]=n[N-2] into the equation for dndt[N-2].
For example, the discrete Laplacian Lx[i] = x[i+1]-2 x[i] +x[i-1] with absorbing boundaries x[n]=x[n-1] can be written as Lx[n-1] = x[n-2] - x[n-1]. The equation for x[n] can then be omitted.