I have an equation of the form
U=T1×S1+T2×S2+T3×S3
where U = utilization
Ti = throughput of i
Si = service time of i
constraint, S1+S2+S3≤R where R is response time U,T1,T2,T3,S1,S2,S3 are single values I know the value of U,Ti and R. From this how will I determine S1,S2,S3.
Please suggest me any mathematical method using which I can solve this.
To transform the inequality into an equality you could add a non negative slack variable to the left hand side of the constraint.
You should expect to get many solutions to this, because you have 2 equations and 3 variables (+1slack).
Related
In traditional Simplex Algorithm notation, we have x at the current basis selection B as so:
xB = AB-1b - AB-1ANxN. How can I compute the AB-1AN term inside a separator in SCIP, or at least iterate over its columns?
I see three helpful methods: getLPColsData, getLPRowsData, getLPBasisInd. I'm just not sure exactly what data those methods represent, particularly the last one, with its negative row indexes. How do I use those to get the value I want?
Do those methods return the same data no matter what LP algorithm is used? Or do I need to account for dual vs primal? How does the use of the "revised" algorithm play into my calculation?
Update: I discovered the getLPBInvARow and getLPBInvRow. That seems to be much closer to what I'm after. I don't yet understand their results; they seem to include more/less dimensions than expected. I'm still looking for understanding at how to use them to get the rays away from the corner.
you are correct that getLPBInvRow or getLPBInvARow are the methods you want. getLPBInvARow directly returns you a of the simplex tableau, but it is not more efficient to use than getLPBInvRow and doing the multiplication yourself since the LP solver needs to also compute the actual tableau first.
I suggest you look into either sepa_gomory.c or sepa_gmi.c for examples of how to use these methods. How do they include less dimensions than expected? They both return sparse vectors.
I want to try to solve a system of ordinary differential equations, perhaps parallelized and came across Julia and DifferentialEquations.jl. the system looks like
x'(t) = f(t)*z(t)
y'(t) = g(t)*z(t)
z'(t) = f(t)*(1-2*x(t))/(2) -g(t)*y(t)
over 10^2 < t < 10^14, but my initial boundary conditions are
x(10^14) == 0
y(10^14) == 0
z(10^14) == 0
Could someone please explain to me how to setup this problem in julia? I checked the documentation and could only find u0 as a parameter, but it doesn't give details on choosing for a right handed set of boundary conditions Many thanks!
You're looking to solve a boundary value problem (BVP). While this area is currently less developed than other areas of DifferentialEquations.jl, there are methods which exist for this which are shown in the tutorial on solving BVPs. The MIRK4 method may be the one to try.
I will note however that your timescale is quite large and may lead to numerical errors. Either using higher precision numbers (BigFloat, ArbFloat, DoubleFloat) may be required for handling that range, or you may want to rescale time in your equations so that way it better fits for standard double precision floating point numbers (Float64).
I am writing web crawler scheduler and have run into problems. First I will describe how I'm trying to find optimal schedule for when my crawler is visiting the page and then I will present my problem.
Scheduler definition
Scheduler is based on this paper "Optimal crawling strategies for web search engines" by J.Wolf. The paper proposes that update times of web pages follow exponential distribution with parameter λ. The problem is finding optimal number of times xi, the page i will be crawled in time interval [0,T]. The function proposed is:
Because this function is convex and its input arguments xi is discrete this kind of problem can be solved using algorithm suggested by Fredrerickson and Johnson in "The Complexity of Selection and Ranking in X + Y and Matrices with sorted columns", that has time complexity O(max{N, log(R/N)}). The optimization algorithm solves the problem by finding N-th element in [RxN] matrix where element at position (i, j) is equal to derivation of j function with input argument x = i, where derivation dj(xi) is equal to:
Because function fi is convex that means that function di has property that is monotonically increasing (matrix has sorted columns).
Problems
I run into problems when evaluating derivation, because of rounding errors d(x+1) - d(x), did not have guarantee to be greater or equal to 0, and I'm not sure that values that I got from optimizer are optional values. Rounding errors happen because value of x can be only positive integers in range of 0 to few billions, therefor exponent in function f is either big negative number or extremely small number (-5000).
Failed Attempts
The first thing I tried, I downloaded arbitrary precision library. This solved my problem but the overhead of library is to big.
The second thing I tried was I expanded d and got function like:
and then I tried to compare dj(xi) and dk(xw) by comparing their terms individually and than try to deduce is dj is bigger or smaller or greater than dk. If I could compare derivation I could solve my problem because optimization algorithm does not need concrete values, instead it only need relations between values. I couldn't find the solution because the term w.
I also tried looking at log(dj(xi)) because log preserves function monotony, but log also had rounding errors and I couldn't compare log(dj) and log(dk) without computing the final values.
If anybody has any other solution that could potentially work I would be most graceful.
I'm modelling an overhead crane and obtained the following equations:
I'm noob when it comes to Scilab and so far I only simullated (using ODE) linear systems with no more than two degrees of freedom, which are simple systems that I can easily convert to am matrix and integrate it using ODE.
But this system in particular I have no clue how to simulate it, not because of the sin and cos functions, but because of the fact that I don't know how to put it in a state space matrix.
I've looked for a few tutorials (listed bellow) but I didn't understand any of those, can somebody tell me how I do it, or at least point where I could learn it?
http://www.openeering.com/sites/default/files/Nonlinear_Systems_Scilab.pdf
http://www.math.univ-metz.fr/~sallet/ODE_Scilab.pdf
Thank you, and sorry about my english
The usual form means writing in terms of first order derivatives. So you'll have relations where the 2nd derivative terms will be written as:
x'' = d(x')/fx
Substitute these into the equations you have. You'll end up with eight simultaneous ODEs to solve instead of four, with appropriate initial conditions.
Although this ODE system is implicit, you can solve it with a classical (explicit) ODE solver by reformulating it this way: if you define X=(x,L,theta,q)^T then your system can be reformulated using matrix algebra as A(X,X') * X" = B(X,X'). Please note that the first order form of this system is
d/dt(X,X') = ( X', A(X,X')^(-1)*B(X,X') )
Suppose now that you have defined two Scilab functions A and B which actually compute their values w.r.t. to the values of Xand X'
function out = A(X,Xprime)
x=X(1)
L=X(2)
theta=X(3)
qa=X(4)
xd=XPrime(1)
Ld=XPrime(2)
thetad=XPrime(3)
qa=XPrime(4);
...
end
function out = B(X,Xprime)
...
end
then the right hand side of the system of 8 ODEs, as it can be given to the ode function of Scilab can be coded as follows
function dstate_dt = rhs(t,state)
X = state(1:4);
Xprime = state(5:8);
out = [ Xprime
A(X,Xprime) \ B(X,Xprime)]
end
Writing the code of A() and B() according to the given equations is the only remaining (but quite easy) task.
I am currently learning Modelica by trying some very simple examples. I have defined a connector Incompressible for an incompressible fluid like this:
connector Incompressible
flow Modelica.SIunits.VolumeFlowRate V_dot;
Modelica.SIunits.SpecificEnthalpy h;
Modelica.SIunits.Pressure p;
end Incompressible;
I now wish to define a mass or volume flow source:
model Source_incompressible
parameter Modelica.SIunits.VolumeFlowRate V_dot;
parameter Modelica.SIunits.Temperature T;
parameter Modelica.SIunits.Pressure p;
Incompressible outlet;
equation
outlet.V_dot = V_dot;
outlet.h = enthalpyWaterIncompressible(T); // quick'n'dirty enthalpy function
outlet.p = p;
end Source_incompressible;
However, when checking Source_incompressible, I get this:
The problem is structurally singular for the element type Real.
The number of scalar Real unknown elements are 3.
The number of scalar Real equation elements are 4.
I am at a loss here. Clearly, there are three equations in the model - where does the fourth equation come from?
Thanks a lot for any insight.
Dominic,
There are a couple of issues going on here. As Martin points out, the connector is unbalanced (you don't have matching "through" and "across" pairs in that connector). For fluid systems, this is acceptable. However, intensive fluid properties (e.g., enthalpy) have to be marked as so-called "stream" variables.
This topic is, admittedly, pretty complicated. I'm planning on adding an advanced chapter to my online Modelica book on this topic but I haven't had the time yet. In the meantime, I would suggest you have a look at the Modelica.Fluid library and/or this presentation by one of its authors, Francesco Casella.
That connector is not a physical connector. You need one flow variable for each potential variable. This is the OpenModelica error message if it helps a little:
Warning: Connector .Incompressible is not balanced: The number of potential variables (2) is not equal to the number of flow variables (1).
Error: Too many equations, over-determined system. The model has 4 equation(s) and 3 variable(s).
Error: Internal error Found Equation without time dependent variables outlet.V_dot = V_dot
This is because the unconnected connector will generate one equation for the flow:
outlet.V_dot = 0.0;
This means outlet.V_dot is replaced in:
outlet.V_dot = V_dot;
And you get:
0.0 = V_dot;
But V_dot is a parameter and can not be assigned to in an equation section (needs an initial equation if the parameter has fixed=false, or a binding equation in the default case).