Stopwatch function in R - matlab

Is there an R timer or stopwatch function similar to MATLAB's tic/toc?

There are plenty of profiling tools in R, as Dirk mentioned. If you want the simplicity of tic/toc, then you can do it in R too.
EDIT: I've cannibalised the garbage collection functionality from the MATLAB package, and tic now lets you choose whether you are interested in total elapsed time or just the user time.
tic <- function(gcFirst = TRUE, type=c("elapsed", "user.self", "sys.self"))
{
type <- match.arg(type)
assign(".type", type, envir=baseenv())
if(gcFirst) gc(FALSE)
tic <- proc.time()[type]
assign(".tic", tic, envir=baseenv())
invisible(tic)
}
toc <- function()
{
type <- get(".type", envir=baseenv())
toc <- proc.time()[type]
tic <- get(".tic", envir=baseenv())
print(toc - tic)
invisible(toc)
}
Usage is, e.g., tic(); invisible(qr(matrix(runif(1e6), nrow=1e3))); toc()

There is a MATLAB emulation package matlab on CRAN. It has implementations of tic and toc (but they look very similar to the functions in Richie Cottons answer; "elapsed" is used instead of "user.self" in proc.time())
> tic
function (gcFirst = FALSE)
{
if (gcFirst == TRUE) {
gc(verbose = FALSE)
}
assign("savedTime", proc.time()[3], envir = .MatlabNamespaceEnv)
invisible()
}
<environment: namespace:matlab>
> toc
function (echo = TRUE)
{
prevTime <- get("savedTime", envir = .MatlabNamespaceEnv)
diffTimeSecs <- proc.time()[3] - prevTime
if (echo) {
cat(sprintf("elapsed time is %f seconds", diffTimeSecs),
"\n")
return(invisible())
}
else {
return(diffTimeSecs)
}
}
<environment: namespace:matlab>

A very simple equivalence with tic and toc that you could have:
tic=proc.time()[3]
...code...
toc=proc.time()[3] - tic
Where the [3] is because we are interested in the third element in the vector returned by proc.time(), which is elapsed time.

Direct equivalents of tic and toc do not exist.
Please see help(system.time) as well as the R Extensions manual about profiling. Discussions of profiling and profiling tools is also in the 'Intro to HPC with R' slides referenced on the High Performance Computing with R taskview

A Closure Approach
A very clean and simple way to do this is by using a closure (which just means having a function within a function):
tic <- function () {
now <- proc.time()
function () {
proc.time() - now
}
}
You start the timer like so:
toc <- tic()
And then you get the time back like this:
toc()
Which outputs a named vector that gets printed like so:
user system elapsed
0.008 0.004 2.055
Even with the simplicity of this version you also get all the functionality of the Matlab and Richie Cotton's versions plus the added feature of being able to run multiple timers:
toc1 <- tic()
toc2 <- tic()

There is a relatively new package tictoc that replicates the features exactly as you would use them in Matlab.
http://cran.r-project.org/web/packages/tictoc/index.html
## Basic use case
tic()
print("Do something...")
Sys.sleep(1)
toc()
# 1.034 sec elapsed

As of the date 2015-03-25,
and possibly earlier,
the pracma
package contains the functions tic() and toc().
Example:
> library(pracma)
> tic()
> for(i in 1:10000) mad(runif(10000)) # kill time
> toc()
elapsed time is 18.610000 seconds

No, but here is a one line solution.
time.it<-function(f) { a<-proc.time(); out<-f(); print(proc.time()-a); out }
And an example for usage:
result<-time.it(function(){ A<-matrix(runif(5000^2),nrow=5000); b<-runif(5000); solve(A,b) } )
user system elapsed
12.788 12.268 8.623
Otherwise, microbenchmark is my favorite in terms of packages.

Just for completeness: you can actually 'simulate' tic
and toc in R, so that you can write
tic
## do something
toc
without parentheses. The trick is to abuse the print
function, as demonstrated in Fun: tic and toc in R:
tic <- 1
class(tic) <- "tic"
toc <- 1
class(toc) <- "toc"
print.tic <- function(x, ...) {
if (!exists("proc.time"))
stop("cannot measure time")
gc(FALSE)
assign(".temp.tictime", proc.time(), envir = .GlobalEnv)
}
print.toc <- function(x,...) {
if (!exists(".temp.tictime", envir = .GlobalEnv))
stop("did you tic?")
time <- get(".temp.tictime", envir = .GlobalEnv)
rm(".temp.tictime", envir = .GlobalEnv)
print(res <- structure(proc.time() - time,
class = "proc_time"), ...)
invisible(res)
}
So typing
tic
Sys.sleep(2)
toc
should results in something like this:
user system elapsed
0.000 0.000 2.002
As I said, it's a trick; system.time, Rprof and
packages such as rbenchmark are the way to measure
computing time in R.

install.packages("tictoc")
library(tictoc)
# Timing nested code.
# The string provided in the call to tic() becomes a prefix to the output of toc()
tic("outer")
Sys.sleep(1)
tic("middle")
Sys.sleep(2)
tic("inner")
Sys.sleep(3)
toc() # inner
# inner: 3.004 sec elapsed
toc() # middle
# middle: 5.008 sec elapsed
toc() # outer
# outer: 6.016 sec elapsed
The tictoc package implements the functionality described by previous answers - thank you for inspiration! The package also adds nested timing, collecting the timings in user-defined variables, custom messages and callbacks.

Related

interrupt a TCP-IP callback function in matlab

Recently wrote code that establishes a connection between two instances of matlab. I can send messages through the TCP-IP connection which will execute code. Now I'm trying to setup the code to be interruptible as I would like to start/stop a function through TCP-IP. Problem though is that sending a second command does nothing until the function is completed. Is there a way to interrupt a TCP-IP callback function?
code:
classdef connectcompstogether<handle
properties
serverIP
clientIP
tcpipServer
tcpipClient
Port = 4000;
bsize = 8;
earlystop
end
methods
function gh = connectcompstogether(~)
% gh.serverIP = '127.0.0.1';
gh.serverIP = 'localhost';
gh.clientIP = '0.0.0.0';
end
function SetupServer(gh)
gh.tcpipServer = tcpip(gh.clientIP,gh.Port,'NetworkRole','Server');
set(gh.tcpipServer,'OutputBufferSize',gh.bsize);
fopen(gh.tcpipServer);
display('Established Connection')
end
function SetupClient(gh)
gh.tcpipClient = tcpip(gh.serverIP,gh.Port,'NetworkRole','Client');
set(gh.tcpipClient, 'InputBufferSize',gh.bsize);
set(gh.tcpipClient, 'BytesAvailableFcnCount',8);
set(gh.tcpipClient, 'BytesAvailableFcnMode','byte');
set(gh.tcpipClient, 'BytesAvailableFcn', #(h,e)gh.recmessage(h,e));
fopen(gh.tcpipClient);
display('Established Connection')
end
function CloseClient(gh)
fclose(gh.tcpipClient);
gh.tcpipClient = [];
end
end
methods
function sendmessage(gh,message)
fwrite(gh.tcpipServer,message,'double');
end
function recmessage(gh,h,e)
Message = fread(gh.tcpipClient,gh.bsize/8,'double');
if Message == 444
gh.Funwithnumbers();
elseif Message == 777
gh.earlystop = 1;
end
end
function Funwithnumbers(gh)
x=1;
while true
if x > 5000, break;end
if gh.earlystop == 1,break;end
x = x+1;
display(x)
end
end
end
end
for ease to understand code.
server
Ser = connectcompstogether;
ser.SetupServer();
ser.sendmessage(333);
Client
cli = connectcompstogether;
cli.SetupClient();
Update:
So after going through the web, I have found out based on this post that the tcpip callback cannot be interrupt. The post was in 2017 which means my 2016a version definitely cannot interrupt a callback.
So An update to my question, Is it possible to start a subprocess in matlab to run the function. I just want to use the callback to start code. If I can start a subprocess from the callback. Than I should be able to free up the main process and use tcpip to start/stop a function on a different computer.
Update 2:
So I tried to utilize parallel processing using the 'spmd' command but the problem still persisted.
function recmessage(gh,h,e)
Message = fread(gh.tcpipClient,gh.bsize/8,'double');
spmd
switch labindex
case 1
if Message == 444
gh.Funwithnumbers();
elseif Message == 777
gh.earlystop = 1;
end
end
end
end
You may use a timer object, which is convenient to delay the execution of some function.
t=timer('ExecutionMode','singleShot', 'StartDelay',0, 'TimerFcn',#myCallback);
start(t);
In this case, the StartDelay is 0, so myCallback will be almost immediately added to the queue of tasks to be processed by Matlab. The execution however will start only after the callback to the tcpip object has been completed. It will block the queue once started, however.
You may try something like:
properties
t=timer('ExecutionMode','singleShot', 'StartDelay',0, 'TimerFcn',#myCallback);
end
function tcpipCallback(gh,tcpObj,~)
message=fread(tcpObj,1,'double');
if message==444
if strcmp(get(t,'Running'),'on')
error('The function is running already');
else
set(gh.t,'UserData',false);
start(gh.t);
end
elseif message==777
set(gh.t,'UserData',true);
end
function myCallback(tObj,~)
ii=0;
while ii<5000
if get(tObj,'UserData'),break,end
ii=ii+1;
pause(.0001); %Pause to interrupt the callback; drawnnow might work too; or perhaps this is not needed at all.
end
end

Plotting 3 variables against 1 variable

I have a code that computes the max value. this code consists of four variables www is the function of a,b, and c labaled xx, yy, and zz respectively, so my question is how can i plot www against xx,yy, and zz? Thanks for helping
objfun file
function f=W4qubit(x,a,b,c,d)
c1=-cos(x(1))*(cos(x(5))*(cos(x(9))*(cos(x(13))-cos(x(15)))-cos(x(11))*(cos(x(13))+cos(x(15))))+...
cos(x(7))*(cos(x(11))*(cos(x(15))-cos(x(13)))-cos(x(9))*(cos(x(13))+cos(x(15)))))-...
cos(x(3))*(cos(x(5))*(cos(x(11))*(cos(x(15))-cos(x(13)))-cos(x(9))*(cos(x(13))+cos(x(15))))-...
cos(x(7))*(cos(x(9))*(cos(x(13))-cos(x(15)))-cos(x(11))*(cos(x(13))+cos(x(15)))));
c2=cos(x(1))*(cos(x(5))*(sin(x(9))*(sin(x(13))*cos(x(10)-x(14))-sin(x(15))*cos(x(10)-x(16)))-...
sin(x(11))*(sin(x(13))*cos(x(12)-x(14))+sin(x(15))*cos(x(12)-x(16))))+...
cos(x(7))*(sin(x(11))*(sin(x(15))*cos(x(12)-x(16))-sin(x(13))*cos(x(12)-x(14)))-...
sin(x(9))*(sin(x(13))*cos(x(10)-x(14))+sin(x(15))*cos(x(10)-x(16)))))+...
cos(x(3))*(cos(x(5))*(sin(x(11))*(sin(x(15))*cos(x(12)-x(16))-sin(x(13))*cos(x(12)-x(14)))-...
sin(x(9))*(sin(x(13))*cos(x(10)-x(14))+sin(x(15))*cos(x(10)-x(16))))-...
cos(x(7))*(sin(x(9))*(sin(x(13))*cos(x(10)-x(14))-sin(x(15))*cos(x(10)-x(16)))-...
sin(x(11))*(sin(x(13))*cos(x(12)-x(14))+sin(x(15))*cos(x(12)-x(16)))));
c3=cos(x(1))*(sin(x(5))*(cos(x(9))*(sin(x(13))*cos(x(6)-x(14))-sin(x(15))*cos(x(6)-x(16)))-...
cos(x(11))*(sin(x(13))*cos(x(6)-x(14))+sin(x(15))*cos(x(6)-x(16))))+...
sin(x(7))*(cos(x(11))*(sin(x(15))*cos(x(8)-x(16))-sin(x(13))*cos(x(8)-x(14)))-...
cos(x(9))*(sin(x(13))*cos(x(8)-x(14))+sin(x(15))*cos(x(8)-x(16)))))+...
cos(x(3))*(sin(x(5))*(cos(x(11))*(sin(x(15))*cos(x(6)-x(16))-sin(x(13))*cos(x(6)-x(14)))-...
cos(x(9))*(sin(x(13))*cos(x(6)-x(14))+sin(x(15))*cos(x(6)-x(16))))-...
sin(x(7))*(cos(x(9))*(sin(x(13))*cos(x(8)-x(14))-sin(x(15))*cos(x(8)-x(16)))-...
cos(x(11))*(sin(x(13))*cos(x(8)-x(14))+sin(x(15))*cos(x(8)-x(16)))));
c4=cos(x(1))*(sin(x(5))*(sin(x(9))*cos(x(6)-x(10))*(cos(x(13))-cos(x(15)))-sin(x(11))*cos(x(6)-x(12))*(cos(x(13))+cos(x(15))))+...
sin(x(7))*(sin(x(11))*cos(x(8)-x(12))*(cos(x(15))-cos(x(13)))-sin(x(9))*cos(x(8)-x(10))*(cos(x(13))+cos(x(15)))))+...
cos(x(3))*(sin(x(5))*(sin(x(11))*cos(x(6)-x(12))*(cos(x(15))-cos(x(13)))-sin(x(9))*cos(x(6)-x(10))*(cos(x(13))+cos(x(15))))-...
sin(x(7))*(sin(x(9))*cos(x(8)-x(10))*(cos(x(13))-cos(x(15)))-sin(x(11))*cos(x(8)-x(12))*(cos(x(13))+cos(x(15)))));
c5=sin(x(1))*(cos(x(5))*(cos(x(9))*(sin(x(13))*cos(x(2)-x(14))-sin(x(15))*cos(x(2)-x(16)))-...
cos(x(11))*(sin(x(13))*cos(x(2)-x(14))+sin(x(15))*cos(x(2)-x(16))))+...
cos(x(7))*(cos(x(11))*(sin(x(15))*cos(x(2)-x(16))-sin(x(13))*cos(x(2)-x(14)))-...
cos(x(9))*(sin(x(13))*cos(x(2)-x(14))+sin(x(15))*cos(x(2)-x(16)))))+...
sin(x(3))*(cos(x(5))*(cos(x(11))*(sin(x(15))*cos(x(4)-x(16))-sin(x(13))*cos(x(4)-x(14)))-...
cos(x(9))*(sin(x(13))*cos(x(4)-x(14))+sin(x(15))*cos(x(4)-x(16))))-...
cos(x(7))*(cos(x(9))*(sin(x(13))*cos(x(4)-x(14))-sin(x(15))*cos(x(4)-x(16)))-...
cos(x(11))*(sin(x(13))*cos(x(4)-x(14))+sin(x(15))*cos(x(4)-x(16)))));
c6=sin(x(1))*(cos(x(5))*(sin(x(9))*cos(x(2)-x(10))*(cos(x(13))-cos(x(15)))-sin(x(11))*cos(x(2)-x(12))*(cos(x(13))+cos(x(15))))+...
cos(x(7))*(sin(x(11))*cos(x(2)-x(12))*(cos(x(15))-cos(x(13)))-sin(x(9))*cos(x(2)-x(10))*(cos(x(13))+cos(x(15)))))+...
sin(x(3))*(cos(x(5))*(sin(x(11))*cos(x(4)-x(12))*(cos(x(15))-cos(x(13)))-sin(x(9))*cos(x(4)-x(10))*(cos(x(13))+cos(x(15))))-...
cos(x(7))*(sin(x(9))*cos(x(4)-x(10))*(cos(x(13))-cos(x(15)))-sin(x(11))*cos(x(4)-x(12))*(cos(x(13))+cos(x(15)))));
c7=sin(x(1))*(sin(x(5))*cos(x(2)-x(6))*(cos(x(9))*(cos(x(13))-cos(x(15)))-cos(x(11))*(cos(x(13))+cos(x(15))))-...
sin(x(7))*cos(x(2)-x(8))*(cos(x(11))*(cos(x(15))-cos(x(13)))-cos(x(9))*(cos(x(13))+cos(x(15)))))+...
sin(x(3))*(sin(x(5))*cos(x(4)-x(6))*(cos(x(11))*(cos(x(15))-cos(x(13)))-cos(x(9))*(cos(x(13))+cos(x(15))))-...
sin(x(7))*cos(x(4)-x(8))*(cos(x(9))*(cos(x(13))-cos(x(15)))-cos(x(11))*(cos(x(13))+cos(x(15)))));
A2=2*a*b;
A3=2*a*c;
A4=2*b*c;
A5=2*a*d;
A6=2*b*d;
A7=2*c*d;
f1=c1+A2*c2+A3*c3+A4*c4+A5*c5+A6*c6+A7*c7;
f=-(f1^2);
my main file of the code
clear
close
clc
%x=[x(1),x(2),x(3),x(4),x(5),x(6),x(7),x(8),x(9),x(10),x(11),x(12),x(13),x(14),x(15),x(16)]; % angles;
lb=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
ub=[pi,2*pi,pi,2*pi,pi,2*pi,pi,2*pi,pi,2*pi,pi,2*pi,pi,2*pi,pi,2*pi];
options = optimoptions(#fmincon,'TolX',10^-12,'MaxIter',1500,'MaxFunEvals',10^8,'Algorithm','sqp','TolFun',10^-8);
a=0:0.1:1;
b=0:0.1:1;
c=0:0.1:1;
w=NaN(length(a),length(b),length(c));
ww=NaN(length(a),length(b),length(c));
www=NaN(length(a),length(c));
for k=1:100
x0=rand([1,16]).*ub*.9986;%7976
for i=1:length(a)
for j=1:length(b)
for l=1:length(c)
dhelp=1-(a(i)^2)-(b(j)^2)-(c(l)^2);
if (dhelp>0 || dhelp==0)
d=sqrt(dhelp);
[~,fval]=fmincon(#(x)W4qubit(x,a(i),b(j),c(l),d),x0,[],[],[],[],lb,ub,[],options);
w(i,j,l)=sqrt(-fval);
else
w(i,j,l)=nan;
end
ww=max(w,ww);
end
end
end
end
www=max(ww,[],3);
yy=b.^2;xx=a.^2;zz=c.^2;
meshc(xx,yy,www)
grid on
zlabel('\fontname{Times New Roman} M_{max}')
xlabel('\fontname{Times New Roman}\alpha^2')
ylabel('\fontname{Times New Roman}\gamma^2')
%title('fontname{Times New Roman} Maximum of the Svetlichny operator. Method 1 (alpha|0001>+beta|0010>+gamma|1000>)')
Not sure, but doesn't
plot(www,[xx;yy;zz]);
do the job for you? I do not have the optimization toolbox, so I can't test your script. But in principle, this should work.

Simpy: How can I represent failures in a train subway simulation?

New python user here and first post on this great website. I haven't been able to find an answer to my question so hopefully it is unique.
Using simpy I am trying to create a train subway/metro simulation with failures and repairs periodically built into the system. These failures happen to the train but also to signals on sections of track and on plaforms. I have read and applied the official Machine Shop example (which you can see resemblance of in the attached code) and have thus managed to model random failures and repairs to the train by interrupting its 'journey time'.
However I have not figured out how to model failures of signals on the routes which the trains follow. I am currently just specifying a time for a trip from A to B, which does get interrupted but only due to train failure.
Is it possible to define each trip as its own process i.e. a separate process for sections A_to_B and B_to_C, and separate platforms as pA, pB and pC. Each one with a single resource (to allow only one train on it at a time) and to incorporate random failures and repairs for these section and platform processes? I would also need to perhaps have several sections between two platforms, any of which could experience a failure.
Any help would be greatly appreciated.
Here's my code so far:
import random
import simpy
import numpy
RANDOM_SEED = 1234
T_MEAN_A = 240.0 # mean journey time
T_MEAN_EXPO_A = 1/T_MEAN_A # for exponential distribution
T_MEAN_B = 240.0 # mean journey time
T_MEAN_EXPO_B = 1/T_MEAN_B # for exponential distribution
DWELL_TIME = 30.0 # amount of time train sits at platform for passengers
DWELL_TIME_EXPO = 1/DWELL_TIME
MTTF = 3600.0 # mean time to failure (seconds)
TTF_MEAN = 1/MTTF # for exponential distribution
REPAIR_TIME = 240.0
REPAIR_TIME_EXPO = 1/REPAIR_TIME
NUM_TRAINS = 1
SIM_TIME_DAYS = 100
SIM_TIME = 3600 * 18 * SIM_TIME_DAYS
SIM_TIME_HOURS = SIM_TIME/3600
# Defining the times for processes
def A_B(): # returns processing time for journey A to B
return random.expovariate(T_MEAN_EXPO_A) + random.expovariate(DWELL_TIME_EXPO)
def B_C(): # returns processing time for journey B to C
return random.expovariate(T_MEAN_EXPO_B) + random.expovariate(DWELL_TIME_EXPO)
def time_to_failure(): # returns time until next failure
return random.expovariate(TTF_MEAN)
# Defining the train
class Train(object):
def __init__(self, env, name, repair):
self.env = env
self.name = name
self.trips_complete = 0
self.broken = False
# Start "travelling" and "break_train" processes for the train
self.process = env.process(self.running(repair))
env.process(self.break_train())
def running(self, repair):
while True:
# start trip A_B
done_in = A_B()
while done_in:
try:
# going on the trip
start = self.env.now
yield self.env.timeout(done_in)
done_in = 0 # Set to 0 to exit while loop
except simpy.Interrupt:
self.broken = True
done_in -= self.env.now - start # How much time left?
with repair.request(priority = 1) as req:
yield req
yield self.env.timeout(random.expovariate(REPAIR_TIME_EXPO))
self.broken = False
# Trip is finished
self.trips_complete += 1
# start trip B_C
done_in = B_C()
while done_in:
try:
# going on the trip
start = self.env.now
yield self.env.timeout(done_in)
done_in = 0 # Set to 0 to exit while loop
except simpy.Interrupt:
self.broken = True
done_in -= self.env.now - start # How much time left?
with repair.request(priority = 1) as req:
yield req
yield self.env.timeout(random.expovariate(REPAIR_TIME_EXPO))
self.broken = False
# Trip is finished
self.trips_complete += 1
# Defining the failure
def break_train(self):
while True:
yield self.env.timeout(time_to_failure())
if not self.broken:
# Only break the train if it is currently working
self.process.interrupt()
# Setup and start the simulation
print('Train trip simulator')
random.seed(RANDOM_SEED) # Helps with reproduction
# Create an environment and start setup process
env = simpy.Environment()
repair = simpy.PreemptiveResource(env, capacity = 1)
trains = [Train(env, 'Train %d' % i, repair)
for i in range(NUM_TRAINS)]
# Execute
env.run(until = SIM_TIME)
# Analysis
trips = []
print('Train trips after %s hours of simulation' % SIM_TIME_HOURS)
for train in trains:
print('%s completed %d trips.' % (train.name, train.trips_complete))
trips.append(train.trips_complete)
mean_trips = numpy.mean(trips)
std_trips = numpy.std(trips)
print "mean trips: %d" % mean_trips
print "standard deviation trips: %d" % std_trips
it looks like you are using Python 2, which is a bit unfortunate, because
Python 3.3 and above give you some more flexibility with Python generators. But
your problem should be solveable in Python 2 nonetheless.
you can use sub processes within in a process:
def sub(env):
print('I am a sub process')
yield env.timeout(1)
# return 23 # Only works in py3.3 and above
env.exit(23) # Workaround for older python versions
def main(env):
print('I am the main process')
retval = yield env.process(sub(env))
print('Sub returned', retval)
As you can see, you can use Process instances returned by Environment.process()
like normal events. You can even use return values in your sub proceses.
If you use Python 3.3 or newer, you don’t have to explicitly start a new
sub-process but can use sub() as a sub routine instead and just forward the
events it yields:
def sub(env):
print('I am a sub routine')
yield env.timeout(1)
return 23
def main(env):
print('I am the main process')
retval = yield from sub(env)
print('Sub returned', retval)
You may also be able to model signals as resources that may either be used
by failure process or by a train. If the failure process requests the signal
at first, the train has to wait in front of the signal until the failure
process releases the signal resource. If the train is aleady passing the
signal (and thus has the resource), the signal cannot break. I don’t think
that’s a problem be cause the train can’t stop anyway. If it should be
a problem, just use a PreemptiveResource.
I hope this helps. Please feel welcome to join our mailing list for more
discussions.

Simpy subway simulation: how to fix interrupt failure of class train while queueing for a resource?

I am working on a train simulation in simpy and have had success so far with a single train entity following the code below.
The trains processes are sections followed by platforms. Each section and platform has a resource of 1 to ensure that only one train utilises at a time.
However I can't find a way to get around the error below:
When I add in a second train to the simulation there is occasionally the situation where one train waits for an unavailable resource and then a failure occurs on that train while it is waiting.
I end up with an Interrupt: Interrupt() error.
Is there a way around these failing queues for resources?
Any help is much appreciated.
import random
import simpy
import numpy
# Configure parameters for the model
RANDOM_SEED = random.seed() # makes results repeatable
T_MEAN_SECTION = 200.0 # journey time (seconds)
DWELL_TIME = 30.0 # dwell time mean (seconds)
DWELL_TIME_EXPO = 1/DWELL_TIME # for exponential distribution
MTTF = 600.0 # mean time to failure (seconds)
TTF_MEAN = 1/MTTF # for exponential distribution
REPAIR_TIME = 120.0 # mean repair time for when failure occurs (seconds)
REPAIR_TIME_EXPO = 1/REPAIR_TIME # for exponential distribution
NUM_TRAINS = 2 # number of trains to simulate
SIM_TIME_HOURS = 1 # sim time in hours
SIM_TIME_DAYS = SIM_TIME_HOURS/18.0 # number of days to simulate
SIM_TIME = 3600 * 18 * SIM_TIME_DAYS # sim time in seconds (this is used in the code below)
# Defining the times for processes
def Section(): # returns processing time for platform 7 Waterloo to 26 Bank
return T_MEAN_SECTION
def Dwell(): # returns processing time for platform 25 Bank to platform 7 Waterloo
return random.expovariate(DWELL_TIME_EXPO)
def time_to_failure(): # returns time until next failure
return random.expovariate(TTF_MEAN)
# Defining the train
class Train(object):
def __init__(self, env, name, repair):
self.env = env
self.name = name
self.trips_complete = 0
self.num_saf = 0
self.sum_saf = 0
self.broken = False
# Start "running" and "downtime_train" processes for the train
self.process = env.process(self.running(repair))
env.process(self.downtime_train())
def running(self, repair):
while True:
# request section A
request_SA = sectionA.request()
########## SIM ERROR IF FAILURE OCCURS HERE ###########
yield request_SA
done_in_SA = Section()
while done_in_SA:
try:
# going on the trip
start = self.env.now
print('%s leaving platform at time %d') % (self.name, env.now)
# processing time
yield self.env.timeout(done_in_SA)
# releasing the section resource
sectionA.release(request_SA)
done_in_SA = 0 # Set to 0 to exit while loop
except simpy.Interrupt:
self.broken = True
delay = random.expovariate(REPAIR_TIME_EXPO)
print('Oh no! Something has caused a delay of %d seconds to %s at time %d') % (delay, self.name, env.now)
done_in_SA -= self.env.now - start # How much time left?
with repair.request(priority = 1) as request_D_SA:
yield request_D_SA
yield self.env.timeout(delay)
self.broken = False
print('Okay all good now, failure fixed on %s at time %d') % (self.name, env.now)
self.num_saf += 1
self.sum_saf += delay
# request platform A
request_PA = platformA.request()
########## SIM ERROR IF FAILURE OCCURS HERE ###########
yield request_PA
done_in_PA = Dwell()
while done_in_PA:
try:
# platform process
start = self.env.now
print('%s arriving to platform A and opening doors at time %d') % (self.name, env.now)
yield self.env.timeout(done_in_PA)
print('%s closing doors, ready to depart platform A at %d\n') % (self.name, env.now)
# releasing the platform resource
platformA.release(request_PA)
done_in_PA = 0 # Set to 0 to exit while loop
except simpy.Interrupt:
self.broken = True
delay = random.expovariate(REPAIR_TIME_EXPO)
print('Oh no! Something has caused a delay of %d seconds to %s at time %d') % (delay, self.name, env.now)
done_in_PA -= self.env.now - start # How much time left?
with repair.request(priority = 1) as request_D_PA:
yield request_D_PA
yield self.env.timeout(delay)
self.broken = False
print('Okay all good now, failure fixed on %s at time %d') % (self.name, env.now)
self.num_saf += 1
self.sum_saf += delay
# Round trip is finished
self.trips_complete += 1
# Defining the failure event
def downtime_train(self):
while True:
yield self.env.timeout(time_to_failure())
if not self.broken:
# Only break the train if it is currently working
self.process.interrupt()
# Setup and start the simulation
print('Train trip simulator')
random.seed(RANDOM_SEED) # Helps with reproduction
# Create an environment and start setup process
env = simpy.Environment()
# Defining resources
platformA = simpy.Resource(env, capacity = 1)
sectionA = simpy.Resource(env, capacity = 1)
repair = simpy.PreemptiveResource(env, capacity = 10)
trains = [Train(env, 'Train %d' % i, repair)
for i in range(NUM_TRAINS)]
# Execute
env.run(until = SIM_TIME)
Your processes request a resource and never release it. That’s why the second trains waits forever for its request to succeed. While it is waiting, the failure process seems to interrupt the process. That’s why you get an error. Please read the guide to resources to understand how SimPy’s resources work and, especially, how to release a resource when you are done.

MATLAB: errorn in butter() command

I wrote the following function:
function [output_signal] = AddDirectivityError (bat_loc_index, butter_deg_vector, sound_matrix)
global chirp_initial_freq ;
global chirp_end_freq;
global sampling_rate;
global num_of_mics;
global sound_signal_length;
for (i=1 : num_of_mics)
normalized_co_freq = (chirp_initial_freq + chirp_end_freq)/ (1.6* sampling_rate);
A=sound_matrix ( i, : ) ;
peak_signal=max(A);
B=find(abs(A)>peak_signal/100);
if (butter_deg_vector(i)==0)
butter_deg_vector(i)=2;
end
[num, den] = butter(butter_deg_vector(i), normalized_co_freq, 'low');// HERE!!!
filtered_signal=filter(num,den, A );
output_signal(i, :)=filtered_signal;
end
This functions runs many-many times without any error. However, when I reach the line: [num, den] = butter ( butter_deg_vector(i), normalized_co_freq, 'low');
And the local variables are: i=3, butter_deg_vector(i)=1, normalized_co_freq=5.625000e-001
MATLAB prompts an error says:
??? Error using ==> buttap Expected N to be integer-valued.
"Error in ==> buttap at 15 validateattributes(n,{'numeric'},{'scalar','integer','positive'},'buttap','N');
Error in ==> butter at 70 [z,p,k] = buttap(n);"
I don't understand why this problem occurs especially in this iteration. Why does this function prompt an error especially in this case?
Try to change the code line for:
[num, den] = butter (round(butter_deg_vector(i)), normalized_co_freq, 'low');