I'm using a select output block to do something if one track has less railcars on it than another track. Right now I'm using the following code in the if condition is true area:
int x
int y
track1.nCars()=x
track2.nCars()=y
(x.intValue() >= y.intValue());
But this isn't working, and I'm wondering how to use the nCars() function in a flowchart block like this?
You can only write 1 line of code in such a condition code box.
As long as the track1 and track2 objects are in the same agent as the SelectOutput, simply turn your code into a 1-line condition:
track1.nCars() >= track2.nCars()
Agents will exit via the true branch if track1 has more/equal cars than track 2
Related
The "Assembler" should stop working for 2 hours after 10 assemblies are done.
How can I achieve that?
There are so many ways to do this depending on what it means to stop working and what the implications are for the incoming parts.. but here's one option
create a resourcePool called Machine, this will be used along with the technicians:
on the "on exit" action of the assembler do this (I use 9 instead of 10 because the out.count() doesn't count until the agent is completely out, so when it counts 9, it means that you have produced 10)
if(self.out.count()==9){
machine.set_capacity(0);
create_MyDynamicEvent(2, HOUR);
}
In your dynamice event (that you have to create) you will add the following code:
machine.set_capacity(1);
A second option is to have a variable countAssembler count the number of items produced... then
on exit you write countAssembler++;
on enter delay you write the following:
if(countAssembler==10){
self.suspend(agent);
create_MyDynamicEvent(2, HOUR,agent);
}
on the dynamic event you write:
assembler.resume(agent);
Don't forget to add the parameter needed in the dynamic event:
Create a variable called countAssembler of type int. Increment this as agents pass through the assembler. Also create a variable called assemblerStopTime. You also record the assembler stop time with assemblerStopTime=time()
Place a selectOutputOut block before the and let them in if countAssembler value is less than 10. Otherwise send to a Wait block.
Now, to maintain the FIFO rule, in the first selectOutputOut condition, you need to check also if there is any agent in the wait block and if the current time - assemblerStopTime is greater than 2. If there is, you free it and send to the assembler with wait.free(0) function. And send the current agent to wait. You also need to reset the countAssembler to zero.
I have the model which I posted before on Stack. I am currently running the iterations through 5 Flow Chart blocks contain enter block and service block. when agent fill service block 5 in flow chart 5, the exit block should start to fill block one and so on. I have used While infinite loop to loop between the five flow chart blocks but it isn't working.
while(true)
{
for (Curing_Drying currProcess : collection) {
if (currProcess.allowedDay == (int)time(DAY)) {
currProcess.enter.take(agent);
}
}
if (queue10.size() <= Throughtput1){
break;
}
}
Image for further illustration 1
Image for further illustration 2
Wondering if someone can tell me what is wrong in the code.
Based on the description and the pictures provided, it isn't clear why the while loop is necessary. The On exit action is executed for each Agent arrival to the Exit block. It seems that the intention is to find the appropriate Curing_Drying block based on number of days since the model start time? If so, then just iterating through the collection is enough.
Also, it is generally a good practice to provide more meaningful names to collections. Using simply collection doesn't say anything about the contents and can get pretty confusing later on.
I am trying to write a code for one state to another state transition where the system will first store the initial model time and then will check a condition (used while loop). It will continue to run the loop until the condition is false and when the condition is false it will record the final model time. So, my main objective is to get the total time that while loop condition is true. The problem is, I don't know how to check the while loop every 1 time step, For example, I tried "wait (1);" in place of "???" section of the below code which is not correct. Can anyone please suggest how I can do this?
My transition code as below:
...
...
initialTime=time();
while ((thisPed.inState(walking) && thisPed.fieldOfVision.contains(pedX, pedY));
{
???
}
finalTime= time();
exposureTime = finalTime - initialTime;
...
...
you can't put while statements in a model that at the same time run with time steps... to do that you have many other ways..
For instance you can generate a transition that goes from that state to the same state (internally) and generate your code there every time step.
Another option is to use a conditional transition in the same way
BUt NOT a while loop
I am working on fire evacuation of a building floor and would like to count the number of people remaining inside the building after 120 seconds? The timer should start once the evacuation process begins, which is by alarm that goes off after a certain amount of time using an event feature.
I know how to count the total number of people inside the building using the function component getPeopleInsideCount and a text with getPeopleInsideCount(). But I don't know what code to use for my problem.
Below is the code:
return pedOffice.countPeds() + pedStudents.countPeds() - pedSink.sink.count();
With this, it will count the people in the building floor and it will stop counting after 120 seconds...
Step 1:
create an event with trigger type timeout, and mode: user control, and timeout=120 seconds.
Step 2:
create a variable called stopCounting as a boolean with initial value false
create a variable called peopleRemaining as an int
Step 3:
when the evacuation begins run the code:
event.restart();
Step 4:
in your event use the following code:
stopCounting=true;
peopleRemaining=getPeopleInsideCount();
Step 5
In your text use the following code instead of getPeopleInsideCount()
stopCounting ? peopleRemaining : getPeopleInsideCount()
Add a dynamic event that returns the count you need.
Once your alarm goes off, you can call that dynamic event 120 seconds later using create_MyDynamicEvent(120, SECOND);
That will execute the event code 120 seconds later.
cheers
I've been given a task to set up an openai toy gym which can only be solved by an agent with memory. I've been given an example with two doors, and at time t = 0 I'm shown either 1 or -1. At t = 1 I can move to correct door and open it.
Does anyone know how I would go about starting out? I want to show that a2c or ppo can solve this using an lstm policy. How do I go about setting up environment, etc?
To create a new environment in gym format, it should have the 5 functions mentioned in the gym.core file.
https://github.com/openai/gym/blob/e689f93a425d97489e590bba0a7d4518de0dcc03/gym/core.py#L11-L35
To lay this down in steps-
Define observation space and action space for your environment, preferably using gym.spaces module.
Write down the step function which performs agent's action and returns a 4 tuple containing - next set of observations from the environment , reward ,
done - a boolean indicating whether the episode is over , and some extra info if you want.
Write a reset function for the environment to reinitialise the episode to a random start state and return a 4 tuple similar to step.
These functions are enough to be able to run an RL agent on your environment.
You can skip the render, seed and close functions if you want.
For the task you have defined,you can model the observation and action space using Discrete(2). 0 for first door and 1 for second door.
Reset would return in it's observation which door has the reward.
Then agent would choose either of the door - 0 or 1.
Then perform a environment step by calling step(action), which will return agent's reward and done flag as true - signifying that the episode is over.
Frankly, the problem you describe seems too simple to accomplish for any reinforcement learning algorithm, but I assume you have provided that as an example.
Remembering for longer horizons is usually harder.
You can read their documentation and toy environments to understand how to create one.