I have the Argo WorkflowTemplate which has n steps.
I want to call last step only if any of the previous step fails.
Example: In a 5 steps templates, if 2nd fails, skip 3 and 4 and only call 5 since its a revert step. If all are passed, don't call 5th because there is no need to revert.
You can define a workflow exit handler to run after all the other steps.
By adding a when condition, you can make sure the exit handler runs if and only if one of the previous steps failed.
So I have a problem where in my SFC it jumps to an inital step but the commands written in the step would not register.
At the end of the SFC a step inputs 5 into A_Status(INT).
The very next transition checks if the value of A_Status is 5.
No problems so far, but after the transition when it jumps to the start of the SFC,
where the first step is supposed to input 0 into A_Status, A_Status stays at 5.
The cycle time of my program is 100ms. I have tried slowing the cycle but it didn't work.
What seems to be the problem here? Maybe the same variable used in such a sequence just doesn't work?
Reply would be greatly appreciated.
You don't mention if you write the values during Entry/Exit or in the SFC step actions. But beware, that on some occasions code from a previous step can be executed later than code in the new step.
Here is a link that explains the call order and why sometimes parts of the code is executed twice:
https://infosys.beckhoff.com/english.php?content=../content/1033/tc3_plc_intro/45035999420423563.html
I've had succes with adding the following code in all the actions to prevent this from happening.
IF STEP_NAME.x THEN // Only execute this while the step is active.
// Insert code here.
END_IF
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
Is there a way to remove only the last snapshot in the CKEditor undo stack or can i replace it with another.Should i implement it on my own?
Example:
Step 1
Step 2 --should be removed and replaced with step 3 (On given situation)
Step 3 -- should become step 2
This feature should be available only if special event occurs.
If your undo snapshots are a result of user actions, following this way:
Step 1.
Step 2.
CKEDITOR.instances.editor.fire( 'lockSnapshot' )
Step 3.
CKEDITOR.instances.editor.fire( 'unlockSnapshot' )
Of course, you have to detect what's going on and fire the right event at the right time.
If changes to the content are done from code, editor#updateSnapshot event would even be better:
function() {
editor.fire( 'saveSnapshot' );
editor.document.body.append(...);
// Makes new changes following the last undo snapshot a part of it.
editor.fire( 'updateSnapshot' );
..
}
I'm trying to understand the FIFO page replacement algorithm, but all the information I can find amounts to what's below. Can you explain how you use a reference string to evaluate a page replacement algorithm, using the particular example of FIFO?
When a page must be replaced, the oldest page is chosen.
In all our examples, the reference string is
1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5
3 frame (9 page faults)
4 frame (10 page faults)
A page fault is when a page is not present in a frame of memory so a trap has to be sent to the OS
and the page has to be added to a frame. If there is no room then something needs to be removed.
FIFO is one method to determine what page will get removed.
The concept is that whatever page got added first to the frame will get removed first. This is what FIFO stands for. Using your first example. I will go down the reference string list and show you what the memory will look like.
1: 1 +1 fault
2: 1,2 +1 fault
3: 1,2,3 +1 fault
4: 2,3,4 +1 fault - 1 gets removed because it was the first added
1: 3,4,1 +1 fault - 2 gets removed because it was the first on the list
2: 4,1,2 +1 fault - 3 got removed..
5: 1,2,5 +1 fault - 4 got removed..
1: 1,2,5 No change - 1 is already present so no page fault was required
2: 1,2,5 No change - 2 is already present in the frame
3: 2,5,3 +1 fault - 1 was removed because it is first
4: 5,3,4 +1 fault - Now 2 got removed
5: 5,3,4 No change - No change because 5 is present in one of the frames.
This gives the total of 9 faults.
You can also think of it as the oldest page gets removed.
Hopefully I made no mistakes :D
------------ * FIRST IN FIRST OUT * -------------
Page no. 1 | 1 +1
Page no. 2 | 1 2 +1
Page no. 3 | 1 2 3 +1
Page no. 4 | 1 2 3 4 +1
Page no. 1 |
Page no. 2 |
Page no. 5 | 2 3 4 5 +1
Page no. 1 | 3 4 5 1 +1
Page no. 2 | 4 5 1 2 +1
Page no. 3 | 5 1 2 3 +1
Page no. 4 | 1 2 3 4 +1
Page no. 5 | 2 3 4 5 +1
Page Faults = 10
page replaced when not availbe in Queue.
go to this link FIFO
here explained all Page repalacemnt algo. very well with example. you understand easily.
1: 1 +1 fault
2: 1,2 +1 fault
3: 1,2,3 +1 fault
4: 2,3,4 +1 fault - 1 gets removed because it was the first added
1: 3,4,1 +1 fault - 2 gets removed because it was the first on the list
2: 4,1,2 +1 fault - 3 got removed..
5: 1,2,5 +1 fault - 1 got removed..
1: 1,2,5 No change - 1 is already present so no page fault was required
2: 1,2,5 No change - 2 is already present in the frame
3: 2,5,3 +1 fault - 1 was removed because it is first
4: 5,3,4 +1 fault - Now 2 got removed
5: 5,3,4 No change - No change because 5 is present in one of the frames.