Duty Cycle Calculation

Question

Below I want to calculate a ratio x:5 meaning for every Xminutes you are on you are off or wait 5 mins to turn back on. On/Off is a complete cycle. Below are three cycles. I am trying to come up with a good way to do this using linq. Basically I am going to return how much time you have to wait or stay off before you start another cycle if that make sense. So below you have a ArrayOnOff[0]= 1:00AM On; ArrayOnOff[1]= 1:01AM Off; ArrayOnOff[2]= 1:02AM On; ArrayOnOff[3]= 1:03AM Off; ArrayOnOff[4]= 1:04AM On; ArrayOnOff[5]= 1:05AM Off; CycleId StateOnOFF 1 1:00AM -on 2 1:01AM -off 3 1:02AM -on 4 1:03AM -off 5 1:04AM -on 6 1:05AM -off

Vulpes · Answer

To avoid using an explicit enumerator I'd probably have done it like this: Cycle[] cycles = new Cycle[ArrayOnOff.Length]; for(int i = 0; i double totalTimeUsed = 0; for(int i = 1; i { double diff = (cycles[i].TimeTag - cycles[i- 1].TimeTag).TotalSeconds; if (i % 2 == 1) totalTimeUsed += diff; else totalTimeUsed -= diff/5.0; if (totalTimeUsed } totalTimeUsed -= (DateTime.Now - cycles[cycles.Length - 1].TimeTag).TotalSeconds / Ratio; This isn't an ideal case for using LINQ but if you must use it then I'd try something like this: double totalTimeUsed = 0; Cycle prev = null; ArrayOnOff.Select(aof => GetCycleInfo(aof)).Select((c i) => { if (i > 0) { double diff = (c.TimeTag - prev.TimeTag).TotalSeconds; if (i % 2 == 1) totalTimeUsed += diff; else totalTimeUsed -= diff/5.0; if (totalTimeUsed } prev = c; return 0.0; // need to return something although we don't use it }).ToArray(); // add ToArray() to execute immediately totalTimeUsed -= (DateTime.Now - prev.TimeTag).TotalSeconds / Ratio;

David Smith · Answer

I am going to try to explain this. If my ration is 1:5. That means for every minute that you are on you have to stay off for five minutes if that make since. I have an input file that will contain a sequence of on/off / on/ off with a timestamp. The input file will populate the array below. The logic I want to simplify and come up with a better way of doing the logic below using linq if possible. ArrayOnOff[0]= 1:00AM On; ArrayOnOff[1]= 1:01AM Off; ArrayOnOff[2]= 1:02AM On; ArrayOnOff[3]= 1:03AM Off; ArrayOnOff[4]= 1:04AM On; ArrayOnOff[5]= 1:05AM Off; Formula: The below formula calculates the amount of time you have left to stay on before exceeding the max. Everything needs to be in seconds Max = 10min Ratio = 5 (off/on) TimeAvailable = Max - [OffTime - OnTime] + [DateTime.Now - OffTime] / Ratio IEnumerator enumerator = ArrayOnOff.GetEnumerator(); enumerator.MoveNext(); previousOn = GetCycleInfo((string)enumerator.Current); enumerator.MoveNext(); previousOff = GetCycleInfo((string)enumerator.Current); double totalTimeUsed = (previousOff.TimeTag - previousOn.TimeTag).TotalSeconds; for (int i = 0; i { enumerator.MoveNext(); Cycle newOn = GetCycleInfo((string)enumerator.Current); en.MoveNext(); Cycle newOff = GetCycleInfo((string)enumerator.Current); double timeWaited = (newOn.TimeTag - previousOff.TimeTag).TotalSeconds; totalTimeUsed -= (newOn.TimeTag - previousOff.TimeTag).TotalSeconds / Ratio; // bought back all time? if (totalTimeUsed { totalTimeUsed = 0.0; } previousOn = newOn; previousOff = newOff; totalTimeUsed += (previousOff.TimeTag - previousOn.TimeTag).TotalSeconds; } totalTimeUsed -= (DateTime.Now - previousOff.TimeTag).TotalSeconds / Ratio;

David Smith · Answer

Ok. Thanks.

Vulpes · Answer

'for' is faster than 'foreach' because the latter has more to do behind the scenes. However the difference is not usually significant. Here I used 'for' because I needed the index of the element in the array to check whether it was odd or even and hence whether it was an 'off' or an 'on'.

David Smith · Answer

Which do you prefer to use in this case a foreach or for? Which is faster in your opinion. It seems the 'for' is faster?

Vulpes · Answer

Personally I find using for or foreach easier to read than using enumerators explicitly.

David Smith · Answer

What's the reason for not using an explicit enumerator.

Kunal Vaishya · Answer

Hello Can you provide more details so i can help you.