I want to make the route timing (arrival) updated with the current traffic in the city.So I fetched Multiple Time Matrix from Google API.Now while routing the Arrival time should be calculated from the Matrix fetched for that time Interval.
For Example : I have fetched 3 Time Matrix. 10:00 {N * N} 12:00 {N * N} and 14:00{N * N}
If previousCustomer's arrival time is 11:00 then for next customer the time should be fetched from Matrix 12:00{}.
So How could I do this using Optaplanner ?
Let me know if anyone one need more explanation. Thanks
Using that matrix from OptaPlanner isn't hard. Just refactor RoadLocation.getDistanceTo(RoadLocation) to something like RoadLocation.getDistanceTo(RoadLocation, LocalDateTime startingTime) and adjust your score rules accordingly.
The big problem is memory while scaling out. If you have 10k locations, then only 1 time interval already costs almost 2GB RAM memory...
Related
I'm new to the search strategies. Does anyone know how to analyze the time complexity of the figure below?
The best way to analyse time complexity for such problems is to check how many times a vertex and edge is accessed. Since in both BFS and DFS a single node is only visited once we can say it would be O(V + E) where v is the number of vertices and E is the number of edges.
I am preparing time series data to build an RNN model (LSTM). The data is collected from sensors installed in a mechanical plant. Consider I have data for input and output temperature of a compressor along with the time stamps.
Like this there is data for around 20 parameters recorded along with their time stamps. Problem is there is a difference in the time stamps at which data is collected.
So how do I ideally match the time stamps to create a single dataframe with all the parameters and a single time stamp?
Since an RNN doesn't know anything about time deltas but only about time steps, you will need to quantify / interpolate your data.
Find the smallest time delta Δt in all of your series
Resample all of your 20 series to Δt/2* or smaller (Nyquist-Theorem)
* Actually you'd need to do a Fourier transform and then use twice the cutoff frequency as sampling rate. Δt/2 might IMHO be a good approximation.
I built an RNN that predicts query execution time for an unseen query. I want to add a timestamp as a feature, as it probably helps to estimate whether the server is busy or not. How can I combine a date/time variable with my query vector and feed it into my RNN model?
Yes, I could calculate the time delta by hand and feed it as a float, but that feels like cheating.
Regardless of the model you are using, your goal is to translate date-time stamps into numerical features that can give some insight into when the server is busy.
If you have periodic server usage, then you might want to create a periodic numerical feature. E.g. Hour # (0-23), or minutes, or maybe even week day # (0-6). If you have a linear trend over time (think server usage is slowly going up on average), then you might want to also translate the date-time stamps into a correctly scaled feature of "time since ...". E.g. number of days since first observation, or # of weeks, etc...
I hope that helps.
How should I calculate the percentage improvement in response time.
I am getting 15306 ms response time for old code and 799 ms response for the updated code. What will be the percentage improvement in response time?
There are two ways to interpret "percentage improvement in response time". One is the classic and ubiquitous formula for computing a percentage change in a data point from an old value to a new value, which looks like this:
(new - old)/old*100%
So for your case:
(799 - 15306)/15306*100% = -94.78%
That means the new value is 94.78% smaller (faster, since we're talking about response time) than the old value.
The second way of interpreting the statement is to take the percentage of the old value that the new value "covers" or "reaches":
new/old*100%
For your case:
799/15306*100% = 5.22%
That means the new value is just 5.22% of the old value, which, for response time, means it takes just 5.22% of the time to respond, compared to the old response time.
The use of the word "improvement" suggests that you want the 94.78% value, as that shows how much of the lag in the old response time was eliminated ("improved") by the new code. But when it comes to natural language, it can be difficult to be certain about precise meaning without careful clarification.
I think the accepted answer suffers from the original question not having
nice round numbers and that there are 3 different ways to state the
result.
Let's assume that the old time was 10 seconds and the new time is 5 seconds.
There's clearly a 50% reduction (or decrease) in the new time:
(old-new)/old x 100% = (10-5)/10 x 100% = 50%
But when you talk about an increase in performance, where a bigger increase is clearly better, you can't use the formula above. Instead, the increase in performance is 100%:
(old-new)/new x 100% = (10-5)/5 x 100% = 100%
The 5 second time is 2x faster than the 10 second time. Said a different way, you can do the task twice (2x) now for every time you used to be able to do it.
old/new = 10/5 = 2.0
So now let's consider the original question
The old time was 15306 ms and the new time is 799 ms.
There is a 94.7% reduction in time.
(old-new)/old x 100% = (15306-799)/15306 x 100% = 94.7%
There is a 1816% increase in performance:
(old-new)/new x 100% = (15306-799)/799 x 100% = 1815.6%
Your new time is 19x faster:
old/new = 15306/799 = 19.16
Actually performance is about how much can be done in the same amount of time.
So the formula is OLD/NEW - 1
In your case your performance increased by 1816% (i.e. you can do 18.16X more in the same time)
15306/799 - 1 = 1816%
Note: before you could do 1/15360, now 1/799 ...
your code's runtime is 94.78% shorter/improved/decreased:
(new - old) / old x 100%
(799 - 15306) / 15306 x 100% =~ -94.78% (minus represents decrease)
your code is 1816% faster:
(old - new) / new x 100%
(15306 - 799) / 799 x 100% =~ 1816%
Couple of responses already answered the question correctly, but let me expand those answers with some additional thoughts and a practical example.
Percentage improvement is usually calculated as ((NEW - OLD)/OLD) * 100
Let us think about it with some practical examples:
If I make $10,000 in my current job and get a new job that offers $12,000 then I am getting 20% increase in salary with this new job ((12000 - 10000)/10000)*100
If Train A travels at 100 miles per hour and Train B travels at 150 miles per hour, Train B is 50% faster than Train A. Simple, right?
It gets tricky when you try to measure a metric using another metric that has an inverse relationship with the metric you are trying to measure. Let me explain what I mean by this.
Let us try the second example again now using "time taken to reach destination". Let us say Train A takes 3 hours to reach the destination and Train B takes 2 hours to reach the same destination. Train B is faster than Train A by what percentage?
If we use the same formula we used in the above examples, we get ((2-3)/3)*100, which is -33%. This simply tells that Train B takes 33% less time to reach the destination than Train A, but that is not what we are trying to determine. Right? We are trying to measure the difference in speed by percentage. If we change the formula slightly and take the absolute value, we get 33%, which may seem right, but not really. (I'll explain why in a minute)
So, what do we do? The first thing we need to do is to convert the metric we have in hand to the metric we want to measure. After that, we should be able to use the same formula. In this example, we are trying to measure difference in speed. So let us first get the speed of each train. Train A travels at 1/3 of the distance per hour. Train B travels at 1/2 distance per hour. The difference in speed in percentage then is: ((1/2 - 1/3)/1/3) * 100 = ((1/2 - 1/3)*3)*100 = (3/2 - 3/3) * 100 = 50%
Which happens to be same as ((3 - 2)/2) * 100.
In short, when the metric we are measuring and the metric we have at hand have an inverse relationship, the formula should be
((OLD - NEW)/NEW) * 100
What is wrong with the original formula? Why can't we use the original formula and conclude that train B is only 33% faster? Because it is inaccurate. The original formula always yields a result that is less than 100%. Imagine a flight reaching the same destination in 15 mins? The first formula tells that flight is 91.6% faster while the second formula tells that the flight is 1100% faster (or 11 times faster), which is more accurate.
Using this modified formula, the percentage improvement for case posted in the original question is ((15306 - 799)/799) * 100 = 1815.6%
((old time - new time) / old time) * 100
This formula will give the Percentage Decreased in New Response time.
In your case, ((15306 - 799)/ 15306) * 100 = 94.78 %
The formula for finding the percentage of reduction is:
P = a/b × 100
Where P is the percentage of reduction, a is the amount of the reduction and b is the original amount that was reduced.
So to calculate a you do: old - new wichi will translate into:
P = ((OLD - NEW)/OLD)*100
I've tried the typical physics equations for this but none of them really work because the equations deal with constant acceleration and mine will need to change to work correctly. Basically I have a car that can be going at a large range of speeds and needs to slow down and stop over a given distance and time as it reaches the end of its path.
So, I have:
V0, or the current speed
Vf, or the speed I want to reach (typically 0)
t, or the amount of time I want to take to reach the end of my path
d, or the distance I want to go as I change from V0 to Vf
I want to calculate
a, or the acceleration needed to go from V0 to Vf
The reason this becomes a programming-specific question is because a needs to be recalculated every single timestep as the car keeps stopping. So, V0 constantly is changed to be V0 from last timestep plus the a that was calculated last timestep. So essentially it will start stopping slowly then will eventually stop more abruptly, sort of like a car in real life.
EDITS:
All right, thanks for the great responses. A lot of what I needed was just some help thinking about this. Let me be more specific now that I've got some more ideas from you all:
I have a car c that is 64 pixels from its destination, so d=64. It is driving at 2 pixels per timestep, where a timestep is 1/60 of a second. I want to find the acceleration a that will bring it to a speed of 0.2 pixels per timestep by the time it has traveled d.
d = 64 //distance
V0 = 2 //initial velocity (in ppt)
Vf = 0.2 //final velocity (in ppt)
Also because this happens in a game loop, a variable delta is passed through to each action, which is the multiple of 1/60s that the last timestep took. In other words, if it took 1/60s, then delta is 1.0, if it took 1/30s, then delta is 0.5. Before acceleration is actually applied, it is multiplied by this delta value. Similarly, before the car moves again its velocity is multiplied by the delta value. This is pretty standard stuff, but it might be what is causing problems with my calculations.
Linear acceleration a for a distance d going from a starting speed Vi to a final speed Vf:
a = (Vf*Vf - Vi*Vi)/(2 * d)
EDIT:
After your edit, let me try and gauge what you need...
If you take this formula and insert your numbers, you get a constant acceleration of -0,0309375. Now, let's keep calling this result 'a'.
What you need between timestamps (frames?) is not actually the acceleration, but new location of the vehicle, right? So you use the following formula:
Sd = Vi * t + 0.5 * t * t * a
where Sd is the current distance from the start position at current frame/moment/sum_of_deltas, Vi is the starting speed, and t is the time since the start.
With this, your decceleration is constant, but even if it is linear, your speed will accomodate to your constraints.
If you want a non-linear decceleration, you could find some non-linear interpolation method, and interpolate not acceleration, but simply position between two points.
location = non_linear_function(time);
The four constraints you give are one too many for a linear system (one with constant acceleration), where any three of the variables would suffice to compute the acceleration and thereby determine the fourth variables. However, the system is way under-specified for a completely general nonlinear system -- there may be uncountably infinite ways to change acceleration over time while satisfying all the constraints as given. Can you perhaps specify better along what kind of curve acceleration should change over time?
Using 0 index to mean "at the start", 1 to mean "at the end", and D for Delta to mean "variation", given a linearly changing acceleration
a(t) = a0 + t * (a1-a0)/Dt
where a0 and a1 are the two parameters we want to compute to satisfy all the various constraints, I compute (if there's been no misstep, as I did it all by hand):
DV = Dt * (a0+a1)/2
Ds = Dt * (V0 + ((a1-a0)/6 + a0/2) * Dt)
Given DV, Dt and Ds are all given, this leaves 2 linear equations in the unknowns a0 and a1 so you can solve for these (but I'm leaving things in this form to make it easier to double check on my derivations!!!).
If you're applying the proper formulas at every step to compute changes in space and velocity, it should make no difference whether you compute a0 and a1 once and for all or recompute them at every step based on the remaining Dt, Ds and DV.
If you're trying to simulate a time-dependent acceleration in your equations, it just means that you should assume that. You have to integrate F = ma along with the acceleration equations, that's all. If acceleration isn't constant, you just have to solve a system of equations instead of just one.
So now it's really three vector equations that you have to integrate simultaneously: one for each component of displacement, velocity, and acceleration, or nine equations in total. The force as a function of time will be an input for your problem.
If you're assuming 1D motion you're down to three simultaneous equations. The ones for velocity and displacement are both pretty easy.
In real life, a car's stopping ability depends on the pressure on the brake pedal, any engine braking that's going on, surface conditions, and such: also, there's that "grab" at the end when the car really stops. Modeling that is complicated, and you're unlikely to find good answers on a programming website. Find some automotive engineers.
Aside from that, I don't know what you're asking for. Are you trying to determine a braking schedule? As in there's a certain amount of deceleration while coasting, and then applying the brake? In real driving, the time is not usually considered in these maneuvers, but rather the distance.
As far as I can tell, your problem is that you aren't asking for anything specific, which suggests that you really haven't figured out what you actually want. If you'd provide a sample use for this, we could probably help you. As it is, you've provided the bare bones of a problem that is either overdetermined or way underconstrained, and there's really nothing we can do with that.
if you need to go from 10m/s to 0m/s in 1m with linear acceleration you need 2 equations.
first find the time (t) it takes to stop.
v0 = initial velocity
vf = final velocity
x0 = initial displacement
xf = final displacement
a = constant linear acceleration
(xf-x0)=.5*(v0-vf)*t
t=2*(xf-x0)/(v0-vf)
t=2*(1m-0m)/(10m/s-0m/s)
t=.2seconds
next to calculate the linear acceleration between x0 & xf
(xf-x0)=(v0-vf)*t+.5*a*t^2
(1m-0m)=(10m/s-0m/s)*(.2s)+.5*a*((.2s)^2)
1m=(10m/s)*(.2s)+.5*a*(.04s^2)
1m=2m+a*(.02s^2)
-1m=a*(.02s^2)
a=-1m/(.02s^2)
a=-50m/s^2
in terms of gravity (g's)
a=(-50m/s^2)/(9.8m/s^2)
a=5.1g over the .2 seconds from 0m to 10m
Problem is either overconstrained or underconstrained (a is not constant? is there a maximum a?) or ambiguous.
Simplest formula would be a=(Vf-V0)/t
Edit: if time is not constrained, and distance s is constrained, and acceleration is constant, then the relevant formulae are s = (Vf+V0)/2 * t, t=(Vf-V0)/a which simplifies to a = (Vf2 - V02) / (2s).