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It is said, "If a negative edge cycle is reachable from the source than the algorithm returns false".
What does this "reachable from source indicates"?
Look at the following image:
Can you give me some example in which this algorithm will return false if there exist a negative edge cycle reachable from source.
Note: I am new to algorithms.
What that means is that if there exists a cycle that has a total weight that is negative, then the algorithm cannot give an answer because repeatedly following the cycle "reduces" the weight of the path. I don't see any negative weight cycles (by inspection) in the graph you show, so the stated limitation shouldn't be a concern in your case.
Edit: "reachable from source" means that the negative weight cycle is only a concern if it is reachable - meaning that a path from the specified source to some node in the negative weight cycle exists - from the start or source node. Bellman Ford finds the shortest path from a distinguished node to all nodes reachable from that node. Does that make sense?
When the algorithm is used to find shortest paths, the existence of negative cycles is a problem, preventing the algorithm from finding a correct answer. However, since it terminates upon finding a negative cycle, the Bellman-Ford algorithm can be used for applications in which this is the target to be sought - for example in cycle-cancelling techniques in network flow analysis.
Please refer this link:- http://evlm.stuba.sk/~partner2/DBfiles/Optimization/Dynamical%20optimization/Optimization_EN_Ford-Belman_algorithm.pdf
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I have a simple network that has several routes from start to end. Vehicles from a transporter fleet will carry agents from the left conveyor to the right conveyor using the moveByTransporter block. What are some syntax i can use to refer to paths/nodes on the network?
Also, what is a sample code line of how i can check the number of vehicles on a specific path?
This is my sample network and idea of trying how to make a new routing instead of just the shortest path (the path i want to follow is via the yellow highlighted one)
The moveTo block will take the route with the shortest distance. If the agent's speed is the same across all choices, then the shortest distance will also be the fastest time.
In the past, I have used Dijkstra's algorithm and manually routed my agents from a to b, then b to c, etc. This way, I could use travel times instead of just distances. This also opens up the possibility of considering congestion by applying a penalty to some segments if there are too many other agents on them. You can also pick a route, but then when you get to the next node, re-calculate the rest of the route for updated congestion considerations.
This is all custom, and I would not recommend it for simple problems. You would be better off to look at other alternatives (assume constant speeds or consider the pedestrian library with walls, etc, depending on your problem).
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I can see how to solve the right side of the equation but when not knowing what C and T are how can I prove it is not necessary and confirm it is sufficient?
I thought about changing the C/T to a variable such as x and starting from N = 1 and solving for x at each step until the RHS is not >= to the LHS. Is this a correct way to go about this?
As the questions stated the bound you provided n(2^(1/n)-1) is sufficient in the sense that if it is true then the set of tasks is schedulable but it is not complete, it will reject task sets that are still schedulable under RMS. A simple example is a harmonic task set. A harmonic task set is one where each period is is an exact multiple of all the tasks with small periods. A trivial example is the task set of two tasks each with a capacity of 1 and a period of 2. It has 100% utilization and is schedulable under RMS but its utilization is greater than the percentage 0.828 listed in the chart for N=2.
In general, to determine if a task set is schedulable under RMS the follow recurrence relationship is solved for each task i:
If this value is less than T_i for each task assuming implicit deadlines) than the task set is schedulable.
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I am confused about a situation which is presented on the following slide:
Last sentences says that:
It is important to note that deterministic does not mean that
xt is non-random. What does this mean? If A and B are random variable, then x must be random right?
I think the point may be that nature may choose randomly among different paths, but once you know which path has been chosen you can predict future values of x_t on the path from past values x_{t-1}, etc. So e.g. nature may flip a coin to choose between the following two paths: x_t=0 for all t, and x_t=1 for all t. Then if you don't know the path, x_t is indeed random. But once you know x_{t-1}, you know x_t.
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I need lots of cables of different small sizes (under 100 meters) and cables are only sold in lenghts of 100 meters.
So, to optimize my purchase, I would like a code where I can input the lengths of all pieces of cables that I need. The code will combine my inputs under the constraint the sum is under 100, while minimizing the total number of 100m-length cables that I need to buy.
If anyone could help with a code in VBA, Matlab or Python I would be very grateful.
This is known as a bin-packing problem, and it's actually very difficult (computationally speaking) to find the optimal solution.
However, it is a problem that is practically useful to solve (as you have seen for yourself) and so there are several approaches that seek to find an approximate solution--one that is "good enough" without guaranteeing that it's the best possible solution. I did a quick search and found this course website, which has some examples that may help you out.
If you are looking for an exact solution, you can ask the related question "will I be able to fit the cables I need into N 100-meter cables?". This feasibility problem can be expressed as a "binary program", which is a special case of a "mixed-integer linear program", for which MATLAB has a solver called intlinprog (requires the optimization toolbox).
I'm sorry that I don't have any code to solve your problem, but I hope that this at least gives you some keywords to help you find more resources!
I believe this is like the cutting stock problem. There are some very good methods to solve this. Here is an implementation and some background. It is not too difficult to write an Excel front-end for this (see here).
If you google for "cutting stock problem" you will find lots of references.
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There is another question with this same title, but the question is asked differently than what's troubling me, and the answer is not sufficient.
The most prominent analogies I hear to explain bandwidth are the highway example, and the pipe example. In the highway example, bandwidth is the amount of cars that can drive on the highway in a given amount of time, and in the pipe its an amount of water that can flow through.
My question is - by measuring by cars per second, or liters per second, does that mean that a longer highway, pipe or copper wire has a higher bandwidth than a shorter one? That seems strange to me.
Wouldn't it make more sense to give the highway bandwidth as the amount of lanes it has - irrespective of a unit of time? It just makes more sense to me and is simpler to say that the pipe is "1 foot in diameter" rather than "it carries 100 litres per second".
Why do we measure bandwidth in bits per second and not just in bits?
"My question is - by measuring by cars per second, or liters per second, does that mean that a longer highway, pipe or copper wire has a higher bandwidth than a shorter one?"
No!
Bandwidth is not about how many cars can fit on the road. It's about how many cars can pass a point on the road during a certain time. How many cars per second can pass under a bridge, for example.
No, it wouldn't. You quote a highway in terms of lanes, because it's more understandable, and a reasonable approximation to assume 4 lanes = 4x as much traffic. But even then, you might have a traffic jam, and then 4 lanes is 'transmitting' fewer cars per minute than it would otherwise.
With a hose pipe, the width of the pipe is the speed of transmission if you assume the same water pressure.
These assumptions don't apply to communications - when I'm transmitting 'a bit' nothing physical is moving *. A 'bit' is the smallest piece that 'information' can be broken down into, and in order to transmit it, something needs to change.
If I turn on my torch and shine it at you, I've sent one 'message' (my torch is on). To send you anything more detailed, I would need to turn it off and on again - morse code is an example of doing this. The pattern of switching it off and on gives you some letters. How fast I can switch it off and on again, is how fast I can send a message.
So it is with bandwidth. I need to change things to communicate. If I can change things faster, I can communicate faster.
"bits" would be a measure of the number of torches I own. Bits per second is how fast I can flick them on and off to send a message.
* Electrons and photons do move, as does air to carry sound. But the signal isn't the thing moving - I don't have to move an atom of air from my mouth to your ear to 'talk' to you, the wave propagates through the medium.