# Tag Info

51

Operations problems like this are tricky, because they always include a large element of uncertainty. As Willie Wong points out, it is possible to construct a scenario in which "closest-first" is a non-optimal strategy. However, proving that there exists a non-optimal strategy does not mean that globally such a strategy is non-optimal. Wait, what? In ...

44

I have little familiarity with ambulances and am not sure if they are stationed when on call, such as a fire truck, but if they are not then you could simply position them such that the distance from any point in the region to the nearest ambulance is minimized. When a call comes in, the nearest ambulance could respond and the entire formation would ...

22

I actually see this problem a lot, believe it or not, in World of Warcraft. The problem is vastly simplified, granted, as it comes in the context of a wargame with fixed points to defend and attack. But the general idea is the same: one may have, say, three points to defend. If one base comes under attack, the natural thing to do is to reinforce ...

22

What I propose is this: Before assigning an ambulance to a call, check the coverage of that ambulance's current vicinity by OTHER ambulances. If you find the coverage to be poor, run the same algorithm on the SECOND nearest ambulance, and so forth. (Fundamentally, what we're trying to do is not just assign an ambulance to a call, but to assign an ambulance ...

12

Let's simplify the situation to the limit so that you'll see clearly what is going on without heavy computations. Assume that you have two stations at $A$ and $B$, which are $1$ mile apart on a straight road and you get calls from random places in between independently with the uniform distribution. Assume also that the total number of calls is exactly the ...

11

Your question is an interesting one, but I think the reason you haven't received an answer to it yet is that it's ill-constrained. That's not a criticism but a technical term. Basically it means that the problem isn't well-defined enough for there to be a clear, distinct answer. (This is what the second part of Rahul Narain's comment is getting at.) To ...

10

If you just want an explicit example, make your second incident happen exactly where C is stationed. Then in the scenario where A picks up incident 2, C picks up incident 1 the total distance traveled is "distance between 1 and C" plus "distance between A and C" (since 2 is where C is stationed). In the scenario where A picks up incident 1, C ...

8

From what I understand there are broadly two distinct applications of mathematics to neuroscience. One uses mathematics to study the biological/chemical/physical aspects of the mechanisms in the brain, such as action potentials and the interactions between neurons. The type of math used here is differential equations/dynamic systems. Relevant wikipedia ...

8

It's entirely analogous to the 15-puzzle. To show that not all initial positions allow you to get to the desired final position, note the following invariant. Assume that in the final position you have the numbers from 1 to 23 written on the pieces starting in the top left corner, and proceeding row-wise to the bottom right corner, where you put 24 to ...

6

From the Chain Rule: $$\frac{\partial\log\lambda}{\partial \log a_{ij}}\;\frac{\partial\log a_{ij}}{\partial t} = \frac{\partial\log\lambda}{\partial t} = \frac{1}{\lambda}\frac{\partial\lambda}{\partial t}.$$ Since $$\frac{\partial\log a_{ij}}{\partial t} = \frac{1}{a_{ij}}\frac{\partial a_{ij}}{\partial t}$$ "solving" for ...

6

"Which is better, graphs or sets" depends crucially on what limitations you're imposing on how you use these. For instance, as a potential foundation for mathematics, sets can be used to do essentially anything if you put your mind to it. For instance, sets can be used to represent a graph. You can represent a graph as a set of labelled nodes, and the edges ...

6

This is a neat problem and can be approached using stochastic calculus and non-linear optimization. I don't believe its explicitly solvable unless trivial assumptions are made. I'll first describe a way to optimize the initial positions of the ambulance. We will then use cost functions developed in from this optimization to discuss how sending a different ...

6

Asaf answered the formal part of your question well. I think the main problem with the theory you're outlining is that your "axioms" are, in fact, not axioms as mathematics uses that word. When you construct a mathematical model of a facet of the world, your axioms are supposed to be things you know will always be true about the thing you're modeling. But ...

5

You could imagine that the medicine gets absorbed by the digestive system at a fast rate $\alpha$ and then consumed by the body at a slower rate $\beta$. Then you have the following system of ordinary differential equations, \begin{align} x' &= -\alpha x, \\ y' &= \alpha x - \beta y, \end{align} where $x$ and $y$ are the amounts of ...

5

Mathematics which works with concrete ideas has some problem with conflicting axioms. Mostly because there is no such thing as a little bit of contradiction. You can only derive contradiction; and from it you can derive anything - which means theories in which there is a contradiction are not useful. There are examples, however, to incompatible axioms ...

5

Scheduling algorithms have only access to information about the past, and are making guesses about what will optimize the unknown future. You have no clue when or where the next emergency will occur. By not sending the closest ambulance to a current emergency, you're risking somebody for a potential gain, which seems unethical. What if you send an ambulance ...

5

There is no right answer to this question. Almost all real-world data is "really" discrete, but it is often more convenient to model it as continuous. You should follow whichever option gives you the easiest model. For example, if the distribution of values appears to be a smooth curve, use a continuous distribution. If you have a huge amount of data and can ...

5

All can be modeled by some $2 \pi$ periodic function $r:\mathbb{R} \to \mathbb{R}$, then the equation is $t \mapsto r(t) \sin t$. For the red wave, use $r(t) = 1$. For the blue wave, use $r(t) = \sqrt{1+\sin^2t}$, for $t \in [-\frac{\pi}{4}, \frac{\pi}{4})$, and let $r$ by $\frac{\pi}{2}$ periodic. For the green wave, use the same formula as for the blue ...

4

Any non-zero velocity will do it, if you assume no friction or other dissipation forces. By conservation of energy it will reach the initial position after one turn with that same speed, and will keep rotating around forever. The maximum speed will be at the bottom, and the smallest speed at the top.

4

Comment to your attempt: instead of reasoning based on the initial positions and conditions of the dog and cat, I advise to sketch a generic intermediate position of both. I tried this approach in the sketch below. I have not checked your second computation whose result differs from mine. I think you should explain it with some text and a sketch. I ...

4

Would L < 1 ? 0 : (int)L+1 be streamlined enough for you? I don't think there's any simpler way to write your function in Java. Yes, this expression uses the ternary conditional operator ?:, which is essentially equivalent to an if–else clause except that it switches between expressions instead of blocks. Depending on how smart your compiler ...

4

I would approach this by making clear assumptions to you model first. if you say that 1) all points in the square are equiprobable for an incident 2) all incidents must be responded to by immediately sending an ambulance 3) the cost (time) can be measured as the simple distance between two points =>you would always try to to have the ambulances ...

4

If you want to model the population with the formula: $$P(t)=P_0 e^{kt},$$ where $P_0$ is the initial population and $k$ is the exponential growth constant, then you must first find the values of $P_0$ and $k$. We are told $P_0=1$, so $$P(t)= e^{kt}$$ Let's now solve for $k$: We know that after 5 minutes, the population is $2$, so $$2 = e^{5k} ... 4 Just a search on google images: If you like the first one, a lot of functions could look like that. For example, get partly cos(t)^3 and partly flat. Of course, with coefficients to adjust it. 4 I think first you need to prove that better assignments are possible with hindsight. Get a log with, for each call, the time the ambulance was sent out, the positions of the ambulances and of the incident, the ambulance sent, and the time of arrival. If the availability of ambulances varies, also log the times that they are in service. If this information ... 4 You try to make a stochastic model corresponding to the deterministic model above. The Master equation of that model is$$\begin{eqnarray}\frac{d}{dt}\mathbb{P}(I,P) & = & \mathbb{T}(I,P|I-1,P)\cdot \mathbb{P}(I-1,P) - \mathbb{T}(I+1,P|I,P)\cdot \mathbb{P}(I,P) \\ & & + \mathbb{T}(I,P|I+1,P)\cdot \mathbb{P}(I+1,P) - ...

4

In the study of dynamical systems there is the concept of 'sensitive dependence on initial conditions' (abbreviated SDIC) which says that two initially nearby trajectories will diverge exponentially quickly. This is known as the 'butterfly effect' in popular terminology. More mathematically, take two trajectories $x_1(t)$, $x_2(t)$ and define the difference ...

3

If you know the interface, then drop perpendiculars from $P$ and $Q$ to the interface. Let the points of intersection be $P'$ and $Q'$. Let $PP' = y_P$ and $QQ' = y_Q$. Now consider the line segment $P'Q'=x$. You need to find a point $O$ inside $P'Q'$ such that $OP' + OQ' = x$. Let $OP' = x_P$ and $OQ' = ... 3 You were given the hint (or command?) to find the time$t$when$s(t)=0$. We will do that, although the relevance to the problem as stated is not entirely clear. So we want to solve the equation $$4t + \frac{2}{t-3} + \frac{2}{3}=0.$$ Multiply through by$3(t-3)$. This is legitimate, since we cannot have$t=3$. We arrive at the equivalent equation ... 3 I'm afraid there is no "good" answer to that. As you said,$X$is a sufficient condition for$Y$, but not vice versa. For example: If$B\Rightarrow B'$and$A'\Rightarrow A$(the assumption is weaker and the conclusion stronger) then$Y$holds but$X$doesn't. So we could generalize$X\$ to $$X_1\,:=\,(B\Rightarrow B') \land (A'\Rightarrow A)$$ This is ...

Only top voted, non community-wiki answers of a minimum length are eligible