## Magento 2 How can I give discount on shipping costs?

I want to give a fixed amount subsidy on shipping cost but not on the product cost, what cart price rule should i use?

## Problem class of assigning N persons to N tasks, zero costs with prefs

I am looking for the general problem class / computational complexity / algorithms for the following problem:

N tasks must be accomplished by N persons. 1 task to be done by exactly 1 person and vice versa. There is a binary preferences matrix whose (i,j) entry is 0 if person i can not do task j, and 1 if it can.

There are no costs involved, no weights and the number of tasks equals the number of people.

I was searching for it but all I could find was the Assignment Problem which has costs/weights associated with each assignment.

As a last resort, perhaps I can transform my preferences matrix to a cost matrix if person can not do job then cost is infinite else cost is zero?

## What are the Costs of Offensive Drugs and Poisons?

What are the costs of offensive drugs and poisons in Cyberpunk 2020? That is, the various types of drugs and poison set out in the Friday Night Firefight section that can be delivered via airgun, dartgun, or squirtgun.

I can find effects but not rules. It is referenced under the Ammunition section in the Gear chapter, but different types are not priced differently (which clearly is not the intention.) If possible, I would prefer rules set out in the Core Rulebook.

## Given complete graph, find optimal path with two costs on each edge

We are given complete graph, such that each edge has two costs $$a \text{ and } b$$. We should find path that passes through each node once and has minimum total cost. Cost of a path is the maximum of both of the sums $$a$$ and $$b$$ of the edges laying on the edges crossed on the path.

One person told me that this is easily solvable by taking the maximum of $$a \text { and } b$$ for each edge and then using the dynamic programming technique storing the final index and the bitmask of visited nodes. I couldn’t find a counter test case, however I’m not sure if this is correct approach.

## The transport costs and you can

The transport costs and you can try the supplement for 0 days So move on and buy it earlier than the stock dries out Maturing men will come upon many detriments among the workout middle and the room lower testosterone stages can severy have an effect on temperament force digestion stamina and healing To assure you gain as a lot as Nerotenze Testo possible from your time on the rec middle boosting unfastened testosterone is…

The transport costs and you can

## Calculate decision boundary of two Gaussians with different missclassification costs

Assuming we have two classes $$C_1$$ and $$C_2$$ represented as two Gaussians with $$(2\mu_2, \sigma)$$ and $$(\mu_2, \sigma)$$. We know further that $$\mu_2 > 0$$ and $$p(C_1) = p(C_2)$$.

We want now calculate the decision boundary assuming that classifying $$x \in C_2$$ as $$C_1$$ is three times more expensive than the opposite. Classifying $$x$$ correctly has no costs.

At first we note that we need some representation of the loss function. Let $$\alpha_1$$ be the decision for $$C_1$$ and $$\alpha_2$$ the decision for $$C_2$$, then we get

\begin{align} R(\alpha_1|x) &= \lambda_{11}p(C_1|x) + \lambda_{12}p(C_2|x)\ R(\alpha_2|x) &= \lambda_{21}p(C_1|x) + \lambda_{22}p(C_2|x) \end{align}

where $$\lambda_{ij} = \lambda(\alpha_i|C_j)$$ is the cost for deciding for class $$i$$ and being in fact class $$j$$.

Hence we have $$\lambda_{11} = \lambda_{22} = 0$$ and with $$\lambda_{21} = c$$ we get $$\lambda_{12} = 3c$$.

We are now searching the the decision boundary, where it holds $$R(\alpha_1|x) = R(\alpha_2|x)$$. We can use the previous equations and substitute the $$\lambda$$‘s: \begin{align} R(\alpha_1|x) = 3c\cdot p(C_2|x) &= c \cdot p(C_1|x) = R(\alpha_2|x)\ \Longleftrightarrow 3c\frac{p(x|C_2)p(C_2)}{p(x)} &= c\frac{p(x|C_1)p(C_1)}{p(x)}\ \Longleftrightarrow 3p(x|C_2) &= p(x|C_1)\ \Longleftrightarrow 3 \exp(-\frac{(x – 2\mu_2)^2}{2\sigma^2}) &= \exp(-\frac{(x – \mu_2)^2}{2\sigma^2})\ \Longleftrightarrow \frac{\exp(-\frac{(x – \mu_2)^2}{2\sigma^2})}{\exp(-\frac{(x – 2\mu_2)^2}{2\sigma^2})} &= 3\ \Longleftrightarrow – \frac{(x – \mu_2)^2}{2\sigma^2} + \frac{(x – 2\mu_2)^2}{2\sigma^2} &= \ln(3)\ \Longleftrightarrow \frac{-x^2 + 2x\mu_2 – \mu_2^2 + x^2 – 4x\mu_2 + \mu_2^2}{2\sigma^2} &= \ln(3)\ \Longleftrightarrow \frac{-2x\mu_2}{2\sigma^2} &= \ln(3)\ \Longleftrightarrow x &= -\ln(3)\frac{\sigma^2}{\mu_2} \end{align}

Sorry for this long equivalent transformation. Is my basic approach correct? And is my result also correct?

Thank you for reading and helping!

## Subtree with minimum sum of nodes’ costs

Let’s consider a tree ( not necessary binary) and to each node $$i$$ we associate a cost $$\sigma(i)$$ that can be non-negative or non-positive. We want to select the set of nodes that minimize $$\sum_i \sigma(i)$$. Of course, the problem until now is easy :

1. If there are only non-negative costs the solution is the empty set.

2. If there are some non-positive costs, we select them all.

But, there is an additional constraint that states: “if a node $$\sigma(i)$$ is selected, it’s predecessor to the root is also selected. Thus the problem is to select the subtree (I am not sure about the vocabulary) that minimize $$\sum_i \sigma(i)$$.

My question is how to design an algorithm that solves this problem and how to prove its correctness and compute its complexity?

(I have a hint that states that we could do some bottom-up traversal, of the tree but I don’t know how to use this information)

Any help will be appreciated. Thanks

## Given a tree find path that maximizes the median of the costs of edges

We have given tree with $$N$$ nodes and $$N-1$$ edges, such that each edges is assigned positive weight. We need to find path of length between $$L$$ and $$R$$ inclusively, with maximum cost. Cost of a path is defined as a median of all costs of edges that lie on the path.

Note that the given numbers $$L$$ and $$R$$ describe the count of edges on the path, not the sum of the costs of all edges.

I have a solution in $$O(N^2 \log N)$$ that tries every possible path and saves the costs of edges in binary search tree, but is is possible to get faster solution?

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