6
votes
Accepted
Families of methods to deal with criterion uncertainties in multicriteria decision analysis
There are many applications of different MCDM (Multi-Criteria Decision Making) method families when there is some kind of uncertainty in weights or amount of objectives or criteria. Mosadeghi et al. (...
- 8,677
5
votes
Accepted
Solving an exponential utility function
The utility function $u(x)=a-be^{-x/20\,000}$ with the conditions $u(0)=0$ and $u(100\,000)=1$ gives $$u(0)=a-be^{-0/20\,000}\implies 0=a-be^0=a-b\implies a=b\tag1$$ and $$u(100\,000)=a-be^{-100\,000/...
- 5,452
4
votes
Accepted
Assignment and scheduling problem with resource constraints
I didn't find a way to express the transition constraints so i give a description what this does and mention what it lacks.
...
- 4,049
3
votes
Accepted
Two criteria in a problem that involves multiple-criteria decision-making
More criteria would not necessarily help, and in fact might make it harder to make a selection (for instance, if every alternative excels on some criterion and does badly on some other criterion). ...
- 40.1k
3
votes
Accepted
How to Use the Weighted Sum Method in R
The GeeksForGeeks page linked in your previous question provides one possible way to scale your attributes. Using that scaling method, your code is correct and two clusters has the highest performance ...
- 40.1k
3
votes
Accepted
Use the Weighted Sum Method in R
The edited version fixed one problem (numeric fields being character strings). Before using the weightedSum function (or doing any other calculations), you also ...
- 40.1k
2
votes
Application with VIKOR multicriteria method
It means that both options have the same ranking and it is up to you what to do with that information.
If you look at the VIKOR paper of Opricovic & Tzeng (2004), you can see that $Q$ is defined ...
- 1,726
2
votes
Accepted
Using VIKOR multicriteria method in R
Yes, lowest $Q$ wins. The weight $v$ is used to form a weighted average of $S$ (multiplied by $v$) and $R$ (multiplied by $1-v$). There is a Wikipedia page that attempts to explain VIKOR.
- 40.1k
1
vote
Weighted sum in the objective function
We typically normalize on seconds, minutes xor dollars, for as far as that is possible. And then leave it to a business stakeholder alignment meeting to tweak the weights. But normalization is not ...
- 5,001
1
vote
How can we choose the right weight to solve multi-objective problem using weighted sum method?
As far as I know, solving the multi-objective optimization by the weighted sum method should give one of the solutions that already exists on the Pareto non-dominance solution frontier. Let us make a ...
- 9,258
1
vote
How can we choose the right weight to solve multi-objective problem using weighted sum method?
I'm going to assume here that bigger is better for all criteria. Let $x_{a,c}$ be the score for alternative $a$ on criterion $c,$ where $a\in \lbrace 1,\dots,A\rbrace$ and $c\in\lbrace 1,\dots,C\...
- 40.1k
1
vote
Normalized VIKOR method multicriteria
Yes! You're absolutely correct.
Basically, whenever normalization is done in MCDM methods, the designer has to see whether the criteria is a benefit criteria (for which the higher the value, the ...
- 529
1
vote
Strengths and weaknesses of some Multiple-criteria decision-making (MCDM) methods
The MCDM methods was created to make better decisions when the decision criteria is usually non-quantitive and is needed to create a decision as fast as possible.
Also, there is really no obvious ...
- 9,258
1
vote
Accepted
Conflicts with weights of the topsis method in R
Normalizing the Weights. According to the topsis() documentation, the argument weights must be "[a] numeric vector with ...
- 1,895
1
vote
MODM to create balanced groups of students to maximise diversity
This is not a direct answer, but I can point you to two papers that are closely related to what you are working on.
[1] Rubin, Paul A. and Lihui Bai (2015). “Forming Competitively Balanced Teams”. IIE ...
- 40.1k
1
vote
Solving an exponential utility function
set k=1 in a=be=1+be^5.
therefore a=k+be^5.
solve for k=a-be^5.
- 11
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