SD Method
Standard Deviation Weighting
SD (Standard Deviation) method is an objective weighting method in MCDM. It assigns higher weights to criteria with greater variation among alternatives, because greater variation provides more information for distinguishing alternatives.
Video Demonstration
Step-by-Step Algorithm
Form the decision matrix
Create the decision matrix with $m$ alternatives and $n$ criteria :
\[ X=\begin{bmatrix} x_{11} & x_{12} & \cdots & x_{1n}\\ x_{21} & x_{22} & \cdots & x_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ x_{m1} & x_{m2} & \cdots & x_{mn} \end{bmatrix} \]
\(x_{ij}\) presents the performance value of ith alternative on jth criterion, m and n are the numbers of alternatives and criteria respectively.
| Alternative | $C_1$ | $C_2$ | $C_3$ |
|---|---|---|---|
| $A_1$ | 10 | 20 | 30 |
| $A_2$ | 15 | 25 | 35 |
| $A_3$ | 25 | 15 | 35 |
Normalize the decision matrix
Normalize the decision matrix to eliminate differences in measurement units and make the criteria comparable.
For benefit criteria :
For cost criteria :
\(x_j^{\max}\) and \(x_j^{\min}\) are respectively the maximum and minimum values of criterion \(C_j\).
C₁ and C₃ are benefit criteria. C₂ is a cost criterion.
| Alternative | $C_1$ | $C_2$ | $C_3$ |
|---|---|---|---|
| $A_1$ | 0.0000 | 0.5000 | 0.0000 |
| $A_2$ | 0.3333 | 0.0000 | 1.0000 |
| $A_3$ | 1.0000 | 1.0000 | 1.0000 |
Calculate the standard deviation
Calculate the arithmetic mean of the normalized values for each criterion.
\(\bar{r}_j\) represents the mean normalized performance of criterion \(C_j\).
Calculate the standard deviation of each normalized criterion to measure its variation across alternatives.
| Metric | $C_1$ | $C_2$ | $C_3$ |
|---|---|---|---|
| Mean ($\bar{r}_j$) | 0.4444 | 0.5000 | 0.6667 |
| Std. Dev. ($\sigma_j$) | 0.4157 | 0.4082 | 0.4714 |
Weights of criteria
Normalize the standard deviation values to obtain the objective weight of each criterion.
The criterion weights satisfy :
Criteria with greater standard deviation receive greater weights.
| Metric | $C_1$ | $C_2$ | $C_3$ |
|---|---|---|---|
| Weight ($w_j$) | 0.3210 | 0.3151 | 0.3639 |
Verification: $w_1 + w_2 + w_3=0.3210 + 0.3151 + 0.3639=1.0000$ ✓
