What to do in these cases where I can't rank by the VIKOR multicriteria method because Q has no value?

    df1<-structure(c(14509.9423478068, 4648.5779693212, 922, 876), .Dim = c(2L, 
    2L), .Dimnames = list(NULL, c("Criteria1", "Criteria2")))

> df1
     Criteria1 Criteria2
[1,] 14509.942       922
[2,]  4648.578       876
      w <- c(0.5,0.5)
      cb <- c('min','max')
      v <- 0.5

> result1
  Alternatives   S   R   Q Ranking
1            1 0.5 0.5 NaN       -
2            2 0.5 0.5 NaN       -
  • $\begingroup$ Related but not a dupe: Using VIKOR multicriteria method in R $\endgroup$ Commented Mar 16, 2022 at 22:36
  • $\begingroup$ It is a different question @SecretAgentMan $\endgroup$
    – Antonio
    Commented Mar 16, 2022 at 23:19
  • $\begingroup$ Yes, which is why I said it is not a duplicate. My comment is intended to help future visitors find a related question. They may have similar questions as you. I hope they upvote both posts. $\endgroup$ Commented Mar 17, 2022 at 12:58

1 Answer 1


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 as follows:

$$Q_j = v(S_j-S^*)/(S^--S^*) + (1-v)(R_j-R^*)/(R^--R^*)$$ where $$S^* = \min_j S_j, \quad S^-=\max_j S_j$$ $$R^* = \min_j R_j, \quad R^-=\max_j R_j$$

In your example, $S_1$ and $S_2$ are both $0.5$, therefore, $S^*=0.5$ and $S^-=0.5$ which leads to a division by zero and consequently, $Q$ is not defined. This holds not only for $S$ but also for $R$.

Opricovic, S., & Tzeng, G.-H. (2004). Compromise solution by MCDM methods: A comparative analysis of VIKOR and TOPSIS. European Journal of Operational Research, 156(2), 445–455. doi:10.1016/s0377-2217(03)00020-1


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