Example 3 Again, we continue with the previous example. Because n$_{1}$ = 3 , n$_{2}$ = 1 , n$_{3}$ = 1 , n$_{4}$ = 1 , and n$_{5}$ = 0 there are in total (3 × 2+1) × (1 × 2+1) × (1 × 2+1) × (1 × 2+1) × (0 × 2+1) = 189 different sets of the form E ( ℓ$_{1}$, ℓ$_{2}$, ℓ$_{3}$, ℓ$_{4}$, ℓ$_{5}$ ) . To illustrate, consider a particular case in which ℓ$_{1}$ = 1 , ℓ$_{2}$ = 2 , ℓ$_{3}$ = 2 , ℓ$_{4}$ = 2 , and ℓ$_{5}$ = 1 . Suppose that we replace the usual constraint N ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 with a more restrictive one (20) in the main optimization problem. Then, the optimization should take the form of (21). As long as constraint (E) holds, constraint (F) can be replaced with those in the last column of min ξ , x ,µ, ζ , ω µ ln ( ∑ 5 i $_{=1}$θ$_{i}$ e − d i µ ) such that (A) ω$_{i}$ ≥ − z$_{i}$e ζ$_{i}$ for ∀ i (B) ζ$_{i}$ ≥ α ∑ 7 j =1 [ x$_{j}$ A$_{ij}$ − Q$_{j}$ ( A T ξ ) j − B$_{j}$b$_{j}$ ( x$_{j}$ − Q$_{j}$ ) ] , ∀ i (C) ζ$_{i}$ ≥ α 7 j $_{=1}$[ x$_{j}$ A$_{ij}$ − B$_{j}$b$_{j}$x$_{j}$ ] , ∀ i (D) µ ≥ 0 (E) ξ ∈ E (1 , 2 , 2 , 2 , 1) (F) x$_{j}$ = 0 if ( A T ξ ) j > B$_{j}$b$_{j}$ x$_{j}$ ∈ [0 , Q$_{j}$ ] if ( A T ξ ) j = B$_{j}$b$_{j}$ x$_{j}$ = Q$_{j}$ if ( A T ξ ) j < B$_{j}$b$_{j}$ for ∀ j (21) E ( ℓ$_{1}$ = 1 , ℓ$_{2}$ = 2 , ℓ$_{3}$ = 2 , ℓ$_{4}$ = 2 , ℓ$_{5}$ = 1) (20) = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 , 0 ≤ ξ$_{1}$ ≤ 0 . 18 , ξ$_{2}$ = ξ$_{3}$ = ξ$_{4}$ = 0 . 18 } E ℓ$_{1}$ 1 = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 , 0 ≤ ξ$_{0}$ ≤ 0 . 18 } E ℓ$_{2}$ 2 = ξ ∈ R $^{5}$| 5 i =1 ξ$_{i}$ = 1 , ξ ≥ 0 ,ξ$_{1}$ = 0 . 18 } E ℓ$_{3}$ 3 = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 ,ξ$_{2}$ = 0 . 18 E ℓ$_{4}$ 4 = ξ ∈ R $^{5}$| 5 i =1 ξ$_{i}$ = 1 , ξ ≥ 0 ,ξ$_{3}$ = 0 . 18 } E ℓ$_{5}$ 5 = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 } (19) 19