Engineering Optimization: Theory and Practice, Fourth Edition

8.9 Primal and Dual Programs in the Case of Less-Than Inequalities 515

λ∗ 03 =

c 03 (x 1 ∗)^13 a^0 (x 2 ∗)^23 a^0 (x∗ 3 )a^330

x 0 ∗

1

3 =

80 (x∗ 1 )(x 2 ∗)

480

=

x 1 ∗x∗ 2

6

(E 7 )

λ∗ 11

λ∗ 11

=c 11 (x∗ 1 )^11 a^1 (x 2 ∗)^21 a^1 (x 3 ∗)a^311

1 = 8 (x 1 ∗)−^1 (x 2 ∗)−^1 (x∗ 3 )−^1 =

8

x∗ 1 x∗ 2 x∗ 3

(E 8 )

Equations(E 5 ) o (Et 8 ) iveg

x∗ 1 = 2 , x∗ 2 = 1 , x∗ 3 = 4

Example8.4 One-degree-of-difficulty Problem

Minimizef=x 1 x 22 x 3 −^1 + 2 x 1 −^1 x 2 −^3 x 4 + 01 x 1 x 3

subject to

3 x 1 −^1 x 3 x 4 −^2 + 4 x 3 x 4 ≤ 1

5 x 1 x 2 ≤ 1

SOLUTION HereN 0 = , 3 N 1 = , 2 N 2 = , 1 N=6,n=4,m=2, and the degree
of difficulty of this problem isN−n− 1 =1. The dual problem can be stated as
follows:

Maximixzev(λ)=

∏m

k= 0

∏Nk

j= 1

(

ckj

λkj

∑Nk

l= 1

λkl

)λkj

subjectto

∑N^0

j= 1

λ 0 j= 1

∑m

k= 0

∑Nk

j= 1

akijλkj= 0 , i= 1 , 2 ,... , n

∑Nk

j= 1

λkj≥ 0 , k= 1 , 2 ,... , m

(E 1 )

Asc 01 = , 1 c 02 = , 2 c 03 = 0, 1 c 11 = , 3 c 12 = , 4 c 21 = , 5 a 011 = , 1 a 021 = , 2 a 031 =
− 1 ,a 041 = , 0 a 012 = − 1 ,a 022 = − 3 ,a 032 = , 0 a 042 = , 1 a 013 = , 1 a 023 = , 0 a 033 =
1 ,a 043 = , 0 a 111 = − 1 ,a 121 = , 0 a 131 = , 1 a 141 = − 2 ,a 112 = , 0 a 122 = , 0 a 132 = , 1