Additional material
5. The different ways to cut the planks and to purchase the precut planks can be given as follows:
2,6m (needed 200)
1
0
1
0
1
0
Type 1 cut
Type 2 cut
Type 3 cut
Type 4 cut
Purchase 5
Purchase 6
2,0m (needed 250)
1
2
0
1
0
1
5m planks (100)
3m planks (150)
Purchase
The variables xi are the type i cuts, i = 1, 2, 3, 4 and x5 is the amount of those 2, 6 meters long
planks that are bought in addition, as well as, x6 is the amount of the 2, 0 meters long planks
bought. We have the following minimization problem
min
2
5 x1
x1 + x3 + x5
≥ 200
x1 + 2x2 + x4 + x6
≥ 250
x1 + x2
≤ 100
x3 + x4
≤ 150
x
+ 25 x2 + 15 x3 + 15 x4 + 2x5 + 23 x6
≥ 0.
The standard form of the above problem is
v> x
min
Ax
= b
x ≥ 0,
 


200
1 0 1 0 1 0 −1
0 0 0
250

1 2 0 1 0 1
0
−1
0
0
 , b =   and
where A = 
100
1 1 0 0 0 0
0
0 1 0
150
0
0
1
1
0
0
0
0
0
1
>
>
2
2
1
1
3
v = /5 /5 /5 /5 2 /2 0 0 0 0 . The x = [0 0 0 0 200 250 0 0 100 150] is a
feasible basic solution. The reduced costs are


1 0 1 0 −1
0
1 2 0 1
0 −1

r>
= [2/5 2/5 1/5 1/5 0 0] − [2 3/2 0 0] 
N
1 1 0 0
0
0
0 0 1 1
0
0
= −31/10 −13/5 −9/5 −13/10 2 3/2 .
Hence the initial tableau is
1
1
1
0
0
2
1
0
-13⁄5
-31⁄10
1
0
0
1
-9⁄5
0
1
0
1
-13⁄10
1
0
0
0
0
0
1
0
0
0
-1
0
0
0
2
0
-1
0
0
3⁄2
0
0
1
0
0
0
0
0
1
0
200
250
100
150
-775
The entering variable is x1 and leaving x9 . The Gaussian elimination leads us to
0
0
1
0
0
-1
1
1
0
1⁄2
1
0
0
1
-9⁄5
0
1
0
1
-13⁄10
1
0
0
0
0
0
1
0
0
0
-1
0
0
0
2
0
-1
0
0
3⁄2
-1
-1
1
0
31⁄10
0
0
0
1
0
100
150
100
150
-465
Now the entering variable is x3 and the leaving is x5 . After Gaussian elimination we obtain
1
0
0
1
0
0
-1
1
1
1
-13⁄10
1
0
0
0
0
0
1
0
1
1
0
0
-1
9⁄5
-13⁄10
0
1
0
0
0
-1
0
0
1
1⁄5
0
-1
0
0
3⁄2
-1
-1
1
1
13⁄10
0
0
0
1
0
100
150
100
50
-285
The entering variable is x2 and leaving x10 . The Gaussian elimination leads to the optimal tableau:
0
0
1
0
0
0
0
0
1
0
1
0
0
0
0
1
0
-1
1
0
0
1
1
-1
1⁄2
0
1
0
0
0
0
-1
-1
1
3⁄2
0
-1
0
0
3⁄2
0
-2
0
1
13⁄5
1
-1
-1
1
13⁄10
150
100
50
50
-220
2
The optimal vertex is then x>
opt = [50 50 150 0 0 100 0 0 0 0] with the minimum costs /5 (50 + 50)+
1/5 · 150 + 3/2 · 100 = 220 euros. That is, to minimize the costs we cut 50 5 meters long planks in
type 1, 50 5 meters long planks in type 2. All 150 3 meters long planks is cut in type 3 and we
purchase 100 2 meters long planks from the store.
6. The gradient and the Hessian matrix for f (x) = e−x1 −x2 + x21 + x22 are
2 + e−x1 −x2
e−x1 −x2
2x1 − e−x1 −x2
.
and Hf =
∇f (x) =
e−x1 −x2
2 + e−x1 −x2
2x2 − e−x1 −x2
The Hessian matrix is positive definite for every x ∈ R2 , and therefore f is strictly convex. The
Newton iterations are:
h
i−1
x(k+1) = x(k) − Hf x(k)
∇f x(k) .
0
The starting point is x(0) =
. Next iteration is
0
1/4
−0, 1065
x(1) = 1
and ∇f x(1) ≈
.
/4
−0, 1065
Moreover,
x
(2)
−0, 0013
0, 2832
(2)
=
and ∇f x
≈
.
0, 2832
−0, 0013
9. From the KKT-conditions we obtain
ex1 −x2 + u1 ex1 − u2
x1 −x2
+ u1 e
x1
x2
−e
u1 ( e
x2
+e
=
0,
=
0,
− 20)
=
0,
u2 x1
=
0,
u1 , u2
≥
0.
The conditions are satisfied when u1 6= 0 and u2 6= 0 (all other possibilities lead to contradiction).
Since u2 6= 0, we must have x1 = 0 and ex1 + ex2 − 20 = 0. Thus, ex2 = 19 and therefore
20
1
and u2 = 361
.
x2 = ln 19. Moreover, u1 = 361
11. Let us denote a> = [a1 a2 a3 ]. The Lagrangian is
L(x, u) = ex1 + x1 x2 + x22 − 2x2 x3 + x23 + u1 x21 + x22 + x23 − 5 + u2 a> x + 2 .
The KKT-conditions are
ex1 + x2 + 2u1 x1 + a1 u2 = 0,
x1 + 2x2 − 2x3 + 2u1 x2 + a2 u2 = 0,
− 2x2 + 2x3 + 2u1 x3 + a3 u2 = 0,
u1 x21 + x22 + x23 − 5 = 0,
a1 x1 + a2 x2 + a3 x3 + 2 = 0,
u1 ≥ 0.
2
The point x̃ = (0, 0, 1) should be a local minimum of the problem, that is, it needs to satisfy the
KKT-conditions. Hence,
1 + a1 u2 = 0,
(0.1)
− 2 + a2 u2 = 0
(0.2)
2u1 + a3 u2 = 0
(0.3)
u1 (1 − 5) = 0
(0.4)
a3 + 2 = 0.
(0.5)
Equations (0.4) and (0.5) yield to u1 = 0 and a3 = −2, respectively. Substituting these to (0.3),
we obtain u2 = 1 and moreover, we derive a1 = −1 and a2 = 2. Thus, the vector a> = [−1 2 − 2].
3