Apresentação de Silvana Nobre, socio-fundadora da Atrium Forest Consulting durante o ciclo de seminários em Pesquisa Florestal Avançada na Universidade Politécnica de Madrid, em 20 de fevereiro de 2013.
Strategies for Landing an Oracle DBA Job as a Fresher
The use of linear programming to integrate forest operations
1. The use of linear programming to
integrate forest operations
Silvana Nobre
Atrium Forest Consulting, Brasil
silvana@atriumforest.com
2. The use of linear programming to
integrate forest operations
1. Overview on Brazilian forest-based companies
2. First-sight Problem: Clonal seedlings allocation
3. Integration A: Plantation Activities Constraints
4. Integration B: Nursery Production process
5. Conclusions
3. 1 . Overview on
Brazilian forest-based
companies
Source: Bracelpa - 2010
179 units
from 400 to 237,000 ha
Total: 4,671,521 ha
Species: Eucalyptus,
Pinus, Teca,
Acácia, Paricá,
Araucária
Industries: Pulp & Paper
Solid & Panels
Steel Mills
Investments
7. 2 - Clonal seedling allocation
Tipical Pulp&Paper Forest Industry Unit
Industry Forest Unit
Productivity
Clear Cut
Annual Replanting Area
Spacing ( 2,5 x 3 )
Trees per area
Plantation eficciency
Seedling Need
95.254 ha
50 m3/ha.year
6 years old
15.876 ha
6 m2
1.667 trees/ha
95%
27.782.475 seedlings/year
Tipical Brazilian Stand
No Stands to plant in a year
Techinical Recomendation Units
Different Clones
20
794
4
6
TRU - 1
......
Fertilization x
---• Clone 1 – 300
• Clone 2 – 260
• Clone 3 – 220
Stand
1
3
5
3
4
3
794
......
Fertilization z
---• Clone 3 – 310
• Clone 1 – 250
• Clone 2 – 210
2
793
......
Fertilization y
---• Clone 2 – 320
• Clone 3 – 270
• Clone 1 – 230
2
4
TRU - 3
1
792
TRU - 2
1
2
ha
stands
Units
clones
TRU
4
8. 2 - Clonal seedling allocation
Maximizar Produção Potencial….
Max Z =
∑ P ij × X ij
•
Pij Probable Productivity of Stand i, if we plant clone j
•
Xij Area of Stand i, we will plant clone j
Exercise….
•
MS Excel ®
•
330 ha
•
5 stands
•
3 Technical Recommendation Units
•
3 Clones
10. 2 - Clonal seedling allocation
Max Z =
∑ P ij ×
X ij
•
Pij Probable Productivity of Stand i, if we plant clone j
•
Xij Area of Stand i, we will plant clone j
Subject to:
∑ X ij ≤ Ij, I =1, 2, 3, 4, 5 (stands)
∑ X ij ≥ Cj, j=1, 2, 3
(clones)
Where:
Ii, area of stand i that can be planted
Cj, at least this area must be planted with clone j
11. 3 - Plantation Activities Constraints
Exercise….
•
MS Excel ®
•
Pre-Planned Harvesting Process
•
Until 3 months after Harvesting
•
Planning Horizon: 12 months
•
Period: 1 month
•
1 team
15. 3 - Plantation Activities Constraints
Max Z = ∑ P ij × X ijk
•
Pij Probable Productivity of Stand i, if we plant clone j
•
Xijk Area of Stand i, we will plant clone j, in period k
Subject to:
∑ X ij ≤ Ij, I =1, 2, 3, 4, 5 (stands)
- Area Constraints
∑ X ijk ≥ Cj, j=1,2, 3
(clones)
- Biological Constraints
∑ X ijk ≤ Tk, k= 1 to 12
(periods)
- Plantation Team Capacity
∑ X ijk = 0 , for some stands i in some periods k - Harvest Constraints
Where:
Ii, area of stand i that can be planted
Cj, at least this area must be planted with clone j
Tj, maximum productivity of a team plantation in each period k
16. 4 - Nursery production process
1st Step :
Clonal Garden
2nd Step : Collect mini-cuttings
from mini-stumps
3rd Step : cuttings production
17. 4 - Nursery production process
4th Step : Green House
5th Step : Open Space Growth
18. 4 - Nursery production process
6th Step : Seedlings Deliver Space
•
MS Excel ®
•
Constraints:
Cutting production team
Green House Occupation
Clonal Garden Capacity
22. Max Z = ∑ P ij × X ijk
• Pij Probable Productivity of Stand i, if we plant clone j
• Xijk Area of Stand i, we will plant clone j, in period k
Subject to:
Area Constraints (stands)
∑ X ij ≤ Ij,
I =1, 2, 3, 4, 5
Biological Constraints (clones)
∑ X ijk ≥ Cj,
j =1, 2, 3
Plantation Team Capacity
∑ X ijk ≤ Tk, k = 1 to 12
Harvest Constraints
∑ X ijk = 0 , for some stands i in some periods k
Cutting Product.Team Capacity
∑ d j X ijk ≤ D k
Greenhouse Max Space
∑ g j X ijk
Mini-Stumps in the Nursery
∑ s j X ijk ≤ S jk , for j =1,2, 3 and k= 1 to 12
+
k= 1 to 12
g j X ijk-1 ≤ G k , for k = 1 to 12
23. Where:
I i,
area of stand i that can be planted
Cj, at least this area must be planted with clone j
Tj, maximum productivity of a team plantation in each period k
dj, nursery efficiency of clone j
Dk cuttings team capacity, in period k
gj, green house occupation factor of clone j
Gk green house maximum occupation in period k
s j,
mini-stumps need factor of clone j
Sjk mini-stumps in the Nursery in production of clone j, in period k
24. 5 - Conclusions
In one year, on 10% of the area you could have
planted the 1st clone, but you planted the 2nd one
10% area
>>>
10% diff produc >>>
Production difference >>>
Minimum Wood value >>>
Prouction Diff Value
Prouction Diff Value
•
>>>
MS Excel ® simplified exercise
•
•
>>>
5% of potential production
10% productivity from first to second clone
1,588 ha
3
30 m /ha
3
47,627 m
50 R$/m3
2,381,355 R$
850,484
€