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.
How to Troubleshoot Apps for the Modern Connected Worker
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
€