UNIVERSIDADE FEDERAL RURAL UNIVERSIDADE FEDERAL RURAL DO DO

RIO DE JANEIRORIO DE JANEIRO

INSTITUTO DE CIÊNCIAS EXATAS INSTITUTO DE CIÊNCIAS EXATAS DEPARTAMENTO DE MATEMÁTICADEPARTAMENTO DE MATEMÁTICA

SET COVERING PROBLEM TO FORMULATE A CROP ROTATION MODEL

Valdomiro Neves Lima

When planning water supply system and distribution network for irrigation, water resource limitation must be pondered

Specific models should be obtained to help making the best choice to overcome this problem, considering as variables:

● the selection and production of crops, ● model of soil exploration in accordance to cultivation, and ● the calendar of water management.

● Frizzone quotes linear programming models which greatly consider the water required for a centre pivot to supply a unit area of a particular cultivation.

● However, the sustainable  rational  use of water, essential  for  today’s economy, is not included in his model.

A work from Food and Agriculture Organization of the United Nations (FAO) in 1979 shows the importance of balancing water  supply  suitability and the biological needs of crop production; which means that an efficient update of water in cultivation can only be achieved when planning, project and the operation of water supply and distribution systems are oriented, including water shortage periods and crop water requirement for optimum growth and high yield.

Therefore, the following parameters are presented to aid the selection of crops according to total water requirement for maximum yields: ● water-use efficiency (Ey in kg/m3 % humidity), ● yield obtained per unit of water volume (kg/m3) and ● the sensitivity to water supply.

Since the water demand must be supplied through the soil by roots absorption, these parameters are a result of the relation between the real evapotranspiration (ETr) and maximum evapotranspiration (ETm), and they generate possible combinations of the irrigation process.

In this project, the combinatorial effect of crop selection and distribution based on the parameters above is dealt with the set covering problem, aiming at a more efficient and rational use of water.Generally, such problem is modeled as (SCP):  Min CT. X

Suj. à A. X X {0, 1}n

with A = a matrix mn, where aij {0, 1} C = cost vector associated with X in n

= vector in m, which has components of the same value: i = 1, i

Particularly, this modeling considers: A = a matrix mn, where aij {0, 1} and the indexes indicate: i = crop type and j = crop i water requirementC = vector associated with X in n, whose cj components obtain their values in the parametrisation of crop i water requirement interval values (mm/per.veget);

X solution vector has its components xj = 1, if j meets crop i needs, and xj = 0, if it does not.

Min CT. XSuj. à A1. X 1 A2. X 2

X {0, 1}n

with A1. X 1 presents special structure

● (non ordered cartesian product = n.o.c.p)● Factor D and E as vertex of bipartite graphs And

in A2. X 2 , Lagragean relaxation is applied. ● constraints in the next n.o.c.p)

x2 + x3 x1 + x4 + x5 x1 + x2 + x4 x4 x3 + x5

Example :

PC1: 3 (2,3) (1,4,5)PC2: 4 (2) (1,4,5) PC3: 4 (3; 1,4) (1,4,5)PC4: 5 (3; 1,4) (4) PC5: 5 (3;1,4) (1,5;ø) PC6: 6 (3;1,4) (4) PC7: 6 (ø;ø) (4;ø) PC8: 6 (ø;1,4;5) (4)PC9: 5 (2) (4) PC10: 5 (2) (1,5;ø)PC11: 6 (ø) (4) PC12: 6 (2;3,5) (4)

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11000000000000000

01100000000000000

00011000000000000

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x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12x13 x14 x15 x16 x17

Matrix A

Graph of Matrix A

Matriz de custo unitário por coluna 18 17 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 5 1 2 3 4 5 4 2 3 4 5 4 2 3 4 5 8 2 3 4 5 6 7 8 9 3 3 4 5 4 4 5 6 7 4 4 5 6 4 5 6 7 8 5 5 6 7 8 9 4 6 7 8 9 6 6 7 8 9 10 11 9 6 7 8 9 10 11 12 13 14 4 7 8 9 10 5 8 9 10 11 12 4 10 11 12 13 2 13 14 2 15 16 2 16 17

Matriz de custo unitário por colunaFASE TEMPO NG NS OBJ OBJR OBJAT COD inicio : 14:14:33 1 ｧ subpr.: 14:14:33 1 1 1.0000 1.0000 17.0000 10 incumb. : 14:14:33 17 17 5.0000 5.0000 17.0000 6 ATUALZ-solucao atual 0 0 2 0 0 2 0 1 0 0 0 0 1 0 0 1 0 1 ｧ c/otim: 14:14:33 17 17 5.0000 5.0000 5.0000 6 solucao = 0 0 2 0 0 2 0 1 0 0 0 0 1 0 0 1 0 incumb. : 14:14:33 135 135 4.0000 4.0000 5.0000 6 ATUALZ-solucao atual 0 0 0 2 0 0 0 2 0 0 0 0 1 0 0 1 0 final : 14:14:33 190 190 4.0000 5 solucao = 0 0 0 2 0 0 0 2 0 0 0 0 1 0 0 1 0 (med,max) = 2.49 6 [subpr.] 71.78 192 [lista ] (min,med,max) = 1 1.35 2 [part D] 1 1.22 2 [part E]

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