Crop Succession Requirements in Agricultural ProductionPlanning

W.K. Klein Haneveld and A.W. Stegeman

University of GroningenDepartment of Econometrics & Operations Research

P.O. Box 8009700 AV GroningenThe Netherlands

W.K.Klein.Haneveld@eco.rug.nlA.W.Stegeman@ppsw.rug.nl

Abstract

A method is proposed to handle crop succession requirements when determining the cultivatedareas of n crops for a planning period of T years. For each year, the decision variables representthe areas where a sequence of m crops is cultivated, m ≥ 1. Here, m + 1 is the length ofthe longest forbidden sequence of crops not containing any forbidden subsequences. The cropsuccession constraints are linear in the decision variables. For each year, the number of nm

decision variables may be reduced by forming suitable combinations of crop sequences. Forthis purpose, an algorithm is presented. Also, multi-year linear programming models for farmproduction planning containing crop succession constraints are considered. It is shown that,under some regularity conditions, a stationary cropping plan is an optimal solution of sucha model. Finally, it is discussed how to determine forbidden crop sequences not containingforbidden subsequences, when a collection of forbidden sequences is given.

AMS 2000 subject classifications. Primary 65K05; secondary 90B10, 90B30, 90C05.Key words and phrases. Agriculture, crop rotation, crop succession requirements, production planning, linearprogramming.

1 Introduction

In agricultural production planning it may happen, and this is often the case, that succession ofcertain crops on the same piece of land is not allowed or not advisable. Otherwise soil fertility willdecrease or the crops may become susceptible to diseases, plagues or weeds. On the other hand,a certain succession of crops may be recommended. We say that in these situations certain cropsuccession requirements have to be fulfilled. For example, in some regions cotton should not begrown after cotton because remaining seeds may cause pests; sorghum after sorghum may causeproblems with the weed striga; potatoes should not succeed potatoes because of the occurrence ofnematodes in the soil. And soya-beans is recommended to alternate with grains in order to restorethe fertility. Often, it is important to leave the land lying fallow so that natural vegetation canrestore the fertility of the soil. Maize and other grains should, for instance, be followed by fallow.Or it might be advisable to let arable land alternate with grassland for cattle. In practice, the cropsuccession requirements usually determine crop rotation cycles. A crop rotation cycle is a sequenceof crops (e.g. soya-beans, maize, fallow) which satisfies the crop succession requirements if appliedcyclically on the same piece of land. A crop rotation cycle is usually implemented in such a way thateach year there is approximately the same acreage for each of the crops in the cycle. Often only fewcrop rotation cycles are taken into consideration, advised by e.g. staff of agricultural experimentalstations. A well-known example is the 8-course rotation cycle cotton-fallow-fallow-cotton-fallow-sorghum-fallow-fallow which has been applied for many years in the Gezira scheme in Sudan. Sincethe introduction of pesticides, insecticides and mechanized cultivation the Gezira rotation schemehas been replaced by the 4-course cycle cotton-wheat-sorghum (or groundnuts)-fallow.

It will be clear, that crop succession requirements have important implications for productionplanning on the farm. Not only is it necessary to take the usage of the land in previous years intoconsideration, but it is also important to be sure, that the production plan is such that also inthe future good (i.e. high yielding) production plans are possible. The implementation of a fixedrecommended crop rotation cycle may take into account these considerations, but as a consequencethere is not much freedom in choosing production plans for future years.

In this paper we will focus instead on feasible production plans, i.e. those satisfying the cropsuccession requirements. We assume the crop succession requirements are given in the form ofcrop sequences not allowed to be cultivated on the same piece of land. In Section 2 we considera planning period of T years. Decision variables are introduced which represent the area of landwhere a certain sequence of m crops is cultivated, the last crop of the sequence being grown inyear t. Here, m + 1 is the length of the longest forbidden crop sequence not containing any shorterforbidden subsequence. It appears that the problem of determining whether the cropping plan ofyear t− 1 is compatible with the cropping plan of year t, can be interpreted as a max-flow problem.In this way, the crop succession requirements may be written as linear contraints in the decisionvariables. In Section 3 we consider linear programming models for farm production planning, whichcontain crop succession constraints. It is shown that, under some regularity conditions, a multi-yearLP-model may be replaced by a one-year model. In this case a stationary cropping plan, i.e. acropping plan not depending on the year index t, is an optimal solution of the multi-year model.Also, it is shown that a stationary cropping plan is composed of several crop rotation cycles. InSection 4 we again consider a planning period of T years, as in Section 2. Here, we focus onreducing the number of decision variables. For n crops, the number of decision variables for year tequals nm. We show that this number may be reduced by forming combinations of some of the cropsequences without losing crop succession information. An algorithm is presented which finds thecrop sequences that may be combined. Finally, in Section 5, we discuss how to determine forbiddencrop sequences not containing forbidden subsequences, when a collection of forbidden sequences is

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given. The results of Section 2.2 and Section 3 are due to Klein Haneveld [2] and were included inSchweigman [3].

2 Crop succession constraints

2.1 Definitions and assumptions

We consider a piece of land of A ha on which n different crops may be grown, where crop nis interpreted as fallow. We consider a planning period of T years, which are numbered t =1, 2, . . . , T , and assume that there is one growing season per year. It is assumed that crop successionrequirements depend only on the types of crops that are grown and not on the methods of cultivationthat are used. Moreover, we assume that the crop succession requirements are given in the formof crop sequences that are not allowed to be cultivated on the same piece of land. We introducethe following definitions. Consider a year t and a series of consequent years t + 1, t + 2, . . . , t + h,with h ≥ 2. Let (j1, j2, . . . , jh), be a sequence of (not necessarily different) crops grown on the samepiece of land, where js, s = 1, 2, . . . , h, refer to the crops grown in year t + s. It is said to be aninadmissible sequence if crop js is not allowed to succeed (j1, j2, . . . , js−1), for some s = 1, 2, . . . , h,on the same piece of land. A sequence of crops that is not forbidden is called admissible. Weassume that the (in)admissibility of a crop sequence does not depend on the year t the sequencewas started. Notice that it may happen that (j1, j2, . . . , jh) is admissible, while (j0, j1, j2, . . . , jh)or (j1, j2, . . . , jh, jh+1) are inadmissible, for certain crops j0 and jh+1. Hence, the admissibility ofa certain sequence may depend on the crops preceding or succeeding the sequence. The sequence(j1, j2, . . . , jh) is regarded as admissible if the succession of the crops j1, j2, . . . , jh on the same pieceof land does not violate the crop succession requirements.

We say that the sequence (i1, i2, . . . , ik) is a subsequence of (j1, j2, . . . , jh) if, for some l ∈[1, h− k + 1], k ≤ h, there holds

(j1, j2, . . . , jh) = (j1, . . . , jl−1, i1, . . . , ik, jl+k, . . . , jh) .

If a sequence is admissible, then all its subsequences are admissible too. Analogously, if a sequence(j1, j2, . . . , jh) is inadmissible, then all crop sequences containing (j1, j2, . . . , jh) are inadmissibletoo. Therefore, the crop succession information may be expressed by all inadmissible sequenceswith the shortest length h. We call such sequences minimal inadmissible sequences. Formally, theyare defined as follows.

Definition 2.1 An inadmissible sequence (j1, j2, . . . , jh) is called minimal if h = 2, or if h ≥ 3 andboth the sequences (j2, j3, . . . , jh) and (j1, j2, . . . , jh−1) are admissible.

As can be seen, a minimal inadmissible sequence does not contain any inadmissible subsequences.Notice that we assume that there are no minimal inadmissible sequences with length 1, i.e. con-sisting of a single crop. Such inadmissible crops are excluded from our analysis. The set of allsequences (j1, j2, . . . , jh) is denoted by {1, 2, . . . , n}h. We denote the length of the longest minimalinadmissible sequence by m + 1, i.e.

m + 1 = max{h : there exists a minimal inadmissible sequence (j) ∈ {1, 2, . . . , n}h } . (2.1)

From Definition 2.1 it follows that m ≥ 1. We assume that all minimal inadmissible sequences areknown. In Section 5 we will discuss how to determine minimal inadmissible sequences from a givencollection of inadmissible sequences.

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To be able to determine all feasible cropping plans for year t, it is sufficient to know:

(a) which crops have been grown in the years t−m, . . . , t− 1 on each partof the available land, and

(b) all admissible sequences of length m + 1.

Notice that by (2.1), (b) is equivalent to knowing all minimal inadmissible sequences.In the sequel we will sometimes use the notation (i) for an arbitrary sequence (i1, i2, . . . , ih) and

(j) for an arbitrary sequence (j1, j2, . . . , jh). Next, we define the set of all admissible sequences oflength m.

S = { (j) ∈ {1, 2, . . . , n}m : (j) is admissible } .

For each year t, we define the following decision variables which indicate the area where a certainsequence (j) ∈ S is applied.

Xt(j) = Xt(j1, j2, . . . , jm) : size of the area (in ha) where in year t− s crop jm−s (2.2)is grown, s = 0, 1, . . . ,m− 1, t = 1, 2, . . . , T ,(j) = (j1, j2, . . . , jm) ∈ S .

For (j) /∈ S, we set Xt(j) = 0. Notice that Xt(j) only indicates the size of the area where cropsequence (j) is grown and not the location of this area on the piece of land under consideration.We will assume that this location is immaterial (see also Remark 3.4). Since we consider n crops,the number of decision variables may be as large as nm for each t. In Section 4 we will reduce thisnumber by combining several sequences in S.

It is required that Xt(j) ≥ 0, (j) ∈ S, and∑(j)∈S

Xt(j) = A , t = 1, 2, . . . , T . (2.3)

This implies that in each year t all land is either cultivated or lying fallow. Notice that the areawhere in year t crop i is grown, is given by∑

(j)∈{1,2,...,n}m−1

Xt((j), i) .

The information under (a) above is now expressed by the values of Xt−1(i), (i) ∈ S, and can beused to determine feasible values for Xt(j), (j) ∈ S. Next, we express the information under (b)by specifying, for each sequence (i) ∈ S, the crops which are allowed to be cultivated after (i) onthe same piece of land. In fact, we consider sequences which are allowed to succeed one another.This notion is defined below. First, we need the following definition.

Definition 2.2 We say that the sequence (j1, j2, . . . , jm) ∈ S is a compatible successor of thesequence (i1, i2, . . . , im) ∈ S if j1 = i2, j2 = i3, . . . , jm−1 = im.

Hence, (j) is a compatible successor of (i) if it is logically possible to cultivate the crops j1, . . . , jm

in the years t − m + 1, . . . , t on the same piece of land where in the years t − m, . . . , t − 1 thecrops i1, . . . , im are cultivated. It can be seen that this is equivalent to the requirement thatj1 = i2, j2 = i3, . . . , jm−1 = im. Notice that for any sequence there are n compatible successors,

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since there are n possible values for jm. For each sequence (i) ∈ S we define a set of all (j) ∈ Swhich are compatible successors of (i).

COMP(i) = { (j) ∈ S : (j) is a compatible successor of (i) } .

If (j) is a compatible successor of (i) this does not imply that (j) is also allowed to succeed (i). Wedefine the last notion as follows.

Definition 2.3 We say that the sequence (j) = (j1, j2, . . . , jm) ∈ S is allowed to succeed thesequence (i) = (i1, i2, . . . , im) ∈ S if (j) is a compatible successor of (i) and the sequence(i1, i2, . . . , im, jm) is admissible.

Notice that Definition 2.3 does not depend on the year t in which the last crop jm is cultivated.For each sequence (i) ∈ S we define a set of all (j) ∈ S that are allowed to succeed (i).

SUC(i) = { (j) ∈ S : (j) is allowed to succeed (i) } . (2.4)

Hence, SUC(i) ⊆ COMP(i) for (i) ∈ S. Since (i) ∈ S is admissible, it follows that SUC(i) 6= ∅ andthat there is a (j) ∈ S such that (i) ∈ SUC(j).

The information under (b) above is now contained in the sets SUC(i), (i) ∈ S. This impliesthat the knowledge of Xt−1(i), (i) ∈ S, and the sets SUC(i), (i) ∈ S, is sufficient to determine thefeasible region of Xt(j), (j) ∈ S.

2.2 Interpretation as a max-flow problem

Here, we focus on determining all feasible cropping plans Xt(j), when the cropping plan Xt−1(i)and the sets SUC(i) are known and (2.3) is satisfied. By interpreting this problem as a max-flowproblem, a system of linear equations in both Xt(j) and Xt−1(i) is obtained. These equations arenecessary and sufficient conditions for the compatibility of Xt−1(i) and Xt(j). We call this systemof equations the crop succession constraints. For the planning period of T years, the crop successionrequirements are satisfied if and only if the crop succession constraints hold for t = 2, 3, . . . , T .

Finding a feasible cropping plan Xt(j) given the cropping plan Xt−1(i) is equivalent to findinga feasible allocation of the crops in year t, given the allocations in years t − m, . . . , t − 1. Thisallocation problem may be formulated as follows. Let Xt−1(i) and Xt(j), (i), (j) ∈ S, be given. Wedefine the transition variables

Zt((i), (j)) : size of the area (in ha) where in years t−m, . . . , t− 1 the cropsequence (i) has been grown and in years t−m + 1, . . . , t the cropsequence (j) is grown, (j) ∈ COMP(i).

For (j) /∈ COMP(i) we set Zt((i), (j)) = 0. The variables Zt((i), (j)) should satisfy:∑(j)∈COMP(i)

Zt((i), (j)) = Xt−1(i) , ∀ (i) ∈ S , (2.5)

∑(i)∈S: (j)∈COMP(i)

Zt((i), (j)) = Xt(j) , ∀ (j) ∈ S , (2.6)

Zt((i), (j)) ≥ 0 , ∀ (i), (j) ∈ S, (j) ∈ COMP(i) , (2.7)

Zt((i), (j)) = 0 if (j) /∈ SUC(i) . (2.8)

We may now define the compatibility of the cropping plans Xt−1(i), (i) ∈ S, and Xt(j), (j) ∈ S,as follows.

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Definition 2.4 We say that the cropping plans Xt−1(i), (i) ∈ S, and Xt(j), (j) ∈ S, are compatibleif and only if (2.3) holds and there exists a feasible solution Zt((i), (j)) for the system (2.5)-(2.8).

By (2.3) the sum of the right-hand side members of (2.5) equals A and the same is true of (2.6). Theright-hand side members of (2.5) and (2.6) being given we recognize in (2.5)-(2.8) the constraintsof a classical transportation problem in which total supply equals total demand and some of thevariables Zt((i), (j)) are restricted to zero values.

For our purposes, however, it is more convenient to interpret (2.5)-(2.8) as a max-flow problem.This is done as follows. Consider a network with a source node, supply nodes (i) ∈ S, demandnodes (j) ∈ S and a sink node. There is a directed arc from the source to every supply node (i)with capacity Xt−1(i). Analogously, there is a directed arc from every demand node (j) to the sinkwith capacity Xt(j). Finally, there is a directed arc from a supply node (i) to a demand node (j)if and only if (j) ∈ SUC(i). Arcs from (i) to (j) have infinite capacity. In this way, Zt((i), (j))represents the flow from (i) to (j) and there exists a feasible solution to (2.5)-(2.8) if and only ifthere exists a flow of A units from the source to the sink.

By the max-flow min-cut theorem (see Theorem 1.1 in Ford and Fulkerson [1]) a flow of A unitsis also the maximum flow through the network, since, by (2.3), the cuts {source}–{supply nodes,demand nodes, sink} and {source, supply nodes, demand nodes}–{sink} both have capacity A.Hence, by the max-flow min-cut theorem, there exists a flow of A units if and only if there are nocuts (separating the source and sink nodes) with capacity less than A. This observation is used toprove the following theorem.

We consider sets I, J ⊂ S with the property that no (j) ∈ J is allowed to succeed any (i) ∈ I,i.e.

(j) /∈ SUC(i) for all (i) ∈ I and all (j) ∈ J . (2.9)

Theorem 2.5 Assume that (2.3) holds. Then the cropping plans Xt−1(i), (i) ∈ S, and Xt(j),(j) ∈ S, are compatible if and only if∑

(i)∈I

Xt−1(i) +∑

(j)∈J

Xt(j) ≤ A , ∀ I, J ⊂ S satisfying (2.9). (2.10)

Proof. It suffices to show that (2.10) is equivalent to the requirement that there are no cutsseparating the source and sink nodes with capacity less than A. Denote the set of all supply nodesby U and the set of all demand nodes by V . Let {U1, U2} and {V1, V2} be disjoint partitions of Uand V respectively. An arbitrary cut separating the source and sink nodes is of the form {source,U1, V1}–{U2, V2, sink}. The capacity of this cut is infinite if there is an arc from U1 to V2. If thisis not the case, then the capacity of the cut equals∑

(i)∈U2

Xt−1(i) +∑

(j)∈V1

Xt(j) . (2.11)

Hence, there are no cuts with capacity less than A if and only if (2.11) is greater than or equal toA for all cuts with no arcs from U1 to V2. By (2.3) this is equivalent to

A−∑

(i)∈U1

Xt−1(i) + A−∑

(j)∈V2

Xt(j) ≥ A . (2.12)

The proof is completed by writing (2.12) in the form∑(i)∈U1

Xt−1(i) +∑

(j)∈V2

Xt(j) ≤ A ,

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and setting I = U1 and J = V2. 2

We call the system of equations (2.10) the crop succession constraints. As can be seen, the cropsuccession constraints are linear in Xt−1(i) and Xt(j). For the whole planning period of T yearsthe crop succession requirements are satisfied if and only if∑

(i)∈I

Xt−1(i) +∑

(j)∈J

Xt(j) ≤ A , t = 2, 3, . . . , T, ∀ I, J ⊂ S satisfying (2.9). (2.13)

However, we may arrive at a situation where production in year T + 1 is impossible because of thecrop succession requirements. Therefore, in (2.13) also t = T + 1 may be included. Additionally,we may want to ensure that the production plan XT+1(j) is a good starting point for the future,and require for example that X1(j) = XT+1(j). This issue will be discussed further in Section 3.1.

2.3 Identifying redundant constraints

The number of crop succession constraints in (2.10) grows exponentially with n (although thenumber of admissible sequences in the set S may be substantially less than nm). However, a lot ofconstraints in (2.10) may be redundant. Here, we discuss a way to identify combinations of I andJ for which the constraint (2.10) is redundant. For I ⊂ S, let J(I) be defined by

J(I) = {(j) ∈ S : (j) /∈ SUC(i) ∀ (i) ∈ I} .

A first observation is that, for a fixed set I ⊂ S, we only need to consider the set J(I) in (2.10),and not all sets J ⊂ S which satisfy (2.9). This is due to the fact that, for a fixed set I ⊂ S, theterm ∑

(j)∈J

Xt(j)

in (2.10), is maximal for J = J(I). Hence, we replace (2.10) by∑(i)∈I

Xt−1(i) +∑

(j)∈J(I)

Xt(j) ≤ A , ∀ I ⊂ S . (2.14)

However, if J(I) = ∅, then (2.14) is implied by (2.3). Therefore, we only need to consider sets Iwith J(I) 6= ∅. We define

S1 = {I ⊂ S : J(I) 6= ∅} .

Suppose that I, I ∈ S1 and I ⊂ I. Then J(I) ⊆ J(I) and it may happen that J(I) = J(I). In thelatter case, the constraint (2.14) for I is implied by the one for I. Hence, we may only considersets I in the following collection S2:

S2 = {I ∈ S1 : there is no I ∈ S1 such that I ⊂ I and J(I) = J(I)} .

When we consider only sets I ∈ S2 there may, however, still be redundant constraints in the system(2.14). The constraint for I1 ∈ S2 is redundant if there exists an I2 ∈ S2, such that

(I1 ∪ J(I1)) ⊂ (I2 ∪ J(I2)) .

It may for example happen that I1 ⊂ I2 and J(I1) ⊂ I2. Or that I1 ⊂ J(I2) and J(I1) ⊂ J(I2).Since excluding all possibilities of redundancy is a rather tedious task, we will end our discussionat this point.

Notice that the roles of I and J are interchangeable, i.e. for J ⊂ S we also could have defineda set I(J) = {(i) ∈ S : (j) /∈ SUC(i) ∀ (j) ∈ J} and similar collections S1 and S2 as above.

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3 Stationary cropping plans in an LP-context

It is well-known that linear programming (LP) is a valuable tool for analysing production plan-ning problems. Also production planning of a farm may be analysed with the use of LP, see e.g.Schweigman [3]. In this section, we consider a generic multi-year LP-model for farm productionplanning containing crop succession constraints. In Section 3.1 it is shown that, under some regu-larity conditions, the multi-year LP-model may be replaced by a one-year LP-model. In this casea stationary cropping plan, i.e. Xt(j) = X(j) for all t, is an optimal solution of the multi-yearLP-model. In Section 3.2 we show that a stationary cropping plan is composed of several croprotation cycles. A method to determine these cycles is presented.

3.1 A stationary linear programming model

We formulate the following generic LP-model for agricultural production planning for a planningperiod of T years. The decision variables are Xt(j), as defined in (2.2). The model is given by:

MaximizeT∑

t=1

∑(j)∈S

ct(j) Xt(j) , (3.1)

subject to∑(i)∈I

Xt−1(i) +∑

(j)∈J

Xt(j) ≤ A , t = 2, 3, . . . , T + 1 , (3.2)

∀ I, J ⊂ S satisfying (2.9),∑(j)∈S

Xt(j) = A , t = 1, 2, . . . , T , (3.3)

∑(j)∈S

at(k, (j)) Xt(j) ≤ bt(k) , k = 1, 2, . . . ,K , t = 1, 2, . . . , T . (3.4)

As mentioned in Section 2.2 we also include the variables XT+1(j) in the crop succession constraints(3.2) to ensure that production in year T + 1 is possible.

The limited availability of land is described by (3.3). For other inputs, the constraints (3.4)are included. The parameters at(k, (j)) indicate, for the crop sequence (j) and year t, how muchof input k is needed to cultivate one ha of crop jm. The inputs may be labour, manure, money,machines, seeds, etc. The parameters bt(k) represent, for year t, the available amount of input k.Also food requirements may be written as (3.4). The parameters ct(j) in the objective functionrepresent, for example, the revenues per ha in year t when crop sequence (j) has been grown. Formore details on the interpretation of the constraints (3.4) and the objective function (3.1) we referto Schweigman [3].

We consider a version of (3.1)-(3.4) in which the parameters do not depend on t, i.e.

at(k, (j)) = a(k, (j)) , bt(k) = b(k) , ct(j) = c(j) , t = 1, 2, . . . , T . (3.5)

We add the following equation to guarantee that the production plan XT+1(j) is a good startingpoint for the future.

X1(j) = XT+1(j) , ∀ (j) ∈ S . (3.6)

The model (3.1)-(3.6) may be called a stationary LP-model. We need the following definition.

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Definition 3.1 We call a cropping plan Xt(j), t = 1, 2, . . . , T , (j) ∈ S, stationary if Xt(j) = X(j),t = 1, 2, . . . , T , (j) ∈ S.

It follows that if an optimal solution of (3.1)-(3.6) is stationary, it is an optimal solution of:

Maximize∑

(j)∈S

c(j) X(j) , (3.7)

subject to ∑(i)∈I

X(i) +∑

(j)∈J

X(j) ≤ A , ∀ I, J ⊂ S satisfying (2.9), (3.8)

∑(j)∈S

X(j) = A , (3.9)

∑(j)∈S

a(k, (j)) X(j) ≤ b(k) , k = 1, 2, . . . ,K . (3.10)

The following proposition shows that if (3.1)-(3.6) is feasible, then it has an optimal solutionwhich is stationary. This result has an important practical implication: in an optimal strategy theproduction plan may be chosen the same in all years of the planning period.

Proposition 3.2

(P1) Problem (3.1)-(3.6) is feasible if and only if problem (3.7)-(3.10) is feasible.

(P2) Suppose (3.1)-(3.6) is feasible. Let X∗(j), (j) ∈ S, be an optimal solution of (3.7)-(3.10).Then Xt(j) = X∗(j), t = 1, 2, . . . , T , (j) ∈ S, is an optimal solution of (3.1)-(3.6).

Proof. Let X be a feasible solution of (3.7)-(3.10). Then Xt = X, t = 1, . . . , T , is a feasiblesolution of (3.1)-(3.6). The proof of (P1) is complete if we show that if (3.1)-(3.6) is feasible, thenthere exists a feasible solution to (3.1)-(3.6), which is stationary.

Suppose (3.1)-(3.6) is feasible. Since the feasible region is compact, it also has an optimalsolution, say X∗

t , t = 1, . . . , T . Define, for h = 1, . . . , T ,

X(h) = (X∗h, X∗

h+1, . . . , X∗T , X∗

1 , . . . , X∗h−1) .

Then each X(h) is a feasible solution to (3.1)-(3.6). Moreover, since X(1) = X∗ and the objectivefunction (3.1) has the same value for all X(h), h = 1, . . . , T , the feasible solutions X(h) are alsooptimal. Define

X =1T

T∑h=1

X(h) =

(1T

T∑t=1

X∗t , . . . ,

1T

T∑t=1

X∗t

).

Since the feasible region of (3.1)-(3.6) is convex X is a feasible solution. This completes the proof of(P1), since X is a stationary solution. Notice that the objective function (3.1) has the same valuefor X as for X∗. Therefore, X is an optimal solution of (3.1)-(3.6) and X1 is an optimal solutionof (3.7)-(3.10).

Next we prove (P2). Suppose (3.1)-(3.6) is feasible. By (P1) also (3.7)-(3.10) is feasible. Sincethe feasible region of (3.7)-(3.10) is compact it has an optimal solution, say X0. The stationarysolution Xt = X0, t = 1, . . . , T , is feasible for (3.1)-(3.6). Moreover, since X1 above has the sameobjective value (3.7) as X0, the stationary solution Xt = X0 has the same objective value (3.1) asX. Hence, Xt = X0 is an optimal solution of (3.1)-(3.6). This completes the proof of (P2). 2

8

Remark 3.3 It is important to note that the requirement (3.5) is of crucial importance in thediscussion above. This implies that we assume that, for example, crop prices and yields per hado not change during the T years of the planning period, which is rather unrealistic. If prices oryields change and the current stationary cropping plan X(j) is not optimal anymore, we still needa multi-year LP-model to transform X(j) into the optimal stationary solution.

It should also be noted that the model (3.1)-(3.6) does not contain crop succession requirementsdue to the last year before the planning period. These can be added by including t = 1 in equation(3.2). However, in this situation Proposition 3.2 is not valid anymore.

Remark 3.4 The cropping plans Xt(j), t = 1, 2, . . . , T , (j) ∈ S, only indicate the sizes of the areaswhere in year t the crop sequence (j) is grown and not the locations of these areas on the A ha ofland that is available. For t = 2, 3, . . . , T , any feasible solution Zt((i), (j)) of (2.5)-(2.8) describes afeasible allocation of the crops in year t. Moreover, for year t = 1 any ordering of X1(j), (j) ∈ S,is a feasible allocation.

We assume that the location of the areas Xt(j) is immaterial in the sense that it has no influenceon the required amount per ha of each input and on the yield per ha. However, the freedom ofchoosing a feasible Zt((i), (j)) and the allocation in year t = 1 may still be used to obtain ‘good’locations for the crops in each year.

Remark 3.5 The amounts of inputs used per ha and the yield per ha usually depends not onlyon the type of crop, but also on the method of cultivation. For example, the yield per ha of maizemay be influenced by the amount of manure that is applied per ha. For this reason, it may beappropriate to define decision variables Xt(f, (j)) indicating the area where crop sequence (j) isgrown and method of cultivation f is used. The model (3.1)-(3.6) may then by changed as follows.Since we assume that the crop succession requirements do not depend on the methods of cultivationthat are used, we may set

Xt(j) =∑

f

Xt(f, (j)) , (j) ∈ S , t = 1, 2, . . . , T ,

and leave the crop succession constraints (3.2) unchanged. Also (3.3) remains the same. Theconstraints (3.4) may be changed into∑

(j)∈S

∑f

at(k, f, (j)) Xt(j) ≤ bt(k) , k = 1, 2, . . . ,K , t = 1, 2, . . . , T .

In this way, it is expressed that the required amounts of inputs and the yield may also depend onthe method of cultivation f . Analogously, the objective function (3.1) may be changed into

MaximizeT∑

t=1

∑(j)∈S

∑f

ct(f, (j)) Xt(j) .

Also, we may replace (3.6) by

X1(f, (j))−XT+1(f, (j)) = 0 , ∀f , ∀ (j) ∈ S .

For this new model we then have an analogous result as Proposition 3.2 in terms of the decisionvariables Xt(f, (j)).

Notice that when m = 1, the decision variables indicate the size of the area where a single cropis grown. Hence, it is not possible to express a dependence of the yield of the current crop on thecrop grown last year on the same piece of land. Therefore, in this case one may prefer to use thedecision variables for the situation m = 2.

9

3.2 Crop rotation cycles

Here we show that a stationary cropping plan X(j) is composed of several crop rotation cycles. Wedefine a crop rotation cycle in terms of sequences in S as follows.

Definition 3.6 A collection of sequences C = {(j)1, (j)2, . . . , (j)h} ⊆ S, h ≥ 1, is called a croprotation cycle if (j)k is allowed to succeed (j)k−1, k = 2, 3, . . . , h, and (j)1 is allowed to succeed(j)h.

Hence, a crop rotation cycle is defined as a collection of sequences in S, that is allowed to beapplied cyclically on the same piece of land. If h = 1 in Definition 3.6, then the single sequence inC consists of m times the same crop. Hence, the same crop is grown permanently. If h = 2, thencycle C consists of two different crops which are alternated. For h = 3 there are two possibilities:either a sequence of three different crops is repeated, or periods of two years the same crop arealternated with one-year periods in which a different crop is grown. In general, the cycle C consistsof at most h different crops.

In Definition 3.6, the number of sequences in C and the length in years of the crop rotationcycle defined by C, are both equal to h. For h ≤ m, this may not seem obvious. However, in thiscase there holds for any sequence (j) in C,

jk = jk+h , k = 1, . . . ,m− h .

The main result of this section is formulated in the following proposition.

Proposition 3.7 Let X(j), (j) ∈ S, satisfy (3.8) and (3.9). Define

SX = {(j) ∈ S : X(j) > 0 } . (3.11)

Then there exist crop rotation cycles Cs ⊆ SX , s = 1, 2, . . . , v, which can be applied such that inevery year the area where sequence (j) is grown, is equal to X(j), (j) ∈ SX .

Proof. Let X(j), (j) ∈ S, satisfy (3.8) and (3.9). Then, by Theorem 2.5, there exists a feasiblesolution Z((i), (j)), (i), (j) ∈ S, (j) ∈ COMP(i), to the following system:∑

(j)∈COMP(i)

Z((i), (j)) = X(i) , ∀ (i) ∈ S , (3.12)

∑(i)∈S: (j)∈COMP(i)

Z((i), (j)) = X(j) , ∀ (j) ∈ S , (3.13)

Z((i), (j)) ≥ 0 , ∀ (i), (j) ∈ S, (j) ∈ COMP(i) , (3.14)

Z((i), (j)) = 0 , if (j) /∈ SUC(i) . (3.15)

We choose a feasible solution Z of (3.12)-(3.15). Next, we construct a network of nodes (j) ∈ SX .There is a directed arc from node (i) to node (j) if and only if Z((i), (j)) > 0. The value ofZ((i), (j)) is interpreted as the flow from node (i) to node (j). The total flow in the network equalsA, since by (3.9), (3.11), (3.12) and (3.13) there holds∑

(i)∈SX

∑(j)∈SX

Z((i), (j)) = A .

10

From (3.12) and (3.13) it can be concluded that for each node (j) in the network, the total flowinto (j) is equal to the total flow out of (j). Because of this flow conservation property, each node(j) in the network is contained in a flow cycle. Suppose node (j) is contained in some flow cycleCs. Let ws be the minimum flow between two consecutive nodes in Cs, i.e.

ws = min{Z((i), (k)) : cycle Cs contains the arc (i) → (k) } .

Because of the flow conservation property we know that ws > 0. The flow in the whole networkcan be described by a finite number of flow cycles Cs, s = 1, 2, . . . , v. Indeed, if a cycle C1 isdeleted from the network, i.e. the flow through the nodes in C1 is decreased by w1, then the flowconservation property still holds for all nodes in the network. Hence, we may delete another cycleC2. In this way, the flow in the network becomes zero after deleting a finite number of cycles.

The flow cycles C1, . . . , Cv determine the values of Z((i), (j)) completely, since

Z((i), (j)) =∑

s: arc (i)→ (j) in Cs

ws . (3.16)

Let hs be given byhs = card(Cs) , s = 1, 2, . . . , v .

The procedure above splits up the transitions Z((i), (j)) into flows ws in cycles Cs of lengths hs.Hence, there holds

v∑s=1

hsws = A . (3.17)

From Definition 3.6 it is clear that a flow cycle Cs may be interpreted as a crop rotation cycle.Moreover, when each rotation cycle Cs is applied on an area of hsws ha, then a rotation schemecan be constructed where each year the area of crop sequence (j) is X(j). This is done as follows.

In the first year, the total area of A ha is divided into v plots with sizes hsws, s = 1, 2, . . . , v.The plot for cycle Cs is divided into hs parts of size ws. On each part a different (j) in Cs is grown.In the following years, each part of the plot for Cs follows the schedule dictated by the rotationcycle Cs. By (3.17), it follows that all land is used. Moreover, by (3.16) and (3.13), the area for(j) ∈ SX is equal to ∑

(i)∈SX

∑s: arc (i)→ (j) in Cs

ws =∑

(i)∈SX

Z((i), (j)) = X(j) .

This completes the proof of Proposition 3.7. 2

Remark 3.8 In the proof of Proposition 3.7 there may be many possible ways of choosing thecycles C1, . . . , Cv. For example, a different feasible solution Z of (3.12)-(3.15) may be taken. Also,the order of picking the cycles may be varied. Or we may choose to consider only cycles Cs whichdo not contain the same sequence (j) more than once. Another possibility is to decrease the flowin cycle Cs by a quantity less than ws.

The cropping plan X is required to satisfy only (3.8) and (3.9). We may for example take Xto be an optimal solution of (3.7)-(3.10). The transitions Z satisfying (3.12)-(3.15) may also bechosen by using preference criteria, e.g. some objective function. If X and Z are chosen optimal(in some sense), then C1, . . . , Cv may be called optimal crop rotation cycles.

11

4 Combining crop sequences

In Section 2.1 it is stated that to determine the feasible cropping plans for year t, it is sufficient toknow (a) which crops have been grown in the years t −m, . . . , t − 1 on each part of the availableland, and (b) all admissble sequences of length m + 1. Here we argue that, given the crop successioninformation (b), in order to satisfy the crop succession requirements it may not be necessary toknow exactly which crops have been grown in the years t−m, . . . , t−1 on each part of the availableland. In some cases, it may be sufficient to know that in a certain year one crop from a certain setof crops has been grown on a certain part of the available land. Hence, given the crop successioninformation (b), we do not need all information under (a) to determine the feasible cropping plansfor year t.

Recall that we expressed (a) in terms of Xt−1(i), (i) ∈ S, and (b) in terms of the sets SUC(i),(i) ∈ S. In this section, we combine certain groups of sequences in S into new sequences and thusobtain a new set of sequences S∗. A sequence (j) ∈ S∗ still has length m. The difference withsequences in S is that now the element jk may represent a set of crops instead of only one crop. Foreach (i) ∈ S∗, the set of successors (j) ∈ S∗ is denoted by SUC∗(i). We define decision variablesXt(j), (j) ∈ S∗, analogous to (2.2). Now Xt−1(i), (i) ∈ S∗, do not contain all information under(a). However, the combinations may be chosen in such a way that the sets SUC∗(i), (i) ∈ S∗, stillcontain all information under (b). Moreover, given Xt−1(i), (i) ∈ S∗, and SUC∗(i), (i) ∈ S∗, it isthen still possible to determine all feasible cropping plans for year t. The combinations of sequencesin S for which this holds are determined by the sets SUC(i), (i) ∈ S.

Notice that forming combinations of groups of sequences in S as above, implies a reductionof the number of decision variables. For each t, the number of decision variables Xt(j) in (2.2)equals card(S) and may be of the order nm. In Section 4.1 we illustrate the process of combiningsequences with a small-scale example. In Section 4.2 a larger example is considered and the needfor a sequence-combining algorithm is expressed. Such an algorithm is presented in Section 4.3. InSection 4.4 we show that when this algorithm is used to combine sequences, the information under(b) above is preserved. Moreover, the cropping plans Xt−1(i), (i) ∈ S∗, and Xt(j), (j) ∈ S∗, arecompatible if and only if crop succession constraints analogous to (2.10) are satisfied.

It should be noted that our aim is to find groups of sequences that can be combined withoutlosing crop succession information. This does not imply that they also have to be combined. Forexample, one may choose not to combine certain sequences (i) and (j) when the yield per ha of thecrop im = jm is not the same for (i) and (j).

4.1 A small-scale example

Consider n = 3 crops and suppose that crop 1 may succeed crop 1 provided that crop 2 is grown onthe land two years before. In this case, (1, 1, 1) and (3, 1, 1) are the minimal inadmissible sequences.Hence, m = 2 and

S = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} .

The sets SUC(i), (i) ∈ S, are given by

SUC(2, 1) = {(1, 1), (1, 2), (1, 3)} ,

SUC(1, 1) = SUC(3, 1) = {(1, 2), (1, 3)} ,

SUC(1, 2) = SUC(2, 2) = SUC(3, 2) = {(2, 1), (2, 2), (2, 3)} ,

SUC(1, 3) = SUC(2, 3) = SUC(3, 3) = {(3, 1), (3, 2), (3, 3)} .

12

As we see, SUC(1, 1) = SUC(3, 1). This implies that if in year t crop 1 is grown on a certain pieceof land, the possibilities for year t + 1 do not depend on whether the crop grown in year t− 1 iscrop 1 or crop 3. Hence, intuitively, we may combine the sequences (1, 1) and (3, 1) into the newsequence (1 or 3, 1). The set SUC(1 or 3, 1) is then equal to SUC(1, 1). The set SUC(2, 1) contains(1, 1) but not (3, 1). In this case, we just replace (1, 1) by the new sequence (1 or 3, 1). Logically,this is not a problem, since a succession of (2, 1) by (1 or 3, 1) implies that (2, 1, 1) is grown. Thesame is true for SUC(1, 3), which contains (3, 1) but not (1, 1). The sets SUC(i) change into

SUC(2, 1) = {(1 or 3, 1), (1, 2), (1, 3)} ,

SUC(1 or 3, 1) = {(1, 2), (1, 3)} ,

SUC(1, 2) = SUC(2, 2) = SUC(3, 2) = {(2, 1), (2, 2), (2, 3)} ,

SUC(1, 3) = SUC(2, 3) = SUC(3, 3) = {(1 or 3, 1), (3, 2), (3, 3)} .

Since SUC(1, 2) = SUC(2, 2) = SUC(3, 2), we may also combine (1, 2), (2, 2) and (3, 2) into thesequence (x, 2), where x denotes all crops. The same holds for (1, 3), (2, 3) and (3, 3). We combinethem into (x, 3). Using the same arguments as above, there holds

SUC∗(2, 1) = {(1 or 3, 1), (x, 2), (x, 3)} ,

SUC∗(1 or 3, 1) = {(x, 2), (x, 3)} ,

SUC∗(x, 2) = {(2, 1), (x, 2), (x, 3)} ,

SUC∗(x, 3) = {(1 or 3, 1), (x, 2), (x, 3)} .

We changed the notation from SUC(i) to SUC∗(i), since all SUC∗(i) are different and no morecombinations can be formed in the same way as above. Hence, the final set of sequences is

S∗ = {(2, 1), (1 or 3), (x, 2), (x, 3)} .

Notice that if (x, 3) is succeeded by (x, 2), then either (1, 3, 2), (2, 3, 2) or (3, 3, 2) is grown. Fromthe sets SUC∗(i) it can be concluded that (1, 1, 1) and (3, 1, 1) are the only inadmissible sequences.Hence, after combining the sequences as we did, we still know all admissible sequences of lengthm + 1 = 3.For (i) ∈ S∗ and t = 1, 2, . . . , T , we define a corresponding variable Xt(i) as follows.

Xt(2, 1) : area where in year t crop 1 is grown and in year t− 1 crop 1 was grown,Xt(1 or 3, 1) : area where in year t crop 1 is grown and in year t− 1 crop 1 or crop 3 was grown,Xt(x, 2) : area where in year t crop 2 is grown,Xt(x, 3) : area where in year t crop 3 is grown.

By forming the combinations as above we have reduced the number of decision variables from 9to 4. With respect to the compatibility of Xt−1(i), (i) ∈ S∗, and Xt(j), (j) ∈ S∗, the analogue ofTheorem 2.5 holds. This result is formulated in Theorem 4.6 below.

4.2 A larger example

In the example of Section 4.1 the combinations that may be formed can be determined by hand.However, for an example with a larger number of crops n or a larger value of m it may be moreconvenient to have a computer program which determines the sequences that may be combined.

13

Here, we consider such a larger example. A sequence-combining algorithm is presented in Section4.3.

Suppose we have the crops cotton (crop 1), sorghum (crop 2), soya beans (crop 3) and fallowland (crop 4). We assume that the following crop succession requirements have to be fulfilled.

(R1) After cotton and sorghum, there should be at least a one-year fallow.

(R2) Cotton may succeed cotton only after a two-year fallow in between.

(R3) Sorghum may succeed sorghum only after a two-year fallow in between.

(R4) Cotton may succeed sorghum only after a two-year fallow in between.

(R5) Soya beans may be succeeded only by sorghum and fallow.

From (R1) it follows that the sequences (1, 1), (1, 2), (1, 3), (2, 1), (2, 2) and (2, 3) are inadmissible.From (R2) it can be concluded that (1, 4, 1) is inadmissible. Requirements (R3) and (R4) indicatethat (2, 4, 2) and (2, 4, 1) are inadmissible. Finally, from (R5) it follows that (3, 1) and (3, 3)are inadmissible. If (R1)-(R5) are the only crop succession requirements, then these inadmissiblesequences are also minimal inadmissible. However, we also include the following requirement.

(R6) At least three out of every five years the land should lie fallow.

This implies that all sequences of length 5 without crop 4, with once crop 4, or with twice crop 4are inadmissible. But not all of these sequences are minimal inadmissible. Determining all minimalinadmissible sequences for (R1)-(R6) is a rather complex task. However, in Section 5 an algorithmwill be presented which determines the minimal inadmissible sequences from a given collectionof inadmissible sequences. Using this algorithm, we obtain the following minimal inadmissiblesequences. The sequences of lengths 2 and 3 are the same as above, i.e. (1, 1), (1, 2), (1, 3), (2, 1),(2, 2), (2, 3), (3, 1), (3, 3), (1, 4, 1), (2, 4, 2) and (2, 4, 1). The minimal inadmissible sequences oflength 4 are (1, 4, 3, 2), (2, 4, 3, 2), (3, 2, 4, 3) and (3, 4, 3, 2). And finally, there are the following 17minimal inadmissible sequences of length 5:

(1, 4, 2, 4, 3) (1, 4, 3, 4, 1) (1, 4, 3, 4, 2) (1, 4, 3, 4, 3) (1, 4, 4, 3, 2) (2, 4, 3, 4, 1) ,

(2, 4, 3, 4, 2) (2, 4, 3, 4, 3) (2, 4, 4, 3, 2) (3, 2, 4, 4, 1) (3, 2, 4, 4, 2) (3, 2, 4, 4, 3) ,

(3, 4, 1, 4, 2) (3, 4, 1, 4, 3) (3, 4, 2, 4, 3) (3, 4, 3, 4, 3) (3, 4, 4, 3, 2) .

Hence, for (R1)-(R6) we have m = 4. Starting with the set S of admissible sequences of length4 and using the same reasoning as in Section 4.1, it can be shown that it suffices to consider theareas where the following sequences are grown, i.e. the sequences in the set S∗ are:

(3, 2, 4, 4) (3, 4, 1, 4) (3, 4, 3, 4) (4, 3, 2, 4) (4, 3, 4, 1) (4, 3, 4, 3) (4, 4, 1, 4) (4, 4, 2, 4) ,

(4, 4, 3, 2) (4, 4, 3, 4) (4, 4, 4, 3) (t, 4, 2, 4) (s, 4, 3, 4) (v, 4, 4, 1) (v, 4, 4, 2) (u, 4, 4, 3) ,

(v, 4, 4, 4) (4, t, 4, 2) (4, s, 4, 3) (4, u, 4, 4) ,

where s denotes crop 1 or crop 2, t denotes crop 1 or crop 3, u denotes crops 1, 2 or 3 and v denotesall four crops. The number of sequences in S is equal to 37, while S∗ contains 20 sequences. Toavoid a lengthy presentation, the sets SUC∗(i), (i) ∈ S∗, are omitted.

14

4.3 An algorithm

Here, we present an algorithm that may be used to combine sequences as in the examples ofSections 4.1 and 4.2. After the combinations are formed, a sequence (i) ∈ S∗ has m elements andeach element is either a single crop or a combined crop. For each crop, the area for the currentyear must be known exactly. Therefore, for (i) ∈ S∗, we want im to be a single crop. The elementsi1, . . . , im−1 may be combined crops. Our algorithm consists of an Initialisation Step followedby Step q, q = 1, 2, . . . ,m − 1. In each Step q, we look for sequences (i) and (j) with identicalsets of successors and which only differ in the q-th element, i.e.

(i1, . . . , iq−1) = (j1, . . . , jq−1) = (r) , iq 6= jq , (iq+1, . . . , im) = (jq+1, . . . , jm) = (k) , (4.1)

for some (r) and (k). For fixed (r) and (k), there are at most n sequences satisfying this prop-erty. They are then combined into the new sequence ((r), y, (k)), where y is the combined cropconsisting of all q-th (single) crops of the sequences that are combined. When no more sequencescan be combined in this way, the new set of admissible sequences S(q) is determined. Step qends by determining the new sets of successors SUC(q)(i), (i) ∈ S(q). A detailed description of theInitialisation Step and a general Step q are given below.

Initialisation Step

Set S(0) = S and SUC(0)(i) = SUC(i), (i) ∈ S(0), and go to Step 1.

Step q (q = 1, . . . ,m− 1)

- For (r) = (r1, . . . , rq−1) and (k) ∈ {1, 2, . . . , n}m−q, define C(q)(r)(k) ⊂ S(q−1) by

C(q)(r)(k) = {(i) ∈ S(q−1) : there is a (j) ∈ S(q−1), such that SUC(q−1)(j) = SUC(q−1)(i)

and (4.1) is satisfied} .

- If C(q)(r)(k) = ∅ for all (r) and (k), then set S∗ = S(q−1) and SUC∗(i) = SUC(q−1)(i), (i) ∈ S∗,

and STOP.

- For any C(q)(r)(k) 6= ∅, combine all sequences in C

(q)(r)(k) into the new sequence ((r), y, (k)), where

y is the combined crop consisting of all q-th (single) crops of the sequences in C(q)(r)(k).

- Define the new set of sequences S(q) by deleting sequences (i) ∈ S(q−1) which were combinedabove and adding the new formed combinations.

- Construct the sets SUC(q)(i), (i) ∈ S(q), as follows. Consider a pair (i), (j) ∈ S(q).

(C1) If (i), (j) ∈ S(q) ∩ S(q−1), then (j) ∈ SUC(q)(i) if and only if (j) ∈ SUC(q−1)(i).

(C2) If (i) ∈ S(q) ∩ S(q−1) and (j) ∈ S(q)\S(q−1), (j) was formed from C(q)(r)(k),

then (j) ∈ SUC(q)(i) if and only if there is a (t) ∈ C(q)(r)(k) with (t) ∈ SUC(q−1)(i).

(C3) If (i) ∈ S(q)\S(q−1), (i) was formed from C(q)(p)(l), and (j) ∈ S(q) ∩ S(q−1),

then (j) ∈ SUC(q)(i) if and only if (j) ∈ SUC(q−1)(s), for any (all) (s) ∈ C(q)(p)(l).

15

(C4) If (i) ∈ S(q)\S(q−1), (i) was formed from C(q)(p)(l), and (j) ∈ S(q)\S(q−1),

(j) was formed from C(q)(r)(k), then (j) ∈ SUC(q)(i) if and only if there is a

(t) ∈ C(q)(r)(k) with (t) ∈ SUC(q−1)(s), for any (all) (s) ∈ C

(q)(p)(l).

- If q ≤ m− 2, then go to Step q + 1. Else, set S∗ = S(q) and SUC∗(i) = SUC(q)(i), (i) ∈ S∗,and STOP.

The construction of the sets SUC(q)(i), (i) ∈ S(q), may need some explanation. In (C1), both (i)and (j) are not combined in Step q and nothing changes. In (C2), sequence (i) is not combined,while (j) is formed in Step q. In this case, (j) is allowed to succeed (i) if and only if some sequence(t), which is combined into (j), is allowed to succeed (i). In Lemma 4.8 below, we will show thatthere is at most one such sequence (t). In (C3), sequence (i) is formed in Step q, while (j) is notcombined. Here, (j) is allowed to succeed (i) if and only if (j) is allowed to succeed some sequence(s), which is combined into (i). Notice that if such a sequence (s) exists, then (j) is allowed tosucceed all (s) which are combined into (i), since, for these sequences (s), the sets SUC(q−1)(s) areidentical. In (C4), both (i) and (j) are formed in Step q. From (C2) and (C3) it follows that (j)is allowed to succeed (i) if and only if some sequence (t), which is combined into (j), is allowed tosucceed some sequence (s), which is combined into (i). As in (C2), for each (s), there is at mostone such sequence (t).

In the algorithm, the order of the steps is Step 1, Step 2, . . ., Step m − 1. In Step q

combinations are formed of sequences in the sets C(q)(r)(k). Since these sequences satisfy (4.1), there

holds that for any (j) ∈ S(q) all elements in (jq+1, . . . , jm) are single crops.It may seem that the order of the steps may be changed. However, this is not true. In Lemma

4.1 below we show that if, after steps 1, . . . , q − 1, we execute Step p with p ∈ {q + 1, . . . ,m− 1},then no sequences will be combined.

The algorithm stops if no sequences are combined in the current step. This is correct, since inLemma 4.2 below it is shown that if in Step q − 1 no sequences are combined, then also in Stepq no sequences will be combined.

Lemma 4.1 Suppose the steps 1, . . . , q − 1 have been executed in this order. If next Step p isexecuted, with p ∈ {q + 1, . . . ,m− 1}, then no sequences will be combined.

Proof. Suppose to the contrary that for some (i), (j) ∈ S(q−1) with

(i1, . . . , ip−1) = (j1, . . . , jp−1) = (r) , ip 6= jp , (ip+1, . . . , im) = (jp+1, . . . , jm) = (k) , (4.2)

for some (r) and (k), there holds SUC(q−1)(i) = SUC(q−1)(j). For any sequence (x) ∈ S(q−1) theelements xq, . . . , xm are single crops. Hence, for any (t) ∈ SUC(q−1)(i), there holds

js = is = ts−1 , s = q + 1, . . . ,m .

But, since p ∈ {q + 1, . . . ,m− 1}, this implies ip = jp which contradicts (4.2). Therefore, in Stepp no sequences will be combined. 2

16

Lemma 4.2 If in Step q − 1 no sequences are combined, then also in Step q no sequences willbe combined.

Proof. The proof is similar to the proof of Lemma 4.1. Suppose to the contrary that for some(i), (j) ∈ S(q−1) satisfying (4.1) for some (r) and (k), there holds SUC(q−1)(i) = SUC(q−1)(j). Sinceno sequences are combined in Step q− 1, for any (x) ∈ S(q−1) the elements xq−1, . . . , xm are singlecrops. Hence, for any (t) ∈ SUC(q−1)(i), there holds

js = is = ts−1 , s = q, . . . , m .

But this implies iq = jq which contradicts (4.1). Therefore, in Step q no sequences will be com-bined. 2

Remark 4.3 An analogous statement as in Lemma 4.2 does not hold for an individual (i) ∈ S(q−1),i.e. when (i) is not combined in Step q − 1, it may still be combined in Step q. An example canbe found in Section 4.2, where the sequences (4, 1, 4, 4), (4, 2, 4, 4) and (4, 3, 4, 4) are not combinedin Step 1, while in Step 2 they are combined into (4, u, 4, 4), where u denotes crops 1, 2 or 3.

Remark 4.4 We have programmed the algorithm using the software package Matlab 5. Thisprogram was also used to obtain the set S∗ in Section 4.2. Our computational experience withthe program learns that on a Pentium 4 PC, the algorithm terminates within approximately 15minutes for values of n and m with nm ≤ 3200 approximately. To our opinion, this covers mostpractical situations.

4.4 No loss of crop succession information

Here, we show that the information under (b), i.e. all admissible sequences of length m + 1, ispreserved when sequences are combined in the algorithm of Section 4.3. First, we introduce somenotation. Let

B(q) = {(i1, . . . , im+1) : (i1, . . . , im) ∈ S(q) , im+1 ∈ {1, 2, . . . , n} , such thatthere is a (j) ∈ SUC(q)(i1, . . . , im) with jm = im+1 } .

Then B(q) contains all admissible sequences of length m + 1 (containing both combined and singlecrops as elements), based on S(q) and the sets SUC(q)(i), (i) ∈ S(q). For any sequence (i) of lengthm, possibly containing combined crops, let

COMB(i) = {(s) ∈ {1, 2, . . . , n}m : (s) is contained in (i)} .

For (i) ∈ S(0) we have COMB(i) = (i). From the definition of the algorithm in Section 4.3it follows that COMB(i) ⊆ S(0), (i) ∈ S(q). Moreover, for any q ∈ {0, 1, . . . ,m − 1}, the setsCOMB(i), (i) ∈ S(q), consitute a disjoint partition of S(0).

To express the information in B(q) in terms of S(0) and SUC(0)(i), (i) ∈ S(0), we define

B(q)0 = {(s1, . . . , sm+1) ∈ {1, 2, . . . , n}m+1 : there is an (i) ∈ B(q) such that

(s1, . . . , sm) ∈ COMB(i1, . . . , im) and im+1 = sm+1} .

Notice that B(0)0 = B(0) contains all admissible sequences of length m + 1 (containing only single

crops as elements), i.e. all information under (b). We have the following result.

17

Proposition 4.5 Suppose that in Step q some sequences are combined. Then there holds

B(q)0 = B

(0)0 .

2

The proof of Proposition 4.5 is presented below. Proposition 4.5 shows that, by combining sequencesin the way the algorithm prescribes, all information under (b) is preserved. As a consequence, theresults of Theorem 2.5, Section 2.3, Proposition 3.2 and Proposition 3.7 are also valid for thedecision variables Xt(j), (j) ∈ S∗ and the sets SUC∗(i), (i) ∈ S∗. The analogue of Theorem 2.5 isas follows.

We consider sets I, J ⊂ S∗ satisfying

(j) /∈ SUC∗(i) for all (i) ∈ I and all (j) ∈ J . (4.3)

Theorem 4.6 Assume that (2.3) holds. Then the cropping plans Xt−1(i), (i) ∈ S∗, and Xt(j),(j) ∈ S∗, are compatible if and only if∑

(i)∈I

Xt−1(i) +∑

(j)∈J

Xt(j) ≤ A , ∀ I, J ⊂ S satisfying (4.3). (4.4)

2

For the whole planning period of T years the crop succession requirements are satisfied if and onlyif (4.4) holds for t = 2, 3, . . . , T . Since the last crop of any sequence in S∗ is a single crop, theknowledge of Xt(j), t = 1, 2, . . . , T , (j) ∈ S∗, implies that for each year 1, 2, . . . , T the area for eachcrop 1, 2, . . . , n is known exactly.

Before we prove Proposition 4.5, we need the following definition and lemma.

Definition 4.7 Let (i) and (j) be combined sequences of length m. We say that (j) is a compatiblesuccessor of (i) if there is an (s) ∈ COMB(i) and a (t) ∈ COMB(j) such that t1 = s2, t2 =s3, . . . , tm−1 = sm.

Notice that a sequence (i) ∈ S(q), q ≥ 1, may have more than n compatible successors (j) ∈ S(q). Anexample is as follows. Consider n = 2 crops and suppose that all crop succession information is givenby the inadmissibility of (2, 2, 2, 2) and (2, 1, 2, 2). Then these sequences are also minimal inadmissi-ble and m = 3. If we execute our algorithm, then S∗ = {(x, 1, 1), (x, 2, 1), (1, 1, 2), (1, 2, 2), (2, x, 2)},where x denotes all crops. In this case (2, x, 2) has three compatible successors: (x, 2, 1), (1, 2, 2)and (2, x, 2).

Lemma 4.8 Suppose (i) ∈ S(q−1) and C(q)(r)(k) 6= ∅. Then there is at most one (t) ∈ C

(q)(r)(k) that is

a compatible successor of (i).

Proof. The proof is similar to the proofs of Lemma 4.1 and Lemma 4.2. Let (s), (t) ∈ C(q)(r)(k), with

(t) a compatible successor of (i). Suppose to the contrary that also (s) is a compatible successorof (i). For any sequence (x) ∈ S(q−1) the elements xq, . . . , xm are single crops. Hence, there holds

sk−1 = tk−1 = ik , k = q + 1, . . . ,m .

18

But this implies sq = tq which contradicts the fact that (s), (t) ∈ C(q)(r)(k). Therefore, if C

(q)(r)(k) 6= ∅,

then, for any (i) ∈ S(q−1), there is at most one (t) ∈ C(q)(r)(k) that is a compatible successor of (i). 2

Proof of Proposition 4.5We will show that, if in Step q some sequences are combined, then B

(q)0 = B

(q−1)0 . Since q ∈

{1, 2, . . . ,m− 1} is arbitrary, this completes the proof.Consider an arbitrary sequence (k1, . . . , km+1) ∈ {1, 2, . . . , n}m+1 and suppose that

(k1, . . . , km) ∈ COMB(s) , (s) ∈ S(q−1) ,

(k2, . . . , km+1) ∈ COMB(t) , (t) ∈ S(q−1) ,

(k1, . . . , km) ∈ COMB(i) , (i) ∈ S(q) ,

(k2, . . . , km+1) ∈ COMB(j) , (j) ∈ S(q) .

Notice that (s), (t), (i) and (j) above are uniquely determined. If in Step q the sequence (i) isformed from some group of sequences, then this group contains (s). The same is true of (j) and(t). For (i) and (j) we distinguish the situations (C1)-(C4) of Section 4.3.

(C1) (i) = (s) and (j) = (t).

(C2) (i) = (s) and (j) is formed from a group containing (t).

(C3) (i) is formed from a group containing (s) and (j) = (t).

(C4) (i) is formed from a group containing (s) and (j) is formed from a group containing (t).

Suppose (k1, . . . , km+1) ∈ B(q−1)0 . Then it follows that (s1, . . . , sm, tm) ∈ B(q−1) and, hence, that

(t) ∈ SUC(q−1)(s). To obtain that (k1, . . . , km+1) ∈ B(q)0 , we need to show that (j) ∈ SUC(q)(i).

But, since (t) ∈ SUC(q−1)(s), this follows for each of (C1)-(C4) from the definition of the algorithmin Section 4.3.

It remains to show that if (k1, . . . , km+1) /∈ B(q−1)0 , then (k1, . . . , km+1) /∈ B

(q)0 . When (k1, . . . , km) /∈

S(0), this holds trivially. We will therefore assume that (k1, . . . , km) ∈ S(0). Suppose that(k2, . . . , km+1) /∈ S(0). Let (i) be as above. If (k1, . . . , km+1) ∈ B

(q)0 , then (i1, . . . , im, km+1) ∈ B(q)

and, hence, there exists a (j) ∈ S(q) with (j) ∈ SUC(q)(i) and jm = km+1. But this implies(k2, . . . , km+1) ∈ COMB(j) which contradicts (k2, . . . , km+1) /∈ S(0). Therefore, if (k2, . . . , km+1) /∈S(0), then (k1, . . . , km+1) /∈ B

(q)0 .

Next, we give the proof for the case (k2, . . . , km+1) ∈ S(0). Let (s), (t), (i) and (j) be asabove. Since (k1, . . . , km+1) /∈ B

(q−1)0 , this implies that (s1, . . . , sm, tm) /∈ B(q−1) and, hence, that

(t) /∈ SUC(q−1)(s). However, since (k1, . . . , km) ∈ COMB(s) and (k2, . . . , km+1) ∈ COMB(t), (t) isa compatible successor of (s). Analogously, (j) is a compatible successor of (i). The proof will becomplete if we show that (j) /∈ SUC(q)(i). Again, we distinguish the cases (C1)-(C4) above.

For (C1) there is nothing to prove. For (C2), it follows from Lemma 4.8 that (t) is the onlycompatible successor of (s) in the group of sequences from which (j) is formed. Hence, from thedefnition of the algorithm in Section 4.3 it follows that (t) /∈ SUC(q−1)(s) implies (j) /∈ SUC(q)(i).Next, we consider (C3). The sequences in the group from which (i) is formed all have the sameset SUC(q−1) as (s). Therefore, (j) = (t) /∈ SUC(q−1)(s) implies (j) /∈ SUC(q)(i). Combining thearguments for (C2) and (C3), it can be seen that also for (C4) there holds (j) /∈ SUC(q)(i). Thiscompletes the proof. 2

19

5 Determining minimal inadmissible sequences

The notion of a minimal inadmissible sequence plays an important role in the analysis above. Thelength of the longest minimal inadmissible sequence indicates how many years we should look intothe past when determining feasible cropping plans for next year. Above, we implicitly assumedthat all minimal inadmissible sequences were known. In practice, however, it may happen thatonly a collection of inadmissible sequences is given. Here, we discuss how the minimal inadmissiblesequences may be determined from such a collection, under the assumption that the given collectionof inadmissible sequences represents all crop succession information. The following example showsthat, in general, this may not be a trivial task.

Example 5.1 Suppose we consider n = 2 crops and all crop succession information is representedby the inadmissibility of the sequences (2, 2, 2, 2) and (1, 2, 2). Then there is only one minimalinadmissible sequence, namely (2, 2). Indeed, suppose we have cultivated (2, 2) in years t− 1 andt. Then we have grown crop 2 in year t− 2, since (1, 2, 2) is inadmissible. And also in year t− 3 wemust have grown crop 2. Otherwise we have a sequence (1, 2, 2) for the years t− 3, t− 2, t− 1. Butnow we have cultivated crop 2 for four successive years, which is also inadmissible. Therefore, (2, 2)is inadmissible. Since (2, 2, 2, 2) and (1, 2, 2) both contain (2, 2), they are not minimal inadmissible.The sequence (2, 2), however, is minimal inadmissible, since the cultivation of crop 2 is not entirelyprohibited.

The example above can still be worked out by hand. However, to determine the minimal inadmis-sible sequences for (R1)-(R6) in the example of Section 4.2 is more difficult. In this case, it is moreconvenient to have a computer program which determines all minimal inadmissible sequences froma given collection of inadmissible sequences. Below, we present such an algorithm. We assume thatan initial collection of inadmissible sequences is given by the sets Mh, h ≥ 1, where

Mh = {(i) ∈ {1, 2, . . . , n}h : (i) is inadmissible } .

Suppose the longest inadmissible sequence has length H + 1, i.e.

H + 1 = max{h : Mh 6= ∅ } . (5.1)

As above, we assume that H ≥ 1. If the collection ∪H+1h=1 Mh represents all crop succession informa-

tion, then its set of minimal inadmissible sequences is unique and the longest minimal inadmissiblesequence has a length of at most H + 1.

Before we determine the minimal inadmissible sequences, we first determine all inadmissiblesequences based on the sets Mh, h = 1, 2, . . . ,H +1. This is done in Step m1 and Step m2 below.

Step m1: For h = 1, 2, . . . ,H, do the following.

1.1 For each (i) ∈ Mh, for each g ∈ {h + 1, . . . ,H + 1}, if there is a (j) ∈ {1, 2, . . . , n}g with(j) /∈ Mg and (i) is a subsequence of (j), then add (j) to Mg.

In Step m1 we enlarge the collection of inadmissible sequences by including all sequences whichcontain an inadmissible subsequence in some set Mh. However, after Step m1 the collection ofinadmissible sequences does not yet contain all inadmissible sequences that can be found, using thesets Mh. We also need to include sequences (not contained in any Mh) of which the inadmissibilityfollows from the sets Mh by logical reasoning. An example is the sequence (2, 2) in Example

20

5.1. Such sequences are identified by making use of similar sets as SUC(i) in (2.4). For (i) ∈{1, 2, . . . , n}h, we define

SUCh(i) = { (j) ∈ {1, 2, . . . , n}h : (j) /∈ Mh , (j1, . . . , jh−1) = (i2, . . . , ih) and ((i), jh) /∈ Mh+1 } .

Hence, (j) ∈ SUCh(i) if and only if (j) is allowed to succeed (i) according to Mh and Mh+1. Noticethat if SUCh(i) = ∅, then (i) is inadmissible. Also, if (i) ∈ {1, 2, . . . , n}h is not contained in anyset SUCh(j), then (i) is inadmissible.

Step m2: For h = H,H − 1, . . . , 1, do the following.

2.1 For all sequences (i) ∈ {1, 2, . . . , n}h with (i) /∈ Mh, determine the sets SUCh(i) based on theinformation in Mh+1.

2.2 Let Dh = {(i) ∈ {1, 2, . . . , n}h\Mh : SUCh(i) = ∅ or (i) /∈ SUCh(j), ∀ (j) /∈ Mh }. If Dh 6= ∅,then go to Step 2.3, else NEXT h.

2.3 Add all (i) ∈ Dh to Mh. For each (i) ∈ Dh, for each g ∈ {h + 1, . . . ,H + 1}, if there is a(j) ∈ {1, 2, . . . , n}g with (j) /∈ Mg and (i) is a subsequence of (j), then add (j) to Mg. Startagain at the beginning of Step m2.

If in Step 2.2 some inadmissible sequences are found, they are added to Mh. Moreover, allsequences containing these sequences are inadmissible too and are added to Mg, g ≥ h + 1. Thismay change the sets SUCg(i), g ≥ h. Therefore, when inadmissible sequences are found, we startagain at the beginning of Step m2. It can be seen that after Step m2 the sets M1,M2, . . . ,MH+1

contain all inadmissible sequences that can be found based on the initial collection of inadmissiblesequences.

In Step m3 below, we determine the minimal inadmissible sequences by deleting inadmissiblesequences that contain an inadmissible subsequence.

Step m3: For h = 1, 2, . . . ,H, do the following.

3.1 For each (i) ∈ Mh, for each g ∈ {h + 1, . . . ,H + 1}, if there is a (j) ∈ Mg that contains (i) asa subsequence, then delete (j) from Mg.

In this way, we end up with all minimal inadmissible sequences in the sets M1,M2, . . . ,MH+1.There holds

m + 1 = max{h : Mh 6= ∅} .

Notice that m may be smaller than H, as can be seen from Example 5.1.There are several ways in which the algorithm above can be made faster. A first observation

is that the sequences which are added to the sets Mh in Step m1 and Step 2.3 (for g ≥ h + 1)are not minimal inadmissible, since they contain an inadmissible subsequence. Hence, in Step m3,these sequences may be deleted without examination. Another change by which the procedure maybecome faster, is by starting with Step m3, followed by a recalculation of H in (5.1). Since Stepm3 excludes sequences from the initial collection which are not minimal inadmissible, after Stepm3 the value of H may have decreased. As a result, the algorithm consisting of Step m3, Stepm1, Step m2 and again Step m3, has more steps but may be faster.

The algorithm above first determines all inadmissible sequences, based on the initial collection,and then determines the minimal inadmissible sequences. To determine all inadmissible sequences,in Step m1 and Step 2.3 inadmissible sequences are added to the current collection. However,

21

the addition in Step 2.3 of sequences to Mg, g ∈ {h + 1, . . . ,H}, is not necessary, since at thispoint they are already contained in the sets Mg, g ∈ {h + 1, . . . ,H}. Moreover, also the additionof sequences to MH+1 is not necessary. In Proposition 5.3 below it is shown that we may replaceStep 2.3 by

2.3* Add all (i) ∈ Dh to Mh and go to Step 2.1.

Remark 5.2 We have programmed a version of the algorithm above using the software packageMatlab 5. We start with Step m3, followed by Step m1, Step m2 and again Step m3, and useStep 2.3* instead of Step 2.3. This program was also used to obtain the minimal inadmissiblesequences for (R1)-(R6) in the example of Section 4.2. Our computational experience with theprogram learns that on a Pentium 4 PC, the algorithm terminates within approximately 20 minutesfor values of n and H with nH ≤ 3400 approximately. To our opinion, this covers most practicalsituations.

Proposition 5.3 Suppose an initial collection Mh, h ≥ 1, is given and Step m1 has been executed.Suppose that in Step 2.2 Dh 6= ∅ for some h. Then there holds for any (i) ∈ Dh,

(F1) If h ≤ H−1, then after Step m2 has been executed for h+1, the sets Mg, g ∈ {h+1, . . . ,H},contain all sequences having (i) as a subsequence.

(F2) In Step 2.3 it is not necessary to add sequences to MH+1, containing (i) as a subsequence.

Proof. First, we prove (F1). Suppose Step m1 has been executed. Observe that there holds forh = 1, 2, . . . ,H

(s) ∈ Mh ⇒ all (t) ∈ {1, 2, . . . , n}g containing (s) are in Mg, g = h + 1, . . . ,H + 1. (5.2)

Suppose that Dh 6= ∅ in Step 2.2 for some h ∈ {1, 2, . . . ,H − 1}. Let (i) ∈ Dh. Suppose thatSUCh(i) = ∅. Then after Step m2 has been executed for h+1, there holds for all x ∈ {1, 2, . . . , n},

(E1) If (i2, . . . , ih, x) /∈ Mh, then ((i), x) ∈ Mh+1,

(E2) If ((i), x) /∈ Mh+1, then (i2, . . . , ih, x) ∈ Mh.

Since the set Mh is still the same as after Step m1 and no sequences have been deleted from Mh+1,it follows from (5.2) that (E2) is impossible. Therefore, if SUCh(i) = ∅, then after Step m1

((i), x)) ∈ Mh+1 , ∀ x ∈ {1, 2, . . . , n} . (5.3)

Notice that this implies that any sequence (y, (i)), y ∈ {1, 2, . . . , n}, not contained in Mh+1 hasan empty set SUCh+1(y, (i)). However, such sequences (y, (i)) do not exist, since they were addedto Mh+1 when Step m2 was executed for h + 1. Therefore, after Step m2 has been executed forh + 1, there holds

(y, (i)) ∈ Mh+1 , ∀ y ∈ {1, 2, . . . , n} . (5.4)

Hence, it follows from (5.3) and (5.4) that after Step m2 has been executed for h+1, all sequencescontaining (i) are contained in Mh+1. Since (5.3) holds after Step m1, it follows from (5.2) thatafter Step m1 all sequences of the form (y, (i), x) and ((i), x, y), x, y ∈ {1, 2, . . . , n}, are containedin Mh+2. But this implies that sequences of the form (x, y, (i)), x, y ∈ {1, 2, . . . , n}, have an emptyset SUCh+2(x, y, (i)). Hence, all sequences of the form (x, y, (i)) were added to Mh+2 when Stepm2 was executed for h + 2. Therefore, after Step m2 has been executed for h + 1, all sequences

22

containing (i) are contained in both Mh+1 and Mh+2. Analogously, it can be shown that after Stepm2 has been executed for h+1, all sequences containing (i) are contained in Mg for g = h+1, . . . ,H.

Suppose next that (i) ∈ Dh due to (i) /∈ SUCh(j) for all (j) /∈ Mh. Then after Step m2 hasbeen executed for h + 1, there holds for all x ∈ {1, 2, . . . , n},

(E3) If (x, i1, . . . , ih−1) /∈ Mh, then (x, (i)) ∈ Mh+1,

(E4) If (x, (i)) /∈ Mh+1, then (x, i1, . . . , ih−1) ∈ Mh.

Since the set Mh is still the same as after Step m1 and no sequences have been deleted from Mh+1,it follows from (5.2) that (E4) is impossible. Therefore, if (i) /∈ SUCh(j) for all (j) /∈ Mh, thenafter Step m1

(x, (i))) ∈ Mh+1 , ∀ x ∈ {1, 2, . . . , n} . (5.5)

Notice that this implies that any sequence ((i), y), y ∈ {1, 2, . . . , n}, not contained in Mh+1 is notcontained in any set SUCh+1. However, such sequences ((i), y) do not exist, since they were addedto Mh+1 when Step m2 was executed for h + 1. Therefore, after Step m2 has been executed forh + 1, there holds

((i), y) ∈ Mh+1 , ∀ y ∈ {1, 2, . . . , n} . (5.6)

Hence, it follows from (5.5) and (5.6) that after Step m2 has been executed for h+1, all sequencescontaining (i) are contained in Mh+1. Since (5.5) holds after Step m1, it follows from (5.2) thatafter Step m1 all sequences of the form (y, x, (i)) and (x, (i), y), x, y ∈ {1, 2, . . . , n}, are contained inMh+2. But this implies that sequences of the form ((i), x, y), x, y ∈ {1, 2, . . . , n}, are not containedin any set SUCh+2. Hence, all sequences of the form ((i), x, y) were added to Mh+2 when Stepm2 was executed for h + 2. Therefore, after Step m2 has been executed for h + 1, all sequencescontaining (i) are contained in both Mh+1 and Mh+2. Analogously, it can be shown that after Stepm2 has been executed for h+1, all sequences containing (i) are contained in Mg for g = h+1, . . . ,H.

Notice that when h is the first (i.e. largest) h for which Dh 6= ∅ in Step 2.2, an analogousproof as above shows that after Step m1 all sequences containing an (i) ∈ Dh are contained in thesets Mg, g = h + 1, . . . ,H. This completes the proof of (F1).

Next, we prove (F2). First, we consider the case h = H. Let (i) ∈ DH . In Step 2.3 sequencescontaining (i) as a subsequence are added to MH+1. As a result, the sets SUCH may change.However, the same changes in SUCH are obtained when (i) is added to MH . This can be seenas follows. The sequences added to MH+1 are of the form (x, (i)) or ((i), x), for x ∈ {1, 2, . . . , n}.Therefore, as a result, only the sets SUCH(j) may change for which either (j) is a compatiblesuccessor of (i), i.e. (j1, . . . , jH−1) = (i2, . . . , iH), or (i) is a compatible successor of (j), i.e.(i1, . . . , iH−1) = (j2, . . . , jH). But these sets SUCH(j) change in the same way when (i) is addedto MH . This completes the proof for the case h = H.

Next, we consider the case h ≤ H − 1. Let (i) ∈ Dh. From the proof of (F1) it follows thatall sequences (k) of length H containing (i) are contained in MH after Step m2 has been exe-cuted for H. Sequences of length H + 1 containing (i) are of the form (x, (k)) or ((k), x), forx ∈ {1, 2, . . . , n}, where (k) has length H and contains (i). From the proof of (F2) for h = Hit follows that adding such sequences to MH+1 will not result in more inadmissible sequences oflength H. This completes the proof of (F2). 2

23

Proposition 5.3 shows that Step 2.3 may be replaced by Step 2.3*. One may wonder whether itis necessary to update the sets SUCh after the sequences in Dh have been added to Mh in Step2.3*. In other words, may we replace Step 2.3* by

Add all (i) ∈ Dh to Mh. NEXT h.

The following example shows that this is not the case.

Example 5.4 Consider n = 3 crops and start with M4 = {(2, 1, 3, 2), (2, 1, 3, 1)} and M3 ={(3, 3, 1), (3, 3, 2), (3, 3, 3)}. Then H = 3 and after Step m1 we have

M4 = {(2, 1, 3, 2), (2, 1, 3, 1), (3, 3, s, t), (u, 3, 3, v), s, t, u, v = 1, 2, 3} .

Next, we execute Step m2. In Step 2.2 we have SUC3(1, 3, 3) = SUC3(2, 3, 3) = ∅. Hence,(1, 3, 3) and (2, 3, 3) are added to M3. When we go to Step 2.1 and change the sets SUC3, wefind that SUC3(2, 1, 3) = ∅. Therefore, also (2, 1, 3) is added to M3. For h = 2, it is found thatSUC2(3, 3) = ∅ and (3, 3) is added to M2. After Step m3, we obtain M4 = ∅, M3 = {(2, 1, 3)} andM2 = {(3, 3)}.

Acknowledgment

We would like to thank Caspar Schweigman for drawing our attention to the subject of cropsuccession requirements in agricultural production planning and for various discussions on thesubject.

References

[1] L.R. Ford Jr. and D.R. Fulkerson (1962) Flows in Networks. Princeton University Press,Princeton NJ.

[2] W.K. Klein Haneveld (1983) Crop rotation and linear programming. Technical Report83-11-OR. Department of Econometrics, University of Groningen.

[3] C. Schweigman (1985) Operations Research Problems in Agriculture in Developing Countries.Khartoum University Press, Khartoum, Sudan.

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