Research article Special Issues

Forecasting arabica coffee yields by auto-regressive integrated moving average and machine learning approaches

  • Received: 23 June 2023 Revised: 26 September 2023 Accepted: 01 November 2023 Published: 11 November 2023
  • Coffee is a major industrial crop that creates high economic value in Thailand and other countries worldwide. A lack of certainty in forecasting coffee production could lead to serious operation problems for business. Applying machine learning (ML) to coffee production is crucial since it can help in productivity prediction and increase prediction accuracy rate in response to customer demands. An ML technique of artificial neural network (ANN) model, and a statistical technique of autoregressive integrated moving average (ARIMA) model were adopted in this study to forecast arabica coffee yields. Six variable datasets were collected from 2004 to 2018, including cultivated areas, productivity zone, rainfalls, relative humidity and minimum and maximum temperatures, totaling 180 time-series data points. Their prediction performances were evaluated in terms of correlation coefficient (R2), and root means square error (RMSE). From this work, the ARIMA model was optimized using the fitting model of (p, d, q) amounted to 64 conditions through the Akaike information criteria arriving at (2, 1, 2). The ARIMA results showed that its R2 and RMSE were 0.7041 and 0.1348, respectively. Moreover, the R2 and RMSE of the ANN model were 0.9299 and 0.0642 by the Levenberg-Marquardt algorithm with TrainLM and LearnGDM training functions, two hidden layers and six processing elements. Both models were acceptable in forecasting the annual arabica coffee production, but the ANN model appeared to perform better.

    Citation: Yotsaphat Kittichotsatsawat, Anuwat Boonprasope, Erwin Rauch, Nakorn Tippayawong, Korrakot Yaibuathet Tippayawong. Forecasting arabica coffee yields by auto-regressive integrated moving average and machine learning approaches[J]. AIMS Agriculture and Food, 2023, 8(4): 1052-1070. doi: 10.3934/agrfood.2023057

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  • Coffee is a major industrial crop that creates high economic value in Thailand and other countries worldwide. A lack of certainty in forecasting coffee production could lead to serious operation problems for business. Applying machine learning (ML) to coffee production is crucial since it can help in productivity prediction and increase prediction accuracy rate in response to customer demands. An ML technique of artificial neural network (ANN) model, and a statistical technique of autoregressive integrated moving average (ARIMA) model were adopted in this study to forecast arabica coffee yields. Six variable datasets were collected from 2004 to 2018, including cultivated areas, productivity zone, rainfalls, relative humidity and minimum and maximum temperatures, totaling 180 time-series data points. Their prediction performances were evaluated in terms of correlation coefficient (R2), and root means square error (RMSE). From this work, the ARIMA model was optimized using the fitting model of (p, d, q) amounted to 64 conditions through the Akaike information criteria arriving at (2, 1, 2). The ARIMA results showed that its R2 and RMSE were 0.7041 and 0.1348, respectively. Moreover, the R2 and RMSE of the ANN model were 0.9299 and 0.0642 by the Levenberg-Marquardt algorithm with TrainLM and LearnGDM training functions, two hidden layers and six processing elements. Both models were acceptable in forecasting the annual arabica coffee production, but the ANN model appeared to perform better.



    The famous Young's inequality, as a classical result, state that: if a,b>0 and t[0,1], then

    atb1tta+(1t)b (1.1)

    with equality if and only if a=b. Let p,q>1 such that 1/p+1/q=1. The inequality (1.1) can be written as

    abapp+bqq (1.2)

    for any a,b0. In this form, the inequality (1.2) was used to prove the celebrated Hölder inequality. One of the most important inequalities of analysis is Hölder's inequality. It contributes wide area of pure and applied mathematics and plays a key role in resolving many problems in social science and cultural science as well as in natural science.

    Theorem 1 (Hölder inequality for integrals [11]). Let p>1 and 1/p+1/q=1. If f and g are real functions defined on [a,b] and if |f|p,|g|q are integrable functions on [a,b] then

    ba|f(x)g(x)|dx(ba|f(x)|pdx)1/p(ba|g(x)|qdx)1/q, (1.3)

    with equality holding if and only if A|f(x)|p=B|g(x)|q almost everywhere, where A and B are constants.

    Theorem 2 (Hölder inequality for sums [11]). Let a=(a1,...,an) and b=(b1,...,bn) be two positive n-tuples and p,q>1 such that 1/p+1/q=1. Then we have

    nk=1akbk(nk=1apk)1/p(nk=1bqk)1/q. (1.4)

    Equality hold in (1.4) if and only if ap and bq are proportional.

    In [10], İşcan gave new improvements for integral ans sum forms of the Hölder inequality as follow:

    Theorem 3. Let p>1 and 1p+1q=1. If f and g are real functions defined on interval [a,b] and if |f|p, |g|q are integrable functions on [a,b] then

    ba|f(x)g(x)|dx1ba{(ba(bx)|f(x)|pdx)1p(ba(bx)|g(x)|qdx)1q+(ba(xa)|f(x)|pdx)1p(ba(xa)|g(x)|qdx)1q} (1.5)

    Theorem 4. Let a=(a1,...,an) and b=(b1,...,bn) be two positive n-tuples and p,q>1 such that 1/p+1/q=1. Then

    nk=1akbk1n{(nk=1kapk)1/p(nk=1kbqk)1/q+(nk=1(nk)apk)1/p(nk=1(nk)bqk)1/q}. (1.6)

    Let E be a nonempty set and L be a linear class of real valued functions on E having the following properties

    L1: If f,gL then (αf+βg)L for all α,βR;

    L2: 1L, that is if f(t)=1,tE, then fL;

    We also consider positive isotonic linear functionals A:LR is a functional satisfying the following properties:

    A1: A(αf+βg)=αA(f)+β A(g) for f,gL and α,βR;

    A2: If fL, f(t)0 on E then A(f)0.

    Isotonic, that is, order-preserving, linear functionals are natural objects in analysis which enjoy a number of convenient properties. Functional versions of well-known inequalities and related results could be found in [1,2,3,4,5,6,7,8,9,11,12].

    Example 1. i.) If E=[a,b]R and L=L[a,b], then

    A(f)=baf(t)dt

    is an isotonic linear functional.

    ii.)If E=[a,b]×[c,d]R2 and L=L([a,b]×[c,d]), then

    A(f)=badcf(x,y)dxdy

    is an isotonic linear functional.

    iii.)If (E,Σ,μ) is a measure space with μ positive measure on E and L=L(μ) then

    A(f)=Efdμ 

    is an isotonic linear functional.

    iv.)If E is a subset of the natural numbers N with all pk0, then A(f)=kEpkfk is an isotonic linear functional. For example; If E={1,2,...,n} and f:ER,f(k)=ak, then A(f)=nk=1ak is an isotonic linear functional. If E={1,2,...,n}×{1,2,...,m} and f:ER,f(k,l)=ak,l, then A(f)=nk=1ml=1ak,l is an isotonic linear functional.

    Theorem 5 (Hölder's inequality for isotonic functionals [13]). Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p1+q1=1. If w,f,g0 on E and wfp,wgq,wfgL then we have

    A(wfg)A1/p(wfp)A1/q(wgq). (2.1)

    In the case 0<p<1 and A(wgq)>0 (or p<0 and A(wfp)>0), the inequality in (2.1) is reversed.

    Remark 1. i.) If we choose E=[a,b]R, L=L[a,b], w=1 on E and A(f)=ba|f(t)|dt in the Theorem 5, then the inequality (2.1) reduce the inequality (1.3).

    ii.) If we choose E={1,2,...,n}, w=1 on E, f:E[0,),f(k)=ak, and A(f)=nk=1ak in the Theorem 5, then the inequality (2.1) reduce the inequality (1.4).

    iii.) If we choose E=[a,b]×[c,d],L=L(E), w=1 on E and A(f)=badc|f(x,y)|dxdy in the Theorem 5, then the inequality (2.1) reduce the following inequality for double integrals:

    badc|f(x,y)||g(x,y)|dxdy(badc|f(x,y)|pdx)1/p(badc|g(x,y)|qdx)1/q.

    The aim of this paper is to give a new general improvement of Hölder inequality for isotonic linear functional. As applications, this new inequality will be rewritten for several important particular cases of isotonic linear functionals. Also, we give an application to show that improvement is hold for double integrals.

    Theorem 6. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p1+q1=1. If α,β,w,f,g0 on E, αwfg,βwfg,αwfp,αwgq,βwfp,βwgq,wfgL and α+β=1 on E, then we have

    i.)

    A(wfg)A1/p(αwfp)A1/q(αwgq)+A1/p(βwfq)A1/q(βwgq) (3.1)

    ii.)

    A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq). (3.2)

    Proof. ⅰ.) By using of Hölder inequality for isotonic functionals in (2.1) and linearity of A, it is easily seen that

    A(wfg)=A(αwfg+βwfg)=A(αwfg)+A(βwfg)A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq).

    ⅱ.) Firstly, we assume that A1/p(wfp)A1/q(wgq)0. then

    A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq)=(A(αwfp)A(wfp))1/p(A(αwgq)A(wgq))1/q+(A(βwfp)A(wfp))1/p(A(βwgq)A(wgq))1/q,

    By the inequality (1.1) and linearity of A, we have

    A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq)1p[A(αwfp)A(wfp)+A(βwfp)A(wfp)]+1q[A(αwgq)A(wgq)+A(βwgq)A(wgq)]=1.

    Finally, suppose that A1/p(wfp)A1/q(wgq)=0. Then A1/p(wfp)=0 or A1/q(wgq)=0, i.e. A(wfp)=0 or A(wgq)=0. We assume that A(wfp)=0. Then by using linearity of A we have,

    0=A(wfp)=A(αwfp+βwfp)=A(αwfp)+A(βwfp).

    Since A(αwf),A(βwf)0, we get A(αwfp)=0 and A(βwfp)=0. From here, it follows that

    A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)=00=A1/p(wfp)A1/q(wgq).

    In case of A(wgq)=0, the proof is done similarly. This completes the proof.

    Remark 2. The inequality (3.2) shows that the inequality (3.1) is better than the inequality (2.1).

    If we take w=1 on E in the Theorem 6, then we can give the following corollary:

    Corollary 1. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p1+q1=1. If α,β,f,g0 on E, αfg,βfg,αfp,αgq,βfp,βgq,fgL and α+β=1 on E, then we have

    i.)

    A(fg)A1/p(αfp)A1/q(αgq)+A1/p(βfq)A1/q(βgq) (3.3)

    ii.)

    A1/p(αfp)A1/q(αgq)+A1/p(βfp)A1/q(βgq)A1/p(fp)A1/q(gq).

    Remark 3. i.) If we choose E=[a,b]R, L=L[a,b], α(t)=btba,β(t)=taba on E and A(f)=ba|f(t)|dt in the Corollary 1, then the inequality (3.3) reduce the inequality (1.5).

    ii.) If we choose E={1,2,...,n}, α(k)=kn,β(k)=nkn on E, f:E[0,),f(k)=ak, and A(f)=nk=1ak in the Theorem1, then the inequality (3.3) reduce the inequality (1.6).

    We can give more general form of the Theorem 6 as follows:

    Theorem 7. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p1+q1=1. If αi,w,f,g0 on E, αiwfg,αiwfp,αiwgq,wfgL,i=1,2,...,m, and mi=1αi=1 on E, then we have

    i.)

    A(wfg)mi=1A1/p(αiwfp)A1/q(αiwgq)

    ii.)

    mi=1A1/p(αiwfp)A1/q(αiwgq)A1/p(wfp)A1/q(wgq).

    Proof. The proof can be easily done similarly to the proof of Theorem 6.

    If we take w=1 on E in the Theorem 6, then we can give the following corollary:

    Corollary 2. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p1+q1=1. If αi,f,g0 on E, αifg,αifp,αigq,fgL,i=1,2,...,m, and mi=1αi=1 on E, then we have

    i.)

    A(fg)mi=1A1/p(αifp)A1/q(αigq) (3.4)

    ii.)

    mi=1A1/p(αifp)A1/q(αigq)A1/p(fp)A1/q(gq).

    Corollary 3 (Improvement of Hölder inequality for double integrals). Let p,q>1 and 1/p+1/q=1. If f and g are real functions defined on E=[a,b]×[c,d] and if |f|p,|g|qL(E) then

    badc|f(x,y)||g(x,y)|dxdy4i=1(badcαi(x,y)|f(x,y)|pdx)1/p(badcαi(x,y)|g(x,y)|qdx)1/q, (3.5)

    where α1(x,y)=(bx)(dy)(ba)(dc),α2(x,y)=(bx)(yc)(ba)(dc),α3(x,y)=(xa)(yc)(ba)(dc),,α4(x,y)=(xa)(dy)(ba)(dc) on E

    Proof. If we choose E=[a,b]×[c,d]R2, L=L(E), α1(x,y)=(bx)(dy)(ba)(dc),α2(x,y)=(bx)(yc)(ba)(dc),α3(x,y)=(xa)(yc)(ba)(dc),α4(x,y)=(xa)(dy)(ba)(dc) on E and A(f)=badc|f(x,y)|dxdy in the Corollary 1, then we get the inequality (3.5).

    Corollary 4. Let (ak,l) and (bk,l) be two tuples of positive numbers and p,q>1 such that 1/p+1/q=1. Then we have

    nk=1ml=1ak,lbk,l4i=1(nk=1ml=1αi(k,l)apk,l)1/p(nk=1ml=1αi(k,l)bqk,l)1/q, (3.6)

    where α1(k,l)=klnm,α2(k,l)=(nk)lnm,α3(k,l)=(nk)(ml)nm,α4(k,l)=k(ml)nm on E.

    Proof. If we choose E={1,2,...,n}×{1,2,...,m}, α1(k,l)=klnm,α2(k,l)=(nk)lnm,α3(k,l)=(nk)(ml)nm,α4(k,l)=k(ml)nm on E, f:E[0,),f(k,l)=ak,l, and A(f)=nk=1ml=1ak,l in the Theorem1, then we get the inequality (3.6).

    In [14], Sarıkaya et al. gave the following lemma for obtain main results.

    Lemma 1. Let f:ΔR2R be a partial differentiable mapping on Δ=[a,b]×[c,d] in R2with a<b and c<d. If 2ftsL(Δ), then the following equality holds:

    f(a,c)+f(a,d)+f(b,c)+f(b,d)41(ba)(dc)badcf(x,y)dxdy12[1baba[f(x,c)+f(x,d)]dx+1dcdc[f(a,y)+f(b,y)]dy]=(ba)(dc)41010(12t)(12s)2fts(ta+(1t)b,sc+(1s)d)dtds.

    By using this equality and Hölder integral inequality for double integrals, Sar\i kaya et al. obtained the following inequality:

    Theorem 8. Let f:ΔR2R be a partial differentiable mapping on Δ=[a,b]×[c,d] in R2with a<b and c<d. If |2fts|q,q>1, is convex function on the co-ordinates on Δ, then one has the inequalities:

    |f(a,c)+f(a,d)+f(b,c)+f(b,d)41(ba)(dc)badcf(x,y)dxdyA|(ba)(dc)4(p+1)2/p[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q4]1/q, (4.1)

    where

    A=12[1baba[f(x,c)+f(x,d)]dx+1dcdc[f(a,y)+f(b,y)]dy],

    1/p+1/q=1 and fst=2fts.

    If Theorem 8 are resulted again by using the inequality (3.5), then we get the following result:

    Theorem 9. Let f:ΔR2R be a partial differentiable mapping on Δ=[a,b]×[c,d] in R2with a<b and c<d. If |2fts|q,q>1, is convex function on the co-ordinates on Δ, then one has the inequalities:

    |f(a,c)+f(a,d)+f(b,c)+f(b,d)41(ba)(dc)badcf(x,y)dxdyA|(ba)(dc)41+1/p(p+1)2/p{[4|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+|fst(b,d)|q36]1/q+[2|fst(a,c)|q+|fst(a,d)|q+4|fst(b,c)|q+2|fst(b,d)|q36]1/q+[2|fst(a,c)|q+4|fst(a,d)|q+|fst(b,c)|q+2|fst(b,d)|q36]1/q+[|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+4|fst(b,d)|q36]1/q}, (4.2)

    where

    A=12[1baba[f(x,c)+f(x,d)]dx+1dcdc[f(a,y)+f(b,y)]dy],

    1/p+1/q=1 and fst=2fts.

    Proof. Using Lemma 1 and the inequality (3.5), we find

    |f(a,c)+f(a,d)+f(b,c)+f(b,d)41(ba)(dc)badcf(x,y)dxdyA|(ba)(dc)41010|12t||12s||fst(ta+(1t)b,sc+(1s))|dtds(ba)(dc)4{(1010ts|12t|p|12s|pdtds)1/p×(1010ts|fst(ta+(1t)b,sc+(1s))|qdtds)1/q+(1010t(1s)|12t|p|12s|pdtds)1/p×(1010t(1s)|fst(ta+(1t)b,sc+(1s))|qdtds)1/q+(1010(1t)s|12t|p|12s|pdtds)1/p×(1010(1t)s|fst(ta+(1t)b,sc+(1s))|qdtds)1/q+(1010(1t)(1s)|12t|p|12s|pdtds)1/p×(1010(1t)(1s)|fst(ta+(1t)b,sc+(1s))|qdtds)1/q}. (4.3)

    Since |fst|q is convex function on the co-ordinates on Δ, we have for all t,s[0,1]

    |fst(ta+(1t)b,sc+(1s))|qts|fst(a,c)|q+t(1s)|fst(a,d)|q+(1t)s|fst(a,c)|q+(1t)(1s)|fst(a,c)|q (4.4)

    for all t,s[0,1]. Further since

    1010ts|12t|p|12s|pdtds=1010t(1s)|12t|p|12s|pdtds=1010(1t)s|12t|p|12s|pdtds (4.5)
    =1010(1t)(1s)|12t|p|12s|pdtds=14(p+1)2, (4.6)

    a combination of (4.3) - (4.5) immediately gives the required inequality (4.2).

    Remark 4. Since η:[0,)R,η(x)=xs,0<s1, is a concave function, for all u,v0 we have

    η(u+v2)=(u+v2)sη(u)+η(v)2=us+vs2.

    From here, we get

    I={[4|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+|fst(b,d)|q36]1/q+[2|fst(a,c)|q+|fst(a,d)|q+4|fst(b,c)|q+2|fst(b,d)|q36]1/q+[2|fst(a,c)|q+4|fst(a,d)|q+|fst(b,c)|q+2|fst(b,d)|q36]1/q+[|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+4|fst(b,d)|q36]1/q}2{[6|fst(a,c)|q+3|fst(a,d)|q+6|fst(b,c)|q+3|fst(b,d)|q72]1/q+[3|fst(a,c)|q+6|fst(a,d)|q+3|fst(b,c)|q+6|fst(b,d)|q72]1/q}
    4{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q16]1/q

    Thus we obtain

    (ba)(dc)41+1/p(p+1)2/pI(ba)(dc)41+1/p(p+1)2/p4{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q16]1/q}(ba)(dc)4(p+1)2/p{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q4]1/q}.

    This shows that the inequality (4.2) is better than the inequality (4.1).

    The aim of this paper is to give a new general improvement of Hölder inequality via isotonic linear functional. An important feature of the new inequality obtained here is that many existing inequalities related to the Hölder inequality can be improved. As applications, this new inequality will be rewritten for several important particular cases of isotonic linear functionals. Also, we give an application to show that improvement is hold for double integrals. Similar method can be applied to the different type of convex functions.

    This research didn't receive any funding.

    The author declares no conflicts of interest in this paper.



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