[R-es] ceiling con max duplica tamaño matriz
Proyecto R-UCA
r-uc@ @end|ng |rom uc@@e@
Jue Ago 25 10:19:16 CEST 2022
Buenas, Sebastián:
No veo qué valor asignas a idx_epsilon.
Un saludo.
El 22/8/22 a las 22:06, Sebastian Kruk escribió:
> Estimados,
>
> Tengo:
> v1
> v_DEM_i_d_m
>
> Donde
>
>> IN_ES_NUM = 469
>> v1 = 1:IN_ES_NUM
>> head(v_DEM_i_d_m)
> Mes 1 Mes 2 Mes 3 Mes 4 Mes 5
> Mes 6 Mes 7 Mes 8 Mes 9 Mes 10
> Mes 11 Mes 12
> ES 1 7.811032258 8.982678571 9.563935484 8.411500000 7.788354839
> 8.976933333 9.517419355 8.902258065 9.328600000 9.391096774
> 9.866800000 10.317838710
> ES 2 0.641419355 1.174714286 1.353838710 1.299900000 1.387096774
> 0.905033333 1.140935484 0.880387097 1.237766667 1.292677419
> 1.327166667 1.387838710
> ES 3 0.003225806 0.003571429 0.003225806 0.003333333 0.003225806
> 0.003333333 0.003225806 0.003225806 0.003333333 0.003225806
> 0.003333333 0.003483871
> ES 4 13.475258065 13.240285714 15.168838710 11.492233333 12.792774194
> 12.989566667 14.996064516 12.911774194 13.835400000 14.635612903
> 14.541266667 15.206000000
> ES 5 18.711258065 23.263535714 24.875967742 23.478000000 27.372548387
> 27.509800000 28.649096774 26.986935484 33.831833333 20.810419355
> 24.669400000 25.797096774
> ES 6 16.112419355 17.903678571 18.250548387 16.712066667 17.353741935
> 17.137233333 18.501000000 17.613258065 19.115766667 20.773354839
> 20.990566667 21.950096774
>
>> class((v_DEM_i_d_m))
> [1] "matrix" "array"
>
>> str(v_DEM_i_d_m)
> num [1:469, 1:12] 7.81103 0.64142 0.00323 13.47526 18.71126 ...
> - attr(*, "dimnames")=List of 2
> ..$ : chr [1:469] "ES 1" "ES 2" "ES 3" "ES 4" ...
> ..$ : chr [1:12] "Mes 1" "Mes 2" "Mes 3" "Mes 4" ...
>
> Si hago:
>
> for (i in idx_epsilon) {
> v1[i] = ceiling(max(v_DEM_i_d_m[i,]))
> }
>
> v1 queda así:
>
>> head(v1)
> 1 2 3 4 5 6
>
>> class(v1)
> [1] "numeric"
>
>> str(v1)
> Named num [1:938] 1 2 3 4 5 6 7 8 9 10 ...
> - attr(*, "names")= chr [1:938] "" "" "" "" ...
>
> No entiendo porque los primeros 469 son los números de 1 a 469.
>
> ¿Que estoy haciendo mal?
>
> Saludos,
>
> Sebastián.
>
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> From: Sebastian Kruk <residuo.solow using gmail.com>
> Date: Wed, 3 Aug 2022 12:22:51 -0300
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> Subject: [R-es] Rglpk_solve_LP
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> Estimados usuarios-R:
>
> Estoy probando el solver GLPK.
>
> Me pasaron un programa en Octave y lo convertí a R pero me da siempre
> que la solución es 0.
>
> En R me quedó así:
>
> ######################################################
>
> #1) Datos
>
> #Período de estudio (días)
> T = 10
>
> #Demanda diaria
> dda = 10*matrix(1,T,1)
>
> #Capacidad de camión (m3)
> K=30
>
> #Capacidad de tanque (m3)
> cap_tqe=40
>
> #Stock inicial (m3)
> s0=15
>
> #Max cantdad diaria de camiones
> x_max=10
>
> ##################################
> #2) Modelo simple, puede entregar todos los días, camiones enteros
> #
> # Variables: z=[x; s], cant camiones x(t) y stock final s(t)
> c = matrix(rep(1:0,each = T))
> vlb = matrix(rep(0,2*T))
> vub = matrix(c(x_max*rep(1,T),cap_tqe*rep(1,T)))
>
> #Rest de balance: s(t)=s(t-1)+K*x(t)-dda(t)
> A1=matrix(0,nrow=T,ncol=2*T)
> A1[,1:T]=diag(K,dim(A1[,1:T]))
> T2 = T+1
> T3 = 2*T
> library(pracma)
> A1[,T2:T3]=-diag(1,T,T)+Diag(rep(1,T-1),-1)
> b1 = dda
> b1[1] = b1[1]-s0
>
> #Tipo de variables y restricciones
> varint=rep("I",T)
> varcont=rep("C",T)
> ctype=rep("S",T)
>
> #vartype=[varint varcont];
> vartype=c(varcont, varcont)
> #-----
> sense=1;
>
> Rglpk_solve_LP(obj = c, mat =A1, dir = rep("<=",nrow(A1)),rhs = b1,max
> = F, types = vartype)
>
> ######################################################
>
> ¿Cuál será el problema?
>
> Gracias,
>
> Sebastián.
>
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