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How can this linear problem be infeasible?
The Next CEO of Stack OverflowPrimal and dual solution to linear programmingQuestions about weak duality theoremHow the dual LP solves the primal LPConstruct a linear programming problem for which both the primal and the dual problem has no feasible solutionHow do you determine which Simplex method you need to use when solving a linear programming problem?What is the significance of an infeasible solution to a network flow problem?Feasibility and boundedness of non-linear programmingLinear programming optimal solutionMultiple optimal solutions for a linear programming problemLinear programming: how to determine if the basic solution is feasible or infeasible?
$begingroup$
I have the following linear problem which is primal:
I have converted this problem to its dual and tried to solve using the lpSolveAPI in R programming language. However, the solve
function returns a 2, which means the problem is infeasible. It is imposible, because the primal problem has an optimal solution, so the dual must have an optimal solution too.
Here is the code that I have written to solve the dual problem:
#Vector de costes
c2<-c(11,10)
#Matriz de restricciones
A2<-matrix(nrow=3,ncol=2)
A2[1,] <- c(1,2)
A2[2,] <- c(2,3)
A2[3,] <- c(3,2)
#Lados derechos
b2<-c(-2,-3,-4)
#Cotas inferiores y superiores
cotasinf<-c(Inf,Inf)
cotassup<-c(0,0)
#Tipos de restricciones
tipores2<-c("<=","<=","<=")
#Se crea problema de programacion lineal
#3 restricciones y 2 variables
ejercicio.2<-make.lp(3,2)
#######Se definen los parametros##########
#Matriz de restricciones
for (i in 1:nrow(A2)) set.row(ejercicio.2,i,A2[i,])
#Funcion objetivo
set.objfn(ejercicio.2,c2)
#Lados derechos
set.rhs(ejercicio.2,b2)
#Tipos de restricciones
set.constr.type(ejercicio.2,tipores2)
#Cotas inferiores y superiores de cada variable
set.bounds(ejercicio.2, lower = cotasinf)
set.bounds(ejercicio.2, upper = cotassup)
#Nombres de variables y restricciones
restrics2 <- c("restriccion.1", "restriccion.2", "restriccion.3")
vars2 <- c("x1", "x2", "x3", "x4")
dimnames(ejercicio.2) <- list(restrics2, vars2)
#Cambiando a problema de minimizacion
lp.control(ejercicio.2,sense="max")
#Se muestra el problema ya cargado
ejercicio.2
#Se resuelve el problema
solve(ejercicio.2)
#Se obtiene el valor de la funcion objetivo en el optimo
get.objective(ejercicio.2)
linear-programming
$endgroup$
add a comment |
$begingroup$
I have the following linear problem which is primal:
I have converted this problem to its dual and tried to solve using the lpSolveAPI in R programming language. However, the solve
function returns a 2, which means the problem is infeasible. It is imposible, because the primal problem has an optimal solution, so the dual must have an optimal solution too.
Here is the code that I have written to solve the dual problem:
#Vector de costes
c2<-c(11,10)
#Matriz de restricciones
A2<-matrix(nrow=3,ncol=2)
A2[1,] <- c(1,2)
A2[2,] <- c(2,3)
A2[3,] <- c(3,2)
#Lados derechos
b2<-c(-2,-3,-4)
#Cotas inferiores y superiores
cotasinf<-c(Inf,Inf)
cotassup<-c(0,0)
#Tipos de restricciones
tipores2<-c("<=","<=","<=")
#Se crea problema de programacion lineal
#3 restricciones y 2 variables
ejercicio.2<-make.lp(3,2)
#######Se definen los parametros##########
#Matriz de restricciones
for (i in 1:nrow(A2)) set.row(ejercicio.2,i,A2[i,])
#Funcion objetivo
set.objfn(ejercicio.2,c2)
#Lados derechos
set.rhs(ejercicio.2,b2)
#Tipos de restricciones
set.constr.type(ejercicio.2,tipores2)
#Cotas inferiores y superiores de cada variable
set.bounds(ejercicio.2, lower = cotasinf)
set.bounds(ejercicio.2, upper = cotassup)
#Nombres de variables y restricciones
restrics2 <- c("restriccion.1", "restriccion.2", "restriccion.3")
vars2 <- c("x1", "x2", "x3", "x4")
dimnames(ejercicio.2) <- list(restrics2, vars2)
#Cambiando a problema de minimizacion
lp.control(ejercicio.2,sense="max")
#Se muestra el problema ya cargado
ejercicio.2
#Se resuelve el problema
solve(ejercicio.2)
#Se obtiene el valor de la funcion objetivo en el optimo
get.objective(ejercicio.2)
linear-programming
$endgroup$
$begingroup$
Perhaps your dual is wrong. You have more chance of getting an answer if you write it as equations and not in R.
$endgroup$
– Michal Adamaszek
Mar 18 at 11:55
$begingroup$
Write the primal program and try $textget.dual.solution(ejercicio.2)$
$endgroup$
– callculus
Mar 18 at 15:36
add a comment |
$begingroup$
I have the following linear problem which is primal:
I have converted this problem to its dual and tried to solve using the lpSolveAPI in R programming language. However, the solve
function returns a 2, which means the problem is infeasible. It is imposible, because the primal problem has an optimal solution, so the dual must have an optimal solution too.
Here is the code that I have written to solve the dual problem:
#Vector de costes
c2<-c(11,10)
#Matriz de restricciones
A2<-matrix(nrow=3,ncol=2)
A2[1,] <- c(1,2)
A2[2,] <- c(2,3)
A2[3,] <- c(3,2)
#Lados derechos
b2<-c(-2,-3,-4)
#Cotas inferiores y superiores
cotasinf<-c(Inf,Inf)
cotassup<-c(0,0)
#Tipos de restricciones
tipores2<-c("<=","<=","<=")
#Se crea problema de programacion lineal
#3 restricciones y 2 variables
ejercicio.2<-make.lp(3,2)
#######Se definen los parametros##########
#Matriz de restricciones
for (i in 1:nrow(A2)) set.row(ejercicio.2,i,A2[i,])
#Funcion objetivo
set.objfn(ejercicio.2,c2)
#Lados derechos
set.rhs(ejercicio.2,b2)
#Tipos de restricciones
set.constr.type(ejercicio.2,tipores2)
#Cotas inferiores y superiores de cada variable
set.bounds(ejercicio.2, lower = cotasinf)
set.bounds(ejercicio.2, upper = cotassup)
#Nombres de variables y restricciones
restrics2 <- c("restriccion.1", "restriccion.2", "restriccion.3")
vars2 <- c("x1", "x2", "x3", "x4")
dimnames(ejercicio.2) <- list(restrics2, vars2)
#Cambiando a problema de minimizacion
lp.control(ejercicio.2,sense="max")
#Se muestra el problema ya cargado
ejercicio.2
#Se resuelve el problema
solve(ejercicio.2)
#Se obtiene el valor de la funcion objetivo en el optimo
get.objective(ejercicio.2)
linear-programming
$endgroup$
I have the following linear problem which is primal:
I have converted this problem to its dual and tried to solve using the lpSolveAPI in R programming language. However, the solve
function returns a 2, which means the problem is infeasible. It is imposible, because the primal problem has an optimal solution, so the dual must have an optimal solution too.
Here is the code that I have written to solve the dual problem:
#Vector de costes
c2<-c(11,10)
#Matriz de restricciones
A2<-matrix(nrow=3,ncol=2)
A2[1,] <- c(1,2)
A2[2,] <- c(2,3)
A2[3,] <- c(3,2)
#Lados derechos
b2<-c(-2,-3,-4)
#Cotas inferiores y superiores
cotasinf<-c(Inf,Inf)
cotassup<-c(0,0)
#Tipos de restricciones
tipores2<-c("<=","<=","<=")
#Se crea problema de programacion lineal
#3 restricciones y 2 variables
ejercicio.2<-make.lp(3,2)
#######Se definen los parametros##########
#Matriz de restricciones
for (i in 1:nrow(A2)) set.row(ejercicio.2,i,A2[i,])
#Funcion objetivo
set.objfn(ejercicio.2,c2)
#Lados derechos
set.rhs(ejercicio.2,b2)
#Tipos de restricciones
set.constr.type(ejercicio.2,tipores2)
#Cotas inferiores y superiores de cada variable
set.bounds(ejercicio.2, lower = cotasinf)
set.bounds(ejercicio.2, upper = cotassup)
#Nombres de variables y restricciones
restrics2 <- c("restriccion.1", "restriccion.2", "restriccion.3")
vars2 <- c("x1", "x2", "x3", "x4")
dimnames(ejercicio.2) <- list(restrics2, vars2)
#Cambiando a problema de minimizacion
lp.control(ejercicio.2,sense="max")
#Se muestra el problema ya cargado
ejercicio.2
#Se resuelve el problema
solve(ejercicio.2)
#Se obtiene el valor de la funcion objetivo en el optimo
get.objective(ejercicio.2)
linear-programming
linear-programming
edited Mar 18 at 11:33
YuiTo Cheng
2,1512937
2,1512937
asked Mar 18 at 10:53
Sergio ReySergio Rey
1
1
$begingroup$
Perhaps your dual is wrong. You have more chance of getting an answer if you write it as equations and not in R.
$endgroup$
– Michal Adamaszek
Mar 18 at 11:55
$begingroup$
Write the primal program and try $textget.dual.solution(ejercicio.2)$
$endgroup$
– callculus
Mar 18 at 15:36
add a comment |
$begingroup$
Perhaps your dual is wrong. You have more chance of getting an answer if you write it as equations and not in R.
$endgroup$
– Michal Adamaszek
Mar 18 at 11:55
$begingroup$
Write the primal program and try $textget.dual.solution(ejercicio.2)$
$endgroup$
– callculus
Mar 18 at 15:36
$begingroup$
Perhaps your dual is wrong. You have more chance of getting an answer if you write it as equations and not in R.
$endgroup$
– Michal Adamaszek
Mar 18 at 11:55
$begingroup$
Perhaps your dual is wrong. You have more chance of getting an answer if you write it as equations and not in R.
$endgroup$
– Michal Adamaszek
Mar 18 at 11:55
$begingroup$
Write the primal program and try $textget.dual.solution(ejercicio.2)$
$endgroup$
– callculus
Mar 18 at 15:36
$begingroup$
Write the primal program and try $textget.dual.solution(ejercicio.2)$
$endgroup$
– callculus
Mar 18 at 15:36
add a comment |
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$begingroup$
Perhaps your dual is wrong. You have more chance of getting an answer if you write it as equations and not in R.
$endgroup$
– Michal Adamaszek
Mar 18 at 11:55
$begingroup$
Write the primal program and try $textget.dual.solution(ejercicio.2)$
$endgroup$
– callculus
Mar 18 at 15:36