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|
R version 4.1.2 (2021-11-01) -- "Bird Hippie"
Copyright (C) 2021 The R Foundation for Statistical Computing
Platform: x86_64-pc-linux-gnu (64-bit)
R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.
R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.
Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.
> library( linprog )
Loading required package: lpSolve
>
> ## Example 1
> ## Steinhauser, Langbehn and Peters (1992)
> cvec <- c(1800, 600, 600) # gross margins
> names(cvec) <- c("Cows","Bulls","Pigs")
> bvec <- c(40, 90, 2500) # endowment
> names(bvec) <- c("Land","Stable","Labor")
> Amat <- rbind( c( 0.7, 0.35, 0 ),
+ c( 1.5, 1, 3 ),
+ c( 50, 12.5, 20 ) )
> result1a <- solveLP( cvec, bvec, Amat, TRUE, verbose = 1 )
> print( result1a )
Results of Linear Programming / Linear Optimization
Objective function (Maximum): 93600
Iterations in phase 1: 0
Iterations in phase 2: 2
Solution
opt
Cows 44
Bulls 24
Pigs 0
Basic Variables
opt
Cows 44.0
Bulls 24.0
S Land 0.8
Constraints
actual dir bvec free dual dual.reg
Land 39.2 <= 40 0.8 0.0 0.8
Stable 90.0 <= 90 0.0 240.0 15.0
Labor 2500.0 <= 2500 0.0 28.8 1375.0
All Variables (including slack variables)
opt cvec min.c max.c marg marg.reg
Cows 44.0 1800 900 2400.000 NA NA
Bulls 24.0 600 450 1200.000 NA NA
Pigs 0.0 600 -Inf 1296.000 -696.0 6.25
S Land 0.8 0 NA 731.092 0.0 NA
S Stable 0.0 0 -Inf 240.000 -240.0 15.00
S Labor 0.0 0 -Inf 28.800 -28.8 1375.00
> # print summary results
> summary( result1a )
Results of Linear Programming / Linear Optimization
Objective function (Maximum): 93600
Solution
opt
Cows 44
Bulls 24
Pigs 0
> # print all elements of the returned object
> print.default( result1a )
$status
[1] 0
$opt
[1] 93600
$iter1
[1] 0
$iter2
[1] 2
$allvar
opt cvec min.c max.c marg marg.reg
Cows 44.0 1800 900 2400.0000 NA NA
Bulls 24.0 600 450 1200.0000 NA NA
Pigs 0.0 600 -Inf 1296.0000 -696.0 6.25
S Land 0.8 0 NA 731.0924 0.0 NA
S Stable 0.0 0 -Inf 240.0000 -240.0 15.00
S Labor 0.0 0 -Inf 28.8000 -28.8 1375.00
$basvar
opt
Cows 44.0
Bulls 24.0
S Land 0.8
$solution
Cows Bulls Pigs
44 24 0
$con
actual dir bvec free dual dual.reg
Land 39.2 <= 40 0.8 0.0 0.8
Stable 90.0 <= 90 0.0 240.0 15.0
Labor 2500.0 <= 2500 0.0 28.8 1375.0
$Tab
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0 0 -0.952 1 -0.28 -0.0056 0.8
Bulls 0 1 3.840 0 1.60 -0.0480 24.0
Cows 1 0 -0.560 0 -0.40 0.0320 44.0
Z-C 0 0 696.000 0 240.00 28.8000 93600.0
$maximum
[1] TRUE
$lpSolve
[1] FALSE
$solve.dual
[1] FALSE
$maxiter
[1] 1000
attr(,"class")
[1] "solveLP"
> # also estimate the dual problem
> result1aD <- solveLP( cvec, bvec, Amat, TRUE, verbose = 1, solve.dual = TRUE )
> result1aD$con
actual dir bvec free dual dual.reg dual.p
Land 39.2 <= 40 0.8 0.0 0.8 0.0
Stable 90.0 <= 90 0.0 240.0 15.0 240.0
Labor 2500.0 <= 2500 0.0 28.8 1375.0 28.8
> all.equal( result1a[-c(8,12)], result1aD[-c(8,10,13)] )
[1] TRUE
>
> # estimation with verbose = TRUE
> result1b <- solveLP( cvec, bvec, Amat, TRUE, verbose = 4 )
[1] "initial Tableau"
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0.7 0.35 0 1 0 0 40
Stable 1.5 1.00 3 0 1 0 90
Labor 50.0 12.50 20 0 0 1 2500
Z-C -1800.0 -600.00 -600 0 0 0 0
Pivot Column: 1 ( Cows )
Pivot Row: 3 ( Labor )
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0 0.175 -0.28 1 0 -0.014 5
Stable 0 0.625 2.40 0 1 -0.030 15
Cows 1 0.250 0.40 0 0 0.020 50
Z-C 0 -150.000 120.00 0 0 36.000 90000
Pivot Column: 2 ( Bulls )
Pivot Row: 2 ( Stable )
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0 0 -0.952 1 -0.28 -0.0056 0.8
Bulls 0 1 3.840 0 1.60 -0.0480 24.0
Cows 1 0 -0.560 0 -0.40 0.0320 44.0
Z-C 0 0 696.000 0 240.00 28.8000 93600.0
> all.equal( result1a, result1b )
[1] TRUE
> # also estimate the dual problem
> result1bD <- solveLP( cvec, bvec, Amat, TRUE, verbose = 4, solve.dual = TRUE )
[1] "initial Tableau"
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0.7 0.35 0 1 0 0 40
Stable 1.5 1.00 3 0 1 0 90
Labor 50.0 12.50 20 0 0 1 2500
Z-C -1800.0 -600.00 -600 0 0 0 0
Pivot Column: 1 ( Cows )
Pivot Row: 3 ( Labor )
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0 0.175 -0.28 1 0 -0.014 5
Stable 0 0.625 2.40 0 1 -0.030 15
Cows 1 0.250 0.40 0 0 0.020 50
Z-C 0 -150.000 120.00 0 0 36.000 90000
Pivot Column: 2 ( Bulls )
Pivot Row: 2 ( Stable )
Cows Bulls Pigs S Land S Stable S Labor P0
Land 0 0 -0.952 1 -0.28 -0.0056 0.8
Bulls 0 1 3.840 0 1.60 -0.0480 24.0
Cows 1 0 -0.560 0 -0.40 0.0320 44.0
Z-C 0 0 696.000 0 240.00 28.8000 93600.0
[1] "initial Tableau"
Land Stable Labor S Cows S Bulls S Pigs P0
Cows -0.70 -1.5 -50.0 1 0 0 -1800
Bulls -0.35 -1.0 -12.5 0 1 0 -600
Pigs 0.00 -3.0 -20.0 0 0 1 -600
Z-C 40.00 90.0 2500.0 0 0 0 0
[1] "initial Tableau for Phase 1"
Land Stable Labor S Cows S Bulls S Pigs M Cows M Bulls M Pigs P0
M Cows 0.70 1.5 50.0 -1 0 0 1 0 0 1800
M Bulls 0.35 1.0 12.5 0 -1 0 0 1 0 600
M Pigs 0.00 3.0 20.0 0 0 -1 0 0 1 600
Z-C 40.00 90.0 2500.0 0 0 0 0 0 0 0
M Z-C -1.05 -5.5 -82.5 1 1 1 0 0 0 -3000
Pivot Column: 3 ( Labor )
Pivot Row: 3 ( M Pigs )
Land Stable Labor S Cows S Bulls S Pigs M Cows M Bulls M Pigs
M Cows 0.70 -6.000 0 -1 0 2.500 1 0 -2.500
M Bulls 0.35 -0.875 0 0 -1 0.625 0 1 -0.625
Labor 0.00 0.150 1 0 0 -0.050 0 0 0.050
Z-C 40.00 -285.000 0 0 0 125.000 0 0 -125.000
M Z-C -1.05 6.875 0 1 1 -3.125 0 0 4.125
P0
M Cows 300
M Bulls 225
Labor 30
Z-C -75000
M Z-C -525
Pivot Column: 1 ( Land )
Pivot Row: 1 ( M Cows )
Land Stable Labor S Cows S Bulls S Pigs M Cows M Bulls
Land 1 -8.571429 0 -1.428571 0 3.571429 1.428571 0
M Bulls 0 2.125000 0 0.500000 -1 -0.625000 -0.500000 1
Labor 0 0.150000 1 0.000000 0 -0.050000 0.000000 0
Z-C 0 57.857143 0 57.142857 0 -17.857143 -57.142857 0
M Z-C 0 -2.125000 0 -0.500000 1 0.625000 1.500000 0
M Pigs P0
Land -3.571429 428.5714
M Bulls 0.625000 75.0000
Labor 0.050000 30.0000
Z-C 17.857143 -92142.8571
M Z-C 0.375000 -75.0000
Pivot Column: 4 ( S Cows )
Pivot Row: 2 ( M Bulls )
Land Stable Labor S Cows S Bulls S Pigs M Cows
Land 1 -2.500000e+00 0 0 -2.857143e+00 1.785714e+00 0
S Cows 0 4.250000e+00 0 1 -2.000000e+00 -1.250000e+00 -1
Labor 0 1.500000e-01 1 0 0.000000e+00 -5.000000e-02 0
Z-C 0 -1.850000e+02 0 0 1.142857e+02 5.357143e+01 0
M Z-C 0 4.440892e-16 0 0 4.440892e-16 -2.220446e-16 1
M Bulls M Pigs P0
Land 2.857143 -1.785714 6.428571e+02
S Cows 2.000000 1.250000 1.500000e+02
Labor 0.000000 0.050000 3.000000e+01
Z-C -114.285714 -53.571429 -1.007143e+05
M Z-C 1.000000 1.000000 -8.526513e-14
[1] "New starting Tableau for Phase II"
Land Stable Labor S Cows S Bulls S Pigs
Land 1 -2.50 0 0 -2.857143 1.785714 642.8571
S Cows 0 4.25 0 1 -2.000000 -1.250000 150.0000
Labor 0 0.15 1 0 0.000000 -0.050000 30.0000
Z-C 0 -185.00 0 0 114.285714 53.571429 -100714.2857
Pivot Column: 2 ( Stable )
Pivot Row: 2 ( S Cows )
Land Stable Labor S Cows S Bulls S Pigs
Land 1 0 0 0.58823529 -4.03361345 1.050420168 731.09244
Stable 0 1 0 0.23529412 -0.47058824 -0.294117647 35.29412
Labor 0 0 1 -0.03529412 0.07058824 -0.005882353 24.70588
Z-C 0 0 0 43.52941176 27.22689076 -0.840336134 -94184.87395
Pivot Column: 6 ( S Pigs )
Pivot Row: 1 ( Land )
Land Stable Labor S Cows S Bulls S Pigs
S Pigs 0.9520 0 0 0.560 -3.840 1 696.0
Stable 0.2800 1 0 0.400 -1.600 0 240.0
Labor 0.0056 0 1 -0.032 0.048 0 28.8
Z-C 0.8000 0 0 44.000 24.000 0 -93600.0
> all.equal( result1aD, result1bD )
[1] TRUE
>
> # estimation with lpSolve
> result1c <- solveLP( cvec, bvec, Amat, TRUE, lpSolve = TRUE, verbose = 4 )
> print( result1c )
Results of Linear Programming / Linear Optimization
(using lpSolve)
Objective function (Maximum): 93600
Solution
opt
Cows 44
Bulls 24
Pigs 0
Constraints
actual dir bvec free
Land 39.2 <= 40 0.8
Stable 90.0 <= 90 0.0
Labor 2500.0 <= 2500 0.0
> # print summary results
> summary( result1c )
Results of Linear Programming / Linear Optimization
Objective function (Maximum): 93600
Solution
opt
Cows 44
Bulls 24
Pigs 0
> # print all elements of the returned object
> print.default( result1c )
$status
[1] 0
$lpStatus
[1] 0
$solution
Cows Bulls Pigs
44 24 0
$opt
[1] 93600
$con
actual dir bvec free
Land 39.2 <= 40 0.8
Stable 90.0 <= 90 0.0
Labor 2500.0 <= 2500 0.0
$maximum
[1] TRUE
$lpSolve
[1] TRUE
$solve.dual
[1] FALSE
$maxiter
[1] 1000
attr(,"class")
[1] "solveLP"
> # also estimate the dual problem
> result1cD <- solveLP( cvec, bvec, Amat, TRUE, lpSolve = TRUE, solve.dual = TRUE )
> result1cD$con
actual dir bvec free dual
Land 39.2 <= 40 0.8 0.0
Stable 90.0 <= 90 0.0 240.0
Labor 2500.0 <= 2500 0.0 28.8
> all.equal( result1c[-c(5,8)], result1cD[-c(5,6,9)] )
[1] TRUE
>
> # using argument const.dir
> const.dir <- c( ">=", ">=", ">=" )
> result1d <- solveLP( cvec, -bvec, -Amat, maximum = TRUE, verbose = 1,
+ const.dir = const.dir )
> print( result1d )
Results of Linear Programming / Linear Optimization
Objective function (Maximum): 93600
Iterations in phase 1: 0
Iterations in phase 2: 2
Solution
opt
Cows 44
Bulls 24
Pigs 0
Basic Variables
opt
Cows 44.0
Bulls 24.0
S Land 0.8
Constraints
actual dir bvec free dual dual.reg
Land -39.2 >= -40 0.8 0.0 0.8
Stable -90.0 >= -90 0.0 240.0 15.0
Labor -2500.0 >= -2500 0.0 28.8 1375.0
All Variables (including slack variables)
opt cvec min.c max.c marg marg.reg
Cows 44.0 1800 900 2400.000 NA NA
Bulls 24.0 600 450 1200.000 NA NA
Pigs 0.0 600 -Inf 1296.000 -696.0 6.25
S Land 0.8 0 NA 731.092 0.0 NA
S Stable 0.0 0 -Inf 240.000 -240.0 15.00
S Labor 0.0 0 -Inf 28.800 -28.8 1375.00
> all.equal( result1a[-8], result1d[-8] )
[1] TRUE
> # also estimate the dual problem
> result1dD <- solveLP( cvec, -bvec, -Amat, TRUE, verbose = 1,
+ const.dir = const.dir, solve.dual = TRUE )
> result1dD$con
actual dir bvec free dual dual.reg dual.p
Land -39.2 >= -40 0.8 0.0 0.8 0.0
Stable -90.0 >= -90 0.0 240.0 15.0 240.0
Labor -2500.0 >= -2500 0.0 28.8 1375.0 28.8
> all.equal( result1aD[-8], result1dD[-8] )
[1] TRUE
>
> # using argument const.dir and lpSolve
> result1e <-solveLP( cvec, -bvec, -Amat, maximum = TRUE, verbose = 1,
+ const.dir = const.dir, lpSolve = TRUE )
> print( result1e )
Results of Linear Programming / Linear Optimization
(using lpSolve)
Objective function (Maximum): 93600
Solution
opt
Cows 44
Bulls 24
Pigs 0
Constraints
actual dir bvec free
Land -39.2 >= -40 0.8
Stable -90.0 >= -90 0.0
Labor -2500.0 >= -2500 0.0
> all.equal( result1c[-5], result1e[-5] )
[1] TRUE
> # also estimate the dual problem
> result1eD <- solveLP( cvec, -bvec, -Amat, TRUE, verbose = 1,
+ const.dir = const.dir, lpSolve = TRUE, solve.dual = TRUE )
> result1eD$con
actual dir bvec free dual
Land -39.2 >= -40 0.8 0.0
Stable -90.0 >= -90 0.0 240.0
Labor -2500.0 >= -2500 0.0 28.8
> all.equal( result1cD[-5], result1eD[-5] )
[1] TRUE
>
>
> ## Example 2
> ## example 1.1.3 of Witte, Deppe and Born (1975)
> cvec <- c(2.5, 2 ) # prices of feed
> names(cvec) <- c("Feed1","Feed2")
> bvec <- c( -10, -1.5, 12)
> names(bvec) <- c("Protein","Fat","Fibre")
> Amat <- rbind( c(-1.6,-2.4 ),
+ c(-0.5,-0.2 ),
+ c( 2.0, 2.0 ) )
> result2a <- solveLP( cvec, bvec, Amat, verbose = 1 )
> print( result2a )
Results of Linear Programming / Linear Optimization
Objective function (Minimum): 10.4545
Iterations in phase 1: 2
Iterations in phase 2: 0
Solution
opt
Feed1 1.81818
Feed2 2.95455
Basic Variables
opt
Feed1 1.81818
Feed2 2.95455
S Fibre 2.45455
Constraints
actual dir bvec free dual dual.reg
Protein -10.00000 <= -10.0 0.00000 0.568182 3.60000
Fat -1.50000 <= -1.5 0.00000 3.181818 1.35000
Fibre 9.54545 <= 12.0 2.45455 0.000000 2.45455
All Variables (including slack variables)
opt cvec min.c max.c marg marg.reg
Feed1 1.81818 2.5 -3.666667 5.000000 NA NA
Feed2 2.95455 2.0 -3.000000 3.750000 NA NA
S Protein 0.00000 0.0 -0.568182 Inf 0.568182 3.60
S Fat 0.00000 0.0 -3.181818 Inf 3.181818 1.35
S Fibre 2.45455 0.0 NA 0.833333 0.000000 NA
> # print summary results
> summary( result2a )
Results of Linear Programming / Linear Optimization
Objective function (Minimum): 10.45455
Solution
opt
Feed1 1.818182
Feed2 2.954545
> # print all elements of the returned object
> print.default( result2a )
$status
[1] 0
$opt
[1] 10.45455
$iter1
[1] 2
$iter2
[1] 0
$allvar
opt cvec min.c max.c marg marg.reg
Feed1 1.818182 2.5 -3.6666667 5.0000000 NA NA
Feed2 2.954545 2.0 -3.0000000 3.7500000 NA NA
S Protein 0.000000 0.0 -0.5681818 Inf 0.5681818 3.60
S Fat 0.000000 0.0 -3.1818182 Inf 3.1818182 1.35
S Fibre 2.454545 0.0 NA 0.8333333 0.0000000 NA
$basvar
opt
Feed1 1.818182
Feed2 2.954545
S Fibre 2.454545
$solution
Feed1 Feed2
1.818182 2.954545
$con
actual dir bvec free dual dual.reg
Protein -10.000000 <= -10.0 0.000000 0.5681818 3.600000
Fat -1.500000 <= -1.5 0.000000 3.1818182 1.350000
Fibre 9.545455 <= 12.0 2.454545 0.0000000 2.454545
$Tab
Feed1 Feed2 S Protein S Fat S Fibre
Feed2 0 1 -0.5681818 1.818182 0 2.954545
Feed1 1 0 0.2272727 -2.727273 0 1.818182
Fibre 0 0 0.6818182 1.818182 1 2.454545
Z-C 0 0 0.5681818 3.181818 0 -10.454545
$maximum
[1] FALSE
$lpSolve
[1] FALSE
$solve.dual
[1] FALSE
$maxiter
[1] 1000
attr(,"class")
[1] "solveLP"
> # also estimate the dual problem
> result2aD <- solveLP( cvec, bvec, Amat, verbose = 1, solve.dual = TRUE )
> result2aD$con
actual dir bvec free dual dual.reg dual.p
Protein -10.000000 <= -10.0 0.000000 0.5681818 3.600000 0.5681818
Fat -1.500000 <= -1.5 0.000000 3.1818182 1.350000 3.1818182
Fibre 9.545455 <= 12.0 2.454545 0.0000000 2.454545 0.0000000
> all.equal( result2a[-c(8,12)], result2aD[-c(8,10,13)] )
[1] TRUE
>
> # estimation with verbose = TRUE
> result2b <- solveLP( cvec, bvec, Amat, verbose = 4 )
[1] "initial Tableau"
Feed1 Feed2 S Protein S Fat S Fibre P0
Protein -1.6 -2.4 1 0 0 -10.0
Fat -0.5 -0.2 0 1 0 -1.5
Fibre 2.0 2.0 0 0 1 12.0
Z-C 2.5 2.0 0 0 0 0.0
[1] "initial Tableau for Phase 1"
Feed1 Feed2 S Protein S Fat S Fibre M Protein M Fat P0
M Protein 1.6 2.4 -1 0 0 1 0 10.0
M Fat 0.5 0.2 0 -1 0 0 1 1.5
Fibre 2.0 2.0 0 0 1 0 0 12.0
Z-C 2.5 2.0 0 0 0 0 0 0.0
M Z-C -2.1 -2.6 1 1 0 0 0 -11.5
Pivot Column: 2 ( Feed2 )
Pivot Row: 1 ( M Protein )
Feed1 Feed2 S Protein S Fat S Fibre M Protein M Fat P0
Feed2 0.6666667 1 -0.41666667 0 0 0.41666667 0 4.1666667
M Fat 0.3666667 0 0.08333333 -1 0 -0.08333333 1 0.6666667
Fibre 0.6666667 0 0.83333333 0 1 -0.83333333 0 3.6666667
Z-C 1.1666667 0 0.83333333 0 0 -0.83333333 0 -8.3333333
M Z-C -0.3666667 0 -0.08333333 1 0 1.08333333 0 -0.6666667
Pivot Column: 1 ( Feed1 )
Pivot Row: 2 ( M Fat )
Feed1 Feed2 S Protein S Fat S Fibre M Protein M Fat
Feed2 0 1 -5.681818e-01 1.818182e+00 0 0.5681818 -1.818182
Feed1 1 0 2.272727e-01 -2.727273e+00 0 -0.2272727 2.727273
Fibre 0 0 6.818182e-01 1.818182e+00 1 -0.6818182 -1.818182
Z-C 0 0 5.681818e-01 3.181818e+00 0 -0.5681818 -3.181818
M Z-C 0 0 -1.804112e-16 4.440892e-16 0 1.0000000 1.000000
P0
Feed2 2.954545e+00
Feed1 1.818182e+00
Fibre 2.454545e+00
Z-C -1.045455e+01
M Z-C 1.110223e-16
[1] "New starting Tableau for Phase II"
Feed1 Feed2 S Protein S Fat S Fibre
Feed2 0 1 -0.5681818 1.818182 0 2.954545
Feed1 1 0 0.2272727 -2.727273 0 1.818182
Fibre 0 0 0.6818182 1.818182 1 2.454545
Z-C 0 0 0.5681818 3.181818 0 -10.454545
> all.equal( result1a, result1b )
[1] TRUE
> # also estimate the dual problem
> result2bD <- solveLP( cvec, bvec, Amat, verbose = 4, solve.dual = TRUE )
[1] "initial Tableau"
Feed1 Feed2 S Protein S Fat S Fibre P0
Protein -1.6 -2.4 1 0 0 -10.0
Fat -0.5 -0.2 0 1 0 -1.5
Fibre 2.0 2.0 0 0 1 12.0
Z-C 2.5 2.0 0 0 0 0.0
[1] "initial Tableau for Phase 1"
Feed1 Feed2 S Protein S Fat S Fibre M Protein M Fat P0
M Protein 1.6 2.4 -1 0 0 1 0 10.0
M Fat 0.5 0.2 0 -1 0 0 1 1.5
Fibre 2.0 2.0 0 0 1 0 0 12.0
Z-C 2.5 2.0 0 0 0 0 0 0.0
M Z-C -2.1 -2.6 1 1 0 0 0 -11.5
Pivot Column: 2 ( Feed2 )
Pivot Row: 1 ( M Protein )
Feed1 Feed2 S Protein S Fat S Fibre M Protein M Fat P0
Feed2 0.6666667 1 -0.41666667 0 0 0.41666667 0 4.1666667
M Fat 0.3666667 0 0.08333333 -1 0 -0.08333333 1 0.6666667
Fibre 0.6666667 0 0.83333333 0 1 -0.83333333 0 3.6666667
Z-C 1.1666667 0 0.83333333 0 0 -0.83333333 0 -8.3333333
M Z-C -0.3666667 0 -0.08333333 1 0 1.08333333 0 -0.6666667
Pivot Column: 1 ( Feed1 )
Pivot Row: 2 ( M Fat )
Feed1 Feed2 S Protein S Fat S Fibre M Protein M Fat
Feed2 0 1 -5.681818e-01 1.818182e+00 0 0.5681818 -1.818182
Feed1 1 0 2.272727e-01 -2.727273e+00 0 -0.2272727 2.727273
Fibre 0 0 6.818182e-01 1.818182e+00 1 -0.6818182 -1.818182
Z-C 0 0 5.681818e-01 3.181818e+00 0 -0.5681818 -3.181818
M Z-C 0 0 -1.804112e-16 4.440892e-16 0 1.0000000 1.000000
P0
Feed2 2.954545e+00
Feed1 1.818182e+00
Fibre 2.454545e+00
Z-C -1.045455e+01
M Z-C 1.110223e-16
[1] "New starting Tableau for Phase II"
Feed1 Feed2 S Protein S Fat S Fibre
Feed2 0 1 -0.5681818 1.818182 0 2.954545
Feed1 1 0 0.2272727 -2.727273 0 1.818182
Fibre 0 0 0.6818182 1.818182 1 2.454545
Z-C 0 0 0.5681818 3.181818 0 -10.454545
[1] "initial Tableau"
Protein Fat Fibre S Feed1 S Feed2 P0
Feed1 1.6 0.5 -2 1 0 2.5
Feed2 2.4 0.2 -2 0 1 2.0
Z-C -10.0 -1.5 12 0 0 0.0
Pivot Column: 1 ( Protein )
Pivot Row: 2 ( Feed2 )
Protein Fat Fibre S Feed1 S Feed2 P0
Feed1 0 0.36666667 -0.6666667 1 -0.6666667 1.1666667
Protein 1 0.08333333 -0.8333333 0 0.4166667 0.8333333
Z-C 0 -0.66666667 3.6666667 0 4.1666667 8.3333333
Pivot Column: 2 ( Fat )
Pivot Row: 1 ( Feed1 )
Protein Fat Fibre S Feed1 S Feed2 P0
Fat 0 1 -1.8181818 2.7272727 -1.8181818 3.1818182
Protein 1 0 -0.6818182 -0.2272727 0.5681818 0.5681818
Z-C 0 0 2.4545455 1.8181818 2.9545455 10.4545455
> all.equal( result2aD, result2bD )
[1] TRUE
>
> # estimation with lpSolve
> result2c <- solveLP( cvec, bvec, Amat, lpSolve = TRUE, verbose = 4 )
> print( result2c )
Results of Linear Programming / Linear Optimization
(using lpSolve)
Objective function (Minimum): 10.4545
Solution
opt
Feed1 1.81818
Feed2 2.95455
Constraints
actual dir bvec free
Protein -10.00000 <= -10.0 0.00000
Fat -1.50000 <= -1.5 0.00000
Fibre 9.54545 <= 12.0 2.45455
> # print summary results
> summary( result2c )
Results of Linear Programming / Linear Optimization
Objective function (Minimum): 10.45455
Solution
opt
Feed1 1.818182
Feed2 2.954545
> # print all elements of the returned object
> print.default( result2c )
$status
[1] 0
$lpStatus
[1] 0
$solution
Feed1 Feed2
1.818182 2.954545
$opt
[1] 10.45455
$con
actual dir bvec free
Protein -10.000000 <= -10.0 0.000000
Fat -1.500000 <= -1.5 0.000000
Fibre 9.545455 <= 12.0 2.454545
$maximum
[1] FALSE
$lpSolve
[1] TRUE
$solve.dual
[1] FALSE
$maxiter
[1] 1000
attr(,"class")
[1] "solveLP"
> # also estimate the dual problem
> result2cD <- solveLP( cvec, bvec, Amat, lpSolve = TRUE, verbose = 4,
+ solve.dual = TRUE )
> result2cD$con
actual dir bvec free dual
Protein -10.000000 <= -10.0 0.000000 0.5681818
Fat -1.500000 <= -1.5 0.000000 3.1818182
Fibre 9.545455 <= 12.0 2.454545 0.0000000
> all.equal( result2c[-c(5,8)], result2cD[-c(5,6,9)] )
[1] TRUE
>
> # using argument const.dir
> const.dir <- c( ">=", ">=", "<=" )
> result2d <- solveLP( cvec, abs( bvec ), abs( Amat ), verbose = 1,
+ const.dir = const.dir )
> print( result2d )
Results of Linear Programming / Linear Optimization
Objective function (Minimum): 10.4545
Iterations in phase 1: 2
Iterations in phase 2: 0
Solution
opt
Feed1 1.81818
Feed2 2.95455
Basic Variables
opt
Feed1 1.81818
Feed2 2.95455
S Fibre 2.45455
Constraints
actual dir bvec free dual dual.reg
Protein 10.00000 >= 10.0 0.00000 0.568182 3.60000
Fat 1.50000 >= 1.5 0.00000 3.181818 1.35000
Fibre 9.54545 <= 12.0 2.45455 0.000000 2.45455
All Variables (including slack variables)
opt cvec min.c max.c marg marg.reg
Feed1 1.81818 2.5 -3.666667 5.000000 NA NA
Feed2 2.95455 2.0 -3.000000 3.750000 NA NA
S Protein 0.00000 0.0 -0.568182 Inf 0.568182 3.60
S Fat 0.00000 0.0 -3.181818 Inf 3.181818 1.35
S Fibre 2.45455 0.0 NA 0.833333 0.000000 NA
> all.equal( result2a[-8], result2d[-8] )
[1] TRUE
> # also estimate the dual problem
> result2dD <- solveLP( cvec, abs( bvec ), abs( Amat ), verbose = 1,
+ const.dir = const.dir, solve.dual = TRUE )
> result2dD$con
actual dir bvec free dual dual.reg dual.p
Protein 10.000000 >= 10.0 0.000000 0.5681818 3.600000 0.5681818
Fat 1.500000 >= 1.5 0.000000 3.1818182 1.350000 3.1818182
Fibre 9.545455 <= 12.0 2.454545 0.0000000 2.454545 0.0000000
> all.equal( result2aD[-8], result2dD[-8] )
[1] TRUE
>
> # using argument const.dir and lpSolve
> result2e <- solveLP( cvec, abs( bvec ), abs( Amat ), verbose = 1,
+ const.dir = const.dir, lpSolve = TRUE )
> print( result2e )
Results of Linear Programming / Linear Optimization
(using lpSolve)
Objective function (Minimum): 10.4545
Solution
opt
Feed1 1.81818
Feed2 2.95455
Constraints
actual dir bvec free
Protein 10.00000 >= 10.0 0.00000
Fat 1.50000 >= 1.5 0.00000
Fibre 9.54545 <= 12.0 2.45455
> all.equal( result2c[-5], result2e[-5] )
[1] TRUE
> # also estimate the dual problem
> result2eD <- solveLP( cvec, abs( bvec ), abs( Amat ), verbose = 1,
+ const.dir = const.dir, lpSolve = TRUE, solve.dual = TRUE )
> result2eD$con
actual dir bvec free dual
Protein 10.000000 >= 10.0 0.000000 0.5681818
Fat 1.500000 >= 1.5 0.000000 3.1818182
Fibre 9.545455 <= 12.0 2.454545 0.0000000
> all.equal( result2cD[-5], result2eD[-5] )
[1] TRUE
>
>
>
> proc.time()
user system elapsed
0.206 0.017 0.214
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