Last data update: 2014.03.03

 R: Solves a linear inverse model using linear programming.
LinpR Documentation

Solves a linear inverse model using linear programming.

Description

Solves a linear inverse model using linear programming

Input presented either as:

  • matrices E, F, A, B, G, H (Linp.double) or

  • as a list (Linp.lim) or

  • as a lim input file (Linp.limfile)

Usage

Linp(...)
## S3 method for class 'lim'
Linp(lim, cost = NULL, ispos = lim$ispos, ...)
## S3 method for class 'limfile'
Linp(file, verbose = TRUE,...)
## S3 method for class 'character'
Linp(...)
## S3 method for class 'double'
Linp(...)

Arguments

lim

a list that contains the linear inverse model specification, as generated by function setup.limfile.

file

name of the inverse input file.

verbose

if TRUE: when reading the file, prints warnings and messages to the screen.

cost

if not NULL, a vector with the coefficients of the cost function (to be minimised).

ispos

if TRUE: all x-values have to be positive.

...

other arguments passed to function linp from packagelimSolve.

Details

Solves the following inverse problem:

min(∑ {Cost_i*x_i})

or

max(∑ {Profit_i*x_i})

subject to

x_i>=0

Ax=B

Gx>=H

and where Cost_i or Profit_i are weighting coefficients

Value

a list containing:

X

vector containing the solution of the linear programming problem.

unconstrained.solution

vector containing the unconstrained solution of the linear programming problem.

residualNorm

scalar, the sum of residuals of equalities and violated inequalities.

solutionNorm

scalar, the value of the quadratic function at the solution.

IsError

logical, TRUE if an error occurred.

Error

linp error text.

type

linp.

Author(s)

Karline Soetaert <karline.soetaert@nioz.nl>

References

Michel Berkelaar and others (2005). lpSolve: Interface to Lpsolve v. 5 to solve linear/integer programs. R package version 1.1.9.

See Also

linp, the more general function from package lpSolve

Ldei, to solve the linear inverse problem by least distance programming

Lsei, to solve the linear inverse problem by lsei (least squares with equality and inequality constraints)

function linp from packagelimSolve

Examples

  # the Blending example
  Linp(LIMBlending)

  # the E coli example: two functions to maximimise
  Linp(LIMEcoli)
 
  # E coli example, but only first function optimised..
  Linp(LIMEcoli, cost = -LIMEcoli$Profit[1,])

  # a foodweb example: need to specify the cost function
  # here just sum of absolute values of flows...
  Linp(LIMRigaAutumn, cost = (rep(1, LIMRigaAutumn$NUnknowns)))

Results


R version 3.3.1 (2016-06-21) -- "Bug in Your Hair"
Copyright (C) 2016 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(LIM)
Loading required package: limSolve
Loading required package: diagram
Loading required package: shape
> png(filename="/home/ddbj/snapshot/RGM3/R_CC/result/LIM/Linp.Rd_%03d_medium.png", width=480, height=480)
> ### Name: Linp
> ### Title: Solves a linear inverse model using linear programming.
> ### Aliases: Linp Linp.lim Linp.limfile Linp.character Linp.double
> ### Keywords: algebra array optimize
> 
> ### ** Examples
> 
>   # the Blending example
>   Linp(LIMBlending)
$residualNorm
[1] 0

$solutionNorm
[1] 31.81818

$X
     PINGREDIENT1 PINGREDIENT2   PFILLER
[1,]    0.5909091    0.1363636 0.2727273

> 
>   # the E coli example: two functions to maximimise
>   Linp(LIMEcoli)
$residualNorm
[1] 1.418687e-11 1.708703e-14

$solutionNorm
[1] 33.11752  0.00000

$X
        GLK1       PGI1       PFKA FBP        FBA       TPIA       GAPA
[1,] 0.00000 807.532745 781.590686   0 781.590686 781.590686 1541.43420
[2,] 5.66169   5.639786   5.553995   0   5.553995   5.553995   11.03607
            PGK       GPMA        ENO PPSA      PYKA         ACEE ZWF PGL GND
[1,] 1541.43420 1492.08909 1492.08909    0 466.65796 1149.2952836   0   0   0
[2,]   11.03607   10.87288   10.87288    0  10.50489    0.1949477   0   0   0
            RPIA          RPE       TKTA1        TKTA2        TALA       GLTA
[1,] 23.62383277 -23.62383277 -5.85076232 -17.77307045 -5.85076232 35.4357492
[2,]  0.07812511  -0.07812511 -0.01934874  -0.05877637 -0.01934874  0.1171877
           ACNA       ICDA SUCA SUCC1 SDHA1 FRDA FUMA  MDH     DLD1
[1,] 35.4357492 35.4357492    0     0     0  100 -100 -100 0.000000
[2,]  0.1171877  0.1171877    0     0     0    0    0    0 9.595867
            ADHE2       PFLA PTA ACKA ACS PCKA         PPC MAEB SFCA ACEA ACEB
[1,] 1.000000e+03 10.0000000   0    0   0    0 194.3849394    0    0    0    0
[2,] 7.228398e-02  0.4041332   0    0   0    0   0.3121354    0    0    0    0
     PPA GLPK GPSA1 RBSK NUOA FDOH GLPD CYOA SDHA2 PNT1A      PNT2A
[1,]   0    0     0    0  140    0    0   40  -100     0 567.965512
[2,]   0    0     0    0    0    0    0    0     0     0   1.878288
              ATPA   GLCUP   GLCPTS GLUP RIBUP ACUP     LACUP       FORUP
[1,] -145.46632929 0.00000 814.1562    0     0    0  0.000000 -10.0000000
[2,]   -0.06247093 5.66169   0.0000    0     0    0 -9.595867  -0.4041332
             ETHUP SUCCUP     PYRUP        PIUP O2TX     CO2TX ATPM ADK
[1,] -1.000000e+03   -100 -27.79634 120.5477822   20 -990.3461 5.87   0
[2,] -7.228398e-02      0   0.00000   0.3986571    0    0.0000 5.87   0
         Growth
[1,] 33.1175226
[2,]  0.1095212

>  
>   # E coli example, but only first function optimised..
>   Linp(LIMEcoli, cost = -LIMEcoli$Profit[1,])
$residualNorm
[1] 1.418687e-11

$solutionNorm
[1] -33.11752

$X
     GLK1     PGI1     PFKA FBP      FBA     TPIA     GAPA      PGK     GPMA
[1,]    0 807.5327 781.5907   0 781.5907 781.5907 1541.434 1541.434 1492.089
          ENO PPSA    PYKA     ACEE ZWF PGL GND     RPIA       RPE     TKTA1
[1,] 1492.089    0 466.658 1149.295   0   0   0 23.62383 -23.62383 -5.850762
         TKTA2      TALA     GLTA     ACNA     ICDA SUCA SUCC1 SDHA1 FRDA FUMA
[1,] -17.77307 -5.850762 35.43575 35.43575 35.43575    0     0     0  100 -100
      MDH DLD1 ADHE2 PFLA PTA ACKA ACS PCKA      PPC MAEB SFCA ACEA ACEB PPA
[1,] -100    0  1000   10   0    0   0    0 194.3849    0    0    0    0   0
     GLPK GPSA1 RBSK NUOA FDOH GLPD CYOA SDHA2 PNT1A    PNT2A      ATPA GLCUP
[1,]    0     0    0  140    0    0   40  -100     0 567.9655 -145.4663     0
       GLCPTS GLUP RIBUP ACUP LACUP FORUP ETHUP SUCCUP     PYRUP     PIUP O2TX
[1,] 814.1562    0     0    0     0   -10 -1000   -100 -27.79634 120.5478   20
         CO2TX ATPM ADK   Growth
[1,] -990.3461 5.87   0 33.11752

> 
>   # a foodweb example: need to specify the cost function
>   # here just sum of absolute values of flows...
>   Linp(LIMRigaAutumn, cost = (rep(1, LIMRigaAutumn$NUnknowns)))
$residualNorm
[1] 5.725698e-13

$solutionNorm
[1] 761.2296

$X
      P1->CO2 Flow(P2,CO2)=ZeroOrder Flow(Z,CO2)=ZeroOrder
[1,] 17.22501               29.95399              27.64922
     Flow(N,CO2)=ZeroOrder Flow(B,CO2)=ZeroOrder Flow(CO2,P1)=ZeroOrder
[1,]              13.40297              247.5388                31.3182
     Flow(CO2,P2)=ZeroOrder Flow(P1,Z)=ZeroOrder Flow(P1,N)=ZeroOrder
[1,]                54.4618             10.49431              4.12297
     Flow(P1,DOC)=ZeroOrder P1->Sedimentation Flow(P2,DOC)=ZeroOrder
[1,]                1.56591               0.1                2.72309
     Flow(P2,Z)=ZeroOrder Flow(P2,D)=ZeroOrder Flow(P2,Sedimentation)
[1,]             21.25472                    0                   0.34
     Flow(N,DOC)=ZeroOrder Flow(N,Z)=ZeroOrder Flow(Z,DOC)=ZeroOrder
[1,]                     0                   0              6.509806
     Flow(Z,D)=ZeroOrder Z->Sedimentation Flow(D,Z)=ZeroOrder
[1,]                   0             0.78                 4.7
     Flow(D,DOC)=ZeroOrder D->Sedimentation Flow(B,N)=ZeroOrder
[1,]                     0            13.92                9.44
     B->Sedimentation Flow(DOC,B)=ZeroOrder
[1,]                0              263.7288

> 
> 
> 
> 
> 
> dev.off()
null device 
          1 
>


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