Difference between revisions of "ApCoCoA-1:GLPK.LPSolve"

Latest revision as of 15:12, 1 November 2020

This article is about a function from ApCoCoA-1. If you are looking for the ApCoCoA-2 version of it, see Package glpk/GLPK.LPSolve.

GLPK.LPSolve

Solving linear programmes.

Syntax

GLPK.LPSolve(Objective_f:POLY, EQ_Poly:LIST, LE_Poly:LIST, GE_Poly:LIST, Bounds:LIST, Method:STRING, MinMax:STRING):LIST

Description

Please note: The function(s) explained on this page is/are using the ApCoCoAServer. You will have to start the ApCoCoAServer in order to use it/them.

@param Objective_f: A linear polynomial which is equivalent to the linear objective function.
@param EQ_Poly: List of linear polynomials, which are equivalent to the equality-part in the list of conditions.
@param LE_Poly: List of linear polynomials, which are equivalent to the lower or equal-part in the list of conditions.
@param GE_Poly: List of linear polynomials, which are equivalent to the greater or equal-part in the list of conditions.
@param Bounds: List of lists with two elements. Each List contains the lower and upper bounds for each variable. You can choose between INT or RAT for the type of each bound, if you type in a (empty) string, then it means minus infinity (first place) or plus infinity (second place).
@param Method: You can choose between the interior-point-method ("InterP") or the simplex-algorithm ("Simplex"). Usually you should use the simplex-algorithm.
@param MinMax: Minimization ("Min") or maximization ("Max"), that's the question.
@return List of linear polynomials, the zeros of the polynomials are the points where the optimal value of the objective function is achieved

Example

-- We want to maximize the Function y = - 1/2x, 
-- with the two conditions y ≤ 6 - 3/4x and y ≥ 1 - x and the bounds 0 ≤ x ≤ 6 and 1/3 ≤ y ≤ 4.

-- We prename the input of GLPK.LPSolve-function.
Use S::=QQ[x,y];
OF := 1/2x + y;
LE := [3/4x + y - 6];
GE := [x + y - 1];
Bounds:=[[0,6], [1/3,4]];

-- Then we compute the solution with
GLPK.LPSolve(OF, [], LE, GE, Bounds, "Simplex", "Max");

-- And we achieve:
------------------------------------- 
Solution Status: OPTIMAL
Value of objective function: 5333333333/1000000000
[x - 266667/100000, y - 4]

Latte.Minimize

Latte.Maximize

@@ Line 1: / Line 1: @@
+{{Version|1|[[Package glpk/GLPK.LPSolve]]}}
 <command>
-<title>LPSolve</title>
+<title>GLPK.LPSolve</title>
-<short_description>solve linear programms</short_description>
+<short_description>Solving linear programmes.</short_description>
 <syntax>
-GLPK.LPSolve(Objective_Function:POLYNOM, EQ_Polynomials:LIST, LE_Polynomials:LIST, GE_Polynomials:LIST, Bounds:LIST, Methode:STRING, MinMax:STRING)
+GLPK.LPSolve(Objective_f:POLY, EQ_Poly:LIST, LE_Poly:LIST, GE_Poly:LIST, Bounds:LIST, Method:STRING, MinMax:STRING):LIST
 </syntax>
 <description>
-{{ApCoCoAServer}}
+<em>Please note:</em> The function(s) explained on this page is/are using the <em>ApCoCoAServer</em>. You will have to start the ApCoCoAServer in order to use it/them.
-The basic idea behind this package is to make the linear optimization program GLPK usable in/with ApCoCoA.
+<itemize>
-The package GLPK contains various functions that let you make use of the GLPK library, rather the stand-alone LP/MIP Solver glpsol.
+<item>@param <em>Objective_f</em>: A linear polynomial which is equivalent to the linear objective function.</item>
+<item>@param <em>EQ_Poly</em>: List of linear polynomials, which are equivalent to the equality-part in the list of conditions.</item>
+<item>@param <em>LE_Poly</em>: List of linear polynomials, which are equivalent to the lower or equal-part in the list of conditions.</item>
+<item>@param <em>GE_Poly</em>: List of linear polynomials, which are equivalent to the greater or equal-part in the list of conditions.</item>
+<item>@param <em>Bounds</em>: List of lists with two elements. Each List contains the lower and upper bounds for each variable. You can choose between INT or RAT for the type of each bound, if you type in a (empty) string, then it means minus infinity (first place) or plus infinity (second place).</item>
+<item>@param <em>Method</em>: You can choose between the interior-point-method ("InterP") or the simplex-algorithm ("Simplex"). Usually you should use the simplex-algorithm.</item>
+<item>@param <em>MinMax</em>:  Minimization ("Min") or maximization ("Max"), that's the question.</item>
+<item>@return List of linear polynomials, the zeros of the polynomials are the points where the optimal value of the objective function is achieved</item>
+</itemize>
-<em>Objective_Function</em>: A linear polynomial which is equivalent to the linear objective Function.
-<em>EQ_Polynomials</em>: List of linear polynomials, which are equivalent to the equality-part in the list of conditions.
-<em>LE_Polynomials</em>: List of linear polynomials, which are equivalent to the lower or equal-part in the list of conditions.
-<em>GE_Polynomials</em>: List of linear polynomials, which are equivalent to the greater or equal-part in the list of conditions.
-<em>Bounds</em>: List of lists with two elements. Each List contains the lower and upper bounds for each variable. You can choose between INT or RAT for the type of each bound, if you type in a (empty) string, then it means minus infinity (first place) or plus infinity (second place).
-<em>Methode</em>: You can choose between the inner-point-method (InnerP) or the simplex-algorithm (Simplex).  Usually you should use the simplex-algorithm.
-<em>MinMax</em>:  Minimization (Min) or maximization (Max), that's the question.
-<em>Specification of the paths</em>: When you execute the GLPK.LPSolve-command, you had to <em>specify the path of the LP solver glpsol</em>, i.e. GlpsolPath:="/usr/bin/";
-and the <em>working path</em>, i.e. WorkingPath:="/home/user/";
-First we want to discuss a rather easy example.
 <example>
-We want to maximize the Function y = - 1/2x,
+-- We want to maximize the Function y = - 1/2x,
-with the two conditions y ≤ 6 - 3/4x and y ≥ 1 - x and the bounds 0 ≤ x ≤ 6 and 1/3 ≤ y ≤ 4.
+-- with the two conditions y ≤ 6 - 3/4x and y ≥ 1 - x and the bounds 0 ≤ x ≤ 6 and 1/3 ≤ y ≤ 4.
-We prename the input of GLPK.LPSol-function.
+-- We prename the input of GLPK.LPSolve-function.
+Use S::=QQ[x,y];
 OF := 1/2x + y;
-LE := 3/4x + y - 6;
+LE := [3/4x + y - 6];
-GE := x + y - 1;
+GE := [x + y - 1];
 Bounds:=[[0,6], [1/3,4]];
-Then we compute the solution with
+-- Then we compute the solution with
-Use S::=Q[x,y];
+GLPK.LPSolve(OF, [], LE, GE, Bounds, "Simplex", "Max");
-$GLPK.LPSolve(OF, LE, GE, Bounds, "Simplex", "Max");
-And we achieve x = 8/3 and y = 4.
-</example>
-[[Image:LinOpt.jpg]]
-<example>
-''Linear programming example 1996 MBA exam''
-A cargo plane has three compartments for storing cargo: front, centre and rear.
-These compartments have the following limits on both weight and space:
-Compartment   Weight capacity (tonnes)   Space capacity (cubic metres)
-Front         10                         6800
-Centre        16                         8700
-Rear          8                          5300
-Furthermore, the weight of the cargo in the respective compartments must be the same
-proportion of that compartment's weight capacity to maintain the balance of the plane.
-The following four cargoes are available for shipment on the next flight:
-Cargo   Weight (tonnes)   Volume (cubic metres/tonne)  Profit (£/tonne)
-C1      18                480                          310
-C2      15                650                          380
-C3      23                580                          350
-C4      12                390                          285
-Any proportion of these cargoes can be accepted. The objective is to determine how much
-(if any) of each cargo C1, C2, C3 and C4 should be accepted and how to distribute each
-among the compartments so that the total profit for the flight is maximised.
-To solve this problem we had to compose a linear program.
-Variables
-We need to decide how much of each of the four cargoes to put in each of the three compartments.
-Hence let:
-xij be the number of tonnes of cargo i (i=1,2,3,4 for C1, C2, C3 and C4 respectively) that is put into
-compartment j (j=1 for Front, j=2 for Centre and j=3 for Rear) where xij >=0 i=1,2,3,4; j=1,2,3
-Note here that we are explicitly told we can split the cargoes into any proportions (fractions) that we like.
-Constraints
-* cannot pack more of each of the four cargoes than we have available
-x11 + x12 + x13 ≤ 18
-x21 + x22 + x23 ≤ 15
-x31 + x32 + x33 ≤ 23
-x41 + x42 + x43 ≤ 12
-* the weight capacity of each compartment must be respected
-x11 + x21 + x31 + x41 ≤ 10
-x12 + x22 + x32 + x42 ≤ 16
-x13 + x23 + x33 + x43 ≤ 8
-* the volume (space) capacity of each compartment must be respected
-x11 + 650x21 + 580x31 + 390x41 ≤ 6800
-x12 + 650x22 + 580x32 + 390x42 ≤ 8700
-x13 + 650x23 + 580x33 + 390x43 ≤ 5300
-* the weight of the cargo in the respective compartments must be the same proportion
-  of that compartment's weight capacity to maintain the balance of the plane
-[x11 + x21 + x31 + x41]/10 = [x12 + x22 + x32 + x42]/16 = [x13 + x23 + x33 + x43]/8
-Objective
-The objective is to maximise total profit, i.e.
-maximise 310[x11+ x12+x13] + 380[x21+ x22+x23] + 350[x31+ x32+x33] + 285[x41+ x42+x43]
-The basic assumptions are:
-* that each cargo can be split into whatever proportions/fractions we desire
-* that each cargo can be split between two or more compartments if we so desire
-* that the cargo can be packed into each compartment (for example if the cargo was spherical it would
-  not be possible to pack a compartment to volume capacity, some free space is inevitable in sphere packing)
-* all the data/numbers given are accurate
+-- And we achieve:
-In ApCoCoa you can solve this LP like this:
+-------------------------------------
+Solution Status: OPTIMAL
-Use S::= Q[x[1..4,1..3]];
+Value of objective function: 5333333333/1000000000
+[x - 266667/100000, y - 4]
-Objective := 310*(x[1,1]+x[1,2]+x[1,3])+310*(x[2,1]+x[2,2]+x[2,3])+350*(x[3,1]+x[3,2]+x[3,3])+285*(x[4,1]+x[4,2]+x[4,3]);
-Balance := [16(x[1,1]+x[2,1]+x[3,1]+x[4,1])-10(x[1,2]+x[2,2]+x[3,2]+x[4,2]), 8(x[1,2]+x[2,2]+x[3,2]+x[4,2])-16(x[1,3]+x[2,3]+x[3,3]+x[4,3])];
-Available := [x[1,1]+x[1,2]+x[1,3]-18, x[2,1]+x[2,2]+x[2,3]-15, x[3,1]+x[3,2]+x[3,3]-23, x[4,1]+x[4,2]+x[4,3]-12];
-Weight := [x[1,1]+x[2,1]+x[3,1]+x[4,1]-10, x[1,2]+x[2,2]+x[3,2]+x[4,2]-16, x[1,3]+x[2,3]+x[3,3]+x[4,3]-8];
-Volume := [480x[1,1]+650x[2,1]+580x[3,1]+390x[4,1]-6800, 480x[1,2]+650x[2,2]+580x[3,2]+390x[4,2]-8700, 480x[1,3]+650x[2,3]+580x[3,3]+390x[4,3]-5300];
-LessEq := Flatten([Available, Weight, Volume]);
-Bounds := [[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"],[0,"Infty"]];
-$GLPK.LPSolve(Objective, Balance, LessEq, [], Bounds, "Simplex", "Max");
-And we receive
-[x[1,1], x[1,2] - 29/5, x[1,3] - 26/5, x[2,1], x[2,2], x[2,3], x[3,1] - 10, x[3,2] - 51/5, x[3,3] - 14/5, x[4,1], x[4,2], x[4,3]]
-This means
-* in the front we put
-  - 0 of cargo one (x[1,1] = 0)
-  - 0 of cargo one (x[2,1] = 0)
-  - 10 tons of cargo three (x[3,3] = 10)
-  - 0 of cargo four (x[4,1] = 0)
-* in the middle we put
-  - 29/5 tons of cargo one (x[1,2] = 29/5)
-  - 0 of cargo one (x[2,2] = 0)
-  - 51/5 tons of cargo three (x[3,2] = 51/5)
-  - 0 of cargo four (x[4,2] = 0)
-* in the rear we put
-  - 26/5 tons of cargo one (x[1,3] = 26/5)
-  - 0 of cargo one (x[2,3] = 0)
-  - 14/5 tons of cargo three (x[3,3] = 14/5)
-  - 0 of cargo four (x[4,3] = 0)
 </example>
 </description>
 <types>
-   <type>cocoaserver</type>
+   <type>apcocoaserver</type>
+  <type>linear_programs</type>
 </types>
+<see>ApCoCoA-1:Latte.Minimize|Latte.Minimize</see>
+<see>ApCoCoA-1:Latte.Maximize|Latte.Maximize</see>
 <key>lpsolve</key>
-<key>solve linear programm</key>
+<key>solve linear program</key>
 <key>solve lp</key>
-<key>jbrandt</key>
+<key>GLPK.LPSolve</key>
-<key>skuehling</key>
+<wiki-category>ApCoCoA-1:Package_glpk</wiki-category>
-<wiki-category>Package_GLPK</wiki-category>
 </command>

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Latest revision as of 15:12, 1 November 2020

GLPK.LPSolve

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Difference between revisions of "ApCoCoA-1:GLPK.LPSolve"

Latest revision as of 15:12, 1 November 2020

GLPK.LPSolve

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