Lingo basic use

Lingo grammar

One 、 summary

1、 brief introduction

LINGO It is a particularly useful software for solving optimization problems , Can quickly solve linear programming 、 Nonlinear programming 、 Linear and nonlinear equations and so on , It is one of the indispensable tools for solving optimization problems in mathematical modeling

（1）LINGO The mathematical programming model of contains the objective function 、 The decision variables 、 Three elements of constraints ;
（2）LINGO In the program , Every sentence must end with a semicolon in English , A statement can be entered in several lines ;
（3）LINGO The comments of are in English ！ Start , It must end with a semicolon in English ;
（4）LINGO Variables of are not case sensitive , Must start with a letter , Can contain numbers and underscores , No more than 32 Characters ;
（5）LINGO In the program , As long as the set is defined , The order of other statements is arbitrary ;
（6）LINGO In the function of “@” start ;
（7）LINGO The program defaults that all variables are nonnegative ;
（8）LINGO In the program "<“ or ”>" Number and "" or " " The number has the same function .

2、 File format

1. suffix “lg4” Express lingo Format of the model file , Only lingo Software can be opened ;
2. suffix “lng” Represents a model file in text format ;
3. suffix “ldt” Express lingo Data files ;
4. suffix “ltf” Express lingo Command script file ;
5. suffix “lgr” Express lingo The report file .

3、 Window details

4、 Optimization model

Usually , An optimization model consists of the following three parts ：

1. Objective function ： It is generally expressed as finding the maximum or minimum value of a mathematical expression
2. The decision variables ： The number of objective functions depends on which variables
3. constraint condition ： Attach some conditional restrictions to variables （ It is usually expressed by equality or inequality ）

Be careful ：Lingo By default, all decision variables are positive , Therefore, the non negative condition of the variable does not need to be entered

Two 、 Basic operators

1、 Logical operators

• #not# Negates the logical value of the operand ,＃not＃ Is a unary operator
• #eq# If two operands are equal , Then for true; Otherwise flase
• #ne# If two operators are not equal , Then for true; Otherwise flase
• #gt# If the operator on the left is strictly greater than the operator on the right , Then for true; Otherwise flase
• #ge# If the operator on the left is greater than or equal to the operator on the right , Then for true; Otherwise flase
• #lt# If the operator on the left is strictly smaller than the operator on the right , Then for true; Otherwise flase
• #le# If the operator on the left is less than or equal to the operator on the right , Then for true; Otherwise flase
• #and# Only if both parameters are true when , The result is true; Otherwise flase
• #or# Only if both parameters are false when , The result is false; Otherwise true

2、 Comparison operator

The comparison between relational operators and logical operators is quite different , The former is the description that the relationship specified by the relational operator in the model is true , The latter only judges whether a relationship is satisfied

Lingo There are three operators in ：=,>=、>, and <=、<,Lingo Strict less than and strict greater than relational operators are not supported in

A < B => A <= B

3、 Arithmetic operator

Operator	describe
-	Take the opposite
^	chengfang
*	ride
/	except
+	Add
-	reduce

Lingo The only unary arithmetic operator is the negation operator

The priority of the operator from high to low is ： Take the opposite -》 reduce

The priority order can be changed by brackets

3、 ... and 、 Common function

1、 Mathematical functions

@abs(x) return x The absolute value of
@sin(x) return x The sine of ,x Use the radian system
@cos(x) return x Cosine of
@tan(x) return x The tangent of
@exp(x) Return constant e Of x Power
@log(x) return x The natural logarithm of
@lgm(x) return x Of gamma The natural logarithm of a function
@sign(x) If x<0 return -1; otherwise , return 1
@floor(x) return x The integral part of . When x>=0 when , Returns no more than x Maximum integer for ; When x<0 when , Return no less than x Maximum integer for .
@smax(x1,x2,…,xn) return x1,x2,…,xn Maximum of
@smin(x1,x2,…,xn) return x1,x2,…,xn Minimum of

2、 Defining function

@bin(x) Limit x by 0 or 1 — be used for 0-1 planning
@bnd(L,x,U) Limit L≤x≤U
@free(x) Cancel the variable x The default lower bound of is 0 The limitation of , namely x You can take any real number
@gin(x) Limit x Is an integer by default ,LINGO Specify that the variable is nonnegative , In other words, the lower bound is 0, The upper bound for +∞[email protected] The default lower bound is cancelled 0 The limitation of , So that the variable can also take a negative value [email protected] Used to set the upper and lower bounds of a variable , It can also cancel the default lower bound of 0 Constraints .

3、 Set cyclic function

@function(setname[(set_index_list)[|conditional_qualifier]]:
expression_list);
@function Corresponding to one of the four set cyclic functions listed below ;setname Is the set to traverse ;set_index_list Is a set index list ;conditional_qualifier It is used to limit the range of set cyclic functions , When the set loop function traverses each member of the set ,LINGO All right conditional_qualifier Evaluate , If it turns out to be true , Execute @function operation , Otherwise, skip , Continue with next cycle .expression_list Is a list of expressions that are applied to each set member , When using @for Function time ,expression_list Can contain multiple expressions , Separated by commas . These expressions will be added to the model as constraints . When using the remaining three set loop functions , expression_list There can only be one expression . If omitted set_index_list , So in expression_list All attributes referenced in are of type setname Set .

1. @for
   This function is used to generate constraints on set members . Scalars based on modeling languages require explicit input of each constraint [email protected] Function allows you to enter only one constraint , then LINGO Automatically generate constraints for each set member .

2. @sum

   This function returns the sum of an expression that traverses the specified set member .

3. @min and @max
   Returns the minimum or maximum value of an expression of a specified set member .

4、 Auxiliary function

1. @if(logical_condition,true_result,false_result)

   @if Function will evaluate a logical expression logical_condition, If it is true to return true_ result, Otherwise return to false_result.

2. @warn(’text’,logical_condition), If the logical condition logical_condition It's true , Then a content is generated as ’text’ Information box

Other functions can be used to query documents

Four 、 model

1、 Assembly section

This part should be based on “SETS:” Start , With “ENDSETS” end , The function is to define the necessary set variables （SET） And its elements （member, The meaning is similar to the subscript of an array ） And attribute （attribute, The meaning is similar to array ）.

To define a primitive set , It must be stated in detail ：

• The name of the set
  • Optional , Set member
  • Optional , Set member properties

Define a primitive set , Use the following grammar ：

setname[/member_list/][:attribute_list]

• The contents in brackets indicate optional

Several ways of listing members

1. Show list set members ： List all members , Comma " , " Or separated by spaces .
2. Implicitly list set members ：setname/member1..memberN/[: attribute_list]; ! use ".." To omit

3. The integrator is not included in the set definition , And in the subsequent data section to define

example ：

!  Common example method , Create a list set 
SETS:
	QUARTERS/1,2,3,4/:DEM,RP,OP,INV;  !  Generate four attributes , The initialization values are  [1 2 3 4]
	factory /1..6/:a, b;  !  Generate a  1 x 6  Matrix 
	! factory  Type name called array ,a, b  Variable names called arrays 
	plant /1..8/: c, d;  !  Generate a  1 x 8  Matrix 
	Cooperation(factory,plant): e, f;  !  Generate a  6 x 8  Matrix , If you exchange a position , be , Generate a  8 x 6  Matrix , You can also use  link(factory, plant)  6 x 8
! Cooperation Big factories are made of factory and plant Two small factories merged , Producibility 6×8 Matrix 
ENDSETS

!  Initialization data 
DATA:
	a = 1, 2, 3, 4, 5, 6
	
ENDDATA:

The matrix to be assigned must be full , You can't give 6 The matrix of elements is only assigned 3 A numerical .
Lingo You can assign an integer to the matrix , You can also assign decimals .

2、 Target and constraint segments

This part actually defines the objective function 、 Constraints, etc , But there are no beginning and end marks for this part , So it's actually, among other things 4 Segments （ There are clear segment marks ） External lingo Model .
Functions are generally used here . for example ：

MIN = @SUM(QUARTERS:400*RP+450*OP+20*INV);  ！ @sum For a summation function , Find 400*RP + 450*OP + 20*INV  And , The first parameter passes in the type of the collection 
@FOR(QUARTERS(I):RP(I)<40);  !  Yes  QUARTERS  Traversal , Pass in the type name , Next is the operation , At the same time, it automatically traverses PR What's in it 
@FOR(QUARTERS(I):I#GT#1:INV(I)=INV(I-1)+RP(I)+OP(I)-DEM(I););
INV(1) = 10+RP(1)+OP(1)-DEM(1);

5、 ... and 、 Case study

1、 Transportation and site selection

A company has 6 A construction site , The position coordinates are (ai, bi) ( Company ： km ), Daily cement consumption di ( Company ： Tons of ）

i 1 2 3 4 5 6

a 1.25 8.75 0.5 5.75 3 7.25

b 1.25 0.75 4.75 5 6.5 7.75

d 3 5 4 7 6 11

existing 2 Stockyard , be located A (5, 1), B (2, 7), remember (xj,yj),j=1,2, / i=1~6 Daily reserves ej Each has 20 Tons of .

Suppose there is a straight road between the stockyard and the construction site , Make a daily supply plan , From A, B How many tons of cement are transported from the two stockyards to each construction site , Minimize the total ton kilometers .

Take the decision variable c_ij Express i The construction site is from j The amount of cement transported from the stockyard . Model （ Linear model ） by ：

$$\begin{gathered} matrix\text{min}\sum _{j=1}^2\sum _{i=1}^6{c}_{ij}\sqrt{(x_j-a_i{)}^2+(y_j-b_i{)}^2} \\ s.t\{\begin{array}{c}{\sum }_{j=1}^2{c}_{ij}=d_i.i=1,2,...,6\\ {\sum }_{i=1}^6{c}_{ij}\le e_j,j=1,2\end{array} \end{gathered}$$ {matrix}
be , Solution available ：

MODEL:
SETS:
    demand/1..6/: a, b, d;
    supply/1,2/:e, x, y;
    link(demand, supply): c;
ENDSETS
DATA:
    a=1.25 8.75 0.5 5.75 3 7.25;
    b=1.25 0.75 4.75 5 6.5 7.75;
    d=3 5 4 7 6 11;
    x=5 2;
    y=1 7;
    e=20 20;
ENDDATA
    [email protected](link(i, j):c(i, j)*@SQRT((a(i)-x(j))^2+(b(i)-y(j))^2));  ! express;
    @FOR(demand(i):@SUM(supply(j):c(i,j))=d(i));  ! condition1;
    @FOR(supply(j):@SUM(demand(i):c(i,j)) < e(j));  ! condition2;
END

2、 The best choice

A drilling crew will start from 10 Among the well locations available for selection 5 Drilling exploration , Minimize the total drilling cost . if 10 The code of the well location is s1,s2…,s10, Corresponding drilling costs c1,c2,…,c10 by 5,8,10,6,9,5,7,6,10,8. And the selection of well location should meet the following restrictions ：
（1） Or select s1 and s7, Or choose to drill s9;
（2） I chose s3 or s4 You can't choose s5, Or vice versa ;
（3） stay s5,s6,s7,s8 You can only choose two at most .

Try to build an integer programming model of this problem , Determine the selected well location .

You can use truth table to establish constraint expression

take 0-1 Variable s_i, if s_i=1, Then it means that the i A well , if s_i=0, Then it means that the i A well . The mathematical model is established as follows ：
$$\begin{gathered} matrix\text{min}\sum _{i=1}^{10}{s}_i{c}_i \\ s.t\{\begin{array}{c}(s_1+s_7-2)(s_9)=0\\ s_3{s}_5+s_4{s}_5=0\\ s_5+s_6+s_7+s_8\le 2\\ {\sum }_{i=1}^{10}{s}_i=5\\ s_i\in 0,1(i=1,2,...,10)\end{array} \end{gathered}$$ {matrix}
be , Solution available ：

MODEL:
SETS:
    var/1..10/:s,c;  !  Create set 
ENDSETS
DATA:
    c=5 8 10 6 9 5 7 6 10 8;  !  Assign a value to a set 
ENDDATA
    MIN = @SUM(var(i):s(i)*c(i));  !  Solution result 
    (s(1)+s(7)-2)*s(9) = 0;  !  Conditions 1
    s(3)*s(5)+s(4)*s(5)=0;  !  Conditions 2
    s(5)+s(6)+s(7)+s(8)<2;  !  Conditions 3
    @SUM(var(i):s(i))=5;  !  Conditions 4
    @FOR(var(i):@BIN(s(i)));  !  Conditions 5, Use for Cycle pair s For each value of ,1 representative true
END