LINGO Part II IE 311 Lab

1
LINGOPart II (IE 311 Lab)
Spring 07
2
Example I (Sets)

• Consider the following sets section
• SETS
• Product / A B /
• Machine / M N /
• Week / 1..2/
• Allowed (Product, Machine, week)
• ENDSETS

Primitive Sets
Derived Set
Index Member 1 (A, M, 1) 2 (A, M, 2) 3 (A,
N, 1) 4 (A, N, 2)
Index Member 5 (B, M, 1) 6 (B, M, 2) 7 (B,
N, 1) 8 (B, N, 2)
Allowed set membership
3
Example II (Explicit member)

• Instead of following statement in Example I
• Allowed (Product, Machine, week)
• Consider the below statement
• Allowed (Product, Machine, week) / A M 1, B N
  2/
• Explicit member list method is used to specify a
  derived sets member list!
• So, Allowed set members would have consisted of
  the two members.

Index Member 1 (A, M, 1) 2 (B, N, 2)
Allowed set membership
4
Example III (Membership Filter)

• Suppose, you have already defined a set called
  TRUCK and each truck has an attribute called
  CAPACITY.
• You would like to derived a subset from TRUCK
  that contain only those trucks capable of hauling
  big load (greater than 5000 kg).
• The TRUCK set is as following
• TRUCK / 1..100 / CAPACITY
• To define a subset of trucks capable of hauling
  big load, you can use a Membership Filter as
  follows
• Heavy_Duty (TRUCK) CAPACITY (1) GT 5000

5
Example III (Membership Filter) (cont.)

• Heavy_Duty (TRUCK) CAPACITY (1) GT 5000

Beginning of the membership filter
Greater than
Derived from
Set name

• Vertical bar character () is used to mark the
  beginning of a membership filter.
• The 1 symbol in the filter is known as a set
  index placeholder.
• When building a derived set that uses a
  membership filter, LINGO generates all the
  combinations of parent set members. Each
  combination is then plugged into the membership
  condition to see if it passes the test. The first
  parent sets value is plugged into 1, the second
  into 2, and so on.

6
LINGO OPERATORS

• ARITHMETIC OPERATORS
• LINGO has five binary (two-operand)
• arithmetic operators
• Power
• Multiplication
• / Division
• Addition
• - Subtraction

7
RELATIONAL LOGICAL OPERATORS

• In LINGO, you use relational and logical
  operators to determine set membership and which
  elements of a set are included in certain
  operation such as summation.
• In an expression involving both relational and
  logical operators, relational operators have
  higher precedence.
• (that is, relational operators are evaluated
  before logical operators).

8
RELATIONAL OPERATORS

• Relational operators have numeric operands and
  return logical results (TRUE or FALSE).
• All relational operators are binary.
• LINGO has six relational operators.
• All relational operators have the same precedence.

9
RELATIONAL OPERATORS
10
LOGICAL OPERATORS

• Logical operators have logical operands (the
  results of relational operations) and return
  logical results.
• Logical operators are used to connect relational
  and logical expressions.
• LINGO has three logical operators, listed here in
  descending order of precedence.

11
LOGICAL OPERATORS
12
GENERAL MATHEMATICAL FUNCTIONS
13
SET-LOOPING FUNCTIONS

• Set-Looping Function Operate over an entire set
  and, with the exception of the _at_FOR operator,
  produce a single result.
• The syntax is as follows

set_operator (set_name condition expression)
14
SET-LOOPING FUNCTIONS

• The condition part in the specification is
  optional. The available functions are listed
  below.

15
VARIABLE DOMAIN FUNCTIONS
16
Example IV(Use of Condition Qualifier )

• Lets use what we have learned till now in an
  example.
• Consider a direct model, which calculates the
  reciprocal of a vector of numbers. Because the
  reciprocal of zero is undefined, we want to
  attempt the calculation only for nonzero numbers.
  The following model accomplishes this by using
  conditional qualifiers in the _at_FOR statements. A
  conditional qualifier is prefaced by the
  symbol.

17

• MODEL
• ! The model calculates reciprocals of numbers
• SETS
• ! Ten numbers are in the vector VAL. The
  reciprocals
• are calculated and stored in the
  corresponding elements
• of RECIP
• NUM / 1..10 / VAL, RECIP
• ENDSETS
• ! Calculate reciprocal only for nonzero
  numbers
• _at_FOR (NUM(I) VAL(I) NE 0 RECIP(I) 1 /
  VAL(I))
• ! If a number is Zero, make its reciprocal
  arbitrarily Zero
• _at_FOR (NUM(I) VAL(I) EQ 0 RECIP(I) 0)
• DATA
• ! Numbers for which to calculate the
  reciprocals
• VAL 2, 0, 8, 40, 1, 0, 0.333333, 50, 3,
  0.2

18

• The bold text in the model means that for all
  members of set NUM whose attribute VAL is not
  equal to zero and proceeded with the calculation
  of reciprocal of that member (I) and assigning it
  to the attribute RECIP(I).

19
Example V (The Sailco Problem (page 99))

• MIN Z 400 (x1 x2 x3 x4)
• 450 (y1 y2 y3 y4)
• 20 (i1 i2 i3 i4)
• ST
• x140
• x240
• x340
• x440
• i1 10 x1 y1 40
• i2 i1 x2 y2 60
• i3 i2 x3 y3 75
• i4 i3 x4 y4 25
• it,yt,xt 0 (t 1,2,3,4)

20

• xt number of sailboards produced during
  regular-time during quarter t.
• yt number of sailboards produced during
  over-time during quarter t.
• it number of sailboards on hand at end of
  quarter t.

21
(No Transcript)
22

• Now, lets write the LP as a LINGO program
• First, we have to define SETS and attributes that
  we will use in the model. As you can variables
  are defined on quarters by index t. We can say
  that our set name is Quarters. Each quarter has
  unique regular time production, over-time
  production, inventory and demand figures. These
  are the attributes of the set Quarters.
• Then, objective function and constraints must be
  written. In this step, we use set looping
  functions which ease our model editing. By this
  way, we can write the same constraints in one
  line by_at_FOR command.
• Its like the FOR loop in PASCAL, with a little
  bit different notation. The last part is DATA
  section, which contains the figures of the
  attributes. Now we are ready to write the model.

23

• MODEL
• !SETS part of the model,
• TIMEPRERIOD NUMBER
• DEM DEMAND
• RP REGULAR TIME PRODUCTION
• OP OVER TIME PRODUCTION
• INV INVENTORY
• SETS
• Quarters/ Q1,Q2,Q3,Q4/TIME,DEM,RP,OP,INV
• ENDSETS
• !OBJECTIVE FUNCTION
• MIN_at_SUM (QUARTERS 400RP 450OP 20INV)
• !CONSTRAINTS
• !PRODUCTION CONSTRAINTS
• _at_FOR(QUARTERS (I) RP(I) lt40)

Set declaration part
Objective Function and Constraints part
Data declaration part
24

• Thats all. The only difficult thing in this
  example is the second constraint set which is
  inventory balance constraint set.
• In that part we try to say that, for all quarters
  greater than 1, current quarters inventory
  figure is equal to the last quarters inventory
  figure plus regular and over time production
  minus demand of current quarter. And if it is
  equal to 1, inventory figure is equal to 10 (on
  hand inventory) plus regular and over time
  production minus demand of 1st quarter.