Transcript IENG-212 Modeling and Optimization

IENG442
LINGO
LAB3
Example -1
A company produces product A and B. Product B
sells for $5 per unit, and product A sells for $3
per unit. Producing a unit of product B requires 5
unit of row material 1 and 2 unit of row material
2. Producing a unit of product A requires 2 unit of
row material 1 and 1 unit of row material 2. 60
units of row material 1 and 25 unit of row
material 2 are available.
Formulate an LP that can be used to maximize
revenue.
Modeling
Variables:
x1: units of product A that should be produced
x2: units of product B that should be produced
Objective Function:
Max Z= 5x2 + 3x1
Subject to (constraints):
5x2 + 2x1 ≤ 60
(Row material 1 constraint)
2x2 + 1x1 ≤ 25
(Row material 2 constraint)
x1 ≥ 0
x2 ≥ 0
Changing model for LINGO:
Solving a LINGO Model
Each model can be solved by:
- Clicking the solve button
- Selecting Solve from the LINGO menu
- Using Ctrl+S keyboard shortcut.
Error (if any) will be reported.
LINGO Solver Status Window
LINGO Solution Report Window
Example -02
Maximize
Subject to
-5/36Q2 + 160/9Q – 125/9
10 ≤Q ≤ 100
LINGO Model
Example -03
Minimize
50S+100F
Subject to
5 + 17≥ 40
20+ 7 ≥ 60
S,F ≥ 0
LINGO Model
Example-04
Consider that each student may eat at most
16 units of food stuff per day and the
amount of protein taken from animal
sources should not be more than the 60%
of the total protein intake.
Formulate as an LP model.
Example-4(Solution):
Objective: To obtain minimum cost daily diet
which satisfying the Nutritional requirements and
eating capacity
Decision Variables: Xj number of units of
food type j into the daily diet of the student
(where j=1 (milk), 2(meat), 3(bread),
4(vegetables)).
MODEL:
Objective Function:
Min Z= 0.42X1+0.68X2+0.32X3+0.17X4
Example-4(Solution):
Subject to:
Calories:
160X1+210X2+120X3+150X4 ≤ 2700
Carbohydrate:
110X1+130X2+110X3+120X4 ≥ 300
Protein:
90X1+190X2+90X3+130X4 ≥ 250
Vitamins:
50X1+50X2+75X3+70X4
≥ 60
Protein balance:
90X1+190X2 ≤ 0.6(90X1+190X2+90X3+130X4)
Or:
36X1+76X2-54X3-78X4
≤0
Eating Capacity: X1+X2+X3+X4
≤ 16
Non-negativity: Xj ≥ 0
; j=1,2,3,4
Lingo Model:
Example-05
A farmer must determine how many acres of corn and wheat
to plant this year. You can find relevant data in the table
below:
Government regulations require that at least 30 bushels of
corn be produced during the current year.
Formulate an LP to maximize the total revenue from wheat
and corn.
Example-05 continuing
a)
X1 = number of acres of corn planted &
X2 = number of acres of wheat planted.
b)
Y1 = number of bushels of corn produced &
Y2 = number of bushels of wheat
Example-05 (Solution)
a)
MAX Z = (10*3) X1 + (25*4) X2
Subject to:
X1 + X2 <= 7
4X1 + 10X2 <= 40
10X1 >= 30
X2 >= 0
Example-05 (LINGO MODEL)
Example-05 (LINGO Solution)
Example-05 (Solution)
b)
MAX Z = 3Y1 + 4Y2
Subject to:
Y1>= 30
Y1/10 + Y2/25 <= 7
Y1/4 + Y2/10 <= 40
Y2 >= 0
Example-05 (LINGO MODEL)
Example-05 (LINGO Solution)
Overview
Using Variable Domain Functions to formulate linear
programming problems
– @BIN(X): Limits the variable X to a binary integer
Value (0 or 1)
– @GIN(X): Limits the variable X to only integer values.
– @BND(L,X,U): Limits the variable X to greater or
equal to L and less than or equal to U.
– @FREE(X): Remove the default lower bound of zero
on a variable, allowing it to take any positive or
negative value.
Integer Programming (IP)
If one or more decision variables must be integer, then
we say that an optimization model is an integer model.
An IP in which all variables are required to be integers
is called a pure integer programming problem.
An IP in which only some of the variables are required
to be integers is called a mixed integer programming
problem.
An IPP in which all the variables must equal 0 or 1 is
called a 0-1 (or binary) IP.
Example 06 (@BIN(X))
A young couple, Evren and Steven, want to divide their
main household chores between them so that each has
two tasks but the total time they spend on household
duties is kept to a minimum. Their efficiencies on these
tasks differ, where the time each would need to perform
the task is given by the following table. Formulate a BIP
model for this problem.
Example 06-Solution
We have 4 types of tasks; marketing, cooking,
dishwashing and laundry
Our decision variables are: Si and Ei (i=1…4).
Si=1 ;if Steven does the ith task
Si=0 ;otherwise
Ei=1 ;if Evren does the ith task
Ei=0 ;Otherwise
Example 06-LINGO Model
Example 06-LINGO Solution
Example 06-LINGO Solution
Example 07 (@GIN(X))
A paper company has machines that produce large
rolls of paper of a given diameter and width, say 100
inches. To fill customer order (200 rolls with 45 inches
width and 110 rolls with 30 inches width), these large
rolls must be cut in to narrow width (using some
patterns). Formulate this problem to minimize the
number of standard rolls cut in filling the order.
Example 07- Solution
We have some patterns to cut the standard rolls:
So our decision variables are:
Xj= number of standard rolls cut by pattern j
(It must be integer)
Example 7-LINGO Model
Example 07-LINGO Solution
Example 07-LINGO Solution
Example 08 (@BND(L,X,U))
A factory has two machines and produces two types of
products. Selling prices, working hours, current orders
amounts and maximum amounts of production are
shown in the table. Formulate this problem to maximize
the profits.
Example 08- Solution
Decision variables are:
Xi =number of product i that should be produced to
maximized the profits (i=1,2)
Example 08-LINGO Model
Example 08-LINGO Solution
Example 08-LINGO Solution
Example 09 (FREE(X))
A baker has 30 ounces of flour and 5 packages of yeast
(additive material). Baking a loaf of bread requires 5
ounces of flour and 1 package of yeast. Each loaf of
bread can be sold for $30. The baker may purchase
additional flour at $4 or sell leftover flour at the same
price.
Formulate this problem to maximize the baker’s profits.
Example 09- Solution
Decision variables:
X1= number of loaves of bread baked
X2=number of ounces by which flour supply is increased by
cash transactions (so, X2 can be positive (purchasing),
negative (selling) or zero)
Note: The baker can not purchase any additional yeast!
Example 09-LINGO Model
Example 09-LINGO Solution
Example 09-LINGO Solution
Brute Production Process (@GIN):
Rylon Corporation manufactures Brute and Chanelle
perfumes. The raw material needed to manufacture each type
of perfume can be purchased for 3$ per pound. Processing 1
lb of raw material requires 1 hour of laboratory time. Each
pound of processed raw material yiels 3 oz of Regular Brute
Perfume and 4 oz of Regular Chanelle Perfume. Regular
Brute can be sold for 7$/oz and Regular Chanelle for $6/oz.
Rylon also has the option of further processing Regular Brute
and Regular Chanelle to produce Luxury Brute, sold at
18$/oz, and Luxury Chanelle,sold at $14/oz. Each ounce of
Regular Brute processed further requires an additional 3
hours of laboratory time and 4$ processing cost and yields 1
oz of Luxury Brute. Each ounce of Regular Chanelle
processed further requires an additional 2 hours of laboratory
time and $4 processing cost and yields 1 oz of Luxury
Chanelle. Each year, Rylon has 6,000 hours of laboratory time
available and can purchase upto 4,000 lb of raw material.
Formulate an LP that can be used to determine how
Rylon can maximize profits. Assume that the cost of
laboratory hours is a fixed cost.
Solution to Example 1:
Determine how much raw material to purchase and mow much of each
type of perfume to produce
X1=number of ounces of Regular Brute sold annually
X2=number of ounces of Luxury Brute sold annuallly
X3=number of ounces of Regular Chanelle sold annually
X4=number of ounces of Luxury Chanelle sold annually
X5=number of pounds of raw material purchased annually
Profit=revenues from perfume sales-processing costs-costs of
purchasing raw material
=7X1+18X2+6X3+14X4-(4X2+4X4)-3X5
Solution to Example 1:
Max z =7X1+14X2+6X3+10X4-3X5
Subject to
X5<=4,000
3X2+2X4+X5<=6,000
X1+X2-3X5=0
X3+X4-4X5=0
Xi>=0 (i=1,2,3,4,5)
Using Sets in LINGO
When and Why?
Whenever you are modeling situations in real life there
will be one or more groups of related objects. Such as
customers, products, trucks, machines or workers.
LINGO allows you to group these related objects
together in to sets. So, sets are groups of related
objects.
Using sets, you can write a series of similar constraints
in a single statement.
Each member in the set may have one or more
characteristics associated with it, which are called
attributes (They can be known in advance or unknown
that LINGO solves for).
Types of Sets in LINGO
Some simple examples:
– Product:
– Truck:
– Warehouse:
Price
Hauling capacity
Storage Capacity
Types of Sets:
1) Primitive Set
2) Derived Set
Primitive Set: Compose only of objects, which can not
be further reduced (e.g.: A set composed of 8 trucks).
Derived set: It made from one or more other sets, or it
is a subset of other sets or it is a combination of
elements from the other sets.
How to use Sets?
Sets are defied in sets section. A sets section may
appear anywhere in a model.
It’s necessary to define sets before using them in a
model.
The sets section begins with “SETS:” and ends with
“ENDSETS”.
Defining Primitive Sets:
Set name / member list / :attribute list ;
Example: machines / m1, m2, m3, m4 / :working hours;
Note: member list may be either explicitly or implicitly.
Defining Primitive Sets
Explicitly:
When the list of members is explicit (cleared and
separated) by entering a unique name for each member.
Example: Trucks / Truck1 Truck2/ Hauling capacity ;
Implicitly:
When there is no list of each member’s name. In fact it is
not necessary to list a name for each set member.
Set Name / Member1 .. Member N / :attribute list ;
Where member1 is the name of the fist member in the
set and member N is the name if the last member.
LINGO automatically generates all the intermediate
member names between member1 and member N.
Defining Primitive Sets
Implicit member list format:
1- 1..n
e.g. 1..5
members: 1, 2, 3, 4, 5
2- day M .. day N e.g. MON..FRI
members: MON, TUE, WED, THU, FRI
3- String M.. String N
e.g. Mach1 .. Mach4
members: Mach1, Mach2, Mach3, Mach4
4- Month M .. Month N
e.g. OCT .. JAN
members: OCT, NOV, DEC, JAN
5- MonthYearM ... MonthYearN e.g. OCT2001 ..JAN2002
members:Oct2001, Nov2001, Dec2001,
Jan2002
Defining Primitive Sets
Set members may have one or more attributes. Each
attribute displays one property of each member.
Example: students / 1..30 / name, surname, ID ;
Example: Suppose that The XYZ company has 5 warehouses
supplying 8 vendors with their products. There is limitation
for capacity and demand for warehouses and vendors,
respectively.
Define the appropriate sets for warehouses and vendors.
SETS:
Warehouse /1..5/ :capacity;
Vendor / 1..8 / demand;
ENDSETS
Defining Derived Sets
A derived set definition had the following syntax:
Set name (parent-set-list) / member list / :attribute list;
The parent-set-list is a list of previously defined sets,
separated by commas.
Without specifying a member list element, LINGO
constructs (builds) all combinations of members from
each parent set to create the members of the new
derived set.
Note: For variable having more than one indices (e.g.:
Xij) they should be represented in a derived set.
Defining Derived Sets
Example1:
SETS:
Product / A B / ;
Machine / M N / ;
Week / 1..2 / ;
Allowed (Product, Machine, Week) ;
ENDSETS
Member
(A, M, 1)
Allowed set
(A, N, 1)
members:
(B, M, 1)
(B, N, 1)
Member
(A, M, 2)
(A, N, 2)
(B, M, 2)
(B, N, 2)
Defining Derived Sets
Example2 (Explicit member):
SETS:
Product / A B / ;
Machine / M N / ;
Week / 1..2 / ;
Allowed (Product, Machine, Week) / A M 1, B N 2 / ;
ENDSETS
Member
Member
(A, M, 1)
(B, N, 2)
Allowed set
members:
Example (defining primitive set)
Define the sets and attributes by using decision variables:
- REGt: Number of cars produced by regular time labor,
during month t (t= 1 … 12)
- OVt: Number of cars produced by over time labor,
during month t (t=1 … 12)
Solution:
SETS:
Month / 1..12 / :REG, OV ;
ENDSETS
Example (defining derived set)
Define the sets and attributes by using decision variables:
- ADVj: dollars spends daily on advertising gas j (j=1,2,3)
- PROD ij: Barrels of crude oil i used daily to produced
gas j (i=1…3)
SETS:
Gas / 1..3 / :ADV ;
Oil / 1..3 / Capacity ;
Production (Oil, Gas) :prod ;
ENDSETS
Membership Filter
Suppose, you have already defined a set called TRUCK
and each truck (There are 100 truck in this set) has an
attribute called Capacity. So, the truck set is as following:
Truck / 1..100 / :Capacity ;
You would like to derived a subset from Truck that
contain only those trucks capable of hauling big load
(greater than 5000 kg). So, you can use a Membership
Filter as follows:
Heavy-Duty (Truck) | Capacity (&1) #GT# 5000 ;
Set
Name
Greater
than
Derived
from
Beginning of the
membership filter
Set index
placeholder
Related to the
first parent
List of Contents
LINGO operators:
– Arithmetic Operators
– Relational Operators
– Logical Operators
General Mathematical Functions
DATA section
Set-Looping Function
–
–
–
–
@SUM
@MIN
@MAX
@FOR
Problem
Arithmetic Operators
LINGO has five binary arithmetic operators:
–
–
–
–
–
^ : Power
* : Multiplication
/ : Division
+ : Addition
- : Subtracting
Relational Operators
Relational operators have numeric operand and return logical
results (True or False).
All relational operators are binary.
LINGO has six relational operators:
– # EQ # : Returns True if the operands are equal, False if not.
– # NE # : Returns True if the operands are not equal, False if the
operands are equal.
– # GT # : Returns True if the left operand is strictly greater than the
right operand, False if not.
– # GE # : Returns True if the left operand is greater than or equal to
the right operand, False otherwise.
– # LT # :Returns True if the left operand is strictly less than the right
operand, False if not.
– # LE # : Returns True if the left operand is less than or equal to the
right operand, False otherwise.
Logical Operators
Logical operators have logical operands and returns
logical results.
Logical operators are used to connect relational and
logical expressions.
Note: In an expression involving both relational and
logical operators, relational operators have higher
precedence (It means that, relational operators are
evaluated before logical operators).
LINGO has three logical operators, listed here in
descending order of precedence.
Logical Operators
# NOT #
– Negates the logical value of its operand. # NOT # is a unary
operator, with its sole argument to the right.
# AND #
– Returns the logical AND of its two operands. # AND # returns
True only if both its arguments are True, otherwise it returns
False.
# OR #
– Returns the logical OR of its two operands. # OR # returns False
only if both its arguments are False, otherwise it returns True.
General Mathematical Functions
@ABS(X): Returns the absolute value of X.
@COS(X): Returns the cosine of X, where X is the angle in
radian.
@EXP(X): Returns the e (2.718281…) to the power X.
@LGM(X): Returns the natural (base e) logarithm of the gamma
function of X. For integral value of X the following is true:
@LGM(X)=@LOG(X-1).
@LOG(X): Returns the natural logarithm of X.
@SIGN(X): Returns -1 if X<0, returns +1 if X>0.
@SIN(X): Returns the sine of X, where X is the angle in radian.
@SIZE (set name): Returns the number of elements in the set.
@SMAX (List): Returns largest value in list of scalars.
@SMIN (List): Returns smallest value in list of scalars.
@TAN(X): Returns the tangent of X, where X is the angle in
radian.
Assigning values to SETS
If you want to assign value to set attributes, you have to
use another optional section which called DATA section.
The DATA section begins with “DATA:” and ends with
“ENDDATA”.
attribute list = value list
Example2: SETS:
Trucks / T1 T2/ :Hauling capacity ;
ENDSET
DATA:
Hauling capacity = 5000 10000 ;
ENDDATA
Applying DATA section
Example2: SETS:
Set1 / s1, s2, s3 / :X, Y;
ENDSETS
DATA:
X = 1 3 5;
Y = 2 4 6;
ENDDATA
OR (for DATA section):
DATA:
X Y=12
34
5 6;
ENDDATA
Set-Looping Function
SET looping functions allow you to iterate through all the
members of a set to perform some operation.
The set looping functions are as follows:
– @FOR: It is used to generate constraints over members of a set.
– @SUM: It computes the sum of an expression over all member
of a set.
– @MIN: Computes the minimum of an expression over all
member of a set.
– @MAX: Computes the maximum of an expression over all
member of a set.
The syntax is as follows:
@function (set name | condition : expression)
@SUM set looping function
Example3: Consider the following model:
SETS:
Trucks /T1 .. T6 / :capacity;
ENDSETS
DATA:
capacity = 5 7 10 10 16 24 ;
ENDDATA
Now, we sum up the values of capacity:
Total capacity= @SUM (Trucks (j) : capacity (j));
OR (in our case):
Total capacity= @SUM (Trucks : capacity);
@SUM / using condition
Example4: In the previous model, how can we sum the
first four elements of the attribute “capacity”?
In this case we have to use conditional qualifier:
Desired capacity = @SUM (Trucks (j) | j # LE # 4 : capacity (j));
@MIN and @MAX set L. Fn.
Example5: In the previous model, how can we compute
the minimum and maximum capacity?
min-capacity = @MIN (Trucks (j) : capacity (j)) ;
max-capacity = @MAX (Trucks (j) : capacity (j)) ;
Example6: And how can we compute the minimum and
maximum capacity among the first four elements?
Again, we have to use conditional qualifier:
Desired min-capacity = @MIN (Trucks (j) | j # LE # 4 : capacity (j));
Desired max-capacity = @MAX (Trucks (j) | j # LT # 5 : capacity (j));
@FOR set looping function
The @FOR function is used to generate constraints
across members of a set. Now, consider the following
example: SETS:
Trucks / T1 .. T5 / :capacity ;
ENDSETS
If we want to limit the capacity of each truck by 10 tons,
what should we do?
We use @FOR function:
@FOR (Trucks (j) : capacity (j) <= 10) ;
Applying @FOR function
Example7: consider the following model:
SETS:
NUM / 1 .. 5 / :value, reciprocal;
ENDSETS
DATA:
value = 3 4 2 8 5;
ENDDATA
How can we compute the reciprocal of these five numbers?
@FOR (NUM (j) : reciprocal (j) = 1/value (j)) ;
Example8: If value = 0 4 2 8 5, what should we do?
@FOR (NUM (j) | value (j) # NE # 0 : reciprocal (j) =1/value (j)) ;
Appling set looping Functions
Problem:
Suppose we have seven songs, each with a
different length, that must be placed on a
cassette tape. The goal is to maximize the
number of songs on one side of the tape without
exceeding half of the total time of the music on
the side.
Problem solution
Lj: Length of the song j
Yj: equal to “1” if song j is selected, “0” otherwise
So, we have:
Max Z= Y1 +Y2 + … +Y7
Subject to:
14Y1+6Y2+… +8Y7 ≤ (1/2)(14+6+ … +8)
Yj: Binary (J=1 … 7)
Converting to LINGO
MODEL:
SETS:
Songs / 1 .. 7 / :Length, Y;
ENDSETS
DATA:
Length = 14 6 4 5 6 4 8 ;
ENDDATA
MAX = @SUM (Songs (j) : Y (j)) ;
@SUM (Songs (j) :Length (j)*Y (j))<= @SUM (Songs (j) : Length (j)) / 2;
@FOR (Songs (j) : @BIN (Y)) ;
END
LINGO answer
Transportation Models
Problem:
A transportation company wants to ship products from 6
warehouses to 5 customers with minimal shipping cost.
Capacity of warehouses and demand of customers are
given. In addition shipping costs from one city to another
are given.
Solution
Ci: Capacity of warehouse i
Dj: Dmand of customer j
Costij: Cost of shipping from warehouse i to customer j
Vij: Amount of shipping from warehouse i to customer j
6
Max Z=
5
 COSTij*Vij
i 1 j 1
Subject to:
6
5
Vij   Dj
i 1
for j=1 … 5
j 1
5
6
j 1
i 1
Vij   Ci
for i=1 ...6
Converting to LINGO
LINGO answer
LINGO answer
LINGO answer
LINGO answer
LINGO answer