View
229
Download
1
Category
Tags:
Preview:
Citation preview
CONTENTS
BASIC CONCEPTS
EXERCISES
DIET PROBLEM
REFERENCES
WHO WE ARE
بسم الله الرحمن الرحيم
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• CONTENTS
DIET PROBLEM
BASIC CONCEPTS
EXERCISES
DIET PROBLEM
Linear programming problems Graphical method of solution
Problem 1 Problem 2 Problem 3 Graphical solution exercise
REFERENCES
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• DIET PROBLEM
DIET PROBLEM
http://www-neos.mcs.anl.gov/CaseStudies/dietpy/WebForms/table.html
Weight is a big problem in our live , but do you know that there are some linear programming online solutions ?
Diet Problem
From here I invite you to visit the following link and have more benefits in your life .
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
WHO WE ARE?
• WHO WE ARE
HANAN HASAN AL-MARHABI
GRADUATED FROM KING ABDULAZIZ UNIVERSITY
WORK AS A TEACHER ASSESTANT IN KING SAUD UNIVERSITY
FOR MORE INFORMATION, WELCOME AT HANAN’S WEBSITEhttp://faculty.ksu.edu.sa/techpen/techhome/Pages/techhome1.aspx
EMAILTO: HAN1.MAR1@HOTMAIL.COM
DIET PROBLEM
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
Basic concepts involve ;
• BASIC CONCEPTS
- Linear programming problems
- Graphical method of solution
DIET PROBLEM
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
LINEAR PROGRAMMING PROBLEMES
• BASIC CONCEPTS
A linear programming problem is one in which we are to find the maximum or minimum value of a linear expression;
(called the objective function), It would be Max z Or Min c
subject to a number of linear constraints of the form
aX1 + bX2 + cX3 + . . .≤ N
aX1 + bX2 + cX3 + . . .
Or aX1 + bX2 + cX3 + . . .≥ N.
X1 , X2 , X3 , . . .> 0
DIET PROBLEM
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
LINEAR PROGRAMMING PROBLEMES
• BASIC CONCEPTS
The largest or smallest value of the objective function is called the optimal value, and
a collection of values of X1, X2, X3, . . . that gives the optimal value constitutes an optimal solution.
The variables X1, X2, X3, . . . are called the decision variables.
EXSERCISE
DIET PROBLEM
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
LINEAR PROGRAMMING PROBLEMES
• BASIC CONCEPTS
If the objective function is
And the constraints are;4X1 + 3X2 – X3 ≥ 3
Max z = 3X1- 2X2 + 4X3
X1+ 2X2 + X3 ≤ 4X1 , X2 , X3 , . . .> 0
Why can't I simply choose, say, X3 to be really large (X3 = 1,000,000 say) and thereby make Max z as large as I want? You can't because;
A - That would make z too large.B – You have to change X1 first.C – It would violate the second constrain .
DIET PROBLEM
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
GRAPHICAL METHOD OF SOLUTION
• BASIC CONCEPTS
The graphical method for solving linear programming problems in two unknowns is as follows;
A- Graph the feasible region by Draw the line ax + by = c ...(1) For each constraint,.
B- Compute the coordinates of the corner points.
C- Substitute the coordinates of the corner points into the objective function to see which gives the optimal value.
D- Optimal solutions always exist when the feasible region is bounded, but may or may not exist when the feasible region is unbounded.
X2
X1
DIET PROBLEM
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Problem 1 Problem 2
problem 3 Graphical solution exercise
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
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 (tones) Space capacity (cubic meters)
Front 10 6800
Center 16 8700
rear 18 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.
Problem 1
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
The following four cargoes are available for shipment on the next flight:
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 maximized.
Cont …
Cargo Weight capacity (tones) Volume (cubic meters/tone)
Profit (£/tone)
C1 18 480 310
C2 15 650 380
C3 23 580 350
C4 12 390 285
Formulate the above problem as a linear program
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM constraints :
here I have 4 cargoes and I need to distribute them among three compartments in the plane,Cargoes : C1 , C2 , C3 , C4 Compartments : front , center , rear My problem is search how can I get best distribution that subject to mentioned constraints.
solution
Here I have tow major types of constraints, Cargoes constraints - compartment constraints
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Cont …First : cargo constraints :
Weights constraints of each cargo respected among each compartment , and as I have 4 cargoes I will have 4 constraints
C1 C2 C3 C4
X11 + X12 +X13
let assume Xij , i =cargo , j = plane compartment
< 18
X21 + X22 +X23 < 15
X31 + X32 +X33 < 23
X41 + X42 +X43 < 12
Weights
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Cont …Second: compartment constraints :
here I have 3 compartments so I will have 4 constraints related to weights and volumes as mentioned.
C1 C2 C3 C4
X11 + X21 +X31
let assume Xij , i =cargo , j = plane compartment
< 10Weight constraints
+X41
X12 + X22 +X32 < 16+X42
X13 + X23 +X33 < 8+X43
480X11 + 650X21+580X31 < 6800
Volume constraints
+390X41
480X12 + 650X22+580X32 < 8700+390X42
480X13 + 650X23+580X33 < 5300+390X43
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Final constraints is Cont…
those proportions should be same , so proportion for any compartment equal to all cargos divided by weight of that compartment , so constraint will be :
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
8 16 10
X11 + X21 +X31 +X41
10= X12 + X22 +X32 +X42
16X13 + X23 +X33 +X43=
8
Xij > 0
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Objective
Cont…
The objective is to maximize total profit :
310(X11 + X12 +X13) + 380(X21+ X22 +X23)
+ 350(X31 + X32 +X33) + 385 (X41 + X42 +X43)
Max z =
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
A company manufactures four products (1,2,3,4) on two machines (X and Y). The time (in minutes) to process one unit of each product on each machine is shown below:
The profit per unit for each product (1,2,3,4) is £10, £12, £17 and £8 respectively. Product 1 must be produced on both machines X and Y but products 2, 3 and 4 can be produced on either machine.
Problem 2
product X Y1 10 27
2 12 19
3 13 33
4 8 23
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
The factory is very small and this means that floor space is very limited. Only one week's production is stored in 50 square meters of floor space where the floor space taken up by each product is 0.1, 0.15, 0.5 and 0.05 (square meters) for products 1, 2, 3 and 4 respectively.
Customer requirements mean that the amount of product 3 produced should be related to the amount of product 2 produced. Over a week approximately twice as many units of product 2 should be produced as product 3.
Cont ….
Machine X is out of action (for maintenance/because of breakdown) 5% of the time and machine Y 7% of the time.
Assuming a working week 35 hours long formulate the problem of how to manufacture these products as a linear program.
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
So variables are :
here 4 products produced by tow machinessolution
Let assume
Xi = the amount of product i(1,2,3,4) produced by machine X Yi = the amount of product (1,2,3,4) produced by machine Y
each product can be produced by any of tow machines so will expressed by tow variables (X2 , Y2). ( X3,Y3), (X4,Y4 ) , just product 1 should produced by both machines so it will expressed on one variable ( X1 )
X1 , X2 , Y2 , X3 , Y3 , X4 , Y4
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Constraints :Cont…
• Floor space :
The factory is very small and this means that floor space is very limited. Only one week's production is stored in 50 square meters of floor space where the floor space taken up by each product is 0.1, 0.15, 0.5 and 0.05 (square meters) for products 1, 2, 3 and 4 respectively.
0.1X1 + 0.15(X2 + Y2) + 0. 5(X3 + Y3)
+ 0.05(X4 + Y4) < 50
• Costumer requirement :
Over a week approximately twice as many units of product 2 should be produced as product 3.
X2 + Y2 = 2(X3 + Y3)
• Available time :
Machine X is out of action (for maintenance/because of breakdown) 5% of the time and machine Y 7% of the time.
10X1 +12 X2 +13X3 < 0.95+8X4 (35) (60)
27Y1 + 19Y2 +33Y3 <+23Y4 0.93 (35) (60)
Xij > 0
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Objective :
Cont…
10X1 + 12(X2 + Y2) + 17(X3 + Y3) + 8(X4 + Y4)
The profit per unit for each product (1,2,3,4) is £10, £12, £17 and £8 respectively.
Max z =
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
A company assembles four products (1, 2, 3, 4) from delivered components. The profit per unit for each product (1, 2, 3, 4) is £10, £15, £22 and £17 respectively. The maximum demand in the next week for each product (1, 2, 3, 4) is 50, 60, 85 and 70 units respectively.
Problem 3
Sages Product 1 Product 2 Prod6uct 3 Product 4
A 2 2 1 1
B 2 4 1 2
C 3 6 1 5
There are three stages (A, B, C) in the manual assembly of each product and the man-hours needed for each stage per unit of product are shown below:
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
The nominal time available in the next week for assembly at each stage (A, B, C) is 160, 200 and 80 man-hours respectively.
Cont…
It is possible to vary the man-hours spent on assembly at each stage such that workers previously employed on stage B assembly could spend up to 20% of their time on stage A assembly and workers previously employed on stage C assembly could spend up to 30% of their time on stage A assembly.
Production constraints also require that the ratio (product 1 units assembled)/(product 4 units assembled) must lie between 0.9 and 1.15.
Formulate the problem of deciding how much to produce next week as a linear program.
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
here 4 products produced solution
Let assume
Xi = the amount of product i(1,2,3,4) produced by machine X
Variables : X1 , X2 , X3 , X4
additional transferred time variables : TBA ,TCA
TBA be the amount of time transferred from B to A
TCA be the amount of time transferred from C to A
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Constraints :Cont…
• Maximum demand :
The maximum demand in the next week for each product (1, 2, 3, 4) is 50, 60, 85 and 70 units respectively.
X1 < 50 X2 < 60
X3 < 85
• Work time :
The nominal time available in the next week for assembly at each stage (A, B, C) is 160, 200 and 80 man-hours respectively.
workers previously employed on stage B assembly could spend up to 20% of their time on stage A assembly and workers previously employed on stage C assembly could spend up to 30% of their time on stage A assembly.
2X1 +2 X2 + X3 < 160+ X4
2X1 + 4X2 +X3 <+2X4 160
X3 < 70
3X1 + 6X2 +X3 <+5X4 80
+ TBA+ TCA
- TBA
- TCA
TBA < 0.2(200)
TCA < 0.3(80)
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Constraints :Cont…
• Ratio constraint :the ratio (product 1 units assembled)/(product 4 units assembled) must lie between 0.9 and 1.15.
0.9 <X1
X4 < 1.15
The profit per unit for each product (1, 2, 3, 4) is £10, £15, £22 and £17 respectively
10X1+15 X2 + 22X3+ 17X4
Objective :
Max z =
Xij > 0
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Max Z = 50x + 18y
Subject to the constraints : 2X + Y < 100 X + Y < 80 X , Y > 0
Graphical solution exercise
Sep 1 : Turn constrains into equations
2X + Y = 100
X + Y = 80
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Cont…
Sep 2 : Draw straight lines for the following equations ,
2X + Y = 100X + Y = 80
To determine tow points on straight line 2X + Y = 100
Put Y = 0 2X = 100
=> X = 50
Put X = 0 y = 100
=> Y = 100 (50, 0) , (0 , 100 ) are the points on the line (1)
To determine tow points on straight line X + Y = 80
Put Y = 0 X = 80
=> X = 80
Put X = 0 y = 80
=> Y = 80 (80, 0) , (0 , 80 ) are the points on the line (2)
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Cont…
(50, 0) , (0 , 100 ) line (1)(80, 0) , (0 , 80 ) line (2)
Y
X
(0 , 100 )
(0 , 80 )
(50,
0)
(80,
0)
A- Draw the feasible .
B- Mark the corner points then determine values of (X,Y ) on each corner’s point.
(0 , 0 )
A
B
C
D
(20 , 60 )
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Cont…Y
X
(0 , 100 )
(0 , 80 )
(50,
0)
(80,
0)
(0 , 0 )
A
B
C
D
(20 , 60 ) Step 3 : Substitute the coordinates of the corner points into the objective function to see which gives the optimal value.
Corner points O.F. O.S.
A ( 0 , 80 )B ( 20 , 60 )
C ( 50 , 0 )
D ( 0 , 0 )
50 (0) + 18 (80)= 1440
50 (20) + 18 (60)= 2080
50 (50) + 18 (0)= 2500
50 (0) + 18 (0)= 0
B ( 20 , 60 )
X = 20 Y = 60 Max z = 2080
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
• EXERCISES
DIET PROBLEM
Cont…
Accurate results : By use the mathematical method
2X + Y = 100X + Y = 80
Then use ( X=20 ) to determine Y on any equation
(- 1 ) ×
X = 20
X + Y = 80 X = 20Y = 80
X + Y = ( 20 , 80 )
Max Z = 50x + 18y
= 50 (20) + 18 (60)= 2080
CONTENTS
BASIC CONCEPTS
EXERCISES
REFERENCES
WHO WE ARE
REFERENCES ;
• REFERENCES
http://people.brunel.ac.uk/~mastjjb/jeb/or/contents.html
DIET PROBLEM http://www.tutorvista.com/content/math/statistics-and-probability/linear-programming/graphical-solution-linear-programming.php
http://people.hofstra.edu/Stefan_Waner/RealWorld/Summary4.html
Recommended