2. بسم الله الرحمن الرحيم CONTENTS EXERCISES DIET PROBLEM REFERENCES
BASIC CONCEPTS
EXERCISES
بسم الله الرحمن الرحيم
DIET PROBLEM
REFERENCES
WHO WE ARE

3. CONTENTS CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
BASIC CONCEPTS
Linear programming problems
BASIC CONCEPTS
Graphical method of solution
EXERCISES
EXERCISES
Problem 1
Problem 2
DIET PROBLEM
Problem 3
Graphical solution exercise
REFERENCES
DIET PROBLEM
REFERENCES
WHO WE ARE

4. DIET PROBLEM Diet Problem CONTENTS EXERCISES DIET PROBLEM REFERENCES
BASIC CONCEPTS
Weight is a big problem in our live , but do you know that there are some linear programming online solutions ?
EXERCISES
DIET PROBLEM
From here I invite you to visit the following link and have more benefits in your life .
REFERENCES
WHO WE ARE

5. WHO WE ARE WHO WE ARE? CONTENTS HANAN HASAN AL-MARHABI EXERCISES
BASIC CONCEPTS
HANAN HASAN AL-MARHABI
GRADUATED FROM KING ABDULAZIZ UNIVERSITY
EXERCISES
WORK AS A TEACHER ASSESTANT IN KING SAUD UNIVERSITY
DIET PROBLEM
REFERENCES
FOR MORE INFORMATION, WELCOME AT HANAN’S WEBSITE
WHO WE ARE
TO:

6. BASIC CONCEPTS Basic concepts involve ; - Linear programming problems
CONTENTS
Basic concepts involve ;
BASIC CONCEPTS
- Linear programming problems
EXERCISES
- Graphical method of solution
DIET PROBLEM
REFERENCES
WHO WE ARE

7. BASIC CONCEPTS LINEAR PROGRAMMING PROBLEMES CONTENTS EXERCISES
A linear programming problem is one in which we are to find the maximum or minimum value of a linear expression;
BASIC CONCEPTS
EXERCISES
aX1 + bX2 + cX
(called the objective function),
It would be Max z Or Min c
DIET PROBLEM
subject to a number of linear constraints of the form
REFERENCES
aX1 + bX2 + cX ≤ N
WHO WE ARE
Or aX1 + bX2 + cX ≥ N.
X1 , X2 , X3 , . . .> 0

8. BASIC CONCEPTS LINEAR PROGRAMMING PROBLEMES EXSERCISE CONTENTS
The largest or smallest value of the objective function is called the optimal value, and
EXERCISES
a collection of values of X1, X2, X3, that gives the optimal value constitutes an optimal solution.
DIET PROBLEM
The variables X1, X2, X3, are called the decision variables.
REFERENCES
EXSERCISE
WHO WE ARE

9. BASIC CONCEPTS LINEAR PROGRAMMING PROBLEMES CONTENTS EXERCISES
If the objective function is
BASIC CONCEPTS
Max z = 3X1- 2X2 + 4X3
And the constraints are;
EXERCISES
4X1 + 3X2 – X3 ≥ 3
X1+ 2X2 + X3 ≤ 4
DIET PROBLEM
X1 , 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;
REFERENCES
WHO WE ARE
A - That would make z too large.
B – You have to change X1 first.
C – It would violate the second constrain .

10. BASIC CONCEPTS GRAPHICAL METHOD OF SOLUTION CONTENTS EXERCISES
The graphical method for solving linear programming problems in two unknowns is as follows;
BASIC CONCEPTS X2 X1
A- Graph the feasible region by Draw the line ax + by = c ...(1)   For each constraint,.
EXERCISES
B- Compute the coordinates of the corner points.
DIET PROBLEM
C- Substitute the coordinates of the corner points into the objective function to see which gives the optimal value.
REFERENCES
WHO WE ARE
D- Optimal solutions always exist when the feasible region is bounded, but may or may not exist when the feasible region is unbounded.

11. EXERCISES Problem 1 Problem 2 problem 3 Graphical solution exercise
CONTENTS
BASIC CONCEPTS
Problem 1
Problem 2
problem 3
Graphical solution exercise
EXERCISES
DIET PROBLEM
REFERENCES
WHO WE ARE

12. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
A cargo plane has three compartments for storing cargo: front, centre and rear. These compartments have the following limits on both weight and space:
BASIC CONCEPTS
EXERCISES
Compartment
Weight capacity (tones)
Space capacity (cubic meters)
Front 10
6800
Center 16
8700
rear 18
5300
DIET PROBLEM
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.
REFERENCES
WHO WE ARE

13. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE Cont …
The following four cargoes are available for shipment on the next flight:
BASIC CONCEPTS
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
EXERCISES
DIET PROBLEM
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.
REFERENCES
WHO WE ARE
Formulate the above problem as a linear program

14. EXERCISES solution CONTENTS EXERCISES constraints : DIET PROBLEM
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.
BASIC CONCEPTS
EXERCISES
DIET PROBLEM
constraints :
Here I have tow major types of constraints,
Cargoes constraints - compartment constraints
REFERENCES
WHO WE ARE

15. EXERCISES Cont … CONTENTS First : cargo constraints : EXERCISES
Weights constraints of each cargo respected among each compartment , and as I have 4 cargoes I will have 4 constraints
BASIC CONCEPTS
EXERCISES C1 C2 C3 C4
let assume Xij , i =cargo , j = plane compartment
DIET PROBLEM
Weights
X11
+ X12
+X13
< 18
REFERENCES
X21
+ X22
+X23
< 15
X31
+ X32
+X33
< 23
WHO WE ARE
X41
+ X42
+X43
< 12

16. EXERCISES Cont … CONTENTS Second: compartment constraints : EXERCISES
here I have 3 compartments so I will have 4 constraints related to weights and volumes as mentioned.
BASIC CONCEPTS
let assume Xij , i =cargo , j = plane compartment
EXERCISES
Weight constraints
X11
+ X21
+X31
+X41
< 10
X12
+ X22
+X32
+X42
< 16
DIET PROBLEM
X13
+ X23
+X33
+X43
< 8 C1 C2 C3 C4
REFERENCES
Volume constraints
480X11
+ 650X21
+580X31
+390X41
<
6800
480X12
+ 650X22
+580X32
+390X42
<
8700
WHO WE ARE
480X13
+ 650X23
+580X33
+390X43
<
5300

17. EXERCISES Cont… CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
Final constraints is
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
BASIC CONCEPTS
EXERCISES 8 10 16
DIET PROBLEM
those proportions should be same , so proportion for any compartment equal to all cargos divided by weight of that compartment , so constraint will be :
REFERENCES
X11
+ X21
+X31
+X41
X12
+ X22
+X32
+X42
X13 =
+ X23
+X33 =
+X43
WHO WE ARE 10 8 16
Xij
>

18. EXERCISES Objective Cont… CONTENTS
BASIC CONCEPTS
The objective is to maximize total profit :
Max z =
EXERCISES
310(X11
+ X12
+X13)
+ 380(X21
+ X22
+X23)
+ 350(X31
+ X32
+X33)
+ 385 (X41
+ X42
+X43)
DIET PROBLEM
REFERENCES
WHO WE ARE

19. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
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:
BASIC CONCEPTS
product X Y 1 10 27 2 12 19 3 13 33 4 8 23
EXERCISES
DIET PROBLEM
REFERENCES
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.
WHO WE ARE

20. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
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.
BASIC CONCEPTS
EXERCISES
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.
DIET PROBLEM
Machine X is out of action (for maintenance/because of breakdown) 5% of the time and machine Y 7% of the time.
REFERENCES
Assuming a working week 35 hours long formulate the problem of how to manufacture these products as a linear program.
WHO WE ARE

21. EXERCISES solution CONTENTS EXERCISES DIET PROBLEM REFERENCES
here 4 products produced by tow machines
Let assume
BASIC CONCEPTS
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
EXERCISES
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 )
DIET PROBLEM
REFERENCES
So variables are :
X1 , X2 , Y2 , X3 , Y3 , X4 , Y4
WHO WE ARE

22. EXERCISES Cont… CONTENTS Constraints : 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.
CONTENTS
Constraints :
Machine X is out of action (for maintenance/because of breakdown) 5% of the time and machine Y 7% of the time.
Floor space :
Over a week approximately twice as many units of product 2 should be produced as product 3.
BASIC CONCEPTS
0.1X1
+ 0.15(X2
+ Y2)
+ 0. 5(X3
+ Y3)
EXERCISES
+ 0.05(X4
+ Y4)
< 50
Costumer requirement :
DIET PROBLEM X2
+ Y2
= 2(X3
+ Y3)
Available time :
REFERENCES
10X1
+12 X2
+13X3
+8X4
<
0.95
(35)
(60)
WHO WE ARE
27Y1
+ 19Y2
+33Y3
+23Y4
<
0.93
(35)
(60)
Xij
>

23. EXERCISES Cont… CONTENTS Objective : EXERCISES DIET PROBLEM REFERENCES
The profit per unit for each product (1,2,3,4) is £10, £12, £17 and £8 respectively.
CONTENTS
BASIC CONCEPTS
Objective :
EXERCISES
Max z =
10X1
+ 12(X2
+ Y2)
+ 17(X3
+ Y3)
+ 8(X4
+ Y4)
DIET PROBLEM
REFERENCES
WHO WE ARE

24. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
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.
BASIC CONCEPTS
EXERCISES
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:
DIET PROBLEM
Sages
Product 1
Product 2
Prod6uct 3
Product 4 A 2 1 B 4 C 3 6 5
REFERENCES
WHO WE ARE

25. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE Cont…
The nominal time available in the next week for assembly at each stage (A, B, C) is 160, 200 and 80 man-hours respectively.
BASIC CONCEPTS
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.
EXERCISES
DIET PROBLEM
Production constraints also require that the ratio (product 1 units assembled)/(product 4 units assembled) must lie between 0.9 and 1.15.
REFERENCES
Formulate the problem of deciding how much to produce next week as a linear program.
WHO WE ARE

26. EXERCISES solution CONTENTS EXERCISES
here 4 products produced
Let assume
BASIC CONCEPTS
Xi = the amount of product i(1,2,3,4) produced by machine X
Variables : X1 , X2 , X3 , X4
EXERCISES
additional transferred time variables : TBA ,TCA
DIET PROBLEM
TBA be the amount of time transferred from B to A
TCA be the amount of time transferred from C to A
REFERENCES
WHO WE ARE

27. EXERCISES Cont… CONTENTS Constraints : EXERCISES DIET PROBLEM + TBA
The nominal time available in the next week for assembly at each stage (A, B, C) is 160, 200 and 80 man-hours respectively.
CONTENTS
The maximum demand in the next week for each product (1, 2, 3, 4) is 50, 60, 85 and 70 units respectively.
Constraints :
Maximum demand :
BASIC CONCEPTS X1
< 50 X2
< 60
< X3 85
<
EXERCISES X3 70
Work time :
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.
DIET PROBLEM
2X1
+2 X2
+ X3
+ X4
<
160
+ TBA
+ TCA
2X1
+ 4X2
+X3
+2X4
<
160
- TBA
REFERENCES
3X1
+ 6X2
+X3
- TCA
+5X4
< 80
WHO WE ARE
TBA
<
0.2(200)
TCA
<
0.3(80)

28. EXERCISES Cont… CONTENTS Constraints : EXERCISES DIET PROBLEM
the ratio (product 1 units assembled)/(product 4 units assembled) must lie between 0.9 and 1.15.
Ratio constraint :
The profit per unit for each product (1, 2, 3, 4) is £10, £15, £22 and £17 respectively
BASIC CONCEPTS X1
0.9
<
<
1.15 X4
EXERCISES
Xij
>
DIET PROBLEM
Objective :
REFERENCES
Max z =
10X1
+15 X2
+ 22X3
+ 17X4
WHO WE ARE

29. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE
Graphical solution exercise
CONTENTS
Max Z = 50x + 18y  
Subject to the constraints :
2X + Y < 100
X + Y < 80  
X , Y > 0     
BASIC CONCEPTS
EXERCISES
DIET PROBLEM
Sep 1 : Turn constrains into equations
2X + Y = 100
REFERENCES
X + Y = 80
WHO WE ARE

30. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE Cont…
Sep 2 : Draw straight lines for the following equations ,
2X + Y = 100
BASIC CONCEPTS
X + Y = 80
To determine tow points on straight line 2X + Y = 100
EXERCISES
Put Y = 0
Put X = 0
2X = 100
y = 100
DIET PROBLEM
=> X = 50
=> Y = 100
(50, 0) , (0 , 100 ) are the points on the line (1)
To determine tow points on straight line X + Y = 80
REFERENCES
Put Y = 0
Put X = 0
X = 80
y = 80
WHO WE ARE
=> X = 80
=> Y = 80
(80, 0) , (0 , 80 ) are the points on the line (2)

31. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE Cont…
(50, 0) , (0 , 100 ) line (1)
BASIC CONCEPTS
(80, 0) , (0 , 80 ) line (2) Y X
EXERCISES
(0 , 100 )
A- Draw the feasible . A
(0 , 80 ) B
DIET PROBLEM
B- Mark the corner points then determine values of (X,Y ) on each corner’s point.
(20 , 60 )
REFERENCES C
(0 , 0 ) D
(50, 0)
(80, 0)
WHO WE ARE

32. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE Cont…Y X
Cont…
(0 , 100 )
CONTENTS A
(0 , 80 ) B
(20 , 60 )
Step 3 : Substitute the coordinates of the corner points into the objective function to see which gives the optimal value.
BASIC CONCEPTS C
EXERCISES
(0 , 0 ) D
(50, 0)
(80, 0)
Corner points
O.F.
O.S.
DIET PROBLEM
A ( 0 , 80 )
50 (0) + 18 (80)= 1440
B ( 20 , 60 )
50 (20) + 18 (60)= 2080
B ( 20 , 60 )
REFERENCES
C ( 50 , 0 )
50 (50) + 18 (0)= 2500
WHO WE ARE
D ( 0 , 0 )
50 (0) + 18 (0)= 0
X = 20
Y = 60
Max z = 2080

33. EXERCISES CONTENTS EXERCISES DIET PROBLEM REFERENCES WHO WE ARE Cont…
Accurate results : By use the mathematical method
2X + Y = 100
BASIC CONCEPTS
(- 1 ) ×
X + Y = 80
X = 20
EXERCISES
Then use ( X=20 ) to determine Y on any equation
X + Y = 80
X = 20
DIET PROBLEM
Y = 80
REFERENCES
X + Y = ( 20 , 80 )
Max Z = 50x + 18y
WHO WE ARE
= 50 (20) + 18 (60)= 2080

34. REFERENCES REFERENCES ; CONTENTS EXERCISES DIET PROBLEM REFERENCES
BASIC CONCEPTS
EXERCISES
DIET PROBLEM
REFERENCES
WHO WE ARE