This document discusses linear programming concepts including decision variables, objective functions, and constraints. It provides examples of linear programming models and asks questions about identifying the standard form, determining extreme points from graphs of the constraint lines, and finding optimal solutions. Several linear programming models are presented and classified as having alternative optimal solutions, an unbounded solution, no feasible solution, or a redundant constraint based on analyzing the graphs of the constraint lines. One example is worked through in detail to find the optimal solution by using simultaneous equations to determine an extreme point and substituting into the objective function.

1. FACULTY OF SCIENCE AND TECHNOLOGY
SEMESTER SEPTEMBER / 2013
SBMP4203
LINEAR PROGRAMMING

2. QUESTION 1 (a)
The three basic components in Linear Programming model is:
i. Decision Variables
ii. Objective Function
iii. Constraints
QUESTION 1 (b)
Maximise
Subject to
(1)
(2)
(3)
Transform the constraint (1) and (2) into sign ≤.
From constraint (1):
The constraint (1) with sign = need to transform into two signs ≥ and ≤ . There are:

3. (i) …………. Constraint (i)
(ii)
Then, the constraint (i) should be transform into sign ≤ .
By multiplying (ii) with (-1)
× (-1)
………… constraint(ii)
From constraint (2):
Also should be convert into sign ≤.
× (-1)
……….. constraint(iii)
The constraint (3) should be the fourth constraint.
(3) ……….constraint(iv)
Then, the LP model is
Maximise
Subject to
…………(1)
……….(2)

4. Next, express the model in standard form:
Maximise
Subject to

5. QUESTION 2 (a)
The LP model given is
Maximize
Subject to
Constraint(1)
Constraint(2)
Convert the constraint (2) into sign ≤ .
From constraint (2):
× (-1)
Express the model in standard form.
Maximize
Subject to

6. QUESTION 2 (b)
Thus, from the standard form;
The initial simplex tableau is
Basis X1 X2 S1 S2 Solution
Z -3 -9 0 0 0
S1 1 6 1 0 12
S2 -1 -2 0 1 -4
QUESTION 2 (c)
The nature of constraints of the Linear Programming model is:
- Inference methods and Constraints

7. QUESTION 3
a) Alternative optimal solutions – Model 2
Maximize
Subject to
(1)
(2)
(3)
(4)
Find all the points at which the lines intersects the and axes. Those constraints are can be
converted into equations or constraint lines.
From constraint (1):
When ;  (0,3)
;  (6,0)
From constraint (2):
When ;  (0,8)
;  (4,0)
From constraint (3):

8. When ;  (0,1)
;  (-1,0)
From constraint (4):
Sketch graphs :
Graphically, several combinations of the decision variables are optimal and we can select the
most desirable optimal solutions. An optimal solution can be found at a corner or extreme point
of the feasible region. Two and more optimal extreme points exist and all the points on the
segment line connecting them are also optimal.
Thus, the model is alternative optimal solutions.

9. b) Unbounded solution – Model 4
Maximize
Subject to
(1)
(2)
Find all the points at which the lines intersects the and axes. Those constraints are can be
converted into equations or constraint lines.
From constraint (1):
From constraint (2):
When ;  (0,0)
;  (0,0)
Sketch graphs :

10. Graphically, the feasible region for this LP problem is open-ended. The objective function is to
maximize the profit of , the line not touch any corner or extreme point because there
is none.
Thus, the model is unbounded solution.
c) No feasible solution – Model 1
maximize
Subject to
(1)
(2)
Find all the points at which the lines intersects the and axes. Those constraints are can be
converted into equations or constraint lines.
From constraint (1):
When ;  (0,1)
;  (1,0)
From constraint (2):

11. When ;  (0,3)
;  (4,0)
Sketch graphs :
Graphically, it means that no feasible region exists; that is no point satisfies all the constraints
and the non-negativity conditions simultaneously.
Thus, the model is no feasible solutions.
d) Redundant constraint – Model 3
Maximize
Subject to
(1)
(2)
(3)
Find all the points at which the lines intersects the and axes. Those constraints are can be
converted into equations or constraint lines.
From constraint (1):
When ;  (0,8)
;  (12,0)

12. From constraint (2):
When ;  (0,6)
;  (3,0)
From constraint (3):
Sketch graphs :
Redundant constraint does not affect the feasible region and it can be removed from the problem.
Thus, the model is redundant constraints.

13. QUESTION 4 (a)
Given LP model is
Subject to
(1)
(2)
Find all the points at which the lines intersects the and axes. Those constraints are can be
converted into equations or constraint lines.
From constraint (1):
When ;
= 3.5  (0,3.5)
;
 (7,0)

14. From constraint (2):
When ;
 (0,8)
;  (4,0)
Sketch graphs :

15. QUESTION 4 (b)
From looking the sketching graph, we can found the extreme points.
The extreme points are:
(0,0) , (0,3.5) , (3,2) and (4,0).
QUESTION 4 (c)
Point B is an intersection point of equation for constraint (1) and constraint (2). Therefore, we
have to solve two simultaneous equations.
By using the simultaneous equations method:
constraint(1) equation
and constraint(2) equation
by multiplying constraint(2) equation by 2:

16. Then, subtraction it to constraint (1) equation to obtain:
----------------------
Now, substitute the value of in constraint (2) equation to get the value of :
Thus, B has the coordinate (3,2) and we can compute the optimum value of the objective
function to complete the analysis.
This gives
Thus,
The optimal solution occurs at (3,2) and its value is Z=13.

17. References
Dr Wan Rosmira Ismail(2010). SBMP4203 Linear Programming. Open University Malaysia.
Meteor Doc. Sdn. Bhd. Selangor Darul Ehsan.