BT1093 Matematik Perniagaan Business Mathematics

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BT1093 Matematik Perniagaan(Business
Mathematics)

• Week 7, Semester 1, 2006/2007
• School of Business Economics, Universiti
  Malaysia Sabah

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Linear Programming
Consider the following situation A factory has
two machines, each able to operate for a maximum
of 40 hours per week.
The factory produces two products A and B.
Product A needs one hours work on machine 1 and
two hours on machine 2.
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Product B needs one hour on machine 1 and one
hour on machine 2.
Product A has profits of 10 per unit and product
B 8 per unit.
How many of each should be produced to maximise
profits?
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Solving linear programming problems requires the
following skills
Creating equations and inequalities
1.
2.
Graphing straight lines
Solving systems of equations
3.
4.
Correct interpretation of mathematical
calculations.
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The profit equation is called an objective
function the machine inequalities are called
constraints.
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2.
The constraints are graphed.
-------------(1)
-------------(2)
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40
Feasible region
40
20
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The area on the graph that is common to both is
called the feasible region.
The solution to the problem lies inside the
feasible region.
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3.
It can be shown that the objective function
(profit) is a maximum at one of the corner points
of the feasible region.
From the diagram, the corner points are (0, 0),
(0, 40), (20, 0).
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4.
A maximum profit of 320 is made when no units of
product A are produced and 40 units of product B
are produced.
Suppose a third machine is introduced into the
production process. It can be used for up to 60
hours per week. Product A needs two hours work
and Product B three hours.
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Adding this equation to the diagram
y
40
G
20
F
x
E
30
40
H
20
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The corner points are marked E to H.
The coordinates of F can be calculated using any
method.
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Under the conditions, maximum profit is achieved
when 15 units of product A and 10 units of
product B are produced.
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How well is machinery being used?
In the first situation where 40 units of B are
produced, both machines are being used to full
capacity since 40 units of B uses the maximum 40
hours on machine 1 and 40 hours on machine 2.
In the second situation, substituting F (15, 10)
into the constraint equations will reveal which,
if any, of the machines are being used to full
capacity.
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Machine 1 15 10 25 (40 max.)
Machine 2 2(15) 10 40 (40 max.)
Machine 3 2(15) 3(10) 60 (60 max.)
Machines 2 and 3 then are being used to
full capacity.
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Example 10-2
Maximise
subject to
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Constraints
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means both must be positive.
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y
180
110
G
70
F
x
H
180
105
73.33
E
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Coordinates
E (73.33, 0) G (0, 70) H (0, 0)
F 3x 2y 220 2x 3y 210
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The function to be maximised is
P attains a maximum value of 468 when x 48 and
y 38.
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So far, the challenge has been to maximise the
objective function, but this is not always the
case. If the objective function represents
COST, for example, then the the aim will be to
minimise the objective function.
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Example
Minimise
subject to
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Constraints
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y
9
G
4
F
H
x
3
I
E
11
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Coordinates
------------------------(1)
-----------(2)
Substitute (1) into (2)
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--------------(1)
---------------(2)
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---------------(1)
---------------(2)
from (2) y 1.71
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Required to minimise Z x y
Z has a minimum value of 3 when x 3 and y 0.
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Tutorial Questions

• 7.1 ? 9, 17, 28
• 7.2 ? 1, 7, 13, 15, 18