# 101878939-Class-4 by sakinajhatial

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```									 Linear Programming is a technique for determining
an optimum
schedule of interdependent activities in view of the
'linear' means that the relationships are represented by
available resources.
straight lines, i.e., the relationships are of the form

A = C + Bx.

The word 'programming' is just an other word for ‘planning’ and
is concerned with optimal allocation of limited resources.

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Basic Terminology
   Linear Function

A linear function contains terms each of which is
composed of only a single, continuous variable raised to (and
only to) the power of 1.
   Objective Function

It is a linear function of the decision variables expressing
the objective of the decision-maker. The most typical forms
of objective functions are:

maximize f(x) or minimize f(x).
   Decision Variables

These are economic or physical quantities whose numerical
values indicate the solution of the linear programming
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   Constraints

These are linear equations arising out of practical limitations.
The mathematical forms of the constraints are:

f(x) ≥ b or f(x) ≤ b or f(x) = b

   Feasible Solution

Any non-negative solution which satisfies all the constraints is
known as a feasible solution.

   Optimal Solution
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General Linear Programming

Optimize ( maximize or minimize)
z = c1x1 + c2x2 + c3x3 + .........+ cnxn
subject to (constraints)
a11x1 + a12x2 + a13x3 + .........+ a1nxn (≥, ≤,
=) b1
a21x1 + a22x2 + a23x3 + .........+ a2nxn (≥, ≤,
=) b2

.................................................................................
..
am1x1 + am2x2 + am3x3 + .........+ amnxn (≥,
≤, =) bm

& (non-negative restrictions)                        x1, x2,....., xn ≥ 0
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Highway Materials (model formulation)
The specification requires
Percent Volumes
that the highway materials
meet the limitations:                     Lower Upper
Limit      Limit
Grav         5         20
el          65         85
Four sources of materials are in the vicinity of the construction
Sand         0         15
required. The composition and
site. 9000 cubic yard material is Silt
Sources
Gravel (% )Sand (%) to (%) Cost (Rs./ cu. yd)
the unit cost of material deliveredSiltthe site are:

River Gr.
15           75     10           60
Sea Gr.        30           70     00           80
Sand           00           100    00           50
On Site        05            30    65           00
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Solution
Decision Variables:
§
Let the material from River gravel = x1
§
Let the material from Sea gravel = x2
§
Let the material from Sand = x3
§
Let the material on Site = x4

Objective Function:
§
Min. Z = 60 * x1 + 80 * x2 + 50 * x3 + 0 * x4

Constraints:
§
x1 + x2 + x3 + x4 ≥ 9000 cu.yd
§
(0.15x1 + 0.3x2 + 0.0x3 + 0.05x4)/(x1 + x2 + x3 +
x4)*100 ≥ 5
§
(0.15x1 + 0.3x2 + 0.0x3 + 0.05x4)/(x1 + x2 + x3 +
x4)*100 ≤ 20
§
(0.75x1 + 0.7x2 + 1.0x3 + 0.3x4)/(x1 + x2 + x3 + x4)*100
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Traffic Engineering (model formulation)
Ø
City Nazim is studying the feasibility of introducing a mass
transit bus system that will reduce the pollution by reducing
in city driving. The initial study seeks the determination of
the minimum number of buses that can handle the
transportation needs

Ø
After collecting the information, the traffic engineer noticed
that the minimum number of buses needed to meet demand
vary with the time of day. He further discovered that the
required number of buses can be assumed constant over
successive interval of 4 hours each. (as shown in fig.)

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Ø
It was decided that, to carry out the required daily
maintenance, each bus could operate only 8 successive hours        a
day. Chalk out the policy to minimize buses.

12
10
8
6
4
2
12:0   4:00       8:00   12:0   4:00   8:00   12:0
0   0      AM         AM     0      PM     PM     0
AM                       Noo                  AM
n

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Solution
12          x              x                x            x
6    x1        2      x         4          x 6
10
3                    5
8
6
4
2
0    12:0       4:00     8:00     12:0     4:00     8:00     12:0
0          AM       AM       0        PM       PM       0
AM                           Noo                        AM
§
Min. Z = x1 + x2 + x3 + x4 + x5 + x6
n

Constraints:
§
x1 + x6 ≥ 4 (12:00 –
4:00)                                      §
x1≥ 0, x2≥ 0, x3≥ 0, x4≥
§
x1 + x2 ≥ 8 (4:00 –                    0,
8:00)
x5≥ 0, x6≥ 0
§
x2 + x3 ≥ 10 (8:00 –
8/2/12
12:00)
Inspection Problem (model formulation)
Ø
A company has two grades of inspectors, 1 and 2 to
undertake quality control inspection.
At least 1500 pieces must be inspected in an 8 hour day.
Grade 1 inspector can check 20 pieces in an hour with an
accuracy of 96%, Grade 2 inspector checks 14 pieces an hour
with an accuracy of 92%

Ø
The daily wages of Grade 1 inspectors are Rs. 200/hour while
by an inspector costs Rs. 100 to the company. If, there are, in
all, 10 Grade 1 and 15 Grade2 inspectors in the company, find
the optimal assignment of inspectors that minimizes the daily
inspection cost.

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Solution
Ø
Let x1, x2 be the nos. of Grade 1 and Grade 2 inspectors.

Ø
Objective function is to minimize the daily cost of inspection
which is equal to wages paid to the inspectors + the cost of
their inspection error.

§
Cost of Grade 1 inspector/hr = 200 +(100*0.04*20)
= Rs. 280
§
Cost of Grade 2 inspector/hr = 150+(100*0.08*14)
=Rs. 262
Therefore,
Z to minimize = 8(280x1+262x2) =2240 x1 + 2096x2

Constraints are x1 ≤ 10     and x2 ≤ 15