Problem 6 Problem Set 2'3A Page 26Modified

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Problem 6 Problem Set 2.3A Page
26(Modified) Electra produces two types of
electric motors, each on a separate assembly
line. The respective daily capacities of the two
lines are 150 and 200 motors. Type I motor uses 2
units of a certain electronic component, and type
II motor uses only 1 unit. The supplier of the
component can provide 400 pieces a day.The
profits per motor of types I and II are 8 and 5
respectively. Formulate the problem as a LPP and
find the optimal daily production.
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Let the company produce x1 type I motors and x2
type II motors per day.
The objective is to find x1 and x2 so as to
Maximize the profit
Subject to the constraints
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We shall use graphical method to solve the above
problem.
Step 1 Determination of the Feasible solution
space
The non-negativity restrictions tell that the
solution space is in the first quadrant. Then we
replace each inequality constraint by an equality
and then graph the resulting line (noting that
two points will determine a line uniquely).
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Next we note that each constraint line divides
the plane into two half-spaces and that on one
half-space the constraint will be and on the
other it will be . To determine the correct
side we choose a reference point and see on which
side it lies. (Normally (0,0) is chosen. If the
constraint line passes through (0,0), then we
choose some other point.) Doing like this we
would get the feasible solution space.
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Step 2 Determination of the optimal solution
The determination of the optimal solution
requires the direction in which the objective
function will increase (decrease) in the case of
a maximization (minimization) problem. We find
this by assigning two increasing (decreasing)
values for z and then drawing the graphs of the
objective function for these two values. The
optimum solution occurs at a point beyond which
any further increase (decrease) of z will make us
leave the feasible space.
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Graphical solution
x2
Maximize z8x15x2
Subject to the constraints
Optimum 1800 at
2x1x2 ? 400 x1 ? 150 x2 ? 200
x1,x2? 0
(100,200)
(0,200)
z1200
(150,100)
z1800
z1000
x1
z1700
z400
(150,0)
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Feed Mix Problem Minimize z 2x1 3x2 Subject
to x1 3 x2 ? 15 2 x1 2 x2
? 20 3 x1 2 x2 ? 24
x1, x2 ? 0
Optimum 22.5 at (7.5, 2.5)
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Graphical Solution of Feed Mix Problem
(0,12)
z 26
z 30
z 36
z 39
(4,6)
z 42
z 22.5
(7.5,2.5)
(15,0)
Minimum at
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Maximize z 2x1 x2 Subject to x1 x2
40 4 x1 x2 100
x1, x2 0
Optimum 60 at (20, 20)
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(0,40)
z maximum at
(20,20)
(25,0)
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Maximize z 2x1 x2 Subject to
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z is maximum at (13, 5) Max z 31
z 28
(5,10)
(10,8)
1
(0,10)
z 20
(13,5)
z 10
z 31
(14.6,0)
2
3
4
z 4
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Exceptional Cases
Usually a LPP will have a unique optimal
solution. But there are problems where there may
be no solution, may have alternative optimum
solutions and unbounded solutions. We
graphically explain these cases in the following
slides. We note that the (unique) optimum
solution occurs at one of the corners of the
set of all feasible points.
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Alternative optimal solutions
Consider the LPP
Maximize
Subject to the constraints
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Graphical solution
x2
Maximize z10x15x2
Subject to the constraints
2x1x2 ? 400 x1 ? 150 x2 ? 200
x1,x2? 0
(100,200)
(0,200)
z maximum 2000 at
z1500
z600
(150,100)
z2000
z1000
x1
z400
(150,0)
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Thus we see that the objective function z is
maximum at the corner (150,200) and also has an
alternative optimum solution at the corner
(100,200). It may also be noted that z is maximum
at each point of the line segment joining them.
Thus the problem has an infinite number of
(finite) optimum solutions. This happens when the
objective function is parallel to one of the
constraint equations.
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Maximize z 5x1 7 x2 Subject to 2 x1 -
x2 -1 - x1 2 x2 -1 x1, x2
0
No feasible solution
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Maximize z x1 x2 Subject to - x1
3 x2 30 - 3 x1
x2 30 x1, x2 0
z70
z50
z30
unbounded solution
z20
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Let x (x1, x2) and y (y1, y2) be two pints in
the x1 x2 plane. Any point t (t1, t2) on the
line segment joining them is of the form
t1 (1 - ?) x1 ? y1 t2 (1 - ?) x2 ? y2 for
some ? 0 ? ? ? 1
? 0 corresponds the left endpoint x ? 1
corresponds the right endpoint y
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More generally, let x(x1, x2, , xn) and y(y1,
y2, , yn) be two points in the n-dimensional
space Vn. The line segment joining x and y is the
set of all points of the form t (1 - ?) x ?
y, for all ? 0 ? ? ? 1 in the sense that
the ith coordinate of t, namely ti is given by
ti (1 - ?) xi ? yi for all i 1, 2,
n.
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Convex sets A subset S in the n-space Vn is
called convex if x, y belong to S implies the
line segment joining them also lies in S.
convex subset
NOT convex
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In Vn, the half spaces
x(x1, x2, , xn) a1x1a2x2 anxn b
x(x1, x2, , xn) a1x1a2x2 anxn b
and the hyperplane
x(x1, x2, , xn) a1x1a2x2 anxn b
are all convex.
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Theorem The intersection of any number of convex
sets is a convex set.
Corollary The set of all feasible solutions of
an LPP is a convex subset.
Extreme Points (Verticescorners) Let S be a
convex set. A point t in S is called an extreme
point of S if it is not strictly between two
distinct points of S.
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In other words, whenever x, y are two points in
S, we cannot find a ? 0 lt ? lt 1 such that t (1
- ?) x ? y
Extreme point
Every point on the boundary is an Extreme point
Extreme point
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Existence of extreme points Theorem Let S be a
nonempty, closed, convex set that is either
bounded from above or bounded from below. Then S
has at least one extreme point.
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Extreme Points and LPP Suppose the set SF of all
feasible solutions of a LPP is nonempty and
bounded. (We already know it is closed and
convex.) Then the optimum solution of the LPP
occurs at a corner (extreme point) of SF.
Proof You may see the book by JC Pant.