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.
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
z
Subject to the constraints
8
5 x x 1 2
150
x
1 2
200
2 x1 x2
400
x
x
1
, x2
0
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).
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.
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.
Graphical solution
x2
Maximize z=8 x1+5 x2 Subject to the constraints Optimum =1800 at (100,200)
0,200) z=1200
(150,100)
z=1800 z=1000 z=400
(150,0)
z=1700
2 x1+ x2 400 x1
150
x2 200 x1, x2 0 x1
Feed Mix Problem
Minimize z = 2 x1 + 3 x2 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)
Graphical Solution of Feed Mix Problem
(0,12) z = 26 z = 30 z= 36 z = 39
(4,6)
z = 22.5
z = 42
Minimum at (7.5,2.5)
(15,0)
Maximize z = 2 x1 + x2 Subject to x1 + x2 ≤ 40 4 x1 + x2 ≤ 100 x1, x2 ≥ 0
Optimum = 60 at (20, 20)
(0,40)
z maximum at (20,20)
(25,0)
Maximize z = 2 x1 + x2 Subject to
10
x
2
2 x1 5 x2
x
3 x1
x
x
1
x
1
, x2
18
2 2
60
0
44
z is maximum at (13, 5)
Max z = 31
z = 28
(5,10)
(10,8)
(0,10) z = 20 z = 10
4
[1]
(13,5) z = 31 (14.6,0) [4]
[3]
[2]
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.
Alternative optimal solutions
Consider the LPP Maximize
z 10 x 5 x 1 2
Subject to the constraints
150
x
1 2
200
2 x1 x2
400
x
x
1
, x2
0
Graphical solution Maximize z=10 x1+5 x2 Subject to the constraints
x2
2 x1+ x2 400 (100,200) 0,200) z=600
z maximum =2000 at z=1500
z=2000
z=1000 z=400
(150,100)
(150,0)
x1
150
x2 200 x1, x2 0 x1
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.
Maximize z = 5 x1 + 7 x2 Subject to 2 x1 -
x2 ≤ -1
- x1 + 2 x2 ≤ -1 x1, x2 ≥ 0
No feasible solution
Maximize z = x1 + x2 Subject to z=70
- x1 + 3 x2 ≥ 30 - 3 x1 + x2 ≤ 30
z=50
x1, x2 ≥ 0 z=30
unbounded solution
z=20
Let x = ( x1, x2) and y = ( y1, y2) be two pints in the x1 x2 plane. Any point t = (t 1, t 2) on the line segment joining them is of the form t 1 = (1 - ) x1 + y1 t 2 = (1 - ) x2 + y2
for some : 0 1
= 0 corresponds the „left‟ endpoint x
= 1 corresponds the „right‟ endpoint y
More generally, let x=( x1, x2, …, xn) and y=( y1, y2, …, yn) be two points in the ndimensional space V n. 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 t i is given by t i = (1 - ) xi + yi for all i = 1 , 2, …n.
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
In V n, the “half spaces” x=( x1, x2, …, xn) | a1 x1+a2 x2 + …+ an xn ≤ b x=( x1, x2, …, xn) | a1 x1+a2 x2 + …+ an xn ≥ b
and the “hyperplane” x=( x1, x2, …, xn) | a1 x1+a2 x2 + …+ an xn = b
are all convex.
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 (Vertices=corners) 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.
In other words, whenever x, y are two points in S, we cannot find a : 0 < < 1 such that t = (1 - ) x + y Extreme point
Extreme point
Every point on the boundary is an Extreme point
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.
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.
