The given problem stated is appropriately mathematical form as LP model is as follows: Determine the member of male and female workers to be employed in order to maximize the monthly total return.įormulate the linear programming – Problem. Data collected on the performance of these workers indicate that a male member contributes Rs 10,000 per month to total return of the industry, while the female worker contribute Rs.8500 per month. 6000 per month, while the female worker doing the same work as the male member gets Rs. The male members of the skilled workers are paid Rs. Due to the limitations of space and finance, the member of workers employed cannot exceed and their salary bill not more than Rs. The problem cannot be solved graphically or by any other method of solving L.P problems i.e., the solution of the problem does not exist.Ī small-scale industry manufactures electrical regulators, the assembly of which being accomplished by a small group of skilled workers, both men and women.
These two shaded regions in the first quadrant do not overlap with the result that there is no point (x, y) common to the two shaded regions. The following figure indicates two shaded regions, one satisfying the constraints -2x + 3y ≤, 9 and the other satisfying the constraints 3x-2y ≤ -20. The problem therefore, has an unbounded solution. As it is not possible to get the farthest corner for the shaded corner region, the maximum value of z cannot be found as it occurs at infinity only. Since we are interested in maximizing z, we increase the value of z till the dotted line passes through the farthest corner of the shaded region from the origin. As z is increased from zero, this dotted line moves to the right, parallel to itself. The objective function, When z = 0 gives the equation 5x 1 + 4x 2 = 0, which is shown by the dotted line passing through the origin 0. This line will represent the equation x 1 + 2x 2 = 2000 as shown in the following figure.Ĭlearly any point P lying on or below the line x 1 + 2x 2 = 2000 will satisfy the inequality x 1 + 2x 2 ≤ 2000 [if x 1 = 500, x 2 = 500 then 500 + 2 x 500 0 is shaded in Fig (1). Similarly mark another Point M such that OM = 1000 To mark the point L such that OL = 2000 by assuming a suitable scales say 100 units = 2 cm
Obviously, any point (x 1, x 2) in the positive quadrant will certainly satisfy non-negativity restrictions: x 1 ≥ 0 x 2 ≥ 0 (Graph the inequality constraints) Consider two mutually perpendicular lines 0x 1, and 0x 2 as axes of coordinates.
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