Linear Programming: Formulations & Graphical Solution. Introduction To Linear Programming.
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(usage of raw material by both paints) ≤(maximum raw material availability)
XE + 2XI ≤ 6 (raw material A)
2XE + XI ≤ 8 (raw material B)
(excess amount of interior over exterior paint) ≤ 1 ton per day
(demand for interior paint) ≤ 2 tons per day
XI - XE ≤ 1
XI ≤ 2
XI ≥ 0
XE ≥ 0
This is a typical optimization problem.
Any values of x1, x2 that satisfy all the constraints of the model is called a feasible solution. We are interested in finding the optimumfeasible solution that gives the maximum profit while satisfying all the constraints.
More generally, an optimization problem looks as follows:
Determine the decision variablesx1, x2, …, xn so as to optimize an objectivefunctionf (x1, x2, …, xn) satisfying the constraints
gi (x1, x2, …, xn) ≤ bi (i=1, 2, …, m).
An optimization problem is called a Linear Programming Problem (LPP) when the objective function and all the constraints are linear functions of the decision variables, x1, x2, …, xn. We also include the “non-negativity restrictions”, namely xj ≥ 0 for all j=1, 2, …, n.
Thus a typical LPP is of the form:
z = c1 x1 + c2 x2+ …+ cn xn
subject to the constraints:
a11 x1 + a12 x2 + … + a1n xn ≤ b1
a21 x1 + a22 x2 + … + a2n xn ≤ b2
. . .
am1 x1 + am2 x2 + … + amn xn ≤ bm
x1, x2, …, xn 0
Aluminum AlloySteel Alloy
Deluxe 2 3
Professional 4 2
Pounds of each alloy needed per frame
The Burroughs garment company manufactures men's shirts and women’s blouses for Walmark Discount stores. Walmark will accept all the production supplied by Burroughs. The production process includes cutting, sewing and packaging. Burroughs employs 25 workers in the cutting department, 35 in the sewing department and 5 in the packaging department. The factory works one 8-hour shift, 5 days a week. The following table gives the time requirements and the profits per unit for the two garments:
Determine the optimal weekly production schedule for Burroughs.
Assume that Burroughs produces x1 shirts and x2 blouses per week.
Profit got = 8 x1 + 12 x2
Time spent on cutting = 20 x1 + 60 x2 mts
Time spent on sewing = 70 x1 + 60 x2 mts
Time spent on packaging = 12 x1 + 4 x2 mts
The objective is to find x1, x2 so as to
maximize the profit z = 8 x1 + 12 x2
satisfying the constraints:
20 x1 + 60 x2≤ 25 40 60
70 x1 + 60 x2 ≤ 35 40 60
12 x1 + 4 x2 ≤ 5 40 60
x1, x2≥ 0, integers