Mathematical Modeling Technique design to optimize the usage of limited recourses. most extensively used in business and economics, but can also be utilized for some engineering problems
Economics-Determination of shadow prices Business application- Maximization of profit Engineering-Design of structures.
Facilitates decision-making process Keep focus on profit under any scenario provides the targets and operating strategies Optimize utilization of assets. Optimize utilization of inventory.
Military Industry Textile industry Agriculture Transportation Energy Telecommunications
Optimize utilization of the assets Optimize inventory management Optimize capacity utilization and shutdown planning Minimize losses 1)Textile industry - 2) Petroleum - Refinery Optimize inventory management Optimize black oil generation and up gradation optimize overall product mix and dispatch
An equation of the form 4x 1 + 5x 2 = 1500 5x 1 + 3x 2 = 1575 x 1 + 2x 2 = 420 defines a straight line in the x 1 -x 2 plane
The optimal solution will be: X 1 = 270 X 2 = 75
Advantages. Easy to Analyze. Disadvantages. Handle only up to 3 variables. Need to draw according scale.
Method:- All the constraints are converted into equal sign by introducing Slack variable and calculate the solution for set of equation to find out the corner points of the feasible region. Substitute all corner point of the feasible region in objective function and thereby determine the corresponding optimal solution. Maximize Z =13x 1 + 11x 2 4x 1 + 5x 2 + x 3 = 1500 5x 1 + 3x 2 + x 4 = 1575 x 1 + 2x 2 + x 5 = 420
Basic Variable Independ ent Columns Basic solution Feasi ble Extreme Point Z value (X1, X2, X3) Yes (270,75,45,0,0) Yes (270,75) 4335 (X1, X2, X4) Yes (300,60,0,-105,0) No (X1, X2, X5) Yes (260,92,0,0,-24) No (X1, X3, X4) Yes (420,0,-180,-525,0) No (X1, X3, X5) Yes (315,0,240,0,105) Yes (315,0) 4095 (X1, X4, X5) Yes (375,0,0,-300,45) No (X2, X3, X4) Yes (0,110,950,1245,0) Yes (0,110) 1210 (X2, X3, X5) Yes (0,525,-1125,0,-630) No (X2, X4, X5) Yes (0,300,0,675,120) Yes (0,300) 3300 (X3, X4, X5) Yes (0,0,1500,1575,420) Yes (0,0) 0
Introducing these slack variables into the inequality constraints and rewriting the objective function such that all variables are on the left-hand side of the equation
Identify the variable that will be assigned a nonzero value in the next iteration so as to increase the value of the objective function. This variable is called the entering variable. Identify the variable, called the leaving variable, which will be changed from a nonzero to a zero value in the next solution.