Solving Linear Programming Problems

How much fruit servings would the family have to consume on a daily basis per person to minimize their cost?Solution: We begin step-wise with the formulation of the problem first.Bananas cost 30 rupees per dozen (6 servings) and apples cost 80 rupees per kg (8 servings).

Tags: Essay Against War DrugsCurrent Hot Topics For Essay WritingGun Control Essay OutlineCheap Christmas Wrapping Paper AustraliaLegalizing Marijuana Research PaperHelp Me With My Essay FreeHoliday Spot EssayPsychopathy EssayShort Essays By ThoreauPierce County Assigned Counsel

The constraint variables – ‘x’ = number of banana servings taken and ‘y’ = number of servings of apples taken. Total Cost C = 5x 10y Constraints: x ≥ 0; y ≥ 0 (non-negative number of servings) Total Vitamin C intake: 8.8x 5.2y ≥ 20 (1) 8.8x 5.2y ≤ 60 (2) To check for the validity of the equations, put x=0, y=0 in (1). Therefore, we must choose the side opposite to the origin as our valid region. Thus we should slide the ruler in such a way that a point is reached, which: 1) lies in the feasible region 2) is closer to the origin as compared to the other points This would be our Optimum Point. It is the one which you will get at the extreme right side of the feasible region here.

Similarly, the side towards origin is the valid region for equation 2) Feasible Region: As per the analysis above, the feasible region for this problem would be the one in between the red and blue lines in the graph! I’ve also shown the position in which your ruler needs to be to get this point by the line in green.

This is used to determine the domain of the available space, which can result in a feasible solution. A simple method is to put the coordinates of the origin (0,0) in the problem and determine whether the objective function takes on a physical solution or not.

If yes, then the side of the constraint lines on which the origin lies is the valid side. The feasible solution region on the graph is the one which is satisfied by all the constraints.

It could be viewed as the intersection of the valid regions of each constraint line as well.

Choosing any point in this area would result in a valid solution for our objective function.Here we are going to concentrate on one of the most basic methods to handle a linear programming problem i.e. In principle, this method works for almost all different types of problems but gets more and more difficult to solve when the number of decision variables and the constraints increases. We will first discuss the steps of the algorithm: We have already understood the mathematical formulation of an LP problem in a previous section.Therefore, we’ll illustrate it in a simple case i.e. Note that this is the most crucial step as all the subsequent steps depend on our analysis here.Now begin from the far corner of the graph and tend to slide it towards the origin. Once you locate the optimum point, you’ll need to find its coordinates.This can be done by drawing two perpendicular lines from the point onto the coordinate axes and noting down the coordinates.The graph must be constructed in ‘n’ dimensions, where ‘n’ is the number of decision variables.This should give you an idea about the complexity of this step if the number of decision variables increases.One must know that one cannot imagine more than 3-dimensions anyway!The constraint lines can be constructed by joining the horizontal and vertical intercepts found from each constraint equation.For the direction of the objective function; let us plot 5x 10y = 50. Now we must calculate the coordinates of this point.Now take a ruler and place it on the straight line of the objective function.