LINEAR AND NON-LINEARPROGRAMING MODELS Vikram Sai Petroleum Engineering ISM Dhanbad Shubham Satyarth Petroleum Engineering ISM Dhanbad OVERVIEW • Firms want to make as much money as possible or maximize their profit. • Firms want their cost to be minimum. Lower the costs, higher the profits. • This process of maximizing or minimizing is known as optimization, or mathematical programming. • Striving to find the optimum solution is a goal in many areas, not just business, or Economics and Finance. • Physics, Chemistry, Biology, Health Care or Engineering are just some of the areas these problems arise in. 1.Decision variables 2.Objective function A mathematical relationship describing an Mathematical symbols objective of the firm, in representing levels of terms of decision OPTIMIZATI activity of a firm OPTIMIZATION variables. ON MODEL MODEL 3.Constraints 4.Parameters Restrictions placed on the firm by the operating environment stated in linear relationships of the decision variables Numerical coefficients and constants used in the objective function and constraint equations. . are linear in terms of the decision variables.LINEAR PROGRAMMING Objective function is to be optimized and all the constraints. . The rest (over half) is used to make things like: Nail Polish Petroleum Jelly Tyres Brushes Plastic Toys .Petroleum Products One 42-gallon barrel of oil creates 19.4 gallons of gasoline. Each Nail polish requires 10 units of Petroleum Product and 3 hours of NonPetroleum Product. Smith has 100 units of Petroleum Product and 60 units of Non-Petroleum Product available.EXAMPLE SMUTCHUCK Fancy Stores Smith makes $200 worth of profit from every Nail polish he sells. and $50 worth of profit from every petroleum Jelly he sells. and each petroleum Jelly requires 5 units of Petroleum Product and 4 units of Non-Petroleum Product. . . EXAMPLE PRODUCT Nail Polish Petroleum Tabular Representation Available Jelly Petroleum Products 10 5 100 NonPetroleum Products 3 4 60 . EXAMPLE Let’s denote The number of Nail polishes produced by X1 The number of petroleum Jellys produced by X2. Smith’s total profit will be Z= 200 X1+50 X2 This is the objective function . Solution Methodology/Implementation GRAPHICAL SOLUTION Limited to models including two variables. . but it is very useful as it provides a better understanding of the solution. and X2 on the vertical axis. We start with a system of coordinates with X1 on the horizontal axis. 10 X1+5 X2≤ 100 .Petroleum Product Constraints Petroleum Product constraint. Non-Petroleum Product Constraint Non-Petroleum Product Constraint: 3 X1+ 4 X2≤ 60 . Feasible Region Area that satisfy all four constraints of the model: 10 X1+5 X2≤ 100 3 X1+ 4 X2≤ 60 X1 ≥ 0 X2 ≥ 0 . Z. and it obviously includes a third variable.Optimization The profit function is Z= 200 X1+50 X2. Z=100. . Let’s just take an arbitrarily selected level of profit. optimum)? . and every combination of Nail polishes and petroleum Jellys on this line will result in a $100 profit.e.Optimization The profit function becomes 100=200 X1+50 X2. Is this the best that the firm can do (i. . and no petroleum Jellys in order to maximize profits.Optimization It follows that the optimum combination of Nail polishes and petroleum Jellys is point C. Smith will need to produce all Nail polishes (10 of them). The profit in this case is Z=2000 . Optimization . . Find the coordinates of the optimum point by solving a system of two equations with two constraints. then move this line out from the origin to find the optimum.Steps of Optimization Plot the model constraints as equations on the graph Find the feasible area Represent the objective function graphically. e. or the amount of Non-Petroleum Product required for each petroleum Jelly). one would have to start from beginning in order to solve the problem. the profitability of each Nail polish. graphically representing the feasible area could be more challenging.DRAWBACKS OF GRAPHICAL APPROACH If the optimization problem is subject to multiple constraints. . If one of the parameters in the problem changes (i. which is often the case in real life. and compare these values. . calculate the value of Z for those coordinates. The point which leads to the highest (in the case of maximization problems) Z or the lowest (in the case of minimization) Z is the optimum solution. one can find the coordinates of the corners.CORNER POINT METHOD Since we know the optimum will be on the boundary of the feasible area. EXCE L . MS.OPTIMIZATION USING EXCEL Another option that is more sophisticated and more simpler is the use of MS-EXCEL to do the Optimization. or some of the constraints may not be linear constraints. the objective function may not be a linear function. real-world problems. Optimization problems that involve nonlinearities are called nonlinear programming (NLP) problem .NON-LINEAR PROGRAMMING • • In many interesting. Optimality Conditions Unconstrained optimization – multivariate calculus problem. the optimum occurs at the point where f '(X) =0 and f''(X) meets second order conditions A relative minimum occurs where f '(X) =0 and f''(X) >0 A relative maximum occurs where f '(X) =0 and f''(X) <0 . For Y=f(X). Concavity and Second Derivative Local max & global max local max f’’(x)<0 f’’(x)>0 local min f’’(x)<0 f‘’(x)>0 Local min & global min . Method The simplest method for solving nonlinear programming model is the method of substitution In substitution method the constraint equation is solved for one variable in terms of another & then substituted into the objective function. . v is the volume p is the price Substituting the value of v in the objective function .6p Where.Example Profit analysis model Z = vp-a-vb Subject to v = 1500-24. 6pb Substituting the value of constants as a=10000$ .8-49.2p Solving for p p=$34.6p2-22000 dz/dp = 1696.8-49.49 .Example Z=1500p-24. b=8$ We get Z=1696.6p2-a-1500b+24.2p 0=1696.8p-24. A curve-fitting procedure was used to estimate the weekly marketing costs required to sustain a production rate of G Gas and O Oil: ◦ Marketing cost for Gas = $25G 2 ◦ Marketing costs for Oil = ($662/3)O 2 The gross profit per unit of Gas sold is about $375. with Marketing Costs Market research indicates that wellspun could sell small numbers of Gas and Oil with no advertising. how much Gas and Oil should wellspun produce? . and the gross profit per unit of Oil is about $700.Wellspun Inc. extensive advertising would be required to sell all that could be produced. the net profits are as follows: ◦ Net profit for Gas = $37G – $25G 2 Net profit for Oil = $700O – ($662/3)O 2 Thus. the revised objective function is Maximize Profit = $375G – 25G 2 + $700 O–($662/3)O 2 Question: Considering the nonlinear marketing costs. However. Therefore. 000 800 800 600 600 400 400 200 200 1.000 1.600 1.800 Weekly profit ($) 1.200 1.Profit Graphs for Gas and Oil Weekly profit ($) 1.400 2 4 0 2 4 6 GD Production rate for doors Production rate for windows Production rate of Oil Production rate of Gas 0 OW .200 1. Wellspun Problem with Nonlinear costs . Production rate of Oil Graphical Display of Nonlinear Formulation Production rate of Gas . Thank You .
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