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# MATH PROGRAM ECON ANL AEB 6592

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This 19 page Class Notes was uploaded by Weldon Rau I on Friday September 18, 2015. The Class Notes belongs to AEB 6592 at University of Florida taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/206596/aeb-6592-university-of-florida in Agricultural Economics And Business at University of Florida.

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Date Created: 09/18/15

Basic Simplex Lecture IV I Some Problem Reformulation A The original problem maxz 50x1 10x2 SI 3x1 2xz g 25 x1 62 g 100 maxzi 50x1 ileZ 0 st 3x1 2x2 sl 25 x1 x2 52 100 In this formulation we turn the inequalities constraints into equalities by adding slack variables In order to maintain consistency s1 s2 2 0 Ifwe had a constraint such as 5x1 4x2 2 20 This equation would be transformed into an equality by 5x1 4xZ isl 20s12 0 In each equation we add a nonnegative slack or surplus variable B Back to the original problem we now have 2 constraint equations and 4 unknowns see notes on Canonical form 1 In general you will have n unknowns and m equations with m lt n a n 7 m variables will be nonzero in any basis or base solution including the slack variables 2 A feasible basic solution is a basic solution satisfying the nonnegativity restrictions 3 Basic variables are those variables in a particular base solution that have nonzero values 4 Nonbasic variables are those variables at zero level in the current solution 5 Each basic solution is associated with an iteration 6 An entering variable at a particular iteration is the variable going from a value of zero to a nonzero value 7 An exiting variable is the variable going from a nonzero value to a zero value 11 Steps in the Simplex Algorithm A Using the standard form compute an initial feasible or starting basic feasible solution AEB 6592 Lecture IV Professor Charles B Moss maxz 50x1 10xl SI 3x1 202 g 25 x1 62 100 max 2750xi710x1 Os1 k2 0 st 3x12x2 SI 25 x1 9 32 100 Is this solution feasible Do the equations hold with an equality and positive values for each variable gt9 x2 1 RHS z 750 71o 0 o o 51 3 2 1 o 25 52 1 1 o 1 100 This is the LP simplex form a canonical form with s1 25 and s2 100 B Select the leaving variable The leaving variable is that variable with the most negative coefficient in the objective function This implies the largest positive change in the objective function In this case x1 has the largest negative coefficient C Select the leaving variable that is select the basic variable that will become nonbasic This is accomplished by finding the most constraining RHS RHS 4 x1 X2 S1 2 750 710 0 3 2 l l l 0 2 0 0 0 25 25325 1 100 100 7100 s1 the slack variable for the labor constraint indicates that the most x1 corn production that can enter the solution is 8333 acres D Generate the next solution using GaussJordon to find the change in basis 1 This operation is called pivoting the whole step is called an iteration We are actually trying to find the next feasible solution in the direction of the optimum 2 The leaving row is called the pivot equation The element at the intersection of the pivot row and the entering column is called the pivot element 3 Question If we had a 3x3 system and the right hand side how would we solve it AEB 6592 Lecture IV Professor Charles B Moss 0 3 2391 l 112i3 311132 01i2 3 2i1Rr3R2 10 1 l 013I2 IZE 2 0 0 7154575 Raw5 I 1031R171 2R3 39 012 arm 0 0152 10 00 010571 0012 x1 x2 2x33 0 71 223 4 Divide the leaving row by the pivot element x1 62 SI sz RHS z 2 3 S1 1 3 10 0 83333 S2 5 Compute the remaining tableau by Old equation or row element 7 Entering column coefficient New pivot equation element AEB 6592 Lecture IV Professor Charles B Moss Example Objective function row cotton column 7504010 x1 9 s1 s2 RHS z 7507750951 7107750 077500 0471020 o5025 x1 1 0 0 83333 s2 1711 171 0710 1 16667 or x1 62 S1 32 RHS z 0 23333 15 0 4166667 x1 1 0 0 83333 52 0 g 7 1 16667 6 Go back to step 2 a Optimality condition If the nonbasic variables have positive coefficients in the maximization problem then you have reached an optimum b Concepts in the tableau i Optimal solution x1 83333 sZ 16667 ii Binding and nonbonding constraints a The rst constraint is binding it constraints the optimal b The second constraint is nonbinding it does not constrain the optimum adding an additional unit will not affect the objective function value iii Finally we will consider the sensitivity of the solution over what ranges will the current solution be optimal What is the value of each unit of additional resource 111 Next Example maxzi 40 720 031 032 033 0 st x1 x2 51 0 6x1 2x2 52 0 36x1 18x 53 0 A In tableau form this problem is expressed as AEB 6592 Lecture IV Professor Charles B Moss x1 x2 s1 s2 s3 RHS z 740 720 0 0 0 12 12 SI 1 1 1 0 0 1 s2 6 2 0 1 0 48 6 8 s3 36 18 0 0 1 360 360 7 A6710 1 Thus s2 is the leaving variable and x1 is the entering variable x1 x2 s1 s2 s3 RHS 720 20 z 0 A 0 A 0 320 12 SI 0 1 7 0 4 A6 24 x1 1 0 8 X 24 Si 0 6 0 76 1 72 7267 2 Thus s1 is the leaving variable and x2 is the entering variable x1 x s1 s2 s3 RHS z 0 10 5 0 360 x2 0 7 4 0 6 x1 1 0 7 2 M 0 6 s3 0 0 3 7 1 36 Graphical Solution and Sensitivity Again Lecture III Problem from Last Time maxz 50x1 10x2 A 31 3cl 2x2 S 25Labor x1 x2 S100Land 1 Labor 3x1 2x2 25 3x1 25 2x2 x 1 3 3 2 x2 03 x1 38333 x1 0 3 x2 125 2 Land x1 x2 100 cl 100 x2 x2 0 3 cl 100 x1 0 3 x2 100 3 Objective 2 50xl 10x2 50xl z 10x2 4 How does the objective function change as X1 and X2 change 2 Ax1 sz dz Aabcl Babc2 B Where do the labor and land constraints intersect 3 3 1 50 x2 3 3 x 50 x2 50 x1100 5050 AEB 6592 Lecture III Professor Charles B Moss C What is the maximum allowable pro t per acre of cotton such that the current solution remains optimal Similarly what is the minimum return on cotton for which this solution remains optimal l Reformulate the objective function z 21x1 22x2 lel Z sz2 Li 21 21 Assuming 22 is xed at 10 what level of 21 would cause the optimum allocation to change 2 Any increase in the value of 2 will result in no change in the optimal solution 3 The range over which the solution is stable in the other direction is bounded by the labor constraint Speci cally if the objective function becomes more steep than the labor constraint then the optimal solution will change To solve for the coef cient at which the solution changes 10 2 amp x1 2 2 2 1 1521 Thus the coef cient on cotton could decline to 15 a decrease of 35acre before the optimal solution changes Similarly for the coef cient on corn the coefficient on corn could be decreased inde nitely without changing the optimal solution However if the coef cient were increased sufficiently corn would enter the solution 4 Z2x 22x 50 2 3 2 322100 253333 Thus if the rate of retum per acre of corn were increased to 3333 acre then corn would enter the optimal solution II New Problem A The lm s problem is how to organize production choose levels of crop plantings to maximize pro ts or net returns over variable costs given 12 acres of land 48 hours of labor and 360 of operating capital AEB 6592 Lecture III Professor Charles B Moss Constraints Corn Oats X1 X2 Land 1 1 Labor 6 2 Capital 3 6 l 8 Retums Acre 40 20 B Speci cation of the mathematical programming model E maX40x1 20x2 SJ cl x2 S l2Land 69cl 2x2 S 48Labor 36xl 18x2 S 360Capital Land clx2 Sl2 x1l2 x2 Jcll23x2 0 990x2 12 Labor 6x12x2 S48 69c148 2x2 x1Sxz 998x2 0 990x2 24 Capital 36x1 18x2 S360 36x1 360 18x2 1 10 9 2X2 cl 103x2 0 cl 03x2 20 Where do these various constraints intersect a Land n Labor l2 x2 8 x2 AEB 6592 Lecture III Professor Charles B Moss b Capital n Labor x2 12x1 4 c Capital n Land 1 l2 x2 10 3x2 l 2 x 2 2 x2 4 x1 8 C Optimal combination of crops 2 40xl 20x2 40x1 z 20x2 i ix x1 40 2 2 This implies an optimal solution at the comer of the Land n Labor constraints 1 lt i lt l 2 3 AK Corn 12 Land 10 8 Capital Objective 20 24 Oats AEB 6592 Lecture III Professor Charles B Moss D Sensitivity Analysis4ver what ranges are the solution stable 1 What if land were increased What constraint would become binding Objective 12 20 24 Oats If land were suf ciently increased then the combination of labor and capital will eventually become constraining The constrained optimal at that point involves X14 and X212 or 16 acres of land an increase of 4 acres The value of the objective function at this point is z 4042012 160 240 400 ShadowValue M E 10 16 12 4 On the down side land can be decreased until the labor constraint is no longer binding before the solution changes At that point X18 and X20 a decrease of 8 acres of land 2 408 200 320 ShadowValue w 10 8 12 4 AEB 6592 Lecture III Professor Charles B Moss 2 What if the labor constraint were increased What constraint would become binding Land Capital Objective 12 20 24 Oats Increasing the labor would cause the Capital n Land intersection to become binding Thus the largest change in labor that would keep the constraint active is X18 and X24 NewLabor 6824 48 8 56 z 408204 320 80 400 ShadowValue M E 5 56 48 8 The labor constraint can likewise be reduced until the point where the land is restricted without change in the active constraints 3 Capital could be decreased until the land and labor intersection is binding at X16 and X26 E Over what range of objective coefficients is the solution stable Z lel sz2 iZ2x2 21 21 Fixingg 20 20 132120Azl 20 40 20 Zl 2 3zl 60Az1 60 40 20 Z 1 AEB 6592 Lecture III Professor Charles B Moss Fixingq40 2 2424 40Azz 40 20 20 40 121 322 Azz 1333 20 667 40 3 3 The Basic Math Programming Model and Graphical Approach Lecture II An Agricultural Example A Assume that the farmer can produce two outputs cotton and corn with 100 acres of land and 25 hours of labor Further assume that each crop has the following input use per acre Cotton Corn Labor 3 2 Land 10 l 0 Finally assume that the pro t per acre is 50 for cotton and 10 for corn B Basic Construction of the Problem The basic formulation of the problem involves answering three questions 1 What does the model seek to determineiWhat are the variables 2 What constrains the variables from taking on any valueiWhar are the constraints 3 What is the objective 7 What is the rule used to determine the best or optimum C In the current model determine each component D Basic Mathematical Formulation maX50x1 10x2 SJ 399 2x2 S 25 x1 x2 S 100 cl x2 2 0 where X1 is cotton and X2 is corn E Graphical solution 1 Basic Formulation of constraints 3xl 2x2 25 3x1 25 2x2 250 2 x x1 3 3 2 x1 x2 100 x1 100 x2 Do these lines intersect 250 2 T gxz x2 1 50 x2 3 3 AEB 6592 Lecture 11 Professor Charles B Moss X1 100 x1 100 x2 Land 8333 50 100 125 2 The region satisfying all constraints is called the feasible region or region of feasibility 2 Choosing the optimal point Construction of the Isoprof1t line Let 2 be the pro t from some allocation of cotton and corn Speci cally Z 50xl 10x2 Question How do we know that the objective value increases as X1 and X2 increase Using basic calculus we differentiate the profit function and set it equal to zero yielding dz 50aml 10alx2 0 abc1 l abc2 5 Thus we know that profit increases as either X101 X2 increases by the first equation Second we know what the effect of increasing X1 on profit is given that we have to decrease X2 For example on the land constraint where we have to trade one unit of X1 for each unit of X2 the effect on profit is 715 Alternatively profit is maximized where the slope of the feasible region is equal to 715 AEB 6592 Lecture 11 Professor Charles B Moss X1 100 8333 3 Two things to note a First the trick is to nd the rightmost isoobjective function line that is still in feasible or solution space b Second notice that the solution is on a comer or extreme point 11 A Slight Reformulation Slack Variables and Matrix Algebra A The complicating factor in the above formulation from simple linear algebra perspective is the inequalities 1 Note that these inequalities can be transformed into equalities if we add nonnegative slack variables max50x1 le2 31 3x1 2x2 31 0 q x2 s2 0 x1 9 s1 32 Z 0 a s1 is the slack or excess variable in the labor constraint while sz is the slack or excess variable in the land constraint b Graphical solution 3x12x2 31 25 3x1 25 2x2 s1 AEB 6592 Lecture 11 Professor Charles B Moss X1 100 8333 50 100 125 2 c Why s15 For a second what if X10 and X2100 Then 250 200 2 10 x2 s 0 1 3 3 3 3 50 10 Sl 3 3 10 50 SI 3 3 T T9 lt 5 200 2 xi T 3 d Two things to note i We can transform an inequality system into an equality system by adding subtracting the appropriate slack surplus variables ii The nonnegativity condition on the slack surplus variable limits the feasibility condition to everything below above the line e The land constraint x1 x2 32 100 cl 100 x2 s2 AEB 6592 Lecture 11 Professor Charles B Moss 2 Combining the two equality representations graphically 8333 6667 8333 100 125 X72 a Thus one alternative is to produce x18333 acres and xz0 acres This exhausts the labor constraint s10 but leaves land in excess szl667 b Another alternative is to use all the land and let labor go slack xz100 x10 s15 and sz0 B Using Matrix Algebra 1 Note that an important part of the solution process was determining the intersection of the two constraints This process can be simpli ed using matrix algebra a Note that the system of equations 3x1 2x2 25 x1 x2 100 can be expressed in matrix form as 3 2 l 25 I l l 100 using row operations we can multiply the rst row by 103 yielding AEB 6592 Lecture 11 Professor Charles B Moss Multiplying the last row by 3 r 0 l 5 Subtracting 23 times the second row from the first yields F1 0 501 I 0 1 50 which is interpreted as x150 xz50 III Stuff from Day A The preceding problem was a simple example of linear programming It contained linear constraints a single linear objective function and continuous variables Many extensions of this framework are now possible 1 The simplest extension is the nonlinear objective function X1 100 8333 50 100 125 2 The nonlinear objective function allows for an optimum not on a comer It may also produce an optimum without bounds

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