+ 5x 2. = x x. + x 2. Transform the original system into a system x 2 = x x 1. = x 1

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1 University of California, Davis Department of Agricultural and Resource Economics ARE 5 Optimization with Economic Applications Lecture Notes Quirino Paris The Pivot Method for Solving Systems of Equations page The Primal Simplex Algorithm Main Ideas of the Primal Simplex Algorithm Technology Measured in Different Units Opportunity Cost Revisited Complementary Slackness Conditions Completing the Final Tableau Linear Programming in Vector Notation The Pivot Method for Solving Systems of Equations (See also chapter 6 of the Linear Programming textbook and chapter of the Symmetric Programming textbook) It is necessary to learn the Pivot method (even if you know other procedures) because it shows how to use the Simplex algorithm (used in all the software) for solving LP problems. The Simplex algorithm uses economic criteria to find the best production plan. This is the only reason why I teach it. Solving systems of equations means transforming the original system that is difficult to read into an equivalent system very easy to read. For example x 6 Transform the original system into a system 0 0 x x that is easy to read. The Pivot method ANSFORMS the original system of linear equations into an identity matrix (or a permutation of the identity matrix) one variable at a time. Each single-variable transformation is called ITERATION. The Rules of the Pivot Method. Define a square transformation matrix. A transformation matrix is an identity matrix with one column replaced by a transformation vector.. Define a transformation vector. A transformation vector is a vector containing all the recipes for eliminating one variable from all but one equation and for normalizing the Pivot element. Choose a Pivot (a pivot is an non-zero element of the matrix of coefficients).

2 . The reciprocal of the pivot element is placed in the same row of the transformation vector.. All the other coefficients of the pivot column are divided by the pivot and placed in the transformation vector with a changed sign.. Rule to locate the transformation vector within the transformation matrix. Place the transformation vector in that column of the transformation matrix whose index corresponds to the row containing the pivot.. Rule for choosing the pivot element. The first pivot can be chosen in any row or column as long as it is not a zero coefficient. Successive pivots must be chosen in a different row and column from those of previous pivots. Example. Consider the following system of equations Add auxiliary variables z and z (their importance variables will become clear later on; they are not slack variables) z + + z 6 A system with four variables and two equations can be solved by choosing (arbitrarily) the value of two variables and solving for the other two. In the example, we choose,, z,z 6. Let us put all these computations in a tableau. The first row of matrices exhibits the pivot element (indicated by the brackets ) chosen in the second row of the original matrix of coefficients. This means that the transformation vector will be the second column of the transformation matrix T. The BI label stands for Basic Indexes, meaning that at this stage variables z, z are associated with the current basis (mathematical ruler, in our intuitive language). All the other variables are equal to zero. Carry out the pre-multiplication of the original matrix by the transformation matrix. Replace z with in the BI column because the pivot was chosen in the second row of the column: this is the computational meaning of pivot. Choose another pivot and continue until all the auxiliary variables have been replaced by original variables in the BI column. T z z sol BI z z 5! 0! 0 z Current Basis: 0 B 0! 0! 6 z 0 T z z sol BI z 0 0!! 8 z Current Basis: B! 0! 6 0 z z sol BI 0!! Current Basis: 5 B 5 0!! 0

3 The solution is exhibited in the sol and BI columns, that is and. Verify the solution. Important information: The coefficients appearing under the z, z headers constitute the inverse of the basis (the mathematical ruler). This can be shown by computing B B I : 5 0 B B yes 5 0 Note the order of the columns of the B matrix. It follows the order of the variables appearing in the last BI column which, in turn, follows the way the pivot elements were chosen during computations. The Primal Simplex Algorithm (See also chapter 7 of the LP textbook and chapter 7 of the Symmetric Programming textbook) To make production decisions an economic agent must consider two criteria:. profitability criterion. producibility criterion. Profitability: A producible commodity must be profitable. In other words, its opportunity cost must be non-positive: OC MC MR 0. A profitable commodity will be produced up to the level where OC MC MR. Producibility: The production plan must satisfy the input constraints and all the components must be nonnegative. We indicate the production plan as x Bi 0 and x NBj where B Basic (produced) and NB Non Basic (not produced). The primal constraints of a linear programming problem indicate producibility (feasibility) of the production plan: Ax b, x 0. Example. The following numerical example will be solved with the Primal Simplex algorithm (PSA). Before tackling the numerical computations it is necessary to have a clear idea of what the various components of the problem mean in a mathematical as well as economic sense. max + 5 subject to all x j 0 and D S wheat corn land labor Add slack (surplus) variables and reorganize all the info in such a way to have all the unknown variables on one side and the constant parameters on the other side of the equal sign Finally, organize the information into a Primal tableau

4 x s sol BI 0 " 0 " 6 0 " 6 0 " 8! "!!!! "! " " 0 x s! Every Primal tableau has six (6) parts. It is necessary to recognize what those parts are in a mathematical sense and what their economic significance is. x s solution 0 " Technology (mrtt) " Basic Production Plan! "... "... " OC MC MR " Algebraically, every Primal tableau assumes the following structure 0 " a ij " x Bi! "... "... " oc x j z j c j " where oc x j z j c j MC j MR j is the algebraic notation for opportunity cost found in the majority of LP books. Main Ideas of the Primal Simplex Algorithm (PSA) Economic Language Mathematical Language 0. Find the do nothing production plan. 0. Find an initial and easy basic feasible solution. Check whether this producible production. Check whether this BFS is an optimal solution plan is optimal If YES, stop If YES, Stop IF NO, continue IF NO, continue. Select another profitable and producible. Select another adjacent BFS that improves the production plan and GO TO. value of the objective function and GO TO. Solution of the numerical example by the Primal Simplex algorithm: Identify the pivot (indicated by the brackets ) using the profitability criterion (Step of the PSA) and the producibility criterion (Step of the PSA). Construct the transformation matrix. Carry out the matrix multiplication.

5 T x s sol BI Ratios 0! 0! 0 " "! " 5 0! 0! 0! 6 0! 6 0! 8 "! " " " "! "! 5 0 0! 0 6 Step of PSA 8 x s 6 " Step : most negative OC MC MR Note: The intersection of the two arrows identifies the pivot in row,. Hence, the transformation vector is located in column of the transformation matrix T (according to rule of the pivot method). A more detailed discussion of the above first iteration (first tableau) of the PSA follows. Step 0 (the initial step) of the PSA identifies (by inspection) the do nothing production plan. This is the current basic feasible solution (BFS). Step 0: Do nothing production plan Current basic feasible solution This is clearly not an optimal solution. 6 x s 8 Step selects the coefficient ( 5) in the third row because it indicates the most negative opportunity cost. Recall that the opportunity cost is defined as OC MC MR. Therefore, at this stage of the decision making process, before deciding which output to produce (hence producing nothing for the time being), the marginal cost is zero. As a consequence, every coefficient in the OC row of the first tableau has the structure of OC (0 ),(0 5),(0 0),(0 0). This is the reason why Step selects ( 5) as the most profitable opportunity cost associated with activity, corn. Step of the PSA selects a better production plan since Step indicated that a new (better) production plan (with corn, > 0 ) could be producible (feasible) and, hopefully, be also optimal. Therefore, the next production plan will include corn at a positive level (if possible) and must adjust the current levels of the other activities by the technical coefficients of corn because the production of corn will deplete the current supply of land and labor. New production plan New basic feasible solution > x s for feasibility for feasibility 5

6 From the two inequalities involving the slack variables (surplus land and surplus labor) we can determine the actual quantitative and feasible level of, corn. 6 x 6 0 minimum ratio 8 x The minimum ratio must be chosen to maintain the producibility (feasibility) of the entire production plan. We can now return to the tableau of the first iteration, carry out the matrix multiplication and continue with the PSA. T x s sol BI Ratios 9! 0 0 9! 0 " "! " 0 9! 0 0!! / !! 6 x s Step of PSA 9/ "! " " " "! " "! 5 0 0! 0 Step : most negative OC MC MR The reading of this second tableau can be summarized as follows: Current feasible New improved production plan production plan > / 6 x s 6 x s / min ratio We can now return to the tableau of the second iteration, carry out the matrix multiplication and continue with the PSA, if necessary. x s sol BI 0 0 " 0 9 " x 0 " "! "!!!! "!! " 7 " y s y s y y TC This third tableau is an optimal tableau because the basic solution is feasible (producible) and all the opportunity costs are nonnegative. The PSA terminates here. The most important aspect to understand the structure of a LP problem is to be able to read and interpret the final (optimal) tableau. This tableau contains all the information useful to the economic agent. Begin with the 6

7 optimal production plan x s 6 Note that x It is remarkable that, without looking explicitly for the dual solution, the PSA produces the dual solution whose economic meaning is that of opportunity costs. optimal opportunity costs y s y s y 6 7 y 9 TC 6 7 Note that TC 6y 6 +8y Therefore, max mintc. The activated technology (optimal feasible basis) is given by the following matrix: B opt 6 It exhibits a permutation of columns because of the pivot selection in the PSA sequence. The coefficients under the slack variables (in the final tableau) constitute the inverse of the optimal basis. This fact can be verified by checking that B opt B opt I : Technology Measured in Different Units Fundamental question: What happens when going from the initial tableau to the final tableau? Answer: All the coefficients of the technology and of the input supplies are measured in different units. Consider again the initial tableau x s sol BI Prices Ruler (Basis) " 0 " 6 y 0 " 6 0 " 8 x s y 0 B! "!!!! "!! 0 " " yes 0 0 x s 7

8 Clearly, the mathematical ruler (basis) is the technology of conserving inputs (do not produce any output). Hence, every coefficient in the initial tableau is measured in input units. Recall, in fact, that every technical coefficient indicates how many units of an input is necessary to produce one unit of output. We conclude that the initial tableau is measured in input units. Analogous discussion can be elaborated for an intermediate tableau. The essential idea to retain is that the basis (mathematical ruler) changes and, therefore, the measurements change accordingly. Consider now the final (optimal) tableau. x s sol BI Prices Ruler (Basis) 0 0 " 0 9 " x c 0 " " c! "!!!! "! B opt! 6 " 7 " In this final tableau, the mathematical ruler (basis) is given by the technology of producing outputs and (recall that the order of the outputs depends on the selection of pivot elements during the PSA execution). Hence, every coefficient of this tableau is measured in output units. Opportunity Costs Revisited The opportunity cost is the single most important notion of microeconomics. Every life decision about every action to be taken is made on the basis of the opportunity cost. The opportunity cost of producing an output is defined as the difference between the marginal cost and the marginal revenue; OC x j MC x j MR x j. In a price-taking environment, the marginal revenue is easy to ascertain. On the contrary, marginal cost may be considered the most difficult notion to measure (correctly) in every economic model. Measure is the crucial word. In the previous discussion, we have seen that economic entities can be measured with different rulers (bases). Hence also marginal costs can be measured with those same rulers (bases). In the previous section we have identified two important rulers (bases): the input ruler and the output ruler. Consider again the initial tableau. The information contained in it can be unpacked to write: input shadow prices b y x s 8 b y 5 Total cost is defined as the sum of the products of input quantities and their prices: TC b y + b y 6y +8y y ( + + ) + y (6 + + x s ) Therefore, marginal costs of outputs and and inputs and are: 8

9 TC MC x y + 6y [y y ] y a 6 TC MC x y + y [y y ] y a TC MC xs y [y y ] x y e s 0 TC 0 MC xs y [y y ] y e x s Now consider the final (optimal) tableau. The information contained in it can be unpacked to write: market output prices x + 9 x s c x x s c x 6 s + 9 x s At the optimal solution of a LP problem total revenue is equal to total cost of the physical plant. Hence, 0 TC c + c c + c c ( x s ) + c ( + 9 x s ) Therefore, marginal costs of outputs and and inputs and are: TC 0 MC x c [c c ] x TC MC x c [c c ] 0 TC MC xs c 6 c [c c ] x s 6 TC 9 MC xs 9 c + 9 c [c c ] x s 9 Finally, it is possible to compute the opportunity cost of all commodities (inputs and outputs) as they appear in the final (optimal) tableau in two different but equivalent ways: Let us recall that, in this example, output market prices (that is, marginal revenues) are c and c 5.. Opportunity cost at market output prices 0 OC x MC x MR x c c [c c ] c y s OC x MC x MR x c c [c c ] c y 0 s OC xs MC xs MR xs c 6 c 0 [5 ] y 6 OC xs MC xs MR xs 9 c + 9 c 0 [5 ] 9 0 y 9 9 9

10 Note that the vertical boxes are the technology vectors appearing in the final tableau.. Opportunity cost at factor cost (shadow input prices) OC x MC x MR x y + 6y c [ 7 6 OC x MC x MR x y + y c [ ] y s 6 9 ] 5 y s 7 OC xs MC xs MR xs y 0 [ 6 9 ] y OC xs MC xs MR xs y 0 [ 6 9 ] 0 9 y Note that the vertical boxes are the technology vectors appearing in the initial tableau. Verify that the two sets of opportunity costs (measured at market output prices and at factor cost) are identical although computed with completely different information. Complementary Slackness Conditions The Economic Equilibrium problem includes primal and dual complementary slackness conditions. (S D)P quantity side (MC MR)Q price side Recall that equilibrium means solution. Therefore, from the final tableau: inputs (land, labor x s ) 7 S D and y 6 y S D x s and y 9 x s y From the final tableau: outputs (wheat, corn ) MC x MR x y s and y s MC x MR x y s and y s When computing the opportunity cost or the quantity level of some commodity, therefore, we can use the appropriate complementary slackness condition. If the commodity is an input we can state directly that if x si > 0 then y i If the commodity is an output if y i > 0 then x si if x j > 0 then y sj if y sj > 0 then x j 0

11 Completing the Final Tableau (When Only the Inverse of the Optimal Basis is Known) To summarize (and verify) all the knowledge accumulated in previous discussions, we now complete the final (optimal) tableau when only knowledge of the initial problem and the inverse matrix of the final (optimal) basis (ruler) are known. Example. Let max x subject to all x j 0 and x 0 x + x x 6 Rearrange (add slack variables and name the vectors) x x 0 s + x s 6 a a a e e b The incomplete final tableau is given as x x s sol BI 0 "??? 7 0 "??? 7 With the available information it is possible to complete the final tableau, that is filling in the missing information. This exercise fulfills the objective of understanding the structure and the meaning of the optimal (final) tableau. To begin with, the coefficients under the slack variables constitute the inverse of the current (optimal, final) basis, say B. Therefore, the solution (basic production plan) is computed as x B B b Recall that a basis (ruler) can measure any point in the given space. Hence w a B a w a B a 7 "? 7 "?! "!!!!! "! "????? "? ??! w a B a Therefore, the partially completed final tableau is: c c x

12 x x s sol BI 0 " " 7 x 5 0 " " 7! "!!!!! "!! " " 0????? 7 The order of the variables under the BI column is identified by the location of the pivot as selected during the PSA. For example, the column under the variable x exhibits a unit value in row. This means that the pivot was selected in that location and, therefore, x replaces (by pivoting) whatever other variable was appearing in row of the BI column. There remains to complete the row of opportunity costs. We have two alternatives:. At factor cost. At market output price Since the factor costs (input shadow prices) are not yet known, it is necessary to compute the opportunity costs at market output prices. Opportunity costs at market output prices: OC MC MR ] mrtt measured OC x MC x MR x [c c 6 c [ 6] y s in x, units 7 OC x : since > 0 it follows that OC x MC x MR x y s by the Economic Equilibrium OC x : since x > 0 it follows that OC x MC x MR x y s by the Economic Equilibrium OC xs MC xs MR xs y 7 OC xs MC xs MR xs y 7 Now that we know the factor costs (input shadow prices), y, y, we can re-compute the Opportunity costs at factor cost 5 OC [ ] 0 7 xs MC xs MR xs 7 7 y OC 0 5 xs MC xs MR xs [ 7 7 ] 7 y ] mrtt messured in input units OC x MC x MR x [y y 5 [ 7 7 ] y s 5 7 OC 7 0 x MC x MR x [ 7 7 ] y s 5 96 OC x MC x MR x 7 5 [ 7 7 ] y s

13 The final tableau can now be completed: TC b y + b y 7 7 x x s sol BI 0 " " 7 x 5 0 " " 7! "!!!!! "!! " 0 5 " y s y s y s y y TC Linear Programming in Vector Notation Partition the technology matrix, A(m n, m < n), in two sub-matrices: a basis (square matrix), B(m m), and the rest of the technology that is not basic, NB, that is, A [B A NB ]. Partition the production plan vector, x(n ), and the market output price vector, c(n ), in similar ways: x x B x NB and c c B c subject to NB. Then, a LP primal problem is stated as x B max c x c B c NB c B x B + c x NB Ax b NB x NB ( m)(m ) + [ (n m)][(n m) ] [B A NB ] x B x b NB Bx B + ANBxNB b Insert surplus variables in the primal constraints Bx B + A NB x NB + Ix s b What follows are the initial and intermediate/final tableaux of the PSA: T x B x NB x s Sol BI B! 0 0! B A NB I! b x s Initial Tableau ""! " "! "" "" "! " " Choose B c B B!! c B c NB 0! 0 as a "pivot" x B x NB x s Sol BI B 0! I B A NB! B b x B Intermediate "! " """""" ""! """ " or Final! 0 ( c c B B! c B B b Tableau B B A NB c NB ) y s,b y s,nb y B TC