Two inputs and three outputs example

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1 University of California, Davis Department of Agricultural and Resource Economics ARE 252 Optimization with Economic Applications Lecture Notes 3 Quirino Paris Price Taker page 1 Two Inputs and Three Outputs Transposition Principle The Dual of Linear Programming by KKT Theory Solution of LP by Graphical Method Reading a System of Equations by Column Anatomy of Linear Programming The Transportation Problem Price Taker (See also chapter 7, Symmetric Programming textbook and chapter 1, Linear Programming textbook) The major hypotheses of the model are that a price-taking economic agent accepts market prices as given (known) parameters in his way to maximize profit using a known technology. The exogenous and the endogenous information of this problem is listed as follows: Exogenous (known) information 1. Technology: Given I inputs and J outputs, a ij is a constant and known technical coefficient (known as Marginal Rate of Technical Transformation MRTT). It measures the quantity of input i required for the production of one unit of output j, i = 1,..., I, j = 1,..., J. 2. Limiting input supplies: known quantities of b i, i = 1,..., I 3. Output market prices: known prices p j, j = 1,..., J Endogenous (unknown information prior to solve the problem): 4. Output quantities: x j, j = 1,..., J 5. Input shadow prices: y i, i = 1,..., I. The Primal and Dual sides of the price-taker s model are stated as Primal: maximize Total Revenue (TR) Demand Supply of any commodity involved. Dual: minimize Total Cost (of physical plant) (TC pp ) Marginal Cost Marginal Revenue of any commodity involved Two inputs and three outputs example x 3 Primal: maxtr = p 1 + p 2 + p 3 x 3 = [ p 1 p 2 p 3 ] = D S p x 1

2 a11 + a 12 + a 13 x 3 b 1 11 a 12 a 13 b 1 a 21 + a 22 + a 23 x 3 b a x 2 = 2 a 21 a 22 a 23 b 2 x 3 x j 0, j = 1,2,3 Ax b, x 0 Note very carefully the arrangement of vectors and matrices. In economic language, the Primal must be read as production maxtr = [mkt output prices] plan limiting technology production input (mrtt) plan supply As stated above, the Dual problem is specified as mintc pp MC MR. Therefore, we must define TC pp, MC and MR. From the Economic Equilibrium, (S D)P = 0. Hence, S P = D P. Therefore, when commodities are inputs, TC pp = S P = D P = b 1 + b 2 = (a 11 + a 12 + a 13 x 3 ) + (a 21 + a 22 + a 23 x 3 ) = b y = x A y two equivalent definitions of total cost based upon the supply(quantity) and demand (quantity) of inputs. Marginal cost of producing output is TC pp ] MC x1 = = a 11 + a 21 = [a 11 a 21 = a 1 y TC pp y MC x2 = = a 12 + a 22 = [a 12 a 22 ] 1 = a 2 y TC pp ] MC x3 = = a 13 + a 23 = [a 13 a 23 x = a y 3 3 and in vector notation: MC = A y. Marginal revenue follows from differentiating total revenue: TR TR TR MR x1 = = p 1, MR x2 = = p 2, MR x3 = = p 3 x 3 Finally: Dual: mintc pp = b 1 + b 2 = [b 1 b 2 ] = b y MC MR 2

3 a p a 21 p 1 a 11 a 21 a 12 + a 22 p 2 a 12 a 22 p 2 y a 2 a p 13 + a 23 p 3 13 a 23 3 y i 0, i = 1,2 A y p, y 0 Note very carefully the arrangement of the vector and matrices. In economic language, the Dual must be read as input limiting input mintc pp = shadow supply prices technology input transposed shadow (mrtt) prices market output prices In compact, vector notation, form: Primal: A x maxtr = p x Ax b x 0 Dual: mintc pp = b y A y p y 0 Transposition Principle A second way to obtain the dual of a linear programming problem. Primal and Dual models contain the same amount of information arranged in different locations. Every column of the Primal problem becomes a row in the Dual problem and vice versa. This is called the Transposition Principle. It is a short cut to state the Dual problem when the Primal problem is specified. Primal: Dual: maxtr = [ p ] x mintc pp = [ b ] y s.t. b s.t. A y p The dual of linear programming by KKT theory There is a third way to obtain the dual of a linear programming problem. This procedure is purely mathematical but produces identical results as those one derived above. 3

4 The Lagrange function of the LP problem is stated as L = p x + y ( b Ax) with KKT conditions L = p A y 0 dual constraints x x L = x [p A y] = 0 dual CSC x L = b Ax 0 y y L y = y [ b Ax] = 0 primal constraints primal CSC By using the information of the dual CSC, p x = y Ax, it is possible to simplify the Lagrange function that becomes the dual objective function: L = p x + b y y Ax = y Ax + b y y Ax = b y Therefore, the dual problem of the LP model becomes mintc pp = b y A y p y 0 Solution of a LP problem by graphical method: Example 2: Two inputs and two outputs Primal: maxtr = D S L 1 Line 1 land L 2 Line 2 labor 0 0 wheat corn Figure 1 presents the solution of this LP problem: 4

5 Figure 1. Graphical solution of a LP problem A, B, C and D are called Extreme Points of the feasible region. The figure below is a reproduction (blow up) of the optimal point C in figure 1 using the gradients of the objective function and of the constraints. Recall that a gradient evaluated at a given point is orthogonal to the corresponding function. The gradient of the objective function falls in the cone defined by the gradients of the constraints and becomes the main diagonal of the associated parallelogram. This graphical structure of the optimal point is a general geometric picture of any optimal point in any mathematical programming problem. 5

6 Slack variables and their meaning Inequalities admit the introduction of an additional variable called Slack variable (algebraic language) Surplus (primal) variable (economic language) Opportunity Cost (dual) variable (economic language) Distance (between a given point and a given line) (geometric language) Primal constraints Dual constraints D S (2 + 3 ) 12 ( + 4 ) 17 MC MR (2 + ) 3 (3 + 4 ) 6 D + Slack = S (2 + 3 ) + x s1 = 12 ( + 4 ) + x s2 = 17 MC = MR + OC (2 + ) = 3+ y s1 (3 + 4 ) = 6 + y s2 See also pages in LP textbook including Table 1.4 that summarizes all the relevant terms in several languages. Reading a system of linear equations by column A system of linear equation can be read by row and by column. The reading by row is familiar (high school). For understanding mathematical programming the reading by column is novel and essential. Example 3. We all agree that = = 3 Now put the information of the above system in boxes (vectors) and name the boxes: = 3 v 1 + v 2 = v 3 Now plot the system by column and construct the parallelogram whose main diagonal is the addition of the two vectors, v 1,v 2 : Figure 2. Vector addition 6

7 Vector subtraction: v 1 v 2 = v 3. See Figure 3. Figure 3. Vector subtraction Now we change only the RHS of the original system of equations, that is = v 1 + v 2 = v 4 In this example we need to find the values of, and using only the diagram of figure 4. Figure 4. Solution of a system of linear equations by parallelogram Conclusion. A system of linear equations can be solved in two geometrical ways: 1. Reading (plotting) the information by row and find the intersection of two lines (three planes, etc.) 2. Reading (plotting) the information by column and construct the parallelogram, (parallelepiped, in a space of dimension greater than 2). Note: From figure 4, v 4 = 2v 3. Hence, multiplying a vector by a scalar can i) Stretch a vector if scalar > 1 ii) Shrink a vector if scalar < 1 iii) Reverse the direction of the vector and either stretch it or shrink it if scalar < 0 7

8 Another important point. In each figure 2, 3 and 4, either v 3 or v 4 is the mathematical object to be measured; v 1 and v 2 form the mathematical ruler (cone); the and constitute the measurement (solution). Anatomy of Linear Programming Example 4. Primal: maxtr = L 1 land L 2 labor wheat Add slack variables. maxtr = x s1 + 0x s x s1 = x s2 = 21 Put the information of the constraints in vector form and name the vectors x 0 + x s x 0 s2 = a 1 a 2 e 1 e 2 b Now graph the output space and the input space as in figure 5. corn OUTPUT SPACE Graphed by row INPUT SPACE Graphed by column Figure 5. Output and input spaces of a LP problem From the output space we identify From the input space we identify Extreme Points: A, B, C, D Feasible Cones (include vector b ) (geometric name) (geometric name) Basic Feasible Solutions (BFS) Feasible Bases (algebraic name) (algebraic name) Basic Production Plans Activated Technologies (economic name) (economic name) 8

9 Question: what is the relation (if any) between Extreme Points and Feasible Cones? To answer this question we must: 1. express the Extreme Points in qualitative terms (coordinates are greater than or equal to zero) 2. identify which vectors constitute the Feasible Cones (those cones that include the RHS vector b ). EXTREME POINTS FEASIBLE CONES A :{ = 0, = 0, x s1 > 0, x s2 > 0} (e 1, e 2 ) B :{ = 0, > 0, x s1 = 0, x s2 > 0} (a 2, e 2 ) C :{ > 0, > 0, x s1 = 0, x s2 = 0} (a 1, a 2 ) D :{ > 0, = 0, x s1 > 0, x s2 = 0} (a 1, e 1 ) Note that each of the Extreme Points contains two coordinates that are equal to zero. Therefore, by combining the information from the Extreme Points and the Feasible Cones we obtain the Basic Systems of Equations that can be solved quantitatively to obtain a precise measure of the production plan at the various Extreme Points A : x s1 + 1 x 0 s2 = 21 e 1 x s1 + e 2 x s2 = b B : C : D : x 1 s2 = 3 21 a 2 + e 2 x s2 = b 6 16 x = a 1 + a 2 = b x s1 = 2 21 a 1 + e 1 x s1 = b Note that each of the four basic systems of equations exhibits the same RHS vector b. This means that the same mathematical object to be measured is being measured by four different mathematical rulers (feasible cones, feasible bases). The fours systems of equations represent four feasible (activated) technologies for transforming inputs into outputs. The LP algorithm, therefore, searches for the best measurement ( x * ) that will maximize the value of the objective function. This discussion illustrates the fact that, in LP, a basic feasible solution occurs always at an extreme point. Since, in general, the extreme points are finite in number, the algorithm to solve a LP problem is also finite (in time). The Simplex Method is such an algorithm. 9

10 Primal problem. The formulation of the transportation problem will be given in general terms. Thus, let Cij = transport cost of one unit of product from origin i to destination j x;; = amount of product shipped from origin i to destination j b; = amount of commodity demanded at destination j a; = amount of commodity supplied by origin i. 1he primal specification of the transportation _ problem, which is reassembled below in a more compact form. Primal minimize TCT m n I::Z:::~jXij i=l j=l m LXij 2: b;, j = 1,...,n i=l j=i i=l,...,m x;; 2: 0 for all i's and j's All the constraints were written with inequalities running in the same direction to facilitate the formulation and interpretation of the dual problem. Before performing this task by way of the Transposition Principle, however, it is important to consider the initial tableau associated with the primal transportation problem which organizes all the information and reveals the special structure of the tram,purlaliun problem. Initial Tableau of Transportation Problem X11 X12 XJn X21 x22 X2n Xml Xm2 Xmn 1 0." 0 1 0". 0". 1 0". 0 2: b l... o... 0 l : b2 o o::: i c, o::: i::: o o. _.- i'''i''"' bn o ". 0 0 " " -1." o o : -am 0 2: 0 2: Dual Vars Pt p~ All the elements of the transportation matrix are either zero or unity. Each column of the initial tableau contains exactly two nonzero coefficients, This tableau exhibits the transportation technology. The symbols pj and pf of the dual variables are chosen for mnemonic reasons to represent prices at destination (d) j and at origin (o) i, respectively. 10

11 The dual of the transportation problem: Dual mrucimfae VA - [ tb;pf - f;; a,pf ] pj - pf ::; Cij for all i's and j's Pd and p 0 > 0 J Economic Interpretation ofthe Transporta- : tion Problem Primal. [ Total Cost of ] min = 'lransportation l [ l [ l Sum Unit cost of Quantity level [ over all transportation of all routes of all routes commodity flows, l Quantity supplied b II.. > Quantity demanded [ ya ongms. t o J th mar k e t - by Jth market [. ] and JQuantity demanded] JQ uan t 1 t y supp 1. 1e dl by all markets ::; b th.. th.. y i - to ongm i GBgffi-- - Dual max l l Value Added Sum over [Shadow]. [ Quantity [ == pnce at demanded of trucking [ all markets] markets at markets firm l Sum over Shadow] [Quantity [. li d - II.. pnce at. supp e a [ on g ms ]. ongms _. by ongms.. l Total revenue ] _ Total cost [ [ of raw ~a~erials at markets at ongms l l Shadow [ price at [ Shadow price at jth market ith origin f l 11

12 L~ A Numerical Transportation Problem De.,1=ription. A trucking comp&ny be..e contracted to supply w~ldy~br~ Lucky supermarkets located in Modesto, Sonoma, &nd Fresno, CpJ.j rnia, with potatoes "ept at two Lucky wa.rchoue:es in Sacra.mcoto and O l and, also in California. The weekly availability (supply) of potatoes at the warehouses is 400 and 600 tons, respectively. The supcrmarket3' weekly neds (; L.:;;:, I L ~ 2- P.f?./.~ IN L WA-R,e tf.~$p.) Sacramento -too ~ ~ p (2 ~T't J..rMi ( m lrt(k f-1 ) ;:;!) ~~ ~ Cu-~;z;l ~ c,, Sonomr. ~'. WO ~1 ~ C-rz =~ ~ ~ Figure The flow of information in a tr&neportat ion problem. )4 j;. \z. ; ~ [_; (demands) a.re 200, 500, and ~ ' ' 300 tone of potatoes, respectively. The tr ~ &- portation unit cos.ts.are given in table 4.1. The task is to ti'. nd. the supply routes that minimize the total tr &- portation coets. for th~ truc" -ing firm.. The.contrad stipulates ~hat all su.- permarkets must re.ce1ve at least. the qu&0t1ty of potatoes specified ab.ve. Solution.., Tlie deiscription of the problem indicates that there fa.re two origins (wa,r.ehoue~). Saera.meoto &nd Oald&Dd, &nd three destinat oos (mar"ets), Mod es:~o. Sonoma, and Fresno. For the purpoee of recog iiing the flow of 'information, it is convenient to represent the linlcing ro tes amoog W arehouse,.and ma.rl.:et.s as in fi.gure 4.1. The formtllat.ion or. the primal problem requires the specification ofiwo sets of constraint.ii: the demand constraint.a and the availability constrai ts. The dem~d constraints etipulate that the~ dem&nd of potatoes.. by ac:b marl.:et be.~atiefied by. the supply Crom all wa.rebouaea: ' ~ ~ J).t11 + z,1. 2: 200 =. b1 Modes.to mark:et xi, +.tn ~ 500 = b, Sonoma marlcet.t %23 ~ 300 = 03 hesno ma.r-1.:et The availability eonstrairits. requi,re th,at ~a.ch warehouse not ship out ~ore Table 4.1. Transportation Unit Costs \ dou«mj.,......, Warehouse Supermarkets Modesto Sonoma hesno Sacramento Oe.lcl&nd of the commodity than i3 available on site :.::D. ~.5" Ztt + Zt, + Z13 :S 400 :: Ot z,1 + z,, + :23 $ 600 = o, The objective is to minimize the tota l transportation Sa.cramenlo warehouse Oakland warehouse coet: minimize TC= 40:::ii + 50: ::1~ + 40: tn + 90.t23 Each variable enters each set of conetraints only once. Therefore, the tr aosporta.tion matrix deaeribing the technology of transportat ion i.& represented in the following initiai tableau where the warehouse toostra fot.s have been reversed in order. to maintain the sam e direction of the inequa lities : Dual Initial Tableau TC z1 1.t12.t 13 J:21.t22.t23 sol 2: 200 l ~ 500 2: : : ', j I t I O I I to,'1 III < I> It I I I I I I O' I I I I' t I ' ' I ' I I'' ' I ' = 0 From the above initial tableau it is eaay to state the dua l problem : 400pf - 600pj maximize VA = 200pf + boop; + 300pg Pt Pi S 40 p~ P!!: 50 P3 Pi $ 80 Pt p; $ 40 p; p; < 30 pg p; :S 90 pf a.nd Pi ~. 0 for all i'e and j 's J The economic inte.rpretation of this dual examp le is similar to that given for lhe gene-ra.l specification. ~ (lt1., vo/'j:1'-- S ~ p I ii 'D fz. f.9 f o '.,c,o r~

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

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