AGBU ANLY & FORECASTING
AGBU ANLY & FORECASTING AGEC 622
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Basic Optimization Problem Notes for AGEC 622 Bruce McCarl Regents Professor of Agricultural Economics Texas AampM University Basic Optimization Problem Optimize FX Subject To st GX 3 S1 X 3 S2 X is a vector of decision variables X is chosen so that the objective FX is optimized FX is called the objective function It is What Will be maximized or minimized In choosing X the choice is made subject to a set of constraints GX 881 and XSS2 must be obeyed Basic Optimization Problem Optimize FX Subject to st GX 3 S1 X 3 S2 A program is a linear programming problem when FX and GX are linear and XS 3 0 When X 8S2 requires X39s to take on integer values you have an integer programming problem It is a quadratic programming problem where GX is linear and FX is quadratic It is a nonlinear programming problem when FX and GX are general nonlinear functions Decision Variables They tell us how much of something to do acres of crops number of animals by type truckloads of oil to move They are generally assumed to be nonnegative They are generally assumed to be continuous Sometimes they are problematic For example When the items modeled can not have a actional part and integer variables are needed They are assumed to be manipulatable in response to the objective This can be problematic also Constraints Restrict how much of a resource can be used What must be done For example acres of land available hours of labor contracts to deliver production requirements nutrient requirements They are generally assumed to be an inviolate limit They can be combined With variables to allow the use of more resources at a speci c price or a buy out at a speci ed level Nature of Objective function A decision maker is assumed to be interested in optimizing a measures of satisfaction by selecting values for the decision variables This measure is assumed to be quanti able and a single item For example Pro t maximization Cost minimization It is the function that When optimized picks the best solution from the universe of possible solutions Sometimes the objective function can be more complicated For example When dealing with pro t risk or leisure Example Applications A rm Wishes to develop a cattle feeding program Objective minimize the cost of feeding cattle Variables quantity of each feedstuff to use Constraint non negative levels of feedstuffs nutrient requirements so the animals don t starve A rm wishes to manage its production facilities Objective maximize pro ts Variables amount to produce inputs to buy Constraints nonneg production and purchase resources available inputs on hand minimum sales per agreements Example Applications A rm Wishes to move goods most effectively Objective Variables Constraints minimize transportation costs amount to move from here to there nonnegative movement available supply by place needed demand by place A rm is researching Where to locate production Objective Variables Constraints facilities minimize production transport cost Where to build amount to move from here to there amount to produce by location nonnegative movement construction production available resources by place products available by place needed demand by place This mixes a transport and a production problem Approach of the Course Users generally know about the problem and are Willing to use solvers as a black box We Will cover appropriate problem formulation results interpretation model use We Will treat the solution processes as a quotblack boxquot Algorithmic details and explanations Will be left to other texts and courses such as industrial engineering Fundamental Types of Uses Mathematical programming is way to develop the optimal values of decision variables However there are a considerable number of other potential usages of mathematical programming Numerical usage is used to determine exact levels of decision variables is probably the least common usage Types of usage problem insight construction numerical usages which nd model solutions solution algorithm development and investigation We discuss the rst two types of use Problem Insight Construction Mathematical programming usage requires a rigorous problem statement One must de ne the objective function the decision variables the constraints complementary supplementary and competitive relationships among variables The data must be consistent A decision maker must understand the problem interacting With the situation thoroughly discovering relevant decision variables and constraining factors in order to select the appropriate option Frequently resultant knowledge outweighs the value of any solutions Numerical Mathematical Programming Three main subclasses prescription of solutions prediction of consequences demonstration of sensitivity Prescriptive usually involves addressing What should I do or normative type questions For example What decision should be made given a particular speci cation of objectives variables and constraints It is probably the model in least common usage over universe of models Do you think that many decision makers yield decision making power to a model Very few circumstances deserve this kind of trust Models are an abstraction of reality that will yield a solution suggesting a practical solution not always one that should be implemented Predictive Most models are used for decision guidance or to predict the consequences of actions They are assumed to adequately and accurately depict the entity being represented They are used to predict in a conditional normative setting In other words if the rm Wishes to maximize pro ts then this is a prediction of What they should do given particular stimulus In business settings models predict consequences of investments acquisition of resources drought management and market price conditions In government policy settings models predict the consequences of policy changes regulations actions by foreign trade partners public service provision weather forecasting environmental change global warming Sensitivity Demonstration Many rms researchers and policy makers would like to know what would happen if an event OCCUIS In these simulations solutions are not always implemented Likewise the solutions may not be used for predictions Rather the model is used to demonstrate what might happen if certain factors are changed In such cases the model is usually speci ed with a quotrealisticquot data set It is then used to demonstrate the implications of alternative input parameters and constraint speci cations Handling IndiVisibilities Notes for AGEC 622 Bruce McCarl Regents Professor of Agricultural Economics Texas AampM University Handling Indivisibilities All or None Integer Programing McCarl and Spreen Chapter 15 Many investment and problems involve cases Where one has to take all or none of an item We cannot build 12 of a plant or buy 34 of a machine We build 1 or 2 or 3 or none but not a actional part This leads to integer programming max C1 W C2 X C3 Y st A1 W A2 X A3 Y Sb W 2 0 X 2 0 and int eger Y equals 0 0r 1 W is a normalcontinuous LP variable X is an integer variable Y is a zero one variable When problems have only X they are called pure integer only Y they are called pure zero one W and X they are called mixed integer other variants exist Handling IndiVisibilities Logical Conditions Integer Programing Integer programming also allows logical conditions to be imposed Suppose I am modeling a bottling plant that runs white milk but can run chocolate If they run any chocolate they encounter cleaning cost of F Let X be amount of chocolate milk processed Then add model component max CX FY s1 X MY S 0 X 2 0 Y equals Oorl Note in this component M is a big number 10 billion and for XgtO this implies Y1 while if FgtO Y0 if XO So if we run any chocolate milk we clean whether it be 1 gallon or one million Y is an indicator variable Handling Indivisibilities Integer Programing Suppose we can buy om k different types of machines and get from them capacity for the ith time period max 2 Cm Xm Z m k s1 2 am Xm Z CAPlk Yk S Ofor MN m k Xm ZOfor allm Yk equals 0 0r lfor all m In this case if they were mutually exclusive we could also add 2 Yk 1 k or if buying one meant we must buy another Y1 Y2 0 or if a machine can only be purchased if we have a minimum volume Zalm Xm ZLLlk YkSOforallz39 m k Handling IndiVisibilities Integer Programming Solution Dif culty All sounds good but problems are hard Let s explore Why Calculus is basis of all continuous optimization but not here because there is no neighborhood around a point in Which a derivative can be de ned 2X1 3X2 s 16 3X1 2X2 s 16 X1 X2 20ampinteger Feasible Region for XY nonnegative integers Figure 15 1 Graph ofFeasible Integer Points for First LP Problem YAxis Note UIbUJNH Handling Indivisibilities Integer Programming Solution Dif culty Figure 151 Graph ofFeasible Integer Points for First LP Problem YAxis Solutions are nite A line between 2 feasible points does not contain all feasible points Moving between points is not always easy Points are on boundary interior and not in general at corners Rounding of LP point may not be bad Handling Indivisibilities Integer Programming Solution Dif culty X1 7X2 23 X1 10X2 54 X1 X2 20ampinteger Consider Figure 15 2 Graph ofFeasible Integer Points for Second Integer Problem 10 c r A c c c o o o o o o o o o 8 o o o o o o o o o o o o o o o o 0 o 6 0 o o o o o o o o N gtlt 4 o o o o o o o o o 2 o o o o o o o 0 o o o o o o o o 0 o 0 A r i 1 r 1 0 2 4 5 8 10 X1 Here Rounding not good and Movement between points hard Handling Indivisibilities Integer Programming Solution Dif culty Mixed Integer Programming Feasible Region X1 integer X2 continuous 3X1 2X2 g 16 2X1 3X2 16 X1 2 0ampinteger X2 20 Figure 153 Mixed Integer Feasible Region 10 X1 X2 Handling Indivisibilities Integer Programming Solution Rounding Solving the problem Max 14X1 X2 s1 2X1 3X2 g 16 3X1 2X2 g 16 X1 X2 20ampinteger as an LP yields X1X232 Obj7 68 which can be rounded to X1X23 Obj7 2 But this may not always be feasible or optimal In this case an objective of 76 arises at X14X22 solintgms Rounding only works well if variable values are large Handling Indivisibilities Integer Programing Solution Branch and Bound Max X2 Solving st 2X1 3X2 s 16 3X1 2X2 s 16 X1 X2 20ampinteger as an LP yields X1X232 obj7 6 We can generate 2 related problems that collectively do not exclude integer variables as follows max 14Xl X2 max 14Xl X2 Prob A 2X1 326 16 or Prob B 226 326 16 3Xl 2X2 g 16 3X1 2X2 6 16 X1 g3 X1 24 X1 X2 20ampinteger X1 X2 20812ir1tegerl Suppose we solveProb A we get Xl3X2333 and again generate 2 more problems the rst with X23 and the other with X2 EI4 Solving these yields an integer solution at X1X23 obj7 2 and another at X12X24 obj68 But optimat solution is Xl4X22 obj7 6 from problem B above Handling Indivisibilities Integer Programing Solution Branch and Bound Now since we are maximizing we choose the 72 as the best solution and call it the incumbent But it is not necessarily optimal in fact it is not at all To verify its optimality we need to go back and investigate the problems we have not yet solved which still have the potential of having an objective function above our current best 72 We would then go back to the right hand problem from the rst setup and eventually nd X14X22 Obj7 6 The above reveals the basic nature of branch and bound It begins by solving an LP then nds a variable that is not integer and generates 2 problems creating a branch It then solves one of these and continues until it nds an integer solution which establishes a bound We then backtrack and try to eliminate all other possible branches either by nding they cannot have a better objective than the incumbent the bounding step or are infeasible This means intermediate solutions are found and we converge to within a tolerance Handling lndivisibilities Integer Programming Solution in EXCEL Integer programming problems can be solved with Excel To do this one adds a constraint restricting the variables to take on integer values This constraint is imposed as one adds any other constrains Namely a variable is chosen with the inquality imposed as one of the last two choices This is portrayed in the following screenshot Add Ennsllaml a eu garments mum mg a j Jmiegev a Namely choosing int restricm the variables to integer values and Bin is chosen to restrict variables to binary or 01 values Handling Indivisibilities Integer Programming Solution in EXCEL Unfortunately the default Excel solver does not reliably get the proper solution for integer programming problems unless you set some options Namely when you call up the solver go into the options box Then check that you are to assume a linear model and set the tolerance smaller Solver Options K Marlims hm seconds OK Solver Parameters x Sgt Target Cell mm W Max 9 Mg 0 game Dr l3 l v v Close By Changing Cells mam 5912am m2 CDnsh alnE sum Linear Maw D ESE mama scam sum U7 gt swss svm k V r I ma sum lt swsa swsis I Show Iterath Results SENEH Derivatives Est mares Ca Tgngem Eurward Newton 9 Quadralt Q Qanlral O CQHJLKJane Add Qhange Delete u Handling Indivisibilities Integer Programing Solution in EXCEL In addition students should realize that EXCEL is not an extraordinarily good choice for the solution of large linear integer or nonlinear programming problems and software like GAMS may be better in a professional setting Excel was taught in this class so as to not require students to learn multiple and initially unfamiliar software packages One can buy improved plugins by looking on the Internet and purchasing an upgraded solver which has this capability For example a Google search for the words excel in solver integer turns up a number of choices Handling Indivisibilities McCarl and Spreen Chapter 16 Integer Programing Solution Knapsack The knapsack problem is a famous 1P formulation Suppose a hiker selects the most valuable items to carry subject to a weight or capacity limit Partial items are not allowed thus choices are depicted by zeroone variables The general problem formulation assuming only one of each item is available is Max Zvaj j st Zdej S W J X Oorl forallj J The decision variables indicate whether the jth alternative item is chosen Each item is worth vi The objective function gives total value of items chosen The capacity used by each Xj is dj The constraint requires total capacity use to be less than or equal to the capacity limit W Handling Indivisibilities Integer Programing Solution Knapsack Suppose an individual is preparing to move Assume a truck is available that can hold at most 250 cubic feet of items Suppose there are 10 items which can be taken and that their names volumes and values implicit below The resultant formulation is Max 17X1 5X2 22X3 12X4 25X5 X6 15X7 21X8 5X9 20X10 st 70X1 10X2 20X3 20X4 15X5 5X6 120x7 5X8 20X9 20X10 S 250 X 0 or 1 for all j The optimal objective function value equals 128 The solution indicates that all items except furniture X7 should be taken There are peculiarities which should be noted First the constraint has 65 units in slack 250185 and no shadow price However for practical purposes the constraint does have a shadow price as the X7 variable would come into the solution if there were 120 more units of capacity but slack is only 65 Further note that each of the variables has a nonzero reduced cost This is because bounds have been added Shadow price and red cost are misleading Handling Indivisibilities Warehouse Location McCarl and Spreen Chapter 16 Warehouse location problems involve location of warehouses within a transportation system so as to minimize overall costs Basic decision involves tradeoffs between xed warehouse construction costs and transportation costs Suppw pomt 1 Suppw pomt 2 Suppw pomt i Handling Indivisibilities Warehouse Location Min 21ka ZZCikXik 212ijij ZZEUZU k k k j i j st Xik ZZU 3 Si foralli k j 2ij Zzij 2 DJ forallj k i ink 2ij g o forallk i j CAPkVk Z ij g 0 for all k j ZAka 3 bm forallm k szoorlj Xikj ijj Zij 2 O forallijk Merges Fixed Charge Capacity and Transportation Problem With transshipments We consider moving goods om supply i to demand j or from i to warehouse k and then on to demand j Handling Indivisibilities Warehouse Location VA VB VC XIA XIB XIC XZA XZB XZC YA1 YAZ YBI YBZ YCl YCZ Z11 Z12 Z21 Z22 506068128 6314634534876 1 l l l l l l l l l l l l l l l l l l l l l l l 1 1 1 1 l l l l 9999 l l 60 l l 70 l 1 VA VB VC 6 01 Xik ij Zij IV 0 for all ij k IV IV Handling Indivisibilities Warehouse Location Variable Value Reduced Cost ROW Activity Shadow Price VA 0 0 l 50 300 VB 0 2 2 75 0 VC 1 0 3 75 700 XIA 0 0 4 50 500 XIB 0 200 5 0 4 ch 0 1000 6 0 300 XZA 0 2 7 0 l00 X23 0 0 8 0 005 ch 70 0 9 0 l00 YA1 0 1052 10 0 l00 YA2 0 5052 11 l 2 YB1 0 0 YB2 0 300 Yc1 20 0 Ycz 50 0 Y 50 0 Y 12 0 600 Y21 5 0 tlt N N O 100 20 Handling Indivisibilities Machinery Selection McCarl and Spreen Chapter 16 The machinery selection problem is a common investment problem In this problem one maximizes pro ts trading off the annual costs of machinery purchase with the extra pro ts obtained by having that machinery A general formulation of this problem is Max ZFkYk 212ijij k jm J st CapikYk ZZAijkajm S 0 for alliandk igDnijjm S bn foralln ZGrkYk J m S er forallr k Yk is a nonnegative integer Xj Z 0 for alljkandm m The decision variables are Yk the integer number of units of the kth type machinery purchased ij the quantity of the jth activity produced using the mth machinery alternative The parameters of the model are Fk the annualized xed cost of the kth machinery type Capik the annual capacity of the kth machinery type to supply the ith resource Gr the usage of the rt machinery restriction when purchasing the kth machinery type ij the per unit net pro t of ij Aijm the per unit use by ij of the ith capacity resource supplied by purchasing machine k D11ij the per unit usage of xed resources of the nth type by ij bn the endowment of the nth resource in the year being modeled and er the endowment of the rt machinery restriction 21 Handling Indivisibilities Machinery Selection Fluwwnh qum 1 Fluw wnh deur 2 dPluwl andFluwZ andFlqu 1nFEnud 1mm Fluw Planter D15 Harvester 1 1 2 1 2 1 1 Obe11vesrrax 75111111 71111111 7121111 72111111 7211111 71111111 7121111 7111111111 712111111 Tamar 1 71511 Capamy VIBE 1nFEnud 721111 121m 2 Capanly m Fanud Capamy ur Plantar Capamy ur D15 Capamy ur Hap Vesta Land Amlable Lmk D150 Flamer 22 ansFluWZ 1nqud 1 Plant D15 8 Tram 1 Planter 1 Cmp Sales 79 33 78 2s 2 s Table 1617 Solution for the Machinery Selection Problem ob 116100 Variable Value Reduced Cost Equation Slack Shadow Price Buy Tractor 1 1 5000 Tractor 1 capacity in Period 1 100 0 Buy Tractor 2 0 0 Tractor 1 capacity in Period 2 130 0 Buy Plow 1 0 0 Tractor 1 capacity in Period 3 50 0 Buy Plow 2 1 1200 Tractor 2 capacity in Period 1 0 12 Buy Planter 1 0 0 Tractor 2 capactiy in Period 2 0 146 Buy Planter 2 1 3300 Tractor 2 capacity in Period 3 0 2226 Buy Disc 1 0 0 Plow 1 capacity in Period 1 0 625 Buy Disc 2 1 0 Plow 1 capacity in Period 2 0 0 Buy Harvester 1 0 0 Plow 2 capacity in Period 1 100 0 Buy Harvestor 2 1 0 Plow 2 capacity in Period 2 180 0 Plow with Tractor 1 and Plow 1 in Period 1 0 245 Planter 1 capacity 0 0 Plow with Tractor 1 and Plow 1 in Period 2 0 120 Planter 2 capacity 130 0 Plow with Tractor 1 and Plow 2 in Period 1 600 0 Disc 1 0 0 Plow with Tractor 1 and Plow 2 in Period 2 0 0 Disc 2 130 0 Plow with Tractor 2 and Plow 1 in Period 1 0 1825 Harvester 1 0 50 Plow with Tractor 2 and Plow 1 in Period 2 0 146 Harvester 2 50 0 Plow with Tractor 2 and Plow 2 in Period 1 0 0 Labor available in Period 1 128 0 Plow with Tractor 2 and Plow 2 in Period 2 0 013 Labor available in Period 2 144 0 Plant with Tractor 1 and Planter 1 0 191 Labor available in Period 3 25 0 Plant with Tractor 1 and Planter 2 600 0 Plow Plant 0 230533 Plant with Tractor 2 and Planter 1 0 1077 Plant Harvester 0 34175 Plant with Tractor 2 and Planter 2 0 0 Land 0 229333 Harvest with Tractor 1 and Harvester 1 0 1775 One Planter 0 0 Harvest with Tractor 1 and Harvester 2 600 0 One Disc 0 0 Harvest with Tractor 2 and Harvester 1 0 2517 Planter 1 to Disc 1 0 0 Harvest with Tractor 2 and Harvester 2 0 5565 Planter 2 to Disc 2 0 0 Sell Crop 84000 0 Yield Balance 0 25 Purchase Inputs 600 0 Input Balance 0 1 10 Basic LP problem formulations Notes for AGEC 622 Bruce McCarl Regents Professor of Agricultural Economics Texas AampM University Basic LP formulations Linear programming formulations are typically composed of a number of standard problem types In these notes we review four basic problems examining their a Basic Structure b Formulation c Example application 1 Answer interpretations The problems examined are the a Transportation problem b Feed mix problem c Joint products problem 1 Disassembly problem Transportation Problem Basic Concept This problem involves the shipment of a homogeneous product om a number of supply locations to a number of demand locations Supply Locations Demand Locations 1 S 2 l 2 2 mgtlt n r Problem given needs at the demand locations how shouldl take limited supply at supply locations and move the goods Further suppose we Wish to minimize cost Objective Minimize cost Variables Quantity of goods shipped from each supply point to each demand point Restrictions Non negative shipments Supply availability at a supply point Demand need at a demand point 3 Transportation Problem Formulating the Problem Basic notation and the decision variable Let us denote the supply locations as i Let us denote the demand locations as j Let us de ne our fundamental decision variable as the set of individual shipment quantities from each supply location to each demand location and denote this variable algebralcally as Movesupplyidemandj Transportation Problem Formulating the Problem The objective function We want to minimize total shipping cost so we need an expression for shipping cost Let us de ne a data item giving the per unit cost of shipments om each supply location to each demand location as costsupplyiademandj Our objective then becomes to minimize the sum of the shipment costs over all supplyi demandj pairs or Minimize Z 2 cost supplyi demand j Move supplyidemandj supplyidemandj Which is the per unit cost of moving om each supply location to each demand location times the amount shipped summed over all possible shipment routes Transportation Problem Formulating the Problem There are three types of constraints 1 2 3 supply availability limiting shipments from each supply point to existing supply so that the sum of outgoing shipments om the supplyith supply point to all possible destinations demandj to not exceed sup plysupplyi 2 Move dem andj supplyidemandj S sup plysupplyi minimum demand requiring shipments into the demandjth demand point be greater than or equal to demand at that point Incoming shipments include shipments from all possible supply points supplyi to the demandjth demand point sgyiMovesupplyiademandj 2 dellnanddemandj nonnegative shipments Move 2 0 supplyidemandj Transportation Problem Formulating the Problem Mm1m1ze Z Z COStSupplymemandjMOVesupplyidemandj supplyi demandj Sit degndj MovesupplyideIrlandj S supplysupplyi for all supplyr gt supplyi MovesupplyideIrlandj demanddemandj for all demand Move 2 0 for all supplyi demandj supplyidemandj Transportation Problem Example Shipping Goods Three plants Four demand markets Quantity availability Miami Houston Supply Available New York 100 Chicago 75 Los Angeles 90 Distances New York 3 Chicago 9 Los Angeles 17 New York Chicago Los Angeles MiamiHoustonMinneapolisPortland Demand Required Miami 30 Houston 7 5 Minneapolis 90 Portland 50 Minneapolis Portland 6 23 3 l3 l3 7 Transportation costs 5 5Distance Miami Houston Minneapolis Portland New York 20 Chicago 50 Los Angeles 90 35 120 20 70 70 40 Transportation Problem Example Shipping Goods costsupplyidemandj Movesupplyi emandj supplyi demandj Sit demand MovesupplyideIrlandj S supplysupplyi for all supplyi supplyi MovesupplyidaIrlandj 2 dernanddemandj for all demandj 2 MovesupplyideIrlandj 0 for all supply1 demand E E E E E E E E E E 2 E 20 40 35 120 50 60 20 70 90 35 70 40 Minimize 1 1 1 1 S 100 1 1 1 1 S 75 1 1 1 1 S 90 1 1 1 2 30 1 1 1 2 75 1 1 1 2 90 1 1 1 2 50 Transportation Problem Example Shipping Goods Solution shadow price represents marginal values of the resources iemarginal value of additional units in Chicago 15 reduced cost represents marginal costs of forcing non basic variable into the solution ie shipments from New York to Portland costs 75 twenty units are left in New York Optimal Solution Objective value 7425 Variable Value Reduced Cost Equation Slack Shadow Price Move 11 3 0 0 l 20 0 Move 12 3 5 0 2 0 l 5 Move13 l 5 0 3 0 5 Move14 0 75 4 0 20 Move21 0 45 5 0 40 Move22 0 3 5 6 0 3 5 Move23 75 0 7 0 45 Move24 0 40 Move31 0 75 Move32 40 0 Move33 0 40 Move34 50 0 Optimal Shipping Pattern Destination Miami Houston Minneapolis Portland Feeding Problem Basic Concept This problem involves composing a minimum cost diet from a set of available ingredients While maintaining nutritional characteristics Within certain bounds Objective Minimize total diet costs Variables how much of each feedstuff is used in the diet Restrictions Non negative feedstuff Minimum requirements by nutrient Maximum requirements by nutrient Total volume of the diet This problem requires two types of indices Type of feed ingredients available from Which the diet can be composed ingredientj com soybeans salt etc Type of nutritional characteristics Which must fall Within certain limits nutrient protein calories etc Feeding Problem Basic Concept Variable Feed 1ngredientj amount of feedstuff ingredientj fed to animal Objective total cost We want to minimize total diet costs across all the feedstuffs so we need an expression for feedstuff costs Let us de ne a data item giving the per unit cost of ingredients as costingrediemj Our objective then becomes to minimize the sum of the diet costs over all feed ingredients Feed 1ngredientj Minimize 2 cost Ingredientj 1ngred1ent which is the per unit cost of ingredients summed over the feed ingredients Feeding Problem Basic Concept Additional parameters representing how much of each nutrient is present in each feedstuff as well as the dietary minimum and maximum requirements for that nutrient are needed Let 1 anutrimingmhentj be the amount of the nutrientth nutrient present in one unit of the ingredientjth feed ingredient 2 ULnutrient and LLnutrient be the maximum and min1mum amount of the nutrientth nutrient in the diet Then the nutrient constraints are formed by summing the nutrients generated from each feedstuff anutrientingredientjFingredientj and requlrlng these to exceed the d1etary m1n1mum andor be less than the maximum Problem then focuses on how much of each feedstuff is used in the diet to maintain nutritional characteristics Within certain bounds 13 Feeding Problem Formulating the Problem The constraints There are four general types of constraints 1 minimum nutrient requirements restricting the sum of the nutrients generated from each feedstuff anutrientingredientjFingredientj to meet the dletary m1n1mum Feed gt minimum anutrientingredientj 1n gredientj nutrient ingredientj 2 maximum nutrient requirements restricting the sum of the nutrients generated om each feedsmff anutrientingredienthingredientj 50 11015 exceed the dietary maximum Feed lt maximum anutrientingredient 1n gredient nutrient ingredientj 3 total volume of the diet constraint requiring the ingredients in the diet equal the required weight of the diet Suppose the weight of the formulated diet and the feedstuffs are the same then ingredientj edingredientj 1 4 nonnegative feedstuff Feed 0 14 1ngredientj Feeding Problem Example cattle feeding Seven nutritional characteristics energy digestible protein fat vitamin A calcium salt phosphorus Seven feed ingredient availability corn hay soybeans urea dical phosphate salt vitamin A New product potato slurry Ingredient costs per kilogram cingrediemj Ingredient Costs per kg Corn 0133 Dical Phosphate 0498 Alfalfa hay 0077 Salt 0110 Soybeans 0300 Vitamin A 0286 Urea 0332 Nutrient requirements per Kilogram Nutrient Unit Minimum Maximum Net energy Mega calories 134 Digestible protein Kilo grams 0071 0 13 Fat Kilograms 005 Vitamin A International Units 2200 Salt Kilograms 0015 002 Calcium Kilograms 00025 001 Phosphorus Kilograms 00035 0012 Weight Kilo grams 1 l 1 5 Feeding Problem Example cattle feeding Nutrient compositions of 1 kg of each feed Nutrient Dical VitaminA Potato Characteristic Corn Hay Soybean Urea Phosrhate Salt Concenrate Slurry Net energy 148 049 129 139 Protein 0075 0127 0438 262 0032 Fat 0357 0022 0013 0009 Vitamin A 600 50880 80 2204600 Salt 1 Calcium 0002 0125 00036 02313 0002 Phosphorus 0035 0023 00075 068 01865 00024 Feeding Problem Example cattle feeding Corn Hay Soybean Urea Dical Salt Vitamin A Slurry Min 133Feedc 077FeedH 3Feeds3 332Feedm 498Feedd 110Feedxu 286FeedVA PFeedSL st 0 75FeedC 12 7FeedH 438FeedSB 262FeedJr 032FeedSL LE 13 protein Max 0357leC 022FeedH 013FeedSB 009EeeulSL LE 05 fat Nutrient FeedXLT LE 02 salt 0002Feedc 0125FeedH 003617eealSB 2313Feedd 002FeedSL LE 01 Calc 0035Feedc 0023FeedH 007517eealSB 68Feedm 1865Feedd 0024FeedSL LE 012 phosp 148FeedC 49FeedH 129FeedSB 139FeedSL GE134 energy 0 75FeedC 12 7FeedH 438FeedSB 262FeedJr 032Feed GE 071 protein Min 600Feedc 50880FeedH 80Eee iSB 2204600FeeulVA GE 2200 VA Nutrient FeedXLT GE 015 salt 0002Feedc 0125FeedH 003617eealSB 2313Feedd 002FeedSL GE0025 Calc 0035Feedc 0023FeedH 007517eealSB 68Feedm 1865Feedd 0024FeedSL GE 0035 phos Volume FeedC FeedH FeedSB FeedUr Feedr1 FeedXLT FeedVA FeedSL 1 Feedc FeedH Feedm Feedur Feedd FeedXLT FeedvA FeedSL GE 0 Feeding Problem Example cattle feeding Solution least cost feed ration is 956 slurry 01 vitamin A 15 salt 02 dicalcium phosphate 14urea 11 soybeans and 01 hay reduced costs of feeding corn is 095 cents shadow prices nonzero values indicate the binding constraints the phosphorous maximum constraint along with the net energy protein salt and calcium minimums and the weight constraint If relaxing the energy minimum we save 0065 Optimal Solution Objective value 0021 Variable Value Reduced Cost Equation Slack Shadow Price Feedc 0 0095 Protein Max 0059 0 FeedH 0001 0 Fat Max 0041 0 FeedSB 0011 0 Salt Max 0005 0 FeedUR 0014 0 Calcium Max 0007 0 Feedd 0002 0 Phosphrs Max 0000 2207 FeedSLT 0015 0 EnergyMin 0000 0065 FeedVA 0001 0 Protein Min 0000 0741 FeedSL 0956 0 Vita Min 0000 0 Salt Min 0000 0218 Calcium Min 0002 4400 Phosphrs Min 0008 0 XI 1 L4 n nnn vv 1511L UUUU Joint Products Problem Basic Concept This problem deals with joint products in Which rms Wish to maximize total pro ts derived om the available production possibilities Each of the production possibilities yields multiple products uses some inputs With a xed market price and uses some resources that are available in xed quantity Problem focuses on maximizing pro t derived om available production possibilities Objective Maximize pro ts Variables 1 How much of quantity of the productth product is produced iii Amount of the inputth input purchased Restrictions Non negativity Supply demand balance Resource availability 19 Joint Products Problem Formulation Basic notation and the decision variable Let us denote the set of the produced products as product the production possibilities as process the purchased inputs as input the available resources as resource Let us define three fundamental decision variables as the set of produced product Salesproduct the set of production poss1b111t1es PI OdUCthnprocess the set of purchased inputs Buylnput 1nput 20 Joint Products Problem Formulation To set up the joint products problem four additional parameter values that give the composite relationship among product process input and resource are needed These parameters are the quantity sold of each produced product product to the quantity yielded by the processth production possibility q productprocess the amount of the inputth input used by the processth production possibility r 1nputprocess the amount of the resourceth resource used by the processth production possibility s resourceprocess the amount of resource availability or endowment by resource b resource 21 Joint Products Problem Formulating the Problem The objective function We want to maximize total pro ts across all of the possible productions To do so three additional required parameters for sale price input purchase cost and other production costs associated With production are needed Let us de ne these parameters as SalePriceproduct InputCost 1nput OtherCost process Then the objective function becomes Maximize Z SalePrice product Sales product product 2 InputCost Buylnput input 1nput 1nput Z OthCost Production process process process 22 Joint Products Problem Formulating the Problem The constraints There are four general types of constraints 1 2 3 demand and supply balance for which quantity sold of each product is less than or equal to the quantity yielded by production 39 lt Salesproduct Z qpmductprocess Productlonpmcess 0 process demand and supply balance for which quantity purchased of each xed price input is greater than or equal to the quantity utilized by the production activities Production Buylnput lt 0 1 1nputprocess process 1nput process resource availability constraint insuring that the quantity used of each xed quantity input does not exceed the resource endowments Z s Production 3 process resourceprocess process resource 4 nonnegatiVity Production gt 0 process Sales Buylnput product 1nput 23 Joint Products Problem Formulating the Problem Maximize SalePrice Sales product product product input InputCostinlput BuyInputinlput process OthCostpmceSS Productlonpmcess st Production lt 0 qproductprocess process salesproduct process process Iinputprocess PIO h lcuonprocess 1Bl l ylnpu tinput S 0 process sresourceprocess bresource Sales BuyInput Production gt 0 product 1nput process 24 Joint Products Problem Example wheat production Two products produced Three inputs Seven production processes Wheat and wheat straw Landfertilizer seed Outputs and inputs per acre Process 1 2 3 4 5 6 7 Wheat yield in bushel 30 50 65 75 80 80 75 Wheat straw yieldbales 10 17 22 26 29 31 32 Fertilizer use in Kg 0 5 10 15 20 25 30 Seed in pounds 10 10 10 10 10 10 10 land 1 1 1 1 1 1 1 Wheat price 4bushel wheat straw price O5bale Fertilizer 2 per kg Seed 02lb 5 per acre production cost for each process Land 500 acres 25 Joint Products Problem Example Wheat Production Maximize mg SalePrlceproduct Saleslproduct gilt InputCostinput BuyInputinput Z OthCostpmceSSProductlonpmcesS process st salesproduct qproductprocess Pr0 h lcuonprocess S 0 process Iinputprocess Productlonprocess Buylnputinput S 0 process 2 sresourceprocess PIO h lctlonprocess Sbresource process Salespmduct BuyInputinput ProductlonProcess 2 O 4Sale15Sale2 5y1 5y2 5y3 5Y4 5Y5 5y6 5Y7 221 ZZZMaximize st Sale1 30y1 50Y2 65Y3 75Y4 80Y5 8034 75y7 LE 0 Salez 10y1 17Y2 22Y3 26Y4 29Y5 3136 32Y7 LE 0 5y2 10Y3 15Y4 20Y5 25y6 30y7 21 LE 0 10y1 10Y2 10Y3 10Y4 10Y5 10Y6 10y7 22 LE 0 Y1Y2Y3Y4Y5Y6Y7 LE 500 SaleISale2Y1Y2Y3Y4Y5Y6Y7Z122GE0 Note that Y refers to Production and Z refers to BuyInput 26 process input Joint Products Problem Example Wheat Production Solution 40000 bushels ofwheat and 14500 bales of straw are produced by 500 acres of the fth production possibility using 10000 kilograms of fertilizer and 5000 lbs of seed reduced cost shows a 16950 cost if the rst production possibility is used shadow prices are values of sales purchase prices of the various outputs and inputs and land values 2875 Optimal Solution Objective value 143450 Variable Value Reduced Cost Equation Slack Shadow Price Salewheat 40000 0 Wheat 0 4 SalestaIW 14500 0 Straw 0 05 Process1 0 16950 Fertilizer 0 2 Process2 0 9600 Seed 0 02 Process3 0 4350 Land 0 2875 Process4 0 1150 Process5 500 0 Process6 0 900 Process7 0 3850 Byinputfm 10000 0 Byinput A 5 000 0 27 Disassembly Problem Basic Concept This problem involves raw product disassembly This problem is common in agricultural processing Where animals are purchased slaughtered and cut into parts steak hamburger etc Which are sold The primal formulation involves the maximization of the product sold revenues less the raw product purchased costs subject to restrictions that relate the amount of component parts to the amount of raw products disassembled Objective Maximize operating pro ts Variables Quantity of the nal product sold Number of units of raW product purchased Restrictions Non negativity Product balance Resource limitation Upper bound on raw disassembly Upper and lower bounds on salesg Disassembly Problem Formulating the Problem Basic notation and the decision variable This problem requires three types of indices Raw products disassembly and let denote it as productj Component parts and let denote it as partk Resource and let denote it as resourcer Let us de ne fundamental decision variables Raw is the number of units of raw product productj purchased Sold is the quantity of component part sold partk 29 Disassembly Problem Formulating the Problem The objective function Because the objective function is to maximize operating pro ts revenue earned by sales minus input costs we need additional expression for l the cost of purchasing one unit of raw product Costlproductj 2 the selling price of component part Pr1cepartk Our objective then becomes to maximize the sum over all nal products sold Soldpa k of the total revenue earned by sales less the costs of all purchased inputs Rawpmductj productj Raw k Sold Max1m1ze 2 Cost productj productj p Prlcepart partk Disassembly Problem Formulating the Problem To set up this problem six additional parameter values that give the relationship among productj partk and resourcer are needed These parameters are the yield component part partk from raw product productj yield partkproductj the use of resource limit resourcer When disassembling raw product productj useraw resourcerproductj the amount of resource limit resourcer used by the sale of one unit of component part partk usesold resourcerpartk the maximum amount of raw product limit resourcer available b resourcer the maximum amount of component part productj available maxavailablepmductj the maximum quantity of component part partk that can be sold maxsoldpartk the minimum amount of the component partk that can be sold minsoldpartk 31 Disassembly Problem Formulating the Problem There are ve types of constraints 1 2 3 4 6 product balance limiting the quantity sold to be no greater than the quantity supplied When the raw product is disassembled 2 yield productj Partkpr0ductj Sold lt 0 Raw partk productj resource limitation constraining on raw product disassembly and product sale USCI39aW productj pgk uSCSOIdresourcerpartk SOldpartk S b Raw resourcerproductj productj TCSOUTCCT upper bound on disassembly Rameductj S maxavaialable productj upper and lower bound on sales Soldpartk s maxsoldpartk Soldpartk 2 m1nsoldpartk Nonnegativity 1avvproductj 7 SOldpartk 2 O 32 Disassembly Problem Formulating the Problem Maximize 2 Cost Raw productj productj productj I Prlcepartk Sold partk st NEW yleldpartkproductj RaWproductj SOldpartk S 0 Z useraw Raw productj resourcerproductj product pgk uSGSOIdresourcerpartk SOldpartk S b TCSOUTCCT Rawpmductj s maxavallablepmductj Soldpartk s maxsoldpartk Soldpartk 2 mmsoldpartk Rawpmductj Soldpartk s O Disassembling 4 type of cars Disassembly Problem Example Junk Yard 5 component parts Escorts 62639s Tbirds Caddy39s Metal seats chrome doors junk Component part yields from each type of car Car Data ESCORTS 626S TBIRDS CADDIES PURCHASE PRICE 85 90 115 140 WEIGHT 2300 2200 3200 3900 DISASSEMBLY COST 100 120 150 170 AVAILABILITY 13 12 20 10 Resource Use to Breakdown Cars CAPACITY 1 1 12 14 LABOR 10 12 15 20 Propctional Breakdown of Cars into Component Parts lb ESCORTS 626S TBIRDS CADDIES METAL 60 55 60 62 SEATS 10 10 06 04 CHROME 05 05 09 14 DOORS 08 10 10 07 JUNK 17 20 15 13 Part Data MAXIMUM PRICE PARTSPACE LABOR METAL 015 0 00010 SEATS 6000 090 0003 00015 CHROME 10000 070 00014 00020 DOORS 5000 100 00016 00025 JUNK 005 0 00001 Disassembly Problem Example Junk Yard Labor endowment 700 hours Junk yard capacity 42 units parts space 60 units Extending the problem by requiring parts to be transformed to other usages if maX sales have been exceed chrome gt metal seat gt junk doors gt 70 metal and 30 junk a m E E m E m m S H o E o D o g 00 o D MIN OBJECTIVE 185 210 265 310 39 015 090 070 100 005 METAL 1380 1210 1920 2418 1 1 07 0 SEATS 230 220 192 156 1 1 0 CHROME 115 110 288 546 1 1 0 DOORS 184 220 320 273 1 1 0 JUNK 391 440 480 507 1 l 03 CAPACITY 1 1 12 14 f 42 LABOR 10 12 15 20 001 0015 0020 0025 0001 S 700 PARTSPACE 003 0014 0016 E 60 Upper Bound 13 12 15 20 6000 10000 5000 Disassembly Problem Example Junk Yard Solution shadow price ie marginal value of additional junk yard capacity is 2476 reduced cost ie selling more 1 more seat costs 095 max sales possibilities on seats chrome doors are exceeded a 320 lbs of seats are converted to junk b 1680 lbs of chrome are converted to metal c 70 of 4866 lbs of doors are converted to metal and the rest is converted to junk Optimal Solution Objective value 183372 Variable Value Reduced Cost Constraint Slack Shadow Price Parts Dlssasemble METAL 0 0150 ESCORTS 4 0 SEATS 0 0050 626s 0 49960 CHROME 0 0150 DOORS 0 0090 TBIRDS 20 JUNK 0 0050 CADDIES 10 Resources Sen CAPACITY 0 24760 LABOR 43512 0 METAL 731862 0 PARTSP 20 0 SEAT S 6000 0 Max Cars CHROME 10000 0 ESCORTS 9 0 DOORS 5000 0 6265 14 0 TBIRDS 0 31688 JUNK 180138 0 CADDIES 0 91356 Max Sell Convert SEATS 0 095 SEAT S 320 0 CHROME 0 0550 DOORS 0 0910 CHROME 1680 0 DOORS 4866 0 36 Including Risk Notes for AGEC 622 Bruce McCarl Regents Professor of Agricultural Economics Texas AampM University Including Risk Where does it occur Maximize CX Subject to AX S b X Z 0 0 Objective function returns Variability in prices production quantities costs and market sales Resource usages A Variability in raw input quality working conditions intermediate product yields and product requirements Resource endowments b Variability in demand rm faces resources available and working conditions Including Risk When incorporating risk there are three big issues 1 What is the nature of risk a What parameters of the model are uncertain b How do we describe their distribution 2 When are risk outcomes revealed Does the producer receive information about uncertain events and will make adaptive decisions 3 How do we model behavioral reaction to risk Is expected pro t maximization not the proper objective but rather some degree of aversion to the variation caused by risk Including Risk Forms of assumed reaction to risk NonRecourse or non adaptive decision making Decisions made now consequences felt later No additional decisions made until consequences felt Example Buy stock now make no decisions for one year Recourse or adaptive decisionmaking Decisions made now consequences arise over time Later time additional decisions made Later one knows what happened between first decision and now Cannot reverse prior actions but can adjust ie can be make adjustments phenomena called irreversibility and recourse Example Buy stock now reView decisions quarterly possibly selling and buying other stocks Including Risk Forms of assumed reaction to risk Expected Value Maximization max EX s1 ZX S I X Z 0 Conservative Fat or thin coef cients Where E E Risk Discount 5 a Risk Discount 5 I Risk Discount E V Maximize Eincome RAP Varianceincome Including Risk Forms of assumed reaction to risk Our risk treatment will be limited because of time constraints see newbookpdf chapter 14 for more extensive treatment 1 Will treat risk in objective function coef cients Will not discuss how to form such probability distributions 2 Will only cover the expected value variance formulation this is the two main one used in practice see newbookpdf chapter 14 for more extensive treatment 3 We will treat non adaptive behavior Including Risk Why model risk Why not just solve for all values of risky parameters Curses of dimensionality and certainty Dimensionality Number of possible plans 3 possible values for 5 parameters 35 243 Certainty Each would be certain of data so we would have 243 different things we could do Which one would we do General Risk Modeling Aim Generate a plan which is Robust in the face of the Uncertainty Not best performer necessarily in any setting but a good performer across many or most of the uncertainty spectrum Including Risk First Risk Model Markowitz meanvariance portfolio choice formulation Given Problem maX suminvest moneyinvestinvestavgreturninvest st suminvestmoneyinvestinvestpriceinvest funds moneyinvestinvest 3 O Markowitz observed not all money is invested in the highest valued stock Inconsistent With LP formulation Why Not a basic solution Markowitz posed the hypothesis that average returns and the variance of returns were important Including Risk Given a linear objective function Z c1X102X2 Where X1 X2 are decision variables c1 c2 are uncertain parameters Then Z is distributed with mean and variance Z chl 62X2 Sz 2 S12X12 522X222gtXlt S12X1X2 Where 621 62 are the means of c1c2 s12 s22 are the variances of c1c2 S12 is the covariance of c1c2 Including Risk De ning terms si2 is the variance of objective function coefficient of X which is calculated using 2k cik ci2N Where cik is the kth observation on the objective value of Xi and N the number of observations sij for i 7 j is the covariance of the objective function coefficients between Xi and Xj calculated by the formula sij Z k cik cicjk cjN Note sij sji r sqrt si2 s12 Where r is correlation ci is the mean value of the objective function coefficient associated with X calculated by Z k cikN Assuming an equally likely probability of occurrence Including Risk EV Model Statistical Background In matrix terms the mean and variance of Z are ZEX a X39 SZ Where in the two by two case Including Risk EV Model Statistical Background Programming Formulation Max E o EXRAP kX39SZ ZEJ39XJ RAPZZSJkaXk j j k SJ 2X S funds J X J 20 for all j or Max B RAP Variance RAP equals dollars of income forgone per dollar of variance reduced It is used by varying RAP from O profit maX solution in nity minimum variance solution Including Risk Data for EV Example Returns by Stock and Event Stock Returns by Stock and Event Stockl Stock2 Stock3 Stock4 Eventl 7 6 8 5 Event2 8 4 l6 6 Event3 4 8 l4 6 Event4 5 9 2 7 Event5 6 7 l3 6 Event6 3 10 ll 5 Event7 2 l2 2 6 Event8 5 4 l8 6 Event9 4 7 l2 5 Eventh 3 9 5 6 Stock Price 22 3O 28 26 Including Risk Mean Returns and Variance Parameters for Stock Exanu e Stockl Stock2 Stock3 Stock4 Mean Returns 470 760 830 580 Variance Covariance Matrix Stockl Stock2 Stock3 Stock4 Stockl 321 352 699 004 Stock2 352 584 1368 012 Stock3 699 1368 6181 164 Stock4 004 012 164 036 Including Risk EV Model Example Max E 5 2 E RAP Variance Max 25196 Zzsijij sl 12X 1 S funds X 20f0rallj J or for the example X1 321 352 699 004 X1 X2 352 584 1368 012 X2 Max 470 760 830 580 X1 X2 X3 X4 X3 699 1368 6181 164 X3 X 004 012 164 036 X4 4 s1 22X1 30X2 28X3 26X4 500 Including Risk EV Model Example X1 321 352 X2 352 584 Max 470 760 830 580 X1X2 X3 X4 X3 699 1368 X4 004 012 s1 22X1 30X2 28X3 26X4 s 500 01 Max 32le2 352X1X2 699X1X3 352X1X2 584X22 1368X2X3 699X1X3 1368X2X3 6181X32 004X1X4 012X2X4 164X3X4 SI 22X1 30X2 28X3 26X4 g 5004 699 1368 6181 164 004 012 164 036 x553 4 004X1X4 012X2X 164X3X4 036X Including Risk Solutions for Alternative Risk Aversion Parameters RAP BUYSTOCK2 BUYSTOCK3 OBJ MEAN VAR STD SHADPRICE RAP BUYSTOCKl BUYSTOCK2 BUYSTOCK3 BUYSTOCK4 OBJ MEAN VAR STD SHADPRICE 17 148 148 125 131 554 23 239 0 857 214 214 19709 140 296 821 392 011 893 043 441 545 929 557 000025 17 143 148 19709 140 0 124 131 547 23 857 287 214 821 392 277 012 960 972 614 459 587 401 236 00005 1 138 16274 127 0 263 16 444 146 764 573 261 504 581 0015 123 430 20 273 420 550 380 129 560 750 234 839 000075 5 12 135 141 7523 86 260 0 324 152 688 331 441 738 025 4372 070 2561 120 125 222 14 230 375 939 576 919 134 I OOLBO 001 355 977 245 138 4460 66 260 705 478 787 050 405 188 753 168 116 121 97 850 224 805 656 026 Including Risk EV Example Frontier EV Frontier 160 Frontier 7 140 i 120 A O 0 Mean Income 0 00 O 0 Mb 00 O I I I I I I I I I 0 2000 4000 6000 8000100001200014000160001800020000 Variance of Income Including Risk Shadow prices Max EX c1 XSX 31 AX Lagrangian EXu EX CDX39SX Ear6X E 2c1gtXs M4 away AX 9 implies roughly u E 2CDXSA So u risk adjusted average returns Including Risk Forming Probability Distributions Probability distributions state the relative frequency of occurrence of a set of mutually exclusive events Desirable Characteristics 1 each of the states of nature must be mutually exclusive 2 probability of occurrence of each of the states of nature must be an unbiased measure of the current probability of that state of nature occurring 3the sum of the probabilities across the states of nature must equal one Second property is the most troubling in When using objective historical data trends events Including Risk Forming Probability Distributions Price F1gur9 1 H QIDHL PI ICE 39 9 9 0 I 9 o o Pncefcwt I Mean ce 1977 1979 1991 1993 1935 1997 1999 1991 1993 Yea Igure 3 Real Price vs Trend 7 quoti Z FA as V 1333 PIECE ka s k 1 3 Prm ecte Prlce a a 1 Price Including Risk Forming Probability Distributions i Figure 4 Real Price vs Trend V5Projected Pricel ED R tc r A VNXW QQ x W iliililllll llllll 1977 1931 1955 1979 1983 1937 Year 1939 1993 1991 i333 PriceCw Pruiected Reai Price PliEE Prubabiiity Distribuiiun Figure 5 Probability Distribution 05 Probability o w i o N i Stationary Price Historic Price Real Price d A 329 372 416 46 503 547 591 634 678 722 Price Including Risk General Lessons Learned on Objective Probabilities Use objective data trends and other systematic effects can bias Use a procedure like regression to develop values expected One may nd residual terms are hetereoskedastic