Basic LP problem formulations

Transcription

1 Basic LP problem formulations Notes for AGEC 622 Bruce McCarl Regents Professor of Agricultural Economics Texas A&M University 1

2 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 d. Answer interpretations The problems examined are the: a. Transportation problem b. Feed mix problem c. Joint products problem d. Disassembly problem 2

3 Transportation Problem McCarl and Spreen Chapter 5 Basic Concept This problem involves the shipment of a homogeneous product from a number of supply locations to a number of demand locations. Supply Locations Demand Locations m n Problem given needs at the demand locations how should I take limited supply at supply locations and move the goods. Further suppose we wish to minimize cost. Objective: Variables: Restrictions: Minimize cost Quantity of goods shipped from each supply point to each demand point Non negative shipments Supply availability at a supply point Demand need at a demand point 3

4 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 define our fundamental decision variable as the set of individual shipment quantities from each supply location to each demand location and denote this variable algebraically as Move supplyi,demandj 4

5 Transportation Problem The objective function: Formulating the Problem We want to minimize total shipping cost so we need an expression for shipping cost Let us define a data item giving the per unit cost of shipments from each supply location to each demand location as cost supplyi,demandj Our objective then becomes to minimize the sum of the shipment costs over all supplyi, demandj pairs or Minimize supply i demand cost supplyi,demandj Move supplyi,demandj j which is the per unit cost of moving from each supply location to each demand location times the amount shipped summed over all possible shipment routes 5

6 Transportation Problem Formulating the Problem There are three types of constraints: 1) supply availability limiting shipments from each supply point to existing supply so that the sum of outgoing shipments from the supplyi th supply point to all possible destinations (demandj) to not exceed supply supplyi demandj Move supplyi,demandj supply supplyi 2) minimum demand requiring shipments into the demandj th demand point be greater than or equal to demand at that point. Incoming shipments include shipments from all possible supply points supplyi to the demandj th demand point. supplyi Move supplyi,demandj 3) nonnegative shipments Move supplyi,demandj 0 demand demandj 6

7 Transportation Problem Formulating the Problem Minimize s.t supplyi demandj supplyi demandj cost supplyi,demandj Move supplyi,demandj Move supplyi,demandj supply supplyi for all supplyi Move supplyi,demandj demand demandj for all demandj Move supplyi,demandj 0 for all supplyi, demandj 7

8 Transportation Problem Example: Shipping Goods Three plants: Four demand markets: New York, Chicago, Los Angeles Miami,Houston,Minneapolis,Portland Quantity availability: Supply Available Demand Required New York 100 Miami 30 Chicago 75 Houston 75 Los Angeles 90 Minneapolis 90 Portland 50 Distances: Miami Houston Minneapolis Portland New York Chicago Los Angeles Transportation costs = 5 +5*Distance: Miami Houston Minneapolis Portland New York Chicago Los Angeles

9 Transportation Problem Example: Shipping Goods Minimize s.t supplyi demandj supplyi demandj cost supplyi,demandj Move supplyi,demandj Move supplyi,demandj supply supplyi for all supplyi Move supplyi,demandj demand demandj for all demandj Move supplyi,demandj 0 for all supplyi, demandj Move 11 Move 12 Move 13 Move 14 Move 21 Move 22 Move 23 Move 24 Move Minimize < < < > > > > 50 Move 32 Move 33 Move 34 9

10 Transportation Problem Example: Shipping Goods Solution shadow price represents marginal values of the resources i.e.marginal value of additional units in Chicago = $15 - reduced cost represents marginal costs of forcing nonbasic variable into the solution i.e. shipments from New York to Portland costs $75 - twenty units are left in New York Optimal Solution: Objective value $7,425 Variable Value Reduced Cost Equation Slack Shadow Price Move Move Move Move Move Move Move Move Move Move Move Move Optimal Shipping Pattern Destination Miami Houston Minneapolis Portland Units Variable Units Variable Units Variable Units Variable New York 30 Move Move Move 13 Chicago 75 Move 23 LA 40 Move Move 10 34

11 Feeding Problem McCarl and Spreen Chapter 5 Basic Concept This problem involves composing a minimum cost diet from a set of available ingredients while maintaining nutritional characteristics within certain bounds. Objective: Variables: Minimize total diet costs 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 = {corn, soybeans, salt, etc.} Type of nutritional characteristics which must fall within certain limits nutrient = {protein, calories, etc.} 11

12 Feeding Problem Basic Concept Variable -- Feed ingredientj amount of feedstuff Objective total cost ingredientj fed to animal We want to minimize total diet costs across all the feedstuffs so we need an expression for feedstuff costs. Let us define a data item giving the per unit cost of ingredients as cost ingredientj. Our objective then becomes to minimize the sum of the diet costs over all feed ingredients Minimize ingredientj cost ingredientj Feed ingredientj which is the per unit cost of ingredients summed over the feed ingredients. 12

13 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). a nutrient,ingredientj be the amount of the nutrient th nutrient present in one unit of the ingredientj th feed ingredient 2). UL nutrient and LL nutrient be the maximum and minimum amount of the nutrient th nutrient in the diet Then the nutrient constraints are formed by summing the nutrients generated from each feedstuff (a nutrient,ingredientj F ingredientj ) and requiring these to exceed the dietary minimum and/or 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

14 The constraints: Feeding Problem Formulating the Problem There are four general types of constraints: 1) minimum nutrient requirements restricting the sum of the nutrients generated from each feedstuff (a nutrient,ingredientj F ingredientj ) to meet the dietary minimum ingredientj a nutrient,ingredientj Feed ingredientj minimum nutrient 2) maximum nutrient requirements restricting the sum of the nutrients generated from each feedstuff (a nutrient,ingredientj F ingredientj ) to not exceed the dietary maximum a nutrient,ingredient Feed ingredient maximum 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 Feed ingredientj = 1 ingredientj 4) nonnegative feedstuff Feed ingredientj 0 14

15 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 (c ingredientj ) Ingredient Costs per kg Corn $0.133 Dical Phosphate $0.498 Alfalfa hay $0.077 Salt $0.110 Soybeans $0.300 Vitamin A $0.286 Urea $0.332 Nutrient requirements per Kilogram: Nutrient Unit Minimum Maximum Net energy Mega calories Digestible protein Kilograms Fat Kilograms Vitamin A International Units Salt Kilograms Calcium Kilograms Phosphorus Kilograms Weight Kilograms

16 Feeding Problem Example: cattle feeding Nutrient compositions of 1 kg of each feed: Nutrient Dical Characteristic Corn Hay Soybean Urea Phosphate Salt Vitamin A Concentrate Potato Slurry Net energy Protein Fat Vitamin A Salt 1 Calcium Phosphorus

17 Feeding Problem Example: cattle feeding Corn Hay Soybean Urea Dical Salt Vitamin A Slurry Min.133Feed C +.077Feed H +.3Feed SB +.332Feed Ur +.498Feed d +.110Feed XLT +.286Feed VA + PFeed SL s.t..075feed C +.127Feed H +.438Feed SB Feed Ur +.032Feed SL LE.13 protein Max Nutrient.0357Feed C +.022Feed H +.013Feed SB +.009Feed SL LE.05 fat Feed XLT LE.02 salt.0002feed C Feed H Feed SB Feed d +.002Feed SL LE.01 Calc.0035Feed C Feed H Feed SB +.68Feed Ur Feed d Feed SL LE.012 phosp 1.48Feed C +.49Feed H Feed SB Feed SL GE 1.34 energy.075feed C +.127Feed H +.438Feed SB Feed Ur +.032Feed SL GE.071 protein Min Nutrient 600Feed C Feed H + 80Feed SB Feed VA GE 2200 VA Feed XLT GE.015 salt.0002feed C Feed H Feed SB Feed d +.002Feed SL GE.0025 Calc.0035Feed C Feed H Feed SB +.68Feed Ur Feed d Feed SL GE.0035 phos Volume Feed C + Feed H + Feed SB + Feed Ur + Feed d + Feed XLT + Feed VA + Feed SL = 1 Feed C, Feed H, Feed SB, Feed Ur, Feed d, Feed XLT, Feed VA, Feed SL GE 0 17

18 Feeding Problem Example: cattle feeding -- Solution - least cost feed ration is 95.6% slurry, 0.1% vitamin A, 1.5% salt, 0.2% dicalcium phosphate, 1.4%urea, 1.1% soybeans, and 0.1% hay - reduced costs of feeding corn is 0.95 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 $ Optimal Solution: Objective value = $0.021 Variable Value Reduced Cost Equation Slack Shadow Price Feed c Protein Max Feed H Fat Max Feed SB Salt Max Feed UR Calcium Max Feed d Phosphrs Max Feed SLT EnergyMin Feed VA Protein Min Feed SL Vita Min Salt Min Calcium Min Phosphrs Min Weight

19 Joint Products Problem McCarl and Spreen Chapter 5 Basic Concept This problem deals with joint products in which firms wish to maximize total profits derived from the available production possibilities. Each of the production possibilities yields multiple products, uses some inputs with a fixed market price, and uses some resources that are available in fixed quantity. Problem focuses on maximizing profit derived from available production possibilities. Objective: Variables: Maximize profits i. How much of quantity of the product th product is produced. ii. Quantity of process th production possibility iii. Amount of the input th input purchased Restrictions: Non negativity Supply demand balance Resource availability 19

20 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, Sales product : the set of production possibilities, Production process : the set of purchased inputs, BuyInput input 20

21 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 process th production possibility, q product,process : the amount of the input th input used by the process th production possibility, r input,process : the amount of the resource th resource used by the process th production possibility, s resource,process : the amount of resource availability or endowment by resource, b resource 21

22 The objective function: Joint Products Problem Formulating the Problem We want to maximize total profits 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 define these parameters as : SalePrice product : InputCost input : OtherCost process Then the objective function becomes Maximize product input process SalePrice product Sales product -InputCost input BuyInput input -OthCost process Production process 22

23 The constraints: Joint Products Problem Formulating the Problem There are four general types of constraints: 1) demand and supply balance for which quantity sold of each product is less than or equal to the quantity yielded by production. Sales product - q product,process Production process 0 2) demand and supply balance for which quantity purchased of each fixed price input is greater than or equal to the quantity utilized by the production activities. process r input,process Production process - BuyInput input 0 3) resource availability constraint insuring that the quantity used of each fixed quantity input does not exceed the resource endowments. process 4) nonnegativity process s resource,process Production process b resource Sales product, BuyInput input, Production process 0 23

24 Joint Products Problem Formulating the Problem Maximize s.t. process process product input process SalePrice product Sales product -InputCost input BuyInput input -OthCost process Production process Sales product - q product,process Production process 0 process r input,process Production process - BuyInput input 0 s resource,process Production process b resource Sales product, BuyInput input, Production process 0 24

25 Joint Products Problem Example: wheat production Two products produced: Three inputs: Wheat and wheat straw Land,fertilizer, seed Seven production processes: Outputs and inputs per acre Process Wheat yield in bushel Wheat straw yield/bales Fertilizer use in Kg Seed in pounds land Wheat price = $4/bushel, wheat straw price =$0.5/bale Fertilizer = $2 per kg, Seed = $0.2/lb. $5 per acre production cost for each process Land = 500 acres 25

26 Maximize s.t. product input process Joint Products Problem Example: Wheat Production SalePrice product Sales product -InputCost input BuyInput input -OthCost process Production process Sales product - q product,process Production process 0 process process process r input,process Production process - BuyInput input 0 s resource,process Production process Sales product, BuyInput input, Production process 0 b resource 4Sale 1 +.5Sale 2-5Y 1-5Y 2-5Y 3-5Y 4-5Y 5-5Y 6-5Y 7-2Z 1 -.2Z 2 Maximize s.t. Sale 1-30Y 1-50Y 2-65Y 3-75Y 4-80Y 5-80Y 6-75Y 7 LE 0 Sale 2-10Y 1-17Y 2-22Y 3-26Y 4-29Y 5-31Y 6-32Y 7 LE 0 + 5Y Y Y Y Y Y 7 - Z 1 LE 0 10Y Y Y Y Y Y Y 7 - Z 2 LE 0 Y 1 + Y 2 + Y 3 + Y 4 + Y 5 + Y 6 + Y 7 LE 500 Sale, 1 Sale 2, Y 1, Y 2, Y 3, Y 4, Y 5, Y 6, Y 7, Z 1, Z 2 GE 0 Note that: Y refers to Production process and Z refers to BuyInput input 26

27 Joint Products Problem Example: Wheat Production -- Solution - 40,000 bushels of wheat and 14,500 bales of straw are produced by 500 acres of the fifth production possibility using 10,000 kilograms of fertilizer and 5,000 lbs. of seed - reduced cost shows a $ cost if the first production possibility is used. - shadow prices are values of sales, purchase prices of the various outputs and inputs, and land values ($287.5). Optimal Solution: Objective value $143,450 Variable Value Reduced Cost Equation Slack Shadow Price Sale wheat 40,000 0 Wheat 0-4 Sale staw 14,500 0 Straw Process Fertilizer 0 2 Process Seed Process Land Process Process Process Process Byinput fert 10,000 0 Byinput seed 5,

28 Disassembly Problem McCarl and Spreen Chapter7 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: Variables: Restrictions: Maximize operating profits : Quantity of the final product sold : Number of units of raw product purchased : Non negativity : Product balance : Resource limitation : Upper bound on raw disassembly : Upper and lower bounds on sales 28

29 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 define fundamental decision variables: Raw productj is the number of units of raw product purchased. Sold partk is the quantity of component part sold. 29

30 The objective function: Disassembly Problem Formulating the Problem Because the objective function is to maximize operating profits (revenue earned by sales minus input costs), we need additional expression for 1). the cost of purchasing one unit of raw product, Cost productj 2). the selling price of component part, Price partk Our objective then becomes to maximize the sum over all final products sold (Sold partk ) of the total revenue earned by sales less the costs of all purchased inputs (Raw productj ). Maximize - productj Cost productj Raw productj + Price partk Sold partk partk 30

31 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 partk,productj : the use of resource limit resourcer when disassembling raw product productj, useraw resourcer,productj : the amount of resource limit resourcer used by the sale of one unit of component part partk, usesold resourcer,partk : the maximum amount of raw product limit resourcer available, b resourcer : the maximum amount of component part productj available, maxavailable productj : the maximum quantity of component part partk that can be sold, maxsold partk : the minimum amount of the component partk that can be sold, minsold partk 31

32 Disassembly Problem Formulating the Problem There are five types of constraints: 1) product balance limiting the quantity sold to be no greater than the quantity supplied when the raw product is disassembled. - yield partk,productj Raw productj + Sold partk 0 productj 2) resource limitation constraining on raw product disassembly and product sale. useraw resourcer,productj Raw productj productj + usesold resourcer,partk Sold partk b resourcer partk 3) upper bound on disassembly Raw productj maxavaialable productj 4) upper and lower bound on sales Sold partk maxsold partk Sold partk minsold partk 6) Nonnegativity Raw productj, Sold partk 0 32

33 Maximize - s.t Disassembly Problem Formulating the Problem Cost productj Raw productj + Price partk Sold partk - yield partk,productj Raw productj + Sold partk 0 productj productj productj partk useraw resourcer,productj Raw productj + usesold resourcer,partk Sold partk b resourcer partk Raw productj maxavailable productj Sold partk maxsold partk Sold partk minsold partk Raw productj, Sold partk 0 33

34 Disassembly Problem Example: Junk Yard Disassembling 4 type of cars: Escorts, 626's, T-birds, Caddy's 5 component parts: Metal, seats, chrome, doors, junk Component part yields from each type of car: Car Data ESCORTS 626S TBIRDS CADDIES PURCHASE PRICE WEIGHT DISASSEMBLY COST AVAILABILITY Resource Use to Breakdown Cars CAPACITY LABOR Propor tional Breakdown of Cars into Component Parts (/lb) ESCORTS 626S TBIRDS CADDIES METAL SEATS CHROME DOORS JUNK Part Data MAXIMUM PRICE PARTSPACE LABOR METAL SEATS CHROME DOORS JUNK

35 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 => metal - seat => junk - doors => 70% metal and 30% junk ESCORTS 626S TBIRDS CADDIES METAL SEATS CHROME DOORS JUNK CONVERT SEATS CONVERT CHROME CONVERT DOORS RHS MIN OBJECTIVE METAL = 0 SEATS = 0 CHROME = 0 DOORS = 0 JUNK = 0 CAPACITY < 42 LABOR < 700 PARTSPACE < 60 Upper Bound

36 Disassembly Problem Example: Junk Yard -- Solution - shadow price: i.e. marginal value of additional junk yard capacity is $ reduced cost: i.e. selling more 1 more seat costs $ max sales possibilities on seats, chrome, doors are exceeded a. 320 lbs. of seats are converted to junk b 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 = $ Variable Value Reduced Cost Constraint Slack Shadow Price Dissasemble: ESCORTS S TBIRDS 20 0 CADDIES 10 0 Sell: METAL SEATS CHROME DOORS JUNK Convert: SEATS CHROME DOORS Parts: METAL SEATS CHROME DOORS JUNK Resources CAPACITY LABOR PARTSP 20 0 Max Cars: ESCORTS S 14 0 TBIRDS CADDIES Max Sell: SEATS CHROME DOORS