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Chapter 5 LP formulations. LP formulations of four basic problem Resource allocation problem Transportation problem Feed mix problem Joint products problem.

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Presentation on theme: "Chapter 5 LP formulations. LP formulations of four basic problem Resource allocation problem Transportation problem Feed mix problem Joint products problem."— Presentation transcript:

1 Chapter 5 LP formulations

2 LP formulations of four basic problem Resource allocation problem Transportation problem Feed mix problem Joint products problem

3 We will examine: Basic Structure Formulation Example application Answer interpretation

4 Resource Allocation Problem The classical LP problem involves the allocation of an endowment of scarce resources among a number of competing products so as to maximize profits. – Objective: Maximize Profits – Competing products index is j; scarce resources index is i – Major decision variable Xj is the number of units of the jth product made – Non negative production (Xj ≥ 0) – Resource usage across all production possibilities is less than or equal to the resource endowment

5 Algebraic Set UP cj: profit per unit of the jth product aij: number of units of the ith resource used when producing one unit of the jth product bi: the endowment of the ith resource

6 Resource Allocation Problem: E-Z Chair Objective: find the number of two types of chairs to produce that will maximize profits. Chair Types: Functional and Fancy Resources: Large & Small Lathe, Chair Bottom Carver, and Labor Profit Contributions: (revenue – material cost - cost increase due to lathe shifts)

7 Information for Problem OBJFunctional$82 - $15 = $67 Fancy$105 - $25 = $80 Resource Requirements When Using The Normal Pattern Hours of Use per Chair Type FunctionalFancy Small Lathe0.81.2 Large Lathe0.50.7 Chair Bottom Carver0.41.0 Labor1.00.8

8 Resource Requirements and Increased Costs for Alternative Methods of Production in Hours of Use per Chair and Dollars Maximum Use of Small LatheMaximum Use of Large Lathe FunctionalFancyFunctionalFancy Small Lathe1.301.700.200.50 Large Lathe0.200.301.301.50 Chair Bottom Carver 0.401.000.401.00 Labor1.050.821.100.84 Cost Increase$1.00$1.50$0.70$1.60 Alternative Production Method

9 Resource Limits Small lathe: 140 hours Large lathe: 90 hours Chair bottom carver: 120 hours Labor: 125 hours

10 Production alternatives and profits Functional, regular method (X1) : $67 (c1) Functional, max small lathe (X2): $66 (c2) Functional, max lg lathe (X3): $66.30 (c3) Fancy, regular method (X4): $80 (c4) Fancy, max small lathe (X5): $78.50 (c5) Fancy, max lg lathe (X6): $78.40 (c6)

11 Empirical Set-UP Max67X 1 +66X 2 +66.3X 3 +80X 4 +78.5X 5 +78.4X 6 s.t.0.8X 1 +1.3X 2 +0.2X 3 +1.2X 4 +1.7X 5 +0.5X 6 ≤ 140 0.5X 1 +0.2X 2 +1.3X 3 +0.7X 4 +0.3X 5 +1.5X 6 ≤ 90 0.4X 1 +0.4X 2 +0.4X 3 +X4X4 +X5X5 +X6X6 ≤ 120 X1X1 +1.05X 2 +1.1X 3 +0.8X 4 +0.82X 5 +0.84X 6 ≤ 125 X1X1,X2X2,X3X3,X4X4,X5X5,X6X6 ≤ 0

12 Solution from Excel Solver McCarl provides GAMS code and solution from GAMS (same as this solution).

13 Interpretation Produce 62 functional chairs using the traditional method, 73 fancy chairs using the traditional method, and 5 fancy chairs using the maximum large lathe method to earn profits of $10,417 Producing functional chairs by the max small lathe method reduces profits by $11.30 per chair made, producing functional chairs by max large lathe reduces profits by $4.08/chair, and producing fancy chairs by max small lathe method reduces profits by $8.40/chair. An hour more small lathe time would increase profits by $33.33; an hour more large lathe time would increase profits by $25.79; an hour more labor would increase profits by 27.44.

14 Dual of the problem (general)

15 Empirical Dual for E-Z Chair Min 140U 1 +90U 2 +120U 3 +125U 4 s.t. 0.8U 1 +0.5U 2 +0.4U 3 +U4U4 ≥67 1.3U 1 +0.2U 2 +0.4U 3 +1.05U 4 ≥66 0.2U 1 +1.3U 2 +0.4U 3 +1.1U 4 ≥66.3 1.2U 1 +0.7U 2 +U3U3 +0.8U 4 ≥80 1.7U 1 +0.3U 2 +U3U3 +0.82U 4 ≥78.5 0.5U 1 +1.5U 2 +U3U3 +0.84U 4 ≥78.4 U1U1, U2U2, U3U3, U4U4 ≥ 0

16 Dual Solution = slack on constraint three in primal Shadow prices in primal answers to primal

17 Transportation Problem This problem involves the shipment of a homogeneous product from a number of supply locations to a number of demand locations. – Objective: Minimize cost – Variables: Quantity of goods shipped from each supply point to each demand point – Restrictions: Non negative shipments – Supply availability at supply point – Demand need at a demand point

18 Diagram of Problem Supply LocationsDemand Locations 12...m12...m AB...nAB...n

19 Formulate the problem Supply locations as supplyi Demand locations as demandj Decision variable = Movesupplyi,demandj – Decision to move a quantity of supply from location i to demand location j costsupplyi,demandj = the cost of moving one unit of product from location i to demand location j

20 Constraints Supply availability: limiting shipments from each supply point so that the sum of outgoing shipments from point supplyi to all possible destinations (demandj) doesn't exceed supplyi 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 to the demandjth demand point. Nonnegative shipments:

21 Algebra

22 Transportation Problem Example: Shipping Goods Three plants:New York, Chicago, Los Angeles Four demand markets:Miami, Houston, Minneapolis, Portland Minimize the cost of shipping product from the three plants to the four demand markets.

23 Quantities

24 Distances

25 Transportation costs = 5 +5*Distance

26 Minimization problem. Note the first three inequalities.

27 Interpretation 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 non-basic variable into the solution i.e. shipments from New York to Portland increase costs by $75. twenty units are left in New York

28 Dual: Note the signs on U1, U2, U3 (the constraints were LE in primal) The dual is a maximization

29 Feeding Problem 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

30 Indices needed Ingredients: How many possible ingredients can be used in the ration? (corn, soybeans, molasses, hay, etc.) Nutrients: What are the essential nutrients to consider? {protein, calories, vitamin A, etc.}

31 Constraints restricting the sum of the nutrients generated from each feedstuff to meet the dietary minimum restricting the sum of the nutrients generated from each feedstuff not to exceed the dietary maximum the ingredients in the diet equal the required weight of the diet. nonnegative feedstuff

32 Algebraic Representation

33 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

34 Information for feeding problem New ingredient: Potato Slurry, initially set cost to $.01/kg

35 More information for problem

36 Nutrient Content per kilogram

37 Minimization problem. Note the LE constraints, and the equality.

38 u1-u5 associated with LE constraints in the min. U12 is associated with an equality constraint. So u12 is unrestricted in sign, which here we work as two parts, u12p and u12n. The dual problem

39 Tracing out the derived demand for slurry Start at $0.01 for slurry (which we've solved) and production is 96% slurry. Sensitivity report – tells you the price at which the answers would change. Look at the "allowable increase" for this price. It is 0.09630319. So at slurry prices greater than.01 + 0.09630319, the optimal solution will change. Plug in.1163, and the slurry falls to 87% of the ration. In the new sensitivity report, the next allowable increase is 0.005930872. At a price of.1222, slurry use falls to 64%. At prices close to.13, use of slurry falls to 0.

40 Joint Products Problem One production process yields multiple products, such as lambs and wool, or wheat grain and straw Maximize profits when each production possibility yields multiple products, uses some inputs with a fixed market price, and uses some resources that are available in fixed quantity.

41 Variables in model the amount of each product produced for sale the production process chosen to produce the products the amount of market inputs to purchase

42 Technical consideration How much of each output will a given production process generate? How much of each market input must be bought for that production process? How much of each fixed resource is used in the production process?

43 Transfer row "Transfer row" is the name given to a certain type of row in an LP problem. These rows take the production generated by a process and distribute it to sales activities. The RHS of a transfer row is generally zero.

44 Example: Wheat-Straw Problem 7 possible processes to produce wheat and straw with different yields. Wheat sells for $4 per bushel, straw for $.50 per small (square) bale. Fertilizer and seed can be purchased and are needed in different amounts for the 7 practices. Fertilizer costs $2/lb and seed costs $.20/lb. Land is limited to 500 acres.

45 Number of decision variables 2 sales variables (wheat and straw) 7 production variables (the 7 possible processes) 2 purchase variables (seed and fertilizer)

46 Number of constraints Transfer row for wheat sales (from processes to sales activity) Transfer row for straw sales (from processes to sales activity) Transfer row for fertilizer purchases (distribute purchased fertilizer to production processes) Transfer row for seed purchases (distribute purchased seed to processes) Limit on land

47 Other information: Yields from Processes

48 Joint Product Problem: Wheat-Straw Note the transfer rows have zero on RHS. It is important to get the inequality in the right direction. Here Sales must be LE production. Use of input must be LE purchases.

49 Excel Solver Solution

50 Dual

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