Lecture 3 – Classic LP Examples

1. Lecture 3 – Classic LP Examples Topics • Employee scheduling problem • Energy distribution problem • Feed mix problem • Cutting stock problem • Regression analysis • Model Transformations

2. Examples of LP Formulations 1. Employee Scheduling Macrosoft has a 24-hour-a-day, 7-days-a-week toll free hotline that is being set up to answer questions regarding a new product. The following table summarizes the number of full-time equivalent employees (FTEs) that must be on duty in each time block. FTEs Interval Time 1 0-4 15 2 4-8 10 3 8-12 40 4 12-16 70 5 16-20 40 6 20-0 35

3. Constraints for Employee Scheduling • Macrosoft may hire both full-time and part-time employees. The former work 8-hour shifts and the latter work 4-hour shifts; their respective hourly wages are $15.20 and $12.95. Employees may start work only at the beginning of 1 of the 6 intervals. • At least two-thirds of the employees working at any one time must be full-time employees. • Part-time employees can only answer 5 calls in the time a full-time employee can answer 6 calls. (i.e., a part-time employee is only 5/6 of a full-time employee.) Formulate an LP to determine how to staff the hotline at minimum cost.

4. Decision Variables xt = # of full-time employees that begin the day at the start of interval t and work for 8 hours yt = # of part-time employees that are assigned interval t (4  12.95) (8  15.20) min 121.6(x1 + • • • + x6) + 51.8(y1 + • • • + y6) 5 ³ s.t. y1 15 x1 + x6 + 6 5 ³ y2 10 x1 + x2 + 6 All shifts must be covered 5 ³ y3 40 x2 + x3 + 6 5 ³ y4 70 x3 + x4 + 6 5 ³ y5 40 x4 + x5 + 6 5 ³ y6 35 x5 + x6 + 6 PT employee is 5/6 FT employee

5. More constraints: 2 x1 + x6 ³ (x6 + x1 + y1) 3 At least 2/3 workers must be full time 2 ³ (x1 + x2 + y2) x1 + x2 3 . . . 2 ³ x5 + x6 (x5 + x6 + y6) 3 Nonnegativity xt ³ 0, yt ³ 0 t =1,2,…,6

6. 2. Energy Generation Problem (with piecewise linear objective) Austin Municipal Power and Light (AMPL) would like to determine optimal operating levels for their electric generators and associated distribution patterns that will satisfy customer demand. Consider the following prototype system Demand requirements 4 MW 1 1 Demand sectors Plants 7 MW 2 2 3 6 MW The two plants (generators) have the following (nonlinear) efficiencies: Plant 1 [ 0, 6 MW] [ 6MW, 10MW] Unit cost ($/MW) $10 $25 Plant 2 [ 0, 5 MW] [5MW, 11MW] Unit cost ($/MW) $8 $28 For plant #1, e.g., if you generate at a rate of 8MW (per sec), then the cost ($) is = ($10/MW)(6MW) + ($25/MW)(2MW) = $110.

7. Problem Statement and Notation Formulate an LP that, when solved, will yield optimal power generation and distribution levels. Decision Variables x = power generated at plant 1 at operating level 1 11 1 x ² 2 ² ² ² ² ² ² 12 2 1 x ² ² ² ² ² ² ² 21 2 x 2 ² ² ² ² ² ² ² 22 y 1 to demand sector 1 = power sent from plant 11 2 y 1 ²²² ² ² ² ² 12 y 3 ² 1 ² ² ² ² ² ² 13 y 1 ² 2 ² ² ² ² ² ² 21 y 2 ² 2 ² ² ² ² ² ² 22 y 3 ² 2 ² ² ² ² ² ² 23

8. Formulation min 10x11 + 25x12 + 8x21 + 28x22 s.t. y y y = x11 + x12 + + 11 12 13 y y y = x21 + x22 + + 21 22 23 y y + = 4 11 21 y y + = 7 12 22 y y = 6 + 13 23 0 £ x11 £ 6, 0 £x12£ 4 0 £ x21 £ 5, 0 £x22£ 6 y11, y12, y13, y21, y22, y32 0 Note that we can model the nonlinear operating costs as an LP only because the efficiencies have the right kind of structure. In particular, the plant is less efficient (more costly) at higher operating levels. Thus the LP solution will automatically select level 1 first.

9. General Formulation of Power Distribution Problem The above formulation can be generalized for any number of plants, demand sectors, and generation levels. Indices/Sets plants i Î I j Î J demand sectors generation levels k Î K Data unit generation cost ($/MW) for plant i at level k Cik = upper bound (MW) for plant i at level k uik = = dj demand (MW) in sector j Decision Variables xik = power (MW) generated at plant i at level k yij = power (MW) sent from plant i to sector j

10. General Network Formulation å å cikxik min iÎI kÎK xik å å s.t. yij = " i Î I jÎJ kÎK å yij = dj " j Î J iÎI • £xik£ uik " i Î I, k Î K 0 £yij" i Î I, j Î J

11. 3. Feed Mix Problem • An agricultural mill produces a different feed for cattle,sheep, and chickens by mixing the following raw ingredients: corn, limestone, soybeans, and fish meal. • These ingredients contain the following nutrients: vitamins, protein, calcium, and crude fat in the following quantities: Nutrient, k Vitamins Protein Calcium Crude Fat Ingredient, i Corn 8 10 6 8 Limestone 6 5 10 6 Soybeans 10 12 6 6 Fish Meal 4 18 6 9 Let aik= quantity of nutrient k per kg of ingredient i

12. Constraints • The mill has (firm) contracts for the following demands. Demand (kg) Cattle Sheep Chicken dj 10,000 6,000 8,000 • There are limited availabilities of the raw ingredients. Supply (kg) Corn Limestone Soybeans Fish Meal si 6,000 10,000 4,0 00 5,000 • The different feeds have “quality” bounds per kilogram. Vitamins Crude fat Protein Calcium min max min max min max min max Cattle 6 -- 6 -- 7 -- 4 8 Sheep 6 -- 6 -- 6 -- 4 8 6 -- Chicken 4 6 6 -- 4 8 The above values represent bounds:Ljk& Ujk

13. Costs and Notation • Cost per kg of the raw ingredients is as follows: Soybeans Fish meal Limestone Corn cost/kg, ci 24¢ 12¢ 20¢ 12¢ Formulate problem as a linear program whose solution yields desired feed production levels at minimum cost. Indices/sets i  I ingredients { corn, limestone, soybeans, fish meal } j  J products { cattle, sheep, chicken feeds } nutrients { vitamins, protein, calcium, crude fat } k  K

14. Data dj demand for product j (kg) si supply of ingredient i (kg) Ljk lower bound on number of nutrients of type k per kg of product j upper bound on number of nutrients of type k per kg of product j Ujk ci cost per kg of ingredient i aik number of nutrients k per kg of ingredient i Decision Variables xij amount (kg) of ingredient i used in producing product j

15. å aikxij£ Ujkdj "j  J, kK iÎI å aikxij ³ Ljk dj "j  J, kK iÎI LP Formulation å å cixij min jÎJ iÎI å xij = dj "j  J s.t. iÎI xij£ si å "i  I jÎJ "i  I, j  J xij³ 0

16. Generalization of feed Mix Problem Gives Blending Problems Blended commodities Raw Materials Qualities corn, limestone, protein, vitamins, feed soybeans, fish meal calcium, crude fat butane, catalytic octane, volatility, gasoline reformate, vapor pressure heavy naphtha pig iron, carbon, metals ferro-silicon, manganese, carbide, various chrome content alloys ³ ³ ³ 2 raw ingredients 1 quality 1 commodity

17. 4. Trim-Loss or Cutting Stock problem • Three special orders for rolls of paper have been placed at a paper mill. The orders are to be cut from standard rolls of 10¢ and 20¢ widths. Length Width Order 1 ¢ 5 10,000¢ 2 ¢ 7 30,000¢ 3 ¢ 9 20,000¢ • Assumption: Lengthwise strips can be taped together • Goal: Throw away as little as possible

18. Problem: What is trim-loss? ¢ 20 ¢ 10 5000' 5' ¢ 5 9' ¢ 7 Decision variables: xj= length of roll cut using pattern, j = 1, 2, … ?

19. Patterns possible ¢ ¢ 10 roll 20 roll x3 x9 x2 x8 x1 x5 x4 x7 x6 5¢ 2 0 0 4 2 2 1 0 0 7¢ 0 1 0 0 1 0 2 1 0 9¢ 0 0 1 0 0 1 0 1 2 Trim loss 0 3 1 0 3 1 1 4 2 min z = 10(x +x +x ) + 20(x +x +x +x +x +x ) 7 9 1 2 3 4 5 6 8 ³ s.t. 2x + 4x + 2x + 2x + x 10,000 1 7 4 5 6 ³ + x x + x + 2x 30,000 8 7 2 5 ³ + x x + x + 2x 20,000 8 3 9 6 xj³ 0, j = 1, 2,…,9

20. Alternative Formulation + 2x9 min z = 3x2 + x3 + 3x5 + x6 + x7 + 4x8 + 5y1 + 7y2 + 9y3 4 s.t. 2x1 + x4 + 2x5 + 2x6 + x7 – y1 = 10,000 x2 + x5 + 2x7 + x8 – y2 = 30,000 x3 + x6 + x8 + 2x9 – y3 = 20,000 xj³ 0, j = 1,…,9; yi³ 0, i = 1, 2, 3 where yi is overproduction of width i

21. 5. Constrained Regression Data (x,y) = { (1,2) , (3,4) , (4,7) } y · 7 6 5 · 4 3 · 2 1 x 1 2 3 4 5 We want to “fit” a linear function y = ax + b to these data points; i.e., we have to choose optimal values for a and b.

22. Objective: Find parameters a and b that minimize the maximum absolute deviation between the data yi and the fitted line yi = axi + b. Ù Ù yi yi and Ù Predicted value observed value In addition, we’re going to impose a priori knowledge that the slope of the line must be positive. (We don’t know about the intercept.) known to be positive a = slope of line Decision variables tive positive or nega b = y-intercept b = b+-b-, b+ 0, b- 0

23. Objective function: Ù min max { |yi-yi| : i = 1, 2, 3 } Ù where yi = axi + b Ù Let w = max { |yi-yi| : i = 1, 2, 3 } Optimization model: min w Ù s.t. w³ |yi-yi|, i = 1, 2, 3

24. Nonlinear constraints: Ù w ³ -  1a + b – 2  y1 = y1 Ù - y2 = y2 w ³  3a + b – 4  Ù w ³ - = y3 y3  4a + b – 7  Convert absolute value terms to linear terms: Note: 2 ³|x| iff 2 ³x and 2 ³ -x Ù Thus w ³ - is equivalent to y y i i w³axi + b-yi and w³ -axi – b + yi

25. so finally … min w - s.t. + - - £ a + b b w 2 - + - - £ - a-b + b w 2 - + - - £ 3a + b b w 4 - - + - - £ - 3a b + b w 4 - + - - £ 4a + b b w 7 - + - + - £ - - 4a b b w 7 a, b+, b-, w³ 0

26. Model Transformations • Direction of optimization: Minimize {c1x1 + c2x2 + … + cnxn} ÛMaximize {–c1x1 – c2x2 – … – cnxn} • Constant term in objective function  ignore • Nonzero lower bounds on variables: • xj>lj replace with xj= yj + lj whereyj 0 • Nonpositive variable: • xj≤ 0 replace with xj= –yj where yj 0 • Unrestricted variables: • xj = y1j– y2j where y1j 0, y2j  0

27. What You Should Know About LP Problems • How to formulate various types of problems. • Difference between continuous and integer variables. • How to find solutions. • How to transform variables and functions into the standard form.