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- 1. Introduction to Operations Research and Linear Programming
- 2. Introduction • Our world is filled with limited resources. • Oil • Land • Human • Time • Business • Resource (Budget) • Manufacturing • m/c , number of worker • Restaurant • Space available for seating
- 3. Introduction • How best to used the limited resources available ? • How to allocate the resource in such a way as to maximize profits or minimize costs ?
- 4. Introduction • Mathematical Programming(MP) is a field of management science or operations research that fines most efficient way of using limited resources/ to achieve the objectives of a business. Optimization
- 5. Characteristics of Optimization Problems • One or more decisions • Restrictions or constraints e.g. Determining the number of products to manufacture a limited amount of raw materials a limited amount of labor • Objective – The production manager will choose the mix of products that maximizes profits – Minimizing the total transportation cost
- 6. Expressing optimization problems mathematically • Decision variables X1,X2 ,X3, … , Xn e.g. the quantities of different products Index n = the number of product types • Constraints – a less than or equal to constraint : f(X1,X2 ,X3, … , Xn) < b – a greater than or equal to constraint : f(X1,X2 ,X3, … , Xn) > b – an equal to constraint : f(X1,X2 ,X3, … , Xn) = b • Objective – MAX(or MIN) : f(X1,X2 ,X3, … , Xn)
- 7. Mathematical formulation of an optimization problem MAX(or MIN) : f(X1,X2 ,X3, … , Xn) Subject to: f(X1,X2 ,X3, … , Xn) < b1 …. f(X1,X2 ,X3, … , Xn) > bk …. f(X1,X2 ,X3, … , Xn) = bm note : n variables , m constraints
- 8. Mathematical programming techniques The function in model : f(x) -Linear -Nonlinear Decision variable -Fractional value e.g. 2.33 -Integer value e.g. 1,2,3,4.. -Binary e.g. 0,1 Test : Linear functions? 1. 2. 3. 5032 321 ≥−+ xxx 602 21 ≥+ xx 75)3/1(4 21 =+ xx 9.0 323 321 321 ≤ ++ −+ xxx xxx 4. 5. 4573 2 2 1 ≤+ xx
- 9. Problem • Blue Ridge Hot Tubs manufactures and sells two models of hot tubs : Aqua Spa and the Hydro-Lux. – Howie Jones, the owner and manager of the company, needs to decide how many of each type of hot tubs to produce – 200 pumps available – Howie expects to have 1,566 production labor hours and 2,880 feet of tubing available. – Aqua-Spa requires 9 hours of labor and 12 feet of tubing – Hydro-Lux require s 6 hours of labor and 16 feet of tubing Assuming that all hot tubs can be sold To maximize profits, how many Aqua-Spas and Hydra-Luxs should be produce?
- 10. Formulating LP Models 1. Understand the problem 2. Indentify the decision variable 3. State the objective as a linear combination of decision variables Max : 350x1 + 300x2 4. State the constraints as linear combinations of the decision variable 4.1 Pumps available 4.2 labor available 4.3 Tubing available 5. Identify any upper or lower bounds on the decision variable x1 > 0 x2 > 0
- 11. Max : 350x1 + 300x2 Subjectto: x1+x2 < 200 , Pumps available 9x1+ 6x2 < 1,566 , Labors available 12x1+ 16x2 < 2,880 , Tubing available x1 > 0 x2 > 0
- 12. General Form of an LP model MAX(or MIN) : C1X1+ C2X2+ , … , CnXn Subject to: a11X1+ a12X2+ , … , a1nXn < b1 ak1X1+ ak2X2+ , … , aknXn > bk am1X1+ am2X2+ , … , amnXn = bm
- 13. Notations a11X1+ a12X2+ , … , a1nXn < b1 X1+ X2+ , … ,+ Xn = b1 X11 + X21+ X31 = b1 X12 + X22+ X32 = b2 1 1 1 bxa i n i i ≤∑= 1 1 bx n i i =∑= j i ij bx =∑= 3 1 j∀,
- 14. Graphical Method: Solving LP Problems
- 15. Graphical Method x1+x2 ≤ 200 (1)
- 16. Graphical Method 9x1+ 6x2 ≤ 1,566 (2) x1+x2 ≤ 200 (1) Note : X1 = 0 , x2 = 1566/6 = 261 X2 = 0 , x1 = 1566/9 =174
- 17. Graphical Method 9x1+ 6x2 ≤ 1,566 (2) x1+x2 ≤ 200 (1) 12x1+ 16x2 ≤ 2,880 (3) Note : X1 = 0 , x2 = 2880/16 = 180 X2 = 0 , x1 = 2880/12 = 240
- 18. Graphical Method MAX 350x1 + 300x2
- 19. Using level curves • MAX 350x1 + 300x2 1. Set a number of objective e.g. Obj = 35,000 2. Finding points (x1,x2) which has obj = 35,000 x1 = 100 , x2 = 0 x1 = 0 , x2 = 116.67
- 20. Graphical Method 9x1+ 6x2 ≤ 1,566 (2) x1+x2 ≤ 200 (1) 12x1+ 16x2 ≤ 2,880 (3) MAX 350x1 + 300x2
- 21. Finding the optimal solution 9x1+ 6x2 = 1,566 (2) x1+x2 = 200 (1) 9x1+ 6(200 –x1) = 1,566 3x1+ 1,200 = 1,566 3x1 = 366 Optimal Solution = x1 = 122 X2 = 78
- 22. Special Conditions in LP Models 9x1+ 6x2 ≤ 1,566 (2) x1+x2 ≤ 200 (1) 12x1+ 16x2 ≤ 2,880 (3) MAX 450x1 + 300x2 1. Alternate Optimal Solutions - having more than one optimal solution
- 23. Special Conditions in LP Models 9x1+ 6x2 ≤ 1,566 (2) x1+x2 ≤ 225 (1) 12x1+ 16x2 ≤ 2,880 (3) MAX 350x1 + 300x2 2. Redundant Constraints A constraint that plays no role in determining the feasible region of the problem.
- 24. Special Conditions in LP Models -x1+ 2x2 ≤ 400 (2) x1+ x2 ≥ 400 (1) x1 ≥ 0 MAX x1 + x2 3. Unbounded Solutions The objective function can be made infinitely large. x2 ≥ 0
- 25. Special Conditions in LP Models 4. Infeasibility x1+ x2 ≤ 150 (1) x1+ x2 ≥ 200 (2) x1 ≥ 0 MAX x1 + x2 x2 ≥ 0 ? – No ways to satisfy all of the constraints
- 26. Using Excel Solver
- 27. Using Excel Solver Model Excel Solver Objective Set Target Cell Decision Variable Changing Cells Constraints Subject to the constraints
- 28. Using Excel Solver 9x1+ 6x2 ≤ 1,566 (2) x1+x2 ≤ 200 (1) 12x1+ 16x2 ≤ 2,880 (3) MAX 350x1 + 300x2 Parameters Profit /unit of each product , Pumps Req /unit, no. Pumps available , Labor Req /unit, no. labor available , Tubing Req /unit, no. Tubing available Decision Variables No. Aqua-Spas produce No. Hydra-Luxs produce Objective Maximize Profit Constraint Pumps available , labor available , Tubing available
- 29. Using Excel Solver
- 30. Using Excel Solver
- 31. Modeling LP Problems 1. Make vs. Buy Decisions - The Electro-Poly corporation received a $750,000 order for various quantities of 3 types of slip rings. - Each slip ring requires a certain amount of time to wire and hardness. Model1 Model2 Model3 Number ordered 3,000 2,000 900 Hours of wiring required per unit 2 1.5 3 Hours of harnessing required per unit 1 2 1 The company has only 10,000 hours of wiring capacity. The company has only 5,000 hours of harnessing capacity.
- 32. Make vs. Buy Decisions (Continue) The company can sub contract to one of its competitors. Determine the number of slip rings to make and the number to buy in order to fill the customer order at the least possible cost Cost($) Model1 Model2 Model3 Cost to make 50 83 130 Cost to buy 61 97 145
- 33. Make vs. Buy Decisions (Continue) • Defining the decision variables mi = number of model i slip rings to make in-house bi = number of model i slip rings to buy from competitor • Defining the objective function – To minimize the total cost Min : 50m1 + 83m2 + 130m3 + 61b1+ 97b2 + 145b3 Subjectto: 2m1 +1.5m2 + 3m3 < 10,000 , wiring constraint 1m1 +2m2 + m3 < 5,000 , harness constraint m1 + b1 = 3,000 , Required order of model1 m2 + b2 = 2,000 , Required order of model2 m3 + b3 = 900 , Required order of model3 m1 ,m2 ,m3 , b1 ,b2 ,b3 , > 0
- 34. Make vs. Buy Decisions Parameters Cost to buy, Cost to make, Number ordered, Hours of wiring , Hours of harnessing Decision Variable - number of model i slip rings to make in-house - number of model i slip rings to buy from competitor Objective Total Cost Constraint wiring constraint harness constraint Required order of each model
- 35. A transportation problem Supply R/M Area Processing Plant Capacity 275,000 200,000 400,000 600,000 300,000 225,000 i1 i2 i3 j1 j2 j3 Distance : (mile) j1 j2 j3 i1 21 50 40 i2 35 30 22 I3 55 20 25
- 36. A transportation problem • How many product should be shipped from r/m area to the processing plant, the trucking company charges a flat rate for every mile. Min the total distance ~ Min total cost of transportation • Defining the decision variable xij = number of R/M to ship from node i to node j • Defining the objective function Min : 21x11+50x12+40x13+35x21+30x22+22x23 +55x31+20x32+25x33 • Defining the constraints x11+x21 + x31 < 200,000 capacity restriction for j1 x12+x22 + x32 < 600,000 capacity restriction for j2 x13+x23 + x33 < 225,000 capacity restriction for j3 x11+x12 + x13 = 275,000 supply restriction for i1 x21+x22 + x23 = 400,000 supply restriction for i2 x31+x32 + x33 = 300,000 supply restriction for i3 x > 0 , for all i and j
- 37. A transportation problem Parameters Distance from I to j, no supply , no capacity Decision Variable number of R/M to ship from node i to node j Objective Min the total distance Constraint capacity restriction supply restriction
- 38. A transportation problem
- 39. A Blending problem • Agri-Pro stocks bulks amounts of four types of feeds that can mix to meet a given customer’s specifications. Nutrient(%) feed1 feed2 feed3 feed4 Corn (i=1) 30% 5% 20% 10% Grain(i=2) 10% 30% 15% 10% Minerals(i=3) 20% 20% 20% 30% Cost per Pound($) 0.25 0.30 0.32 0.15
- 40. A Blending problem • Agri-Pro has just received an order from a local chicken farmer for 8,000 pounds of feed. • The farmers wants this feed to contain at least 20% corn, 15% grain , and 15% minerals. • What should Agri-Pro do to fill this order at minimum cost? • Defining the decision variable xi = pounds of feed (j) to use in the mix • Defining the objective function Min : 0.25x1+0.30x2+0.35x3+0.15x4 , total cost • Defining the constraints x1+x2 + x3 + x4 = 8,000 0.3x1+0.05x2 + 0.20x3 + 0.10x4 > 0.20 × 8,000 --- Corn 0.1x1+0.30x2 + 0.15x3 + 0.10x4 > 0.15× 8,000 --- Grain 0.2x1+0.20x2 + 0.20x3 + 0.30x4 > 0.15 × 8,000 --- Mineral x1 , x2, x3 , x4 > 0
- 41. A Blending problem Parameters Cost per Pound($), Nutrient(%) of each feed Order quantity , Required Nutrient(%) Decision Variable pounds of feed (j) to use in the mix Objective Minimize Cost Constraint Satisfy the Required Nutrient(%)
- 42. A Blending problem
- 43. A production and inventory planning problem • Upton Corporation is trying to plan its production and inventory levels for the next 6 months • Given the size of Upton’s warehouse, a maximum of 6,000 units • The owner like to keep at least 1,500 units in inventory any months as safety stock • The company wants to produce at no less than half of its maximum production capacity each month. • Estimate that the cost of carrying a unit in any given month is equal to 1.5% of the unit production cost in the same month. m1 m2 m3 m4 m5 m6 Unit production cost (Cm) 240 250 265 280 280 260 Unit Demanded(Dm) 1000 4500 6000 5500 3500 4000 Maximum Production(Maxp m) 4000 3500 4000 4500 4000 3500
- 44. A production and inventory planning problem • The number of units carried in inventory using the averaging the beginning and ending inventory for each month • Current inventory 2,750 units • To minimize production and inventory costs, the company wants to identify the production and inventory plan. • Defining the decision variable pm = the number of units to product in month m bm = the beginning inventory for month m
- 45. A production and inventory planning problem • Defining the objective function Min : , Subject to 2/))(015.0( 1 6 1 6 1 + == +×+ ∑∑ mmm mm mm bbcpC mm Maxpp ≤ m∀, m∀, m∀, mmmm Dpbb −+=+1 m∀, m∀, mm Maxpp )2/1(≥ 000,6≤−+ mmm Dpb 500,1≥−+ mmm Dpb
- 46. A production and inventory planning problem Parameters Maximum Production(Maxp), Unit production cost(Cm) , Unit Demanded(Dm) Decision Variable the number of units to product in month m , pm the beginning inventory for month m, bm Objective Minimize Cost Constraint Upper bound and Lower bound of the production and inventory , Balance constraint
- 47. A production and inventory planning problem

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