管理数学
- 1. 管理數學 Chapter 2: System of Linear EquationsXX. XX, XXX by XXXX
- 2. AgendaLinear Systems as Mathematical Models Linear Systems Having One or No Solutions Linear Systems Having Many Solutions
- 3. Linear <-- Linea1 x + a2 y = b or a1 x1 + a2 x2 +....+ an xn = b
- 4. 2x + 3y = 4 x2 + y2 = 1 x1 - x2 + x3 - x4 = 6 z = 5 - 3x + y/2 sin x + ey = 1 xy = 2 7x1 + 3x2 + 9/x3 + 2x4 = 1 x1 + 2x2 + 3x3 +....+ nxn = 1Which one is linear?
- 5. Example 1A firm produces bargain and deluxe TV sets by buying the components, assembling them, and testing the sets before shipping.
- 6. ResourcesThe bargain set requires 3 hours to assemble and 1 hour to test. The deluxe set requires 4 hours to assemble and 2 hours to test. The firm has 390 hours for assembly and 170 hours for testing each week.
- 7. QuestionUse a system of linear equations to model the number of each type of TV set that the company can produce each week while using all of its available labor.
- 8. Problem FormulationDefine decision variables (unit of scale) Define the linear relation between variables (write the linear equations)
- 9. Example 2A dietitian is to combine a total of 5 servings of cream of mushroom soup, tuna, and green beans, among other ingredients, in making a casserole.
- 10. Ingredient NutritionsEach serving of soup has 15 calories and 1 gram of protein, each serving of tuna has 160 calories and 12 gram of protein, and each serving of green beans has 20 calories and 1 gram of protein.
- 11. QuestionIf these three foods are to furnish 380 calories and 27 grams of protein in casserole, how many servings of each should be used?
- 12. Example 3A retailer has warehouses in Lima and Canton, from which two stores—one in Tiffin and one in Danville—place orders for bicycles. Tiffin orders 38 and Danville orders 46.
- 13. Limitations and QuestionEach warehouse has enough to supply all orders but twice as many are to be shipped from Lima to Danville as from Canton to Tiffin. Write the linear equations.
- 14. AgendaLinear Systems as Mathematical Models Linear Systems Having One or No Solutions Linear Systems Having Many Solutions
- 15. Solving a System of Linear EquationsProblem formulation: variable definition and equations Algorithms or formula Interpretation of solutions
- 16. x + 3y = 9 -2x + y = -4{(1)-2x + y = 3 -4x + 2y = 2{(2)4x - 2y = 6 6x - 3y = 9{(3)System of 2 Linear Equations
- 17. x1 + x2 + x3 = 2 --------(1) 2x1 + 3x2 + x3 = 3 ------(2) x1 - x2 - 2x3 = -6 --------(3)(1)(2)(3)System of 3 Linear Equations
- 18. A System of Linear Equations ( A Linear System ) A finite collection of linear equations a11 x1 + a12 x2 +....+ a1n xn = b1 a21 x1 + a22 x2 +....+ a2n xn = b2 .... am1 x1 + am2 x2 +....+ amn xn = bm
- 19. A SolutionTo a equation: a1x1 + a2x2 + .... + anxn = b ( t1, t2,...., tn ) To a linear system : A solution to each of linear equation simultaneously ps. Solution Set
- 20. Elementary Transformations1. Interchange the position of two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a multiple of one equation to another equation.
- 21. x1 + x2 + x3 = 2 --------(1) 2x1 + 3x2 + x3 = 3 ------(2) x1 - x2 - 2x3 = -6 --------(3)Interchange (1) and (3). Multiply (2) by 1/2. Add a -1 multiple (2) to (1).Continuous Operations
- 22. x1 + x2 + x3 = 2 -------------(1) x1 + 1.5x2 + 0.5x3 = 3 -----(2) x1 - x2 - 2x3 = -6 ------------(3)Ax1 - x2 - 2x3 = -6-----------(3) x1 +1.5x2 + 0.5x3 = 1.5--(2) -0.5x2 + 0.5x3 = 0.5---(1)BResult of operations
- 23. x1 + x2 + x3 = 2 --------(1) 2x1 + 3x2 + x3 = 3 ------(2) x1 - x2 - 2x3 = -6 --------(3)A x1 = ? ------(1) x2 = ? ------(2) x3 = ? ------(3)DThe Objective
- 24. Solve the problem by elementary transformationx1 + x2 + x3 = 2 ------(1) 2x1 + 3x2+ x3 = 3 ------(2) x1 - x2 - 2x3= -6 ------(3)Add a -2 multiple (1) to (2). Add a -1 multiple (1) to (3).
- 25. Iteration 1x1 + x2 + x3 = 2 ------(1) x2 - x3 = -1 ------(2) - 2x2 -3x3 = -8 ------(3)Add a -1 multiple (2) to (1). Add a 2 multiple (2) to (3).
- 26. Iteration 2x1 + 2x3= 3 ------(1) x2 - x3 = -1 ------(2) -5x3 = -10 ------(3)Multiply (2) by -1/5. Add a -2 multiple (3) to (1). Add a 1 multiple (3) to (2).
- 27. Final Iterationx1 = -1 ------(1) x2 = 1 ------(2) x3 = 2 ------(3)Final answer: (x1, x2, x3) = (-1, 1, 2)
- 28. DifficultyIt is very hard to carry variables, xi’s, through the calculation process when applying elementary transformation
- 29. 2 3 -4 7 5 -17 1 0 5 -8 33 5 6 0 -2 5 8 9 128 9 123 5 0 -2 8 9 5 -2 93 5 0 -2Matrix
- 30. Transfer to be : AX = BMatrix Notationa11x1 + a12x2 +....+ a1nxn = b1 a21x1 + a22x2 +....+ a2nxn = b2 am1x1 + am2x2 +....+ amnxn = bm................
- 31. x1 + x2 + x3 = 2 ------(1) 2x1 + 3x2 + x3 = 3 ------(2) x1 - x2 - 2x3 = -6 ------(3)Example 1Transfer (I) to matrix format.(I)
- 32. x1 x2 x3 X =1 3 -2B =1 -1 -2 2 -3 -5 -1 3 5A =Matrix RepresentationAX = B
- 33. 3x1 + 2x2 - 5x3 = 7 x1 - 8x2 + 4x3 = 9 2x1 + 6x2 - 7x3 = -2Transfer (I) to matrix format.(I)3 2 -5 1 -8 4 2 6 -7A=x1 x2 x3 X=7 9 -2B=Example 2
- 34. Matrix of Coefficients1 1 1 2 3 1 1 -1 -2x1+x2+x3=2 2x1+3x2+x3=3 x1-x2-2x3=-6Coefficients of the system or Matrix A
- 35. Augmented Matrixx1+x2+x3=2 2x1+3x2+x3=3 x1-x2-2x3=-61 1 1 2 2 3 1 3 1 -1 -2 -6Coefficients and RHS, or [ A | B ].
- 36. Reduced Echelon Form1. Any rows with all zeros are at the bottom. 2. Leading 1. 3. Leading 1 to the right. 4. All other elements in a leading 1 column are zeros.
- 37. 1 0 8 0 1 2 0 0 01 2 0 4 0 0 0 0 0 0 1 31 0 0 2 0 0 1 4 0 1 0 31 2 3 0 0 0 0 1 0 0 0 0Examples I
- 38. 1 2 0 3 0 0 0 3 4 0 0 0 0 0 11 7 0 8 0 1 0 3 0 0 1 2 0 0 0 0Examples II
- 39. Elementary Row Operations1. Interchange two rows 2. Multiply the elements of a row by a nonzero constant 3. Add a multiple of the elements of one row to the corresponding elements of another row.
- 40. x1 + x2 + x3 = 2 --------(1) 2x1 + 3x2 + x3 = 3 ------(2) x1 - x2 - 2x3 = -6 --------(3)1 1 1 2 2 3 1 3 1 -1 -2 -6Continuous Operations
- 41. 1 1 1 2 2 3 1 3 1 -1 -2 -6Interchange r.1 and r.3. Multiply r.2 by 1/2. Add a -1 multiple r.2 to r.3.1 -1 -2 -6 1 1.5 0.5 1.5 0 -0.5 0.5 0.5Result
- 42. 1 1 1 2 2 3 1 3 1 -1 -2 -6Through Elementary Row operationsThe ObjectiveReduced echelon Form ? ? ?
- 43. Equivalent SystemsSuppose that A and B are both systems of linear equations. A and B are equivalent if they are related through elementary transformations. A and B has the same solution if they are equivalent.
- 44. Solving a system of linear equationsGauss-Jordan Elimination Gauss Elimination
- 45. Gauss-Jordan Elimination1. Write the augmented matrix. 2. Derive the reduced echelon form of the augmented matrix 3. Write the system of equations corresponding to the reduced echelon form.
- 46. x1 + x2 + x3 = 2 --------(1) 2x1 + 3x2 + x3 = 3 ------(2) x1 - x2 - 2x3 = -6 --------(3)1 1 1 2 2 3 1 3 1 -1 -2 -6Perform J-G Elimination
- 47. 1 1 1 2 2 3 1 3 1 -1 -2 -61 1 1 2 0 1 -1 -1 0 -2 -3 -8+*(-2)+*(-1)Pivoting Step 1
- 48. 1 1 1 2 0 1 -1 -1 0 -2 -3 -81 0 2 3 0 1 -1 -1 0 0 -5 -10+*(-1) +*(2)Pivoting Step 2
- 49. 1 0 2 3 0 1 -1 -1 0 0 -5 -10++*(-2)1 0 0 -1 0 1 0 1 0 0 1 2Pivoting Step 4*(-1/5)
- 50. 1 0 0 -1 0 1 0 1 0 0 1 2Stop?Reduced Echelon Form?Write the linear equations: x1 = -1, x2 = 1, x3 = 2
- 51. AgendaLinear Systems as Mathematical Models Linear Systems Having One or No Solutions Linear Systems Having Many Solutions
- 52. Conditions of Solution SetsConsistent : A linear system with at least one solution Inconsistent : A linear System with no solution
- 53. Conditions of Solution SetEmpty--infeasible One--feasible with unique solutions Many--feasible with infinite many solutions
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