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Department of Material Science and Engineering University of Ghana, Legon First Semester 2018 Numerical Methods FAEN 301 Course Project Submitted by ID 10607444 10616366 10620620 10625002 Students Jacqueline Abeasi Jonathan Selorm Degbedzui Gabriel Opoku Okempa Philip Contents 1 Introduction 1 2 Solution Approach and Algorithm 2.1 General Problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Specific Questions . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Question 1 - Investment In An Entrepreneurial Venture 2.2.2 Question 2 - Clock Production Company . . . . . . . . 2.2.3 Question 3 - Is Aid an Invariably Fatal Disease? . . . . 2 2 24 24 25 27 3 Conclusion 30 4 Appendix 31 4.1 Question 1 - Investment In An Entrepreneurial Venture . . . . 31 4.2 Question 2 - Clock Production Company . . . . . . . . . . . . 32 References 36 i Chapter 1 Introduction Numerical analysis is the study of algorithms that use numerical approximation (as opposed to general symbolic manipulations) for the problems of mathematical analysis (as distinguished from discrete mathematics). Numerical analysis naturally finds applications in all fields of engineering and the physical sciences, but in the 21st century, the life sciences and even the arts have adopted elements of scientific computations. Ordinary differential equations appear in the movement of heavenly bodies (planets, stars and galaxies); optimization occurs in portfolio management; numerical linear algebra is important for data analysis; stochastic differential equations and Markov chains are essential in simulating living cells for medicine and biology. The overall goal of the field of numerical analysis is the design and analysis of techniques to give approximate but accurate solutions to hard problems. This project employees the use of software such as Microsoft Excel, Visual Basic and MATLAB. Solutions obtained can be used as models to solve similar problems in many fields of study 1 Chapter 2 Solution Approach and Algorithm 2.1 General Problems 2 Question 1 8 4 A= 2 1 5 9 4 5 7 8 3 7 2 4 1 2 3 4 1 5 8 7 6 , B = 7 8 5 9 , C = 3 4 9 , D = 1 7 5 6 1 4 2 7 6 9 4 5 2 8 2 6 9 3 8 (a) Matrix multiplication between AB will not be possible as the their dimensions do not agree. A has a dimension of (4 x 4) and B (3 x 4). Two matrices can be multiplied if and only if the column size of the first matrix is equal to the row size of the second matrix. (b) Matrix A has 4 columns while second Matrix C has 3 rows. Hence matrix multiplication between AB will not be possible as the their dimensions do not agree.Two matrices can be multiplied if and only if the column size of the first matrix is equal to the row size of the second matrix. (c) 8 4 AD = 2 1 5 9 4 5 7 8 3 7 2 4 6 7 ∗ 7 1 9 4 5 2 8 2 6 82 110 130 9 = 111 114 177 67 56 113 3 8 82 89 144 (d) Matrix multiplication between BC will not be possible as the their dimensions do not agree. B has a dimension of (3 x 4) and C (3 x 3). Two matrices can be multiplied if and only if the column size of the first matrix is equal to the row size of the second matrix. (e) 4 1 2 3 4 7 BD = 7 8 5 9 ∗ 1 5 6 1 4 4 5 2 8 2 6 37 41 65 9 = 125 109 201 3 79 53 119 8 (f) Matrix C has 3 columns while second Matrix D has 4 rows. Hence matrix multiplication between CD will not be possible as the their dimensions do not agree.Two matrices can be multiplied if and only if the column size of the first matrix is equal to the row size of the second matrix. 3 10 (a) 8 4 A= 2 1 5 9 4 5 7 8 3 7 2 6 7 9 |A − λI| = 0 8 5 7 2 λ 0 0 0 4 9 8 6 − 0 λ 0 0 | = 0 | 2 4 3 7 0 0 λ 0 1 5 7 9 0 0 0 λ 8−λ 5 7 2 4 9−λ 8 6 | = 0 | 2 4 3−λ 7 1 5 7 9−λ 21.6363 7 λ= 2.6114 −2.2477 The correspondent eigenvectors are: for λ = -2.2477 for λ = 7 −0.3614 −0.2611 V = 0.8189 −0.3614 for λ = 21.6363 −0.8566 −0.2109 V = 0.1450 0.4481 for λ = 2.6114 −0.4890 −0.6208 V = −0.3694 −0.4890 4 11 −0.3609 0.8329 V = −0.2139 −0.3609 Matrix C 1 5 8 C = 3 4 9 2 7 6 |C − λI| = 0 The eigenvalues are : 1 5 8 λ 0 0 | 3 4 9 − 0 λ 0 | = 0 2 7 6 0 0 λ 1−λ 5 8 4−λ 9 | = 0 | 3 2 7 6−λ for λ = -1.6792 15.2221 λ = −1.6792 −2.5430 forλ = 15.2221 −0.8721 V = −0.2266 0.4337 for λ = -2.5430 (b) determinant of Matrix A −0.5405 V = −0.6096 −0.5799 −0.5428 V = −0.5829 0.6047 5 12 det(A) = λ1 × λ2 × λ3 × λ4 det(A) = 21.6363 × 7 × 2.6114 × −2.2477 det(A) = −888.9816 . Matrix C det(C) = λ1 × λ2 × λ3 det(C) = 15.2221 × −1.6792 × −2.5430 det(C) = 65.0015 (c) The inverse of matrix C using Gauss Jordan method 1 0 0 1 0 0 1 5 8 C = 3 4 9 2 7 6 1 5 8 1 0 0 3 4 9 0 1 0 2 7 6 0 0 1 5 8 1 0 0 −11 −15 −3 1 0 −3 −10 −2 0 1 5 8 1 0 0 −11 −15 3 1 0 0 65 13 3 −11 −65 −325 0 39 24 −88 0 −715 0 0 110 −165 0 0 65 13 3 −11 −46475 0 0 1 0 0 0 0 27885 −18590 −9295 232375 0 0 −35750 5625 0 65 13 3 −11 0 0 −0.6 0.4 0.2 1 0 0 −0.1538 0.2308 0 1 0.2 0.04615 −0.1692 6 13 C −1 (d) 8 0 0 0 −0.6 0.4 0.2 −0.1538 0.2308 = 0 0.2 0.04615 −0.1692 8 5 7 2 4 9 8 6 A= 2 4 3 7 1 5 7 9 8 5 7 2 1 0 0 0 4 9 8 6 0 1 0 0 2 4 3 7 0 0 1 0 1 5 7 9 0 0 0 1 8 5 7 2 1 0 0 0 0 −13 −9 −10 1 −2 0 0 0 −11 −5 −26 1 0 −4 0 0 −35 −49 −70 1 0 0 −8 8 5 7 2 1 0 0 0 0 −13 −9 −10 0 1 −2 0 0 0 34 −228 2 22 −52 0 0 0 −322 −560 −22 70 0 −104 5 7 2 1 0 0 0 −13 −9 −10 −2 0 0 1 0 34 −228 2 22 −52 0 0 0 −92456 −104 9464 −16744 −3536 1 0.625 0.875 0.25 0.125 0 0 0 0 −1 −0.6923 −0.7692 0.07692 −0.1538 0 0 0 0 1 −6.7059 0.05882 0.6471 −1.5294 0 −3 0 0 0 −1 −1.1249 × 10 0.1024 −0.1811 −0.03824 0.1253 0.0256 −0.04528 −9.56 × 10−3 1 0.625 0.875 0 0 −1 −0.6923 0 0.07779 −0.2325 0.1393 0.02941 0 0.06636 −0.03958 −0.315 0.2564 0 1 0 0.1024 −0.1811 −0.03824 0 0 0 −1 −1.1249 × 10−3 −0.06724 −0.06023 −0.2303 0.23391 −1 −0.625 0 0 0 −1 0 0 0.1237 −0.2599 −0.0788 0.2069 0 0 1 0 0.06636 −0.03958 −0.315 0.2564 0 0 0 −1 −1.1249 × 10−3 0.1024 −0.1811 −0.03824 7 14 −1 0 0 0 −0.1440 0.1022 −0.1811 0.1046 0 −1 0 0 0.1237 −0.2599 −0.0788 0.2069 0 0 1 0 0.06636 −0.03958 −0.315 0.2564 0 0 0 −1 −1.1249 × 10−3 0.1024 −0.1811 −0.03824 1 0 0 0 0.1440 −0.1022 0.1811 −0.1046 0 1 0 0 −0.1237 0.2599 0.0788 −0.2069 0 0 1 0 0.06636 −0.03958 −0.315 0.2564 0 0 0 1 1.1249 × 10−3 −0.1024 0.1811 0.03824 The inverse of the matrix A is therefore 0.1440 −0.1022 −0.1237 0.2599 A−1 = 0.06636 −0.03958 1.1249 × 10−3 −0.1024 (e) LU decomposition of matrix C 5 8 4 9 7 6 0 0 1 0 L32 1 U12 U13 U22 0 U32 U33 U11 U12 U13 L21 U12 + U22 L21 U13 + U23 LU = L21 U11 L31 U11 L31 U12 + L32 U22 L31 U13 + L32 U23 + U33 . 1 C= 3 2 1 L = L21 L31 U11 U = U21 U31 0.1811 −0.1046 0.0788 −0.2069 −0.315 0.2564 0.1811 0.03824 U11 = 1; U12 = 5; U13 ; U14 = 9 L21 U11 = 3 ⇒ L21 = 3 L21 + U12 + U22 = 4 ⇒ U22 = −11 L21 U13 + U13 = 9 U23 = −15 L31 U11 = 2 8 15 L31 = 2 L31 U12 + L32 U22 = 7 3 L32 = 11 L31 U13 + L32 U32 + U33 = 6 −65 U33 = 11 . 1 L = 3 2 . 0 1 3 11 0 0 1 1 5 8 U = 0 −11 −15 −65 0 0 11 14 b = 16 15 let the vector X be the solution 1 3 2 1 3 2 5 8 X1 14 4 9 X2 = 16 7 6 X3 15 Z1 14 0 0 1 0 Z2 = 16 3 1 Z3 15 11 Z1 = 14, Z2 = −26, −65 11 1 5 8 X1 14 0 −11 −15 X2 = −26 −65 −65 0 0 X3 11 11 Z3 = 9 16 −65 −65 X3 = 11 11 X3 = 1 −11X2 − 15X3 = −26, −11X2 = 11 X2 = 1 X1 + 5X2 + 8X3 = 14 X1 = 14 − 13 X1 = 1 Therefore the solution is 1 X = 1 1 10 17 Question 2 Forward Elimination of Unknowns Since there are four equations, there will be three steps of forward elimination of unknowns. First step Divide Row 1 by 4.2857 10 7 and then multiply it by 4.2857 10 7 , that is, multiply Row 1 by 4.2857 107 4.2857 107 = 1 . Row 1 (1) = 4.2857 107 − 9.2307 105 0 0 − 7.88710 3 Subtract the result from Row 2 to get 4.2857 107 0 − 6.5 0 − 9.2307 105 3.7688 105 − 0.15384 0 0 − 4.2857 107 6.5 4.2857 107 c1 − 7.887 103 0 5.4619 105 c 2 7.887 103 = 0.15384 c3 0.007 − 3.6057 105 c 4 0 Divide Row 1 by 4.2857 10 7 and then multiply it by − 6.5 , that is, multiply Row 1 by − 6.5 4.2857 107 = −1.5167 10 −7 . ( ) Row 1 − 1.5167 10−7 = − 6.5 0.14000 0 0 1.196210 −3 Subtract the result from Row 3 to get 4.2857 107 0 0 0 − 9.2307 105 3.7688 105 − 0.29384 0 0 − 4.2857 107 6.5 4.2857 107 c1 − 7.887 103 0 5.4619 105 c 2 7.887 103 = 0.15384 c3 5.8038 10−3 − 3.6057 105 c 4 0 Divide Row 1 by 4.2857 10 7 and then multiply it by 0, that is, multiply Row 1 by 0 4.2857 107 = 0 . Row 1 (0) = 0 0 0 0 0 Subtract the result from Row 4 to get 4.2857 107 0 0 0 − 9.2307 105 3.7688 105 − 0.29384 0 0 − 4.2857 107 6.5 4.2857 107 11 c1 − 7.887 103 0 5.4619 105 c 2 7.887 103 = 0.15384 c3 5.8038 10−3 − 3.6057 105 c 4 0 Second step We now divide Row 2 by 3.7688 105 and then multiply it by − 0.29384, that is, multiply Row 2 by − 0.29384 3.7688 105 = −7.7966 10−7 . Row 2 (− 7.7966 10−7 ) = 0 − 0.29384 33.414 − 0.42584 − 6.1492 10 −3 Subtract the result from Row 3 to get 4.2857 107 0 0 0 − 9.2307 105 3.7688 105 0 0 0 − 4.2857 107 − 26.914 4.2857 107 c1 − 7.887 103 0 5.4619 105 c 2 7.887 103 = 0.57968 c3 1.1953 10 − 2 − 3.6057 105 c 4 0 Divide Row 2 by 3.7688 105 and then multiply by 0 , that is, multiply Row 2 by 0 3.7688 105 = 0 . Row 2 (0) = 0 0 0 0 0 Subtract the result from Row 4 to get 4.2857 107 0 0 0 − 9.2307 105 3.7688 105 0 0 0 − 4.2857 107 − 26.914 4.2857 107 c1 − 7.887 103 0 5.4619 105 c 2 7.887 103 = 0.57968 c3 1.1953 10 − 2 − 3.6057 105 c 4 0 Third step We now divide Row 3 by − 26.914 and then multiply by 4.2857 10 7 , that is, multiply Row 2 by 4.2857 107 − 26.914 = −1.5924 106 . ) ( Row 3 − 1.5924 106 = 0 0 4.2857 107 − 9.2307 105 − 1.903410 4 Subtract the result from Row 4 to get 4.2857 107 0 0 0 − 9.2307 105 3.7688 105 0 0 0 − 4.2857 107 − 26.914 0 Back substitution From the fourth equation, 12 c1 − 7.887 103 0 5.4619 105 c 2 7.887 103 = 0.57968 c3 1.1953 10− 2 5.625 105 c 4 1.9034 104 5.625 105 c4 = 1.9034 104 1.9034 104 5.625 105 c4 = = 3.3837 10−2 Substituting the value of c4 in the third equation, − 26.914c3 + (0.57968)c4 = 1.195310−2 1.1953 10 −2 − (0.57968)c 4 − 26.914 c3 = = 1.1953 10 −2 − (0.57968) 3.3837 10 −2 − 26.914 = 2.8469 10 −4 Substituting the value of c3 and c 4 in the second equation, ( ) 3.7688105 c2 + − 4.2857107 c3 + 5.4619105 c4 = 7.887 103 c2 = ( ) ( ) ( 7.887 103 − − 4.2857 107 c3 − 5.4619 105 c 4 3.7688 105 ) ( ) ( 7.887 103 − − 4.2857 107 2.8469 10−4 − 5.4619 105 3.3837 10−2 = 3.7688 105 = 4.2615 10−3 Substituting the values of c2 , c3 and c 4 in the first equation, ( ) 4.2857107 c1 + − 9.2307105 c2 + (0)c3 + (0)c4 = −7.887 103 c1 = = ( ) ( ) ( − 7.887 103 − − 9.2307 105 c 2 − (0)c3 − (0)c 4 4.2857 107 − 7.887 103 − − 9.2307 105 4.2615 10−3 4.2857 107 = −9.2244 10 −5 Hence the solution vector is 13 ) ) c1 − 9.2244 10−5 c −3 2 = 4.2615 10 c3 2.8469 10− 4 −2 c4 3.3837 10 The stress on the inside radius of the outer cylinder is then given by 1 − c3 (1 + ) + c4 r 2 30 106 1 − 0.3 = 2.8469 10− 4 (1 + 0.3) + 3.3837 10− 2 2 2 1 − 0.3 6.5 = E 1 − 2 = 30683psi 14 Question 3 1. 2 -1 A= 0 0.5 -1 2 -1 0 0 -1 2 -1 0.5 0 -1; 1 0.5 5 b= −5 1.5 a. U= 1.4142 -0.7071 0 0.3536 0 1.2247 -0.8165 0.2041 0 0 1.1547 -0.7217 0 0 0 0.5590 0 0 1.4142 -0.7071 1.2247 0 U' = 0 -0.8165 1.1547 0.3536 0.2041 -0.7217 U U x = b T U x y U y = b T b. 0 0 0 y1 0.5 1.4142 -0.7071 1.2247 0 0 y 2 5 = 0 -0.8165 1.1547 0 y3 −5 0.3536 0.2041 -0.7217 0.5590 y 4 1.5 -0.3536 3.8784 y= -1.5877 -0.5590 c. U x = y 15 0 0 0 0.5590 1.4142 -0.7071 0 0.3536 x1 -0.3536 x 2 3.8784 0 1.2247 -0.8165 0.2041 = x3 -1.5877 0 0 1.1547 -0.7217 0 0 0 0.5590 x 4 -0.5590 1 2 x= -2 -1 2. D= 2.0000 0 0 0 0 1.5000 0 0 0 0 1.3333 0 0 0 0 0.3125 16 Question 4 (a) 00 mu (t) + ku(t) = F0 coswt w0 = u(t) = r k m F0 [coswt − cosw0 t] − w2 ) m(w02 Figure 6: Graph of u(t), m= 1, k = 9;F0 =1; w = 2 and t [0,2π] 1 u(t) = [cos2t − cos3t] 5 The integral, I I= Z2π u(t)dt 0 Exact value, Z2π u(t)dt = 0 0 Let the n = 20; i. Trapezoidal Rule i.e. Approximating integral using Trapezoidal rule I = −8.7197 × 10−17 ii. Simpson’s 1/3 i.e. Approximating integral using Simpson’s 1/3 I = −8.4290 × 10−17 n = 18 17 6 iii. Simpson’s 3/8 i.e. Approximating integral using Simpson’s 3/8 I = −8.7646 × 10−16 iv. Romberg n = 20 i.e. Approximating integral using Romberg I = −2.9587 × 10−15 18 7 (b) 0 Adding the term cu (t) to the left side of the eqn in a. u(t) = C1 er1 t + C2 er2 t + c2 w 2 F0 [cwsin(wt) + m(w02 − w2 )cos(wt)] ...(1) + m2 (w02 − w2 ) p p −c − (c2 − 4w0 m2 ) (c2 − 4w0 m2 ) r1 = andr2 = 2m 2m for k =9; m = 1; F0 = 1; c = 10; w = 2; calculating for r1 and r2 ; −c + r1 = -1 r2 = -9 Subtituting the value of r1 and r2 into eqn (1) u(t) = C1 e−t + C2 e−9t + 1 [20sin(2t) + 5cos(3t)] 425 taking u(0) = 0 and u(1) = 0; u(0) = C1 + C2 + 5 = 0 ...(1) 425 u(1) = C1 e−1 + C2 e−9 + 16.1052 = 0 ...(2) 425 solving (1) and (2) simultaneously yields: C1 = -0.1030 C2 = 0.09127 Figure 7: Resulting Graph of solving the above The integral, I I= Z 2π u(t)dt 0 Taken n = 20, 19 8 i. Trapezoidal rule I = - 0.0875 ii. Simpson’s 1/3 rule I = - 0.0909 iii. Simpson’s 3/8 rule let n = 18 I = - 0.09258 iv. Romberg I = - 0.09266 (c) i. Z1.5 x2 lnxdx 1 For n = 2 - The exact integral is I = 0.192259 - The Gauss quadrature I = 0.192269 Relative error for Gauss Quadrature is very small hence gives the best estimate. ii. Z0.35 0 x2 2 dx, n = 3 −4 - The exact integral is I = - 0.176820 - The Gauss quadrature gives I = - 0.176820 The ralative error = −0.176820 + 0.176820 =0 −0.176820 Both methods yield the same result and hence the accuracy is 100% Z 3.5 x √ dx, n = 4 x2 − 4 3 - The exact integral is I = 0.636213 - The Gauss quadrature yields I = 0.636213 The two answers are the same so the accuracy is 100% 20 9 Question 5 a) f ( xi +1 ) − f ( xi ) x f ( xi ) a = 0.24 xi = 0.1 x = 0.05 xi+1 = x + x = 0.1 + 0.05 = 0.15 f (0.1) = 1 − e( −0.240.1) = 0.023714 f (0.15) = 1 − e( −0.240.15) = 0.035360 f (0.1) = f (0.15) − f (0.1) 0.05 0.035360− 0.023714 0.05 = 0.23291 The exact value of f (0.1) can be calculated by differentiating f (x ) = 1 − e − ax , x 0 as f ( x ) = d f (x ) dx Knowing that d − ax e = −ae − ax dx 21 we find f ( x ) = d (1 − e −ax ) dx = ae− ax = 0.24e −0.24 x ( f (0.1) = (0.24) e − ( 0.240.1) ) = 0.23431 The absolute relative true error is True Value − Approximat e Value 100 True Value t = = 0.23431− 0.23291 100 0.23431 = 0.59761% b) a = 0.12 f (0.1) = 1 − e − ( 0.120.1) = 0.011928 f (0.15) = 1 − e − ( 0.120.15) = 0.017839 f (0.1) = f (0.15) − f (0.1) 0.05 0.017839− 0.011928 0.05 = 0.11821 f ( x ) = d (1 − e −ax ) dx = ae− ax = 0.12e −0.12 x f (0.1) = (0.12)(e −( 0.120.1) = 0.11856 22 The absolute relative true error is t = = True Value − Approximat e Value 100 True Value 0.11857− 0.11821 100 0.11857 = 0.29940% The estimate of the derivative decreased. 23 2.2 2.2.1 Specific Questions Question 1 - Investment In An Entrepreneurial Venture (a). With the Wyndor Glass Co problem, the optimal solution for two activities had to be found for a certain limited amount of resources. An optimal combination of this two activities had to be found. Let x be the fraction of purchase of the ownship in the first friends company Let y be the fraction of purchase of the ownship in the second friends company (b) maximize 4500x + 4500y Subject to : 5000x + 4000y ≤ 6000 400x + 500y ≤ 600 x ≤1 y ≤1 xG , x W ≥ 0 (c) Optimal Solution:(x,y) = (2/3, 2/3) and P = 6000 24 2.2.2 Question 2 - Clock Production Company Three friends, David, LaDeana, and Lydia are the sole partners and workers in a company which produces fine clocks. David and LaDeana work a maximum of 40 hours each per week while Lydia works a maximum of 20 hours per week The company makes two different types of clocks, namely: • A grandfather clock and a wall clock • A wall clock Each grandfather clock built and shipped yields a profit of 300 dollars, while each wall clock yields a profit of 200 dollars. Problem: The three partners now want to determine how many clocks of each type should be produced per week to maximize the total profit Solution Approach The Following is the linear programming model for the Clock Production Company: a maximize 300xG + 200xW Subject to : 6xG + 4xW ≤ 40 8xG + 4xW ≤ 40 3xG + 3xW ≤ 20 xG , x W ≥ 0 e C(G): $300 -¿ $375 % of allowable increase = 100(375-300/100) = 75% The optimal solution will remain the same since $75 is within the allowable increase. f By the 100% rule for simultaneous chnages in the objective function, the optimal solution may or may not chnage. C(G): $300 -¿ $375 % of allowable increase = 100(375-300/100) = 75% C(W): $200 -¿ $175 % of allowable decrease = 100(200-175/50) = 50% 25 g ”Graph” When the profit for grandfather clock increases to $375, the optimal solution is still (G,W) = (x1,x2) = (3.333.3.333) and P = 1666.67 if both profit estimates change thus grandfather clock increases to $375 and wall clock decreases to $175, then the optimal solution changes to: (G,W) = (x1,x2) = (5,0) and P = 1875 h Lydia should increase her hours slightly since she has the highest shadow price connected to her i The shadow price for David is zero because all of his available hours are not being used and consequently an increase in his hours would not impact total profit j Yes, this increase is within the allowable increase. The increase in total will be $33.33 x 5 = $166.65 k By the 100the optimal solution may or may not chnage. The shadow prices may not be used to determine the effect on profit. C(L): 20 -¿ 25 % of allowable increase = 100(25-20/100) = 25% C(D): 40 -¿ 35 % of allowable increase = 100(40-35/100) = 75% sum = 125% 26 2.2.3 Question 3 - Is Aid an Invariably Fatal Disease? Solution Approach QUESTION 1 s(t) = e−kt show thats Tavg = Tavg = Z ∞ 1 k s(t) dt 0 Tavg = Z ∞ e−kt dt 0 lim b→∞ lim [ b→∞ QUESTION 2 Z b e−kt dt 0 e−kt e−k∗0 + ] −k −k Tavg = 6.4 months s(t), where t = 5 years Find the fraction remaining after 5 years −t s(t) = e Tavg −60 s(t) = e 6.4 s(t) = 8.4818 ∗ 10−5 QUESTION 3a s(t) = e−kt show that S(t) = 2 27 −t T1 2 0.5 = e −kt 1 2 −kt 1 ln(0.5) = ln(e 2 ) ln(0.5) = −kt 1 2 ln(0.5) −k t1 = 2 t1 = ln(2) k k =, ln(2) t1 2 2 − 0.5 = e ln(2) ∗t t1 2 0.5 = 2 −t T1 2 QUESTION 3b show thats t 1 = Tavg ln(2) 2 t1 = 2 k = ln(2) k 1 Tavg t 1 = Tavg ln(2) 2 QUESTION 4 Incurable f raction that survives af ter 5 years = 0.14 − 0.10 = 0.04 = 4% s(t) = 0.04 = e−k5 28 k = 0.644/yr s(t) = e−0.644t F raction surviving af ter 2 years = e−0.644∗2 = 0.2757 = 27.58 29 Chapter 3 Conclusion Advanced numerical analysis are essential in making numerical weather predictions feasible, computing the trajectory of spacecraft using numerical solution of a solution of differential equations. 30 Chapter 4 Appendix 4.1 Question 1 - Investment In An Entrepreneurial Venture 31 4.2 Question 2 - Clock Production Company a 32 b c 33 d 34 35 References [1] W. L. Winston, et. al, Developing Spreadsheet- Based Decision Support Systems Using Excel and VBA for Excel, Dynamic Ideas, 2007. [2]Hillier and Lieberman,Introduction to Operations Research, 7th edition. [3]D. G. Zill, A First Course in Differential Equations with Modeling Applications, 10th ed. New York: Brooks/Cole, 2013 36