MCSE-004. Dec 2013 Solutions Manual IGNOUUSER

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1 MCSE-004 Dec 2013 Solutions Manual IGNOUUSER

2 1 1. (a) Verify the distributive property of floating point numbers i.e. prove : a(b-c) ab ac a=.5555e1, b=.4545e1, c=.4535e1 Define : Truncation error, Absolute Error and Relative Error. { } Addition/Subtraction of floating point numbers To add/subtract 2 floating point numbers, the exponents of the two numbers must be equal. If not the mantissa is shifted appropriately. Then the Mantissas are added or subtracted. After that the result is normalized and the exponent is appropriately adjusted. Multiplication of floating point numbers To multiply 2 floating point numbers, the mantissas are multiplied and exponents are added. After that the resulting mantissa is normalized and the exponent is appropriately adjusted. LHS (b-c) =.0010E1 =.1000E-1 a(b-c) =.5555E1 *.1000E-1 =.05555E0 =.5555E-1 RHS ab =.5555E1 *.4545E1 =.2524E2 ac =.5555E1 *.4535E1 =.2519E2 ab ac =.0005E0 =.5000E-3 Clearly LHS RHS For definitions, please refer block. 1. (b)find the real root of the equation x = e -x using Newton-Raphson Method. List the cases where Newton s Method fail. Given x=e -x multiplying both sides by e x and arranging, we get xe x 1 = 0 f(x) = xe x 1 f (x) = (1 + x)e x

3 2 Let x 0 = 1 By Newton Raphson method we have, x n+1 = x n - x 1 = 1 - = f(x 1 ) = , f (x 1 ) = x 2 = = f(x 2 ) = , f (x 2 ) = x 3 = = f(x 3 ) = , f (x 3 ) = x 4 = = Since x3 and x4 are same upto 3 decimal places, is the required root. Method fails in the cases listed below i) Division by zero occurs in some cases. let f(x) = x 3 2x = 0 f (x) = 3x 2 4x By Newton Raphson method we have x n+1 = x n - If we choose initial value x 0 = 0 then we get division by zero error. ii) Infinite loop case In some cases, say x 0 is the initial value and 1 st approximation gives x 1 which inturn gives 2 nd approximation as x 0 again and the loop continues.

4 3 1. (c) Solve by Gauss-Seidel method. 2x 1 x 2 + x 3 = -1 x 1 + 2x 2 x 3 = 6 x 1 x 2 + 2x 3 = -3 correct to 3 decimal places The given system of equations can we written as x 1 = (1/2)(-1 + x 2 - x 3 ) x 2 = (1/2)(6 - x 1 + x 3 ) x 3 = (1/2)(-3 - x 1 + x 2 ) Let initial approximation x 0, y 0, z 0 is 0,0,0 First approximation x 1 (1) = (1/2)( ) = x 2 (1) = (1/2)( ) = 3.25 x 3 (1) = (1/2)( ) = Second approximation x 1 (2) = (1/2)( ) = x 2 (2) = (1/2)( ) = x 3 (2) = (1/2)( ) = Similarly, we have [x 1 (3), x 2 (3), x 3 (3) ] = [ , , ] [x 1 (4), x 2 (4), x 3 (4) ] = [ , , ] [x 1 (5), x 2 (5), x 3 (5) ] = [ , , ] [x 1 (6), x 2 (6), x 3 (6) ] = [ , , ] [x 1 (7), x 2 (7), x 3 (7) ] = [ , , ] [x 1 (8), x 2 (8), x 3 (8) ] = [ , , ] [x 1 (9), x 2 (9), x 3 (9) ] = [ , , ] [x 1 (10), x 2 (10), x 3 (10) ] = [ , , ] [x 1 (11), x 2 (11), x 3 (11) ] = [ , , ] Since last 2 approximations are same upto 3 decimal places. Hence the required solution is x 1 = x 2 = x 3 =

5 4 1. (d)let f(x) = ln(1+x), x 0 = 1 and x 1 = 1.1. Use linear interpolation to calculate an approximate value of f(1.04) and obtain a bound on the truncation error. We have x 0 = 1, x 1 = 1.1 f(x 0 ) = ln(1+x 0 ) = f(x 1 ) = ln(1+x 1 ) = Lets compute Lagranges fundamental polynomials, l 0 (x) = (x-x 1 ) / (x 0 x 1 ) = (x 1.1)/(1 1.1) = (x 1.1)/(-0.1) l 1 (x) = (x-x 0 ) / (x 1 x 0 ) = (x 1)/(1.1 1) = (x 1)/( 0.1) Hence Lagranges Interpolating polynomial is given by P 1 (x) = l 0 (x)* f(x 0 ) + l 1 (x)* f(x 1 ) =- 10(x 1.1)* (x 1)* f(1.04) = P 1 (1.04) = -10( )* (1.04 1)* = The truncate error in linear interpolation is given by E 1 (f;x) = (1/2)(x-x 0 )(x-x 1 )f ( ) where lies between 1 and 1.1 = (1/2)(x 1)(x 1.1)(-1/(1+x) 2 ) The max value of is and maximum value of (x-1)(x-1.1) occurs at (1+1.1)/2=1.05 * ( ) = (e) Consider initial value problem = x+y; y(0) = 1 Find y(0.2) using Runge-Kutta Method of fourth order. Also compare it with exact solution y = -(1+x) + 2e x to find the error. We have, x 0 =0, y 0 =1, h=0.2 Then we get, k 1 =hf(x 0,y 0 ) = 0.2(0+1) = 0.2 k 2 = hf(x 0 + h/2,y 0 + k 1 /2) = 0.2( ) = 0.24 k 3 = hf(x 0 + h/2,y 0 + k 2 /2) = 0.2( ) =0.244 k 4 = hf(x 0 + h,y 0 + k 3 ) = 0.2( ) = Therefore y = y 0 + 1/6(k 1 +2 k k 3 + k 4 ) = =

6 5 Given, y = -(1+x) + 2e x Putting x = 0.2, we get y = = Error = = (a) Find the interval in which the smallest positive root of the following equation lies using Bisection Method x 3 x 4 = 0 Given f(x) = x 3 x 4 f(0) = = -4 f(1) = = -4 f(2) = = 2 Clearly f(1)*f(2)<0, hence root lies between 1 and 2 First approximation of root, x 1 = (1+2)/2 = 1.5 f(1.5) = Clearly f(1.5)*f(2)<0, hence root lies between 1.5 and 2. Second approximation of root, x 2 = (1.5+2)/2 = 1.75 f(1.75) = Clearly f(1.75)*f(2)<0, hence root lies between 1.75 and 2. Third approximation of root, x 3 = f(1.875) = Now, f(1.875)*f(2) > 0, Hence the interval in which smallest root lie is [1.75,2] and the root is (b) Solve the following linear system of equations using Gauss Elimination Method. x 1 + x 2 + x 3 = 3 4x 1 + 3x 2 + 4x 3 = 8 9x 1 + 3x 2 + 4x 3 = 7

7 6 Given system of equations x 1 + x 2 + x 3 = 3 4x 1 + 3x 2 + 4x 3 = 8 9x 1 + 3x 2 + 4x 3 = 7...(1) Eliminate x 1 from second equation of (1), multiply first equation by 4 and subtract it from second equation (3 4) x 2 + (4 4) x 3 = 8 4*3 - x 2 = 8 12 x 2 = 4 Eliminate x 1 from third equation of (1), multiply first equation by 9 and subtract it from third equation. (3 9) x 2 + (4 9) x 3 = 7 9*3-6 x 2 + (-5 x 3 ) = x x 3 = 20 Hence given system of equations become x 1 + x 2 + x 3 = 3 x 2 = 4 6 x x 3 = 20...(2) Substituting value of x 2 in third equation we get, x 3 = -4 /5 = -0.8 Substituting value of x 2, x 3 in the first equation we get, x 1 = 3 4 (-0.8) = -0.2 Hence the required solution is x 1 = x 2 = 4 x 3 = (c) Give properties of polynomial equations. i) A polynomial equation of degree n has n solutions. ii) Non-real complex solution with real coefficient, if exist, occur in conjugate pairs. iii) Descarte s rule of signs The number of positive real solutions of the given polynomial equation is equal to the number of variations in signs of the polynomial.

8 7 3. (a) The table below gives the values of tanx for 0.10 x x y=tanx Find i) tan0.12 ii) tan0.26 First lets construct the difference table x y=tanx Δ Δ2 Δ3 Δ i) tan0.12 here, h=0.05, a=0.10, x=0.12 u=( )/0.05 = 0.4 By Newton s Forward Interpolation formula we have tan(0.12) = f(a) + u Δ f(a) + Δ 2 f(a) + Δ 3 f(a) +... = * = = ii) tan0.26 here, h=0.05, a=0.10, x=0.26 u=( )/0.05 = 3.2 By Newton s Forward Interpolation formula we have tan(0.26) = f(a) + u Δ f(a) + Δ 2 f(a) + Δ 3 f(a) +... = * = =

9 8 3. (b) Evaluate I = dx, correct to three decimal places using i) Trapezoidal, and ii) Simpson s rule with h=0.5 and h=0.25 Here, f(x) = The value of f(x) at x=0,0.25,0.5,0.75,1 are tabulated below. x f(x) f 0 f 1 f 2 f 3 f 4 i) Trapezoidal rule Taking h=0.25, we have I = (0.25/2)[ f 0 + f 4 + 2(f 1 + f 2 + f 3 )] = 0.125*( *2.04) = ii) Simpsons rule h = 0.25, I = (0.25/3)[ f 0 + f 4 + 4(f 1 + f 3 ) + 2 f 2 ] = * ( * *0.67) = h=0.5 we have to consider only f 0, f 2, f 4 I = (0.5/3)[ f 0 + f 4 + 4* f 2 ] = *( *0.67) = (c) Determine the value of y when x=0.1 given that y(0) = 1 and y = x 2 + y. let h = 0.1/5 = 0.02 x 0 = 0, y 0 = 1 By Euler s method we have y 1 = y 0 + hf(x 0,y 0 ) = *(0+1) = 1.1 y 2 = y 1 + hf(x 1,y 1 ) = *(0.02* ) = y 3 = y 2 + hf(x 2,y 2 ) = *(0.04* ) =

10 9 y 4 = y 3 + hf(x 3,y 3 ) = *(0.06* ) = y 5 = y 5 + hf(x 4,y 4 ) = *(0.08* ) = Hence the value of y at 0.1 is (a) A problem in statistics is given to the three students A,B and C whose chances of solving it are 1/2, 3/4 and 1/4 respectively. What is the probability that the problem will be solved. Given, probability that A will solve the problem P(A) = 1/2 probability that B will solve the problem P(B) = 3/4 probability that C will solve the problem P(C) = 1/4 Hence, probability that A will not solve the problem P( ) = 1 1/2 = 1/2 probability that B will not solve the problem P( ) = 1 3/4 = 1/4 probability that C will not solve the problem P( ) = 1 1/4 = 3/4 Since all events are independent, so P(no one will solve the problem) = P( )* P( )* P( ) = (1/2)*(1/4)*(3/4) = (3/32) P(that prob will be solved) = 1 (3/32) = 29/ (b) Calculate the correlation coefficient for the following heights (in inches) of fathers (x) and their sons (y) : x: y: The various calculations on the given data are tabulated below. x y x-x' y-y' (x-x')*(y-y') (x-x')*(x-x') (y-y')*(y-y')

11 10 where x = mean value of x = y = mean value of y = We have the correlation coefficient r = = = (c) Three identical bags have the following proportion of balls. First bag: 2 black 1 white Second bag: 1 black 2 white Third bad: 2 black 2 white One of the bag is selected and one ball is drawn. It turns out to be white. What is the probability of drawing a white ball again. The first one not been returned. Let E = event of selection of ith bag(i=1,2,3) A = event of drawing white ball Then P(E 1 ) = P(E 2 ) = P(E 3 ) = 1/3 Also, P(A E 1 ) = C(1,1)/C(3,1) = 1/3 P(A E 2 ) = C(2,1)/C(3,1) = 2/3 P(A E 3 ) = C(2,1)/C(4,1) = 1/2 Let C denote the further event of drawing another white ball. P(C E 1 A) = 0/2 = 0 P(C E 2 A) = 1/2 P(C E 3 A) = 1/3 Hence by Bayes theorem P(C A) = = = = 1/6

12 11 5. (a) Evaluate [ ( )] using Simpsons rule with 11 points. Here, h = (6-1)/11 = 5/11 x f(x) (5/11) (10/11) (15/11) (20/11) (25/11) (30/11) (35/11) (40/11) (45/11) (50/11) By Simpsons one-third rule, [ ( )] = (5/33) [( ) + 4( ) + 2( )] = (5/33) [ ( ) + 2( ) ] =

13 12 5. (b) Estimate the sale of a particular quantity for 1966 using the following table. Year Sale in thousands First lets construct the difference table. Year Sale here, h= =10 a=1981 x=1966 hence, u=( )/10 = By Newton s Backward Interpolation formula we have f(1966) = f(a) + u f(a) + 2 f(a) + 3 f(a) + 4 f(a) +... = 52 + (-1.5)* = = (-7) + (-10)

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