Solutions to the midterms
===========================================================================

 Any answer that was mostly correct got at least 1/2 the marks, and
 any answer that was mostly wrong should got at most 1/2 the marks.  Do
 No .5 marks were given; instead, I gave you a benefit of doubt
 and assumed that you know what you are doing.



QUESTION 1 

Part A

 The largest exponent that can be represented is +99  and the
 largest mantissa is .999999 (since it must first be normalized).  
 Hence, the largest positive floating-point number in this system is
 0.999999 x 10^{+99}; in unnormalized form with no significant bit after 
 the dot, it is 999999^{93}.


Part B

 The largest round-off error occurs when the exponent has the largest value
 (99) and the number to be represented is exactly halfway between two
 floating-point numbers.  For example, the real number
 2.32425 x 10^{98} is represented using either of the floating-point
 numbers .23242 x 10^{99} or .23243 x 10^{99}, with a round-off
 error in both cases of .0000005 x 10^{99} = 5 x 10^{92}.

Part C

 After normalizing, we do: 
   0.0353472 x 10^{+2} (+) 0.35347222 x 10^{+2} = 
                                     0.38881922 x 10^{+2} , and   
   0.38881922 x10^{+2}  (+) 0.00000011353472 x 10^{+2} = 
                  0.38881922 x 10^{+2} = 0.388819 x 10^{+2} .
 Last number doesn't affect the result. In general, add the smallest numbers  
 first. 

 The result of the subtraction is 0.00000022 x 10^{+2} = 0.000000 x 10^{+2} 
 which is .22 x 10^{-4}. This number is representable !. So no problem with 
 it. A couple of other possible answers were accepted: In general, don't 
 subtract two too close numbers; or double the precision; extend the length 
 of the mantissa. These answers are not quite correct (why ?), but I 
 accepted them.



QUESTION 2

Part A

   [a_0,b_0] = [0,1].

   m_0 = (a_0+b_0)/2 = 1/2 and
   f(m_0) = 16^{1/2}-10 = sqrt{16}-10 = 4-10 = -6 < 0 so
   [a_1,b_1] = [m_0,b_0] = [1/2,1].

   m_1 = (a_1+b_1)/2 = 3/4 and
   f(m_1) = 16^{3/4}-10 = (sqrt[4]{16})^3-10 = 2^3-10 = -2 < 0 so
   [a_2,b_2] = [m_1,b_1] = [3/4,1].


Part B

 From part (b), [a_2,b_2] = [3/4,1], so the bisection method will return
 the approximate root (a_2+b_2)/2 = 7/8 = 0.875.  The bound on the error 
 is half the size of the interval [a_2,b_2] which is (b_2-a_2)/2 = 1/8 = 0.125.

 Note: these values may be different depending on the initial values guessed in
 Part A. Important is to show your work.


Part C

 The error at step i is no more than {1/2}/{2^i} = 1/2^{i+1}.  
 To guarantee that the error at step i is less than 0.005, we pick i 
 such that 1/2^{i+1} < 0.005, i.e., 1/2^i < 0.01.  By solving this 
 for i, we get i > log_2(100) > 6, which is almost 7. 




QUESTION 3

Part A

float secant(float x0, float x1, int it, float tol) {

    float f0 = f(x0);
    float f1 = f(x1);
    int k = 1;
    float x2;

    do {
       /* Compute the secant and update the approximation. */
     x2 = x1 - ((x1-x0)/(f1-f0))*f1; 
     x0 = x1;
     f0 = f1;
     x1 = x2;
     f1 = f(x2);
     k = k+1; 
    } while ((fabs(x2 - x1) >= tol) && (k < it)) 
    /* Error = |x2-x1|, so stop when |x2-x1| < tol. */
    return (x2);

 } /* END secant() */




QUESTION 4

Part A

  1. 14
  2. 14
  3. 54
  4. false


Part B 

   int (*a)[13]  -- Pointer to an array holding 13 integers


   int *a[13] -- Array of 13 pointers to integers 


   char (*f)() -- pointer to a function returning chars


   (*p)++ -- value of item at which p is pointing is first evaluated 
             and then incremented


Part C 

   p->b = ' ';    -- illegal
  
   p->e[3] = 10;  -- legal

   (*p).d.a = '*';  -- legal
   
   p->d->c = 20;  -- illegal


Part D

  %-8d  456 .....    

  %08d  00000456