Practice Problems

The following problems are drawn from CSE 330 coursework. Each problem includes a complete worked solution. Try solving each one yourself before expanding the solution.

Problem 1: Form Types and Representable Ranges

Given $\beta = 2$ , $m = 3$ , and $-2 \le e \le 1$ , find the minimum, maximum, and total count of representable positive numbers for each of the three forms below.

(a) Standard Form

$x = \pm\,(0.d_1\,d_2\,d_3)_2 \times 2^e, \qquad d_1 \ne 0 \;\Rightarrow\; d_1 = 1$

Free digits: $d_2, d_3 \in \{0,1\}$ → $2^2 = 4$ mantissas
Exponents: $e \in \{-2,-1,0,1\}$ → 4 values


Minimum	$(0.100)_2 \times 2^{-2} = \frac{1}{2} \times \frac{1}{4} = \dfrac{1}{8}$
Maximum	$(0.111)_2 \times 2^{1} = \frac{7}{8} \times 2 = \dfrac{7}{4}$
Total (positive)	$2^2 \times 4 = 16$
Total (with sign)	$16 \times 2 = 32$

(b) Normalized Form

In this form the leading digit after the radix point is 1 and no other digit is fixed. All three digits $0.1d_1\,d_2\,d_3$ vary, giving a wider set of representable mantissas.

$x = \pm\,(0.1d_1\,d_2\,d_3)_2 \times 2^e$


Minimum	$(0.1000)_2 \times 2^{-2} = \frac{1}{2} \times \frac{1}{4} = \dfrac{1}{8}$
Maximum	$(0.1111)_2 \times 2^{1} = \frac{15}{16} \times 2 = \dfrac{15}{8}$
Total (positive)	$2^3 \times 4 = 32$
Total (with sign)	$32 \times 2 = 64$

(c) Denormalized Form

$x = \pm\,(1.d_1\,d_2\,d_3)_2 \times 2^e$

The leading 1 is explicit; all three digits $d_1, d_2, d_3 \in \{0,1\}$ are free → $2^3 = 8$ mantissas.

Equivalently in the 0. style: $(0.1\,d_1\,d_2\,d_3)_2 \times 2^{e+1}$ .


Minimum	$(1.000)_2 \times 2^{-2} = 1 \times \frac{1}{4} = \dfrac{1}{4}$
Maximum	$(1.111)_2 \times 2^{1} = \frac{15}{8} \times 2 = \dfrac{15}{4}$
Total (positive)	$8 \times 4 = 32$
Total (with sign)	$32 \times 2 = 64$

Problem 2: Converting Decimal to Standard Floating-Point

Convert $(6.25)_{10}$ to Standard floating-point representation with $\beta = 2$ , $m = 3$ , and $-1 \le e \le 3$ .

Step 1: Convert $6.25$ to binary:

$6.25 = 4 + 2 + 0.25 = (110.01)_2$

Step 2: Write in F-format (shift radix point left so the integer part becomes 0):

$110.01_2 = 0.11001 \times 2^3$

Step 3: Round to $m = 3$ mantissa digits.

The mantissa so far: $0.11001$ . We keep $d_1 d_2 d_3 = 110$ . The next digit is $d_4 = 0$ , so we truncate:

$fl(6.25) = (0.110)_2 \times 2^3$

Step 4: Verify:

$(0.110)_2 \times 2^3 = \left(\frac{1}{2}+\frac{1}{4}\right) \times 8 = \frac{3}{4} \times 8 = 6.0$

Rounding Error:

$\text{R.E.} = |6.25 - 6.0| = 0.25$

Relative Error:

$\delta = \frac{|6.25 - 6.0|}{6.25} = 0.04$

Problem 3: Floating-Point Multiplication

Given $x = \dfrac{3}{8}$ and $y = \dfrac{5}{8}$ , find $fl(x \times y)$ using $m = 4$ (Convention 1, $\beta = 2$ ), and compute the rounding error.

Step 1: Convert to Standard Form:

$x = \frac{3}{8} = (0.375)_{10} = (0.011)_2 = (0.11)_2 \times 2^{-1}$

$y = \frac{5}{8} = (0.625)_{10} = (0.101)_2 \times 2^0$

As $m = 4$ , so there is no need to round. Therefore, $fl(x) = x = (0.11)_2 \times 2^{-1} = \frac{3}{8}$ and $fl(y) = y = (0.101)_2 \times 2^0 = \frac{5}{8}$ .

Step 2: Compute the exact product:

$x \times y = fl(x) \times fl(y) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64} = 0.234375$

Step 3: Convert to binary F-format:

$\frac{15}{64} = (0.001111)_2 = (0.1111)_2 \times 2^{-2}$

The mantissa $0.1111$ has exactly $m = 4$ digits. Therefore, no rounding needed.

$fl(x \times y) = (0.1111)_2 \times 2^{-2} = \frac{15}{16} \times \frac{1}{4} = \frac{15}{64}$

Rounding Error:

$\text{R.E.} = \left|\frac{15}{64} - \frac{15}{64}\right| = 0$

No rounding error in this case since the exact result fits within $m = 4$ digits.

Problem 4: Finding Machine Epsilon

Given $\beta = 2$ , $m = 4$ , $-100 \le e \le 100$ , find the machine epsilon $\xi_M$ using the IEEE Normalized Form.

For the Normalized Form:

$\xi_M = \frac{1}{2}\,\beta^{-m}$

Substituting $\beta = 2$ , $m = 4$ :

$\xi_M = \frac{1}{2} \times 2^{-4} = \frac{1}{2} \times \frac{1}{16} = \boxed{\frac{1}{32}}$

Problem 5 — Quadratic Roots and Loss of Significance

Compute the roots of $x^2 - 12x + 5 = 0$ keeping four significant figures throughout. Show the Loss of Significance and apply the workaround.

Exact Roots (high precision)

$x = \frac{12 \pm \sqrt{144 - 20}}{2} = \frac{12 \pm \sqrt{124}}{2} = 6 \pm \sqrt{31}$

$x_1 = 6 + \sqrt{31} \approx 11.5678 \qquad x_2 = 6 - \sqrt{31} \approx 0.43218$

With 4 Significant Figures

A toy computer approximates $\sqrt{31} \approx 5.568$ (4 s.f., rounding $5.5677...$ ).

$x_1 = 6 + 5.568 = 11.568 \;\to\; \mathbf{11.57} \qquad\text{✓ close to } 11.5678$

$x_2 = 6 - 5.568 = 0.4320 \;\to\; \mathbf{0.4320} \qquad\text{✗ actual: } 0.43218$

The subtraction $6 - 5.568$ causes cancellation (1 leading digit lost), giving 3–4 digits of accuracy instead of 4.
(The error grows worse with more extreme cases(see Problem 2 analogue with 56x).)

Workaround Using Vieta’s Formulas

For $x^2 - 12x + 5 = 0$ :

$x_1 + x_2 = 12, \qquad x_1 \cdot x_2 = 5$

Step 1 — compute the stable root first (addition, no cancellation):

$x_1 = 6 + 5.568 = 11.57$

Step 2 — recover $x_2$ from the product:

$x_1 \cdot x_2 = 5 \;\Longrightarrow\; x_2 = \frac{5}{x_1} = \frac{5}{11.57} = \mathbf{0.4322}$

This matches the accurate value $0.43218$ to 4 significant figures. The division completely avoids the catastrophic cancellation.