The following problem is used to demonstrate the limitation of FLP in computers.

A computer uses 8-bit for FLP (1bit for sign, 3 bit for exponent with excess-3 rep., rest for fraction). Assume Excess-3 is used to represent exponents, and 000 and 111 in exponent field are reserved.

EXCESS_3

Decimal	0	1	2	3	4	5	6	7
Unsigned	000	001	010	011	100	101	110	111
Excess-3	reserved	-2	-1	0	1	2	3	reserved

Only 12*16 + 31 (+ 3 special) = 223 numbers can be represented

	REPRESENTATION			Decimal Value
[Special Exponent 000] Denormal numbers	0	000	0000	+0
	0	000	0001	.0001 = 0.0625
	0	000	0010	.0010 = 0.125
	0	000	1111	0.1111 = 0.5+0.25+0.125+0.625 = 0.9375
			16 combinations
	1	000	0000	-0
	1	000	0001	-0.0001 = - 0.0625
	1	000	0010	-0.0010 = -0.125
	1	000	1111	-0.1111 = 0.5+0.25+0.125+0.625 = -0.9375
			16 combinations
[Special Exponent 111]	0	111	0000	+ ?
	1	111	0000	- ?
	X	111	X 1 X	NaN
				3 numbers
NORMALIZED POSITIVE FLPs
min +ve fra	0	001	0000	1 * 2 -2 = 0.25
Next	0	001	0001	(1+ 1/16) * 2 -2 = 0.25+1/64
	0	001	0010	(1+ 1/8) * 2 -2 = 0.25+ 2/64
	0	001	0011	(1+ 1/8+ 1/16) * 2 -2 = 0.25+3/64
	0	001	1111	(1+ + 1/4+1/8+ 1/16) * 2 -2 = 0.25+15/64
			16 combinations
	0	010	0000	1 * 2 -1 = 0.5 [ = 0.25+ 16/64]
	0	010	0001	(1+ 1/16) * 2 -1 = 0.5+1/32
	0	010	0010	(1+ 1/8) * 2 -1 = 0.5+ 2/32
	0	010	0011	(1+ 1/8+ 1/16) * 2 -1 = 0.5+3/32
	0	010	1111	(1+ + 1/4+1/8+ 1/16) * 2 -1 = 0.5 +15/32
			16 combinations
	0	011	0000	1 * 2 0 = 1 [ =0.5+ 16/32]
	0	011	0001	(1+ 1/16) * 2 0 = 1 + 1/16
	0	011	0010	(1+ 1/8) * 2 0 = 1 + 2 /16
	0	011	0011	(1+ 1/8+ 1/16) * 2 0 = 1 + 3/16
	0	011	1111	(1+ + 1/4+1/8+ 1/16) * 2 -1 = 1 +15/16
			16 combinations
	0	100	0000	1 * 2 1 = 2 [ =1 + 16/16]
	0	100	0001	(1+ 1/16) * 2 1 = 2 + 1/8
	0	100	0010	(1+ 1/8) * 2 1 = 2 + 2/8
	0	100	0011	(1+ 1/8+ 1/16) * 2 1 = 2 + 3/8
	0	100	1111	(1+ + 1/4+1/8+ 1/16) * 2 1 = 2 + 15/8
			16 combinations
	0	101	0000	1 * 2 2 = 4 [= 2 + 16/8]
	0	101	0001	(1+ 1/16) * 2 2 = 4 +
	0	101	0010	(1+ 1/8) * 2 2 = 4 + 2/4
	0	101	0011	(1+ 1/8+ 1/16) * 2 2 = 4 +
	0	101	1111	(1+ + 1/4+1/8+ 1/16) * 2 2 = 4 + 15/4
			16 combinations
	0	110	0000	1 * 2 3 = 8 [ 4 + 16/16]
	0	110	0001	(1+ 1/16) * 2 3 = 8 +
	0	110	0010	(1+ 1/8) * 2 3 = 8 + 2/2 = 9
	0	110	0011	(1+ 1/8+ 1/16) * 2 3 = 8 + 3/2 = 9.5
nextLargest	0	110	1110	(1+ + 1/4+1/8+ 0) * 2 3 = 8+ 14/2 =15
Largest	0	110	1111	(1+ + 1/4+1/8+ 1/16) * 2 3 = 8+ 15/2 =15.5
			16 combinations
	Positive numbers- 6 * 16
NORMALIZED NEGATIVE FLPs
min ve fra	1	001	0000	-1 * 2 -2 = - 0.25
next	1	001	0001	-(1+ 1/16) * 2 -2 = - (0.25 + 1/64)
nextLargest	1	110	1110	-(1+ + 1/4+1/8+ 0) * 2 3 = 8+ 14/2 =15
Largest	1	110	1111	-(1+ + 1/4+1/8+ 1/16) * 2 3 = 8+ 15/2 =15.5
	Negative - 6 * 16 numbers

Observation: Big gap (0.25=1/4) between 0 and next numbers(+ve and ve)

The gap (precision) is 1/64 in between 0.25 and 0.5 (16 numbers)

The precision is 1/32 from 0.5 to 1 (16 numbers)

The precision continuously decreases, and finally it will be 0.5 in between 8 and 15.5.

	-1		-0.5	-0.25	0	0.25	0.5		1
	16 numbers		16 numbers	No normalized numbers. Denormalized numbers are used		16 numbers	16 numbers

e. What is 11000000?

1

100

0000

-1 .0 * 2 1 = -2

f. How 1 will be represented?

0

011

0000

1.0 * 2 0 = 1

g. What is the smallest positive number in this representation? (without denormal)

min +ve

0

001

0000

1 * 2 -2 = 0.25

h. What is the maximum number?

Max

0

110

1111

1 .1111 * 2 3 = - 1.9375 * 8 = 15.5

i. What is the smallest negative number (in absolute value)? (without denormal)

max -ve

0

001

0000

1 * 2 -2 = 0.25

j. What is the minimum number?

Min

1

110

1111

-1 .1111 * 2 3 = - 1.9375 * 8 = -15.5

k. How 0.2 will be represented?

0.2 = 0.00110011 = 1.110011 * 2 ^-3 ; cannot be represented- need to be rounded

Closest possible number is 0.01 = 0.25 (similar to rounding 0.0099, = 0.01)

~0.2=0.25

0

001

0000

1 * 2 -2 = 0.25

l. How 0 will be represented.