Module: TM112
TMA: 01
Page 1 of 14
,Question 1:
a
i) Unsigned binary representation may have been chosen for the
purpose that is can only contain magnitude of number, thus only
containing positive numbers. It is stated that the minimum
number in a group is zero and therefore the use of 32-bit
unsigned binary meets the requirements.
ii) The given range is from 0 to 32 – bit unsigned binary, so the
largest number that can belong to a group can be found using
the formula
( 2n −1 ),
32
2 −1
232−1 is the expression using the power of 2 to find the largest
group number.
iii) Using the expression from answer 1) ii), 232−1=4294967295, we can
see that the group has reached capacity, if a new member
intends to join then a problem of overflow may occur as the value
of the number of group members is too large for the storage
space.
b
i) 0.1011100 02 converted to decimal place;
( 0 × 20 ) + ( 1 ×2−1 ) + ( 0 × 2−2 ) + ( 1 ×2−3 ) + ( 1 ×2−4 ) + ( 1 ×2−5 )
+ ( 0 × 2−6 ) + ( 0 × 2−7 ) + ( 0 × 2−8 )
¿ 1 ∕ 21 + 1 ∕ 23 + 1 ∕ 24 + 1 ∕ 25
¿ 0.5+0.125+ 0.0625+0.03125
¿ 0.71875
10
Page 2 of 14
, ii) 0.1011100 12 converted to decimal place.
( 0 × 20 ) + ( 1 ×2−1 ) + ( 0 × 2−2 ) + ( 1 ×2−3 ) + ( 1 ×2−4 ) + ( 1 ×2−5 )
+ ( 0 × 2−6 ) + ( 0 × 2−7 ) + ( 1 ×2−8 )
¿ 1 ∕ 21 + 1 ∕ 23 + 1 ∕ 24 + 1 ∕ 25+ 1 ∕ 28
¿ 0.5+0.125+ 0.0625+0.03125+ 0.00390625
¿0.72265625
10
iii) The binary number from parts b)i) and b)ii) closest to 0.72:
Part i) : ( 0.71875 )10
¿ ( 0.72 )− ( 0.71875 )
¿ 0.00125
Part ii) : ( 0.72265625 )10
¿ ( 0.72 )− ( 0.72265625 )
¿ 0.00265625
Thus, the closest number to value 0.72 is from part a)i) at 0.00125.
iv) If a customer changes $100,000 into GBP and at 0.1011100 12 (clerk
error) instead of 0.1011100 02:
At rate 0.1011100 02 : 100 000 x 0.71875 = £718,750
at rate 0.1011100 12 : 100 000 x 0.72265625 = £72265.625
72265 – 718750 = £390.625
The total the client would receive with the clerical error is £390.63 (to
nearest penny).
Page 3 of 14
TMA: 01
Page 1 of 14
,Question 1:
a
i) Unsigned binary representation may have been chosen for the
purpose that is can only contain magnitude of number, thus only
containing positive numbers. It is stated that the minimum
number in a group is zero and therefore the use of 32-bit
unsigned binary meets the requirements.
ii) The given range is from 0 to 32 – bit unsigned binary, so the
largest number that can belong to a group can be found using
the formula
( 2n −1 ),
32
2 −1
232−1 is the expression using the power of 2 to find the largest
group number.
iii) Using the expression from answer 1) ii), 232−1=4294967295, we can
see that the group has reached capacity, if a new member
intends to join then a problem of overflow may occur as the value
of the number of group members is too large for the storage
space.
b
i) 0.1011100 02 converted to decimal place;
( 0 × 20 ) + ( 1 ×2−1 ) + ( 0 × 2−2 ) + ( 1 ×2−3 ) + ( 1 ×2−4 ) + ( 1 ×2−5 )
+ ( 0 × 2−6 ) + ( 0 × 2−7 ) + ( 0 × 2−8 )
¿ 1 ∕ 21 + 1 ∕ 23 + 1 ∕ 24 + 1 ∕ 25
¿ 0.5+0.125+ 0.0625+0.03125
¿ 0.71875
10
Page 2 of 14
, ii) 0.1011100 12 converted to decimal place.
( 0 × 20 ) + ( 1 ×2−1 ) + ( 0 × 2−2 ) + ( 1 ×2−3 ) + ( 1 ×2−4 ) + ( 1 ×2−5 )
+ ( 0 × 2−6 ) + ( 0 × 2−7 ) + ( 1 ×2−8 )
¿ 1 ∕ 21 + 1 ∕ 23 + 1 ∕ 24 + 1 ∕ 25+ 1 ∕ 28
¿ 0.5+0.125+ 0.0625+0.03125+ 0.00390625
¿0.72265625
10
iii) The binary number from parts b)i) and b)ii) closest to 0.72:
Part i) : ( 0.71875 )10
¿ ( 0.72 )− ( 0.71875 )
¿ 0.00125
Part ii) : ( 0.72265625 )10
¿ ( 0.72 )− ( 0.72265625 )
¿ 0.00265625
Thus, the closest number to value 0.72 is from part a)i) at 0.00125.
iv) If a customer changes $100,000 into GBP and at 0.1011100 12 (clerk
error) instead of 0.1011100 02:
At rate 0.1011100 02 : 100 000 x 0.71875 = £718,750
at rate 0.1011100 12 : 100 000 x 0.72265625 = £72265.625
72265 – 718750 = £390.625
The total the client would receive with the clerical error is £390.63 (to
nearest penny).
Page 3 of 14
