Essay TT257 Cisco Networking (TT257)

Module: TM257
Assignment: TMA 01

,QUESTION 1
Grades from Cisco NetAcad gradebook;

M1>3: 98.1

M4>7: 86

M8>10: 61.2

 GradeAverage is (98.1 + 86 + 61.2) / 3 = 81.76 to 2d.p.
 QuestionScore is (0.2 x 81.76) = 16.35 = 17 (rounded up).



QUESTION 2
a. Converting binary MAC address to hexadecimal form, the most
memorable way is to split the binary up into 4 bits then attach
the placeholder value then calculate, as follows; Dec He
x

23 22 21 20 20 21 22 23
9 9

8 4 2 1 8 4 2 1 10 A

8+4 = 12 1 1 0 0 0 0 0 1 =1 11 B

4+1 = 5 0 1 0 1 1 1 0 0 8+4 =12 12 C

=0 0 0 0 0 1 1 0 1 8+4+1=13 13 D


=1 0 0 0 1 0 1 0 1 4+1=5 14 E


8+4=12 1 1 0 0 0 0 0 0 =0 15 F


4+1=5 0 1 0 1 0 1 1 1 4+2+1=7




Binary 1100-0001 0101-1100 0000-1101 0001-0101 1100-0000 0101-0111

Decimal 12 -1 5 - 12 0 -13 1–5 12 - 0 5-7

Hexadecimal C1 5C 0D 15 C0 57




Hexadecimal MAC address – C1-5C-0D-15-C0-57

2

, b. Converting binary IP address
01100011.01000010.00100001.00000001 to dotted decimal;


Convert first octet;
Positional Value 128 64 32 16 8 4 2 1
Binary Number 0 1 1 0 0 0 1 1
Add Together 0 +64 +32 +0 +0 +0 +2 +1
Result 99


Convert second octet;
Positional Value 128 64 32 16 8 4 2 1
Binary Number 0 1 0 0 0 0 1 0
Add Together 0 +64 +0 +0 +0 +0 +2 +0
Result 66


Convert third octet;
Positional Value 128 64 32 16 8 4 2 1
Binary Number 0 0 1 0 0 0 0 1
Add Together 0 +0 +32 +0 +0 +0 +0 +1
Result 33


Convert fourth octet;
Positional Value 128 64 32 16 8 4 2 1
Binary Number 0 0 0 0 0 0 0 1
Add Together 0 +0 +0 +0 +0 +0 +0 +1
Result 1


Converted decimal address: 97.64.33.1.




2