Q1.  What is the difference between a unicast, multicast and broadcast address?
-	Unicast
o	LSB = 0
o	From one source to one destination – One-to-One
-	Multicast
o	LSB = 1
o	From one source to all possible destinations – One-to-All
-	Broadcast
o	From one source to multiple destinations stating an interest in receiving the traffic i.e. One-to-Many

Q2. Why is there no need for CSMA/CD on a full-duplex Ethernet LAN?
-	There are no collisions, because data is transmitted in both directions alongside each other.
-	Collisions can only occur in a shared medium, such as WiFi and older Ethernet technologies such as thinnet and 10/100bT networks, implemented with hubs instead of switches. Hubs are half-duplex.

Q3. Compare the data rates of Standard Ethernet, Fast Ethernet, Gigabit Ethernet and Ten-Gigabit Ethernet.
-	Standard Ethernet – 10 Mbps
-	Fast Ethernet – 100 Mbps
-	Gigabit Ethernet – 1,000 Mbps (1 Gbps)
-	Ten-Gigabit Ethernet – 10,000 Mbps (10 Gbps)


Q4. What are the common Standard Ethernet implementations?
-	10Base5
o	Thick coax
o	500m
-	10Base2
o	Thin coax
o	185m
-	10Base-T
o	2 UTP
o	100m
-	10Base-F
o	2 Fibre
o	2000m
Q5. What are the common Fast Ethernet implementations?
-	100Base-TX
o	STP
o	2 wires
o	100m
-	100Base-FX
o	Fibre
o	2 wires
o	100m
-	100Base-T4
o	UTP
o	4 wires
o	100m
Q6. What are the common Gigabit Ethernet implementations?
-	1000Base-SX
o	Fibre – short-wave
o	2 wires
o	550m
-	1000Base-LX
o	Fibre – long-wave
o	2 wires
o	5000m
-	1000Base-CX
o	STP
o	2 wires
o	25m
-	1000Base-T4
o	Cat 5 UTP
o	4 wires
o	100m
Q7. What are the common Ten-Gigabit Ethernet implementations?
-	10GBase-S
o	Multi-mode Fibre
o	2 wires
o	300m
-	10GBase-L
o	Single-mode Fibre
o	2 wires
o	10,000m
-	10GBase-E
o	Single-mode Fibre
o	2 wires
o	40,000m

Q13. What is the ratio of useful data to the entire packet for the smallest Ethernet frame? What is the ratio for the largest frame?
-	Smallest frame
o	The minimum length of an Ethernet frame is 64 Bytes. The useful data in that Ethernet frame is 46 Bytes.
o	46 / 64 = 0.7185 = 71.2%
-	Largest frame
o	The maximum frame length = 1518 Bytes
o	Maximum payload length = 1500 Bytes
o	1500 / 1518 = 0.9881 = 98.9%

Q14. Suppose the length of a 10Base5 cable is 2500 m. If the speed of propagation in a thick coaxial cable is 200,000,000 m/s, how long does it take for a bit to travel from the beginning to the end of the network? Assume there are 10 µs delay in the equipment.
-	____2500m_____ = 0.0125 ms = 12.5 µs
(200,000,000 m/s)
-	12.5 µs + 10 µs = 22.5 µs