<h3><u>CSMA/CD Protocol:
</u></h3>
Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.
But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.
There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.
<u>Example:
</u>
, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.
12:00 AM A will see collisions
Pocket Size to detect the collision.
![\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26T_%7Bt%7D%20%5Cgeq%202%20T_%7BP%7D%5C%5C%26%5Cfrac%7BL%7D%7BB%7D%20%5Cgeq%202%20T_%7BP%7D%5C%5C%26L%20%5Cgeq%202%20%5Ctimes%20T_%7BP%7D%20%5Ctimes%20B%5Cend%7Baligned%7D)
CSMA/CD is widely used in Ethernet.
<u>Efficiency of CSMA/CD:</u>
- In the previous example we have seen that in worst case
time require to detect a collision.
- There could be many collisions may happen before a successful completion of transmission of a packet.
We are given number of collisions (contentions slots)=4.
Distance = 1km = 1000m
![\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Speed%20%7D%3D2%20%5Ctimes%2010%5E%7B8%7D%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bsec%7D%5C%5C%20%26T_%7BP%7D%3D%5Cfrac%7B1000%7D%7B2%20%5Ctimes%2010%5E%7B8%7D%7D%3D%280.5%29%20%5Ctimes%2010%5E%7B-5%7D%3D5%20%5Ctimes%2010%5E%7B-6%7D%5C%5C%20%26T_%7Bt%7D%3D5%20%5Cmu%20%5Cmathrm%7Bsec%7D%5Cend%7Baligned%7D)
Answer:
Air mass sensors is the right answer i think
Explanation:
Answer:
Amount of concrete need to make slab = 1,500 feet³
Explanation:
Given:
Length of slab = 50 feet
Width of slab = 30 feet
Height of slab = 1 feet
Find:
Amount of concrete need to make slab
Computation;
Amount of concrete need to make slab = Volume of cuboid
Volume of cuboid = (l)(b)(h)
Amount of concrete need to make slab = (50)(30)(1)
Amount of concrete need to make slab = 1,500 feet³