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1 answer:
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Answer:
d. The system of water spray pipes did not have sufficient water supply
Explanation:
The Union Carbide India Ltd. was chemical factory situated at Bhopal that produces various pesticides, batteries, plastics, welding equipment, etc. The plant in Bhopal produces pesticides. In the year 1984, on the night of 2nd December, a devastating disaster occurred on the Bhopal plant which killed millions of people due to the leakage of the poisonous, methyl isocyanate. This disaster is known as Bhopal Gas tragedy.
Soon after the gas pipe exploded, the fire brigade started spraying water into the air to the cloud of the gases in the air but there was not sufficient amount of water in the water sprays and so it was not effective.
Thus the correct answer is option (d).
Answer:
the average arrival rate \lambda in units of packets/second is 15.24 kbps
the average number of packets w waiting to be serviced in the buffer is 762 bits
Explanation:
Given that:
A single channel with a capacity of 64 kbps.
Average packet waiting time in the buffer = 0.05 second
Average number of packets in residence = 1 packet
Average packet length r = 1000 bits
What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?
The Conservation of Time and Messages ;
E(R) = E(W) + ρ
r = w + ρ
Using Little law ;
r = λ × T_r
w = λ × T_w
r / λ = w / λ + ρ / λ
T_r =T_w + 1 / μ
T_r = T_w +T_s
where ;
ρ = utilisation fraction of time facility
r = mean number of item in the system waiting to be served
w = mean number of packet waiting to be served
λ = mean number of arrival per second
T_r =mean time an item spent in the system
T_w = mean waiting time
μ = traffic intensity
T_s = mean service time for each arrival
the average arrival rate \lambda in units of packets/second; we have the following.
First let's determine the serving time T_s
the serving time T_s
= 0.015625
now; the mean time an item spent in the system T_r = T_w +T_s
where;
T_w = 0.05 (i.e the average packet waiting time)
T_s = 0.015625
T_r = 0.05 + 0.015625
T_r = 0.065625
However; the mean number of arrival per second λ is;
r = λ × T_r
λ = r / T_r
λ = 1000 / 0.065625
λ = 15238.09524 bps
λ ≅ 15.24 kbps
Thus; the average arrival rate \lambda in units of packets/second is 15.24 kbps
b) Determine the average number of packets w waiting to be serviced in the buffer.
mean number of packets w waiting to be served is calculated using the formula
w = λ × T_w
where;
T_w = 0.05
w = 15238.09524 × 0.05
w = 761.904762
w ≅ 762 bits
Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits