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Nataliya [291]
3 years ago
15

Your baby brother has found the hammer and is eagerly eyeing one of the boxes. Describe and analyze an algorithm to determine if

it is possible to retrieve all the keys without smashing any box except the one your brother has chosen.
Computers and Technology
1 answer:
erastovalidia [21]3 years ago
7 0

Answer:

Question is incomplete.

Assuming the below info to complete the question

You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.

Detailed answer is written in explanation field.

Explanation:

We have to find the reachability using the directed graph G = (V, E)

In this V are boxes are considered to be non empty and it may contain key.

Edges E will have keys .

G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.

Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.

Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.

If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.

After first search from B we can start marking all other vertex of graph G.

So algorithm will be O ( V +E ) = O (n+m) time.

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Consider the following statements: Statement A: In Duplex transmission, either node can transmit while the other node can receiv
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Answer:

<u>d. Statement A is true and Statement B is false</u>

Explanation:

Indeed, when using duplex transmission either node can transmit while the other node can receive data from the network. Also, in half-duplex transmission, both the nodes can transmit as well as receive data.

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Levi's an experienced marketing strategist for a software company that sells a learning platform to public schools. He developed
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Answer:

B. He segments data using his company’s CRM dashboards, giving his organization access to data that powers their decision-making.

E. He encourages learning from failure, which is necessary for testing the possibilities and for learning what does not work.

Explanation:

He seems to have realized that learning from mistake is important, and that can be done through the research of the company's past. Also, he has also learnt to analyze the data as well, as he is able to constantly evolve company.s marketing strategies to fit to school's training needs. And this is impossible without analysis and research. It also looks like that he is a good learner, and loves exploring new things. And he must be using analytic software like Tableau OR Power BI, and he might be using Machine learning as well, and definitely the latest. And he is definitely a good manager.

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Computer Networks - Queues
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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 T_w 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  = \dfrac{1000}{64*1000}

= 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

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