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omeli [17]
3 years ago
13

An online conference room reservation system lets a user reserve rooms. The user is required to enter contact information and re

servation detail before saving the room reservation. Which of the following identifies the use cases?
a)
Reserve Room.

b)
Reserve Room, Enter Contact Information, Enter Reservation Detail.

c)
Reserve Room, Enter Contact Information and Reservation Detail.

d)
Reserve Room, Save Reservation.
Computers and Technology
1 answer:
lisov135 [29]3 years ago
7 0

Answer:b)Reserve Room, Enter Contact Information, Enter Reservation Detail

Explanation: The use case is described as actors , actors goal, usage of use cases, units of scope and initial point and terminating point of use cases .

The situation mentioned in the question will take place for the reservation of the room through the instances in the form like filling the contact details for the identification , filling out reservation information as well for getting the knowledge about the room requirement by the person.

These details help in reaching out the particular person if required, number of days he wants to book the room,any extra facility he would like to have etc. After these detail filling activity the room can be reserved successfully.Thus , the most appropriate option is option (b).

Other options are incorrect because room reservation cannot be done without any formality as mentioned in the question and entering contact detail and reservation detail one by one.

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Assume that passwords are selected from four-character combinations of 26 alphabeticcharacters. Assume that an adversary is able
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Answer:

for a) the time required is 228488.5 seconds= 63.469 hours

for b) the time required is 54 seconds

Explanation:

for a) since each combination is equally probable , then the number of possible combinations is

CT=Combinations = number of characters^length of password = 26⁴

then the number of combinations at time t will be the total , less the ones already tried:

Ct = CT - (n-1) , since n=α*t  → Ct=CT-α*t  

since each combination is equally probable , then the probability to succeed  

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but the probability of having a success in time t , means also not succeeding in the previous trials , then

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Pt = 1/ (CT- α*t +1) П[1-1/ (CT- α*k +1)] = 1/ (CT- α*t +1) П[(CT- α*k )] /(CT- α*k +1)]  

since α=1 ,  

Pt = 1/ (CT- t +1) П[(CT- k )] /(CT- k +1)] = 1/ (CT- t +1)  * [CT- (t-1) ]/CT  =  1/CT

then the expected value of the random variable t= time to discover the correct password is

E(t) = ∑ t* Pt = ∑ t *1/CT , for t=1 until t=CT/α =CT

E(t) = ∑ t *(1/CT) = (1/CT) ∑ t = (1/CT) * CT*(CT+1)/2 = (CT+1)/2

therefore

E(t) = (CT+1)/2 = (26⁴ +1)/2 = 228488.5 seconds = 63.469 hours

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time required = time to find character 1 + time to find character 2 +time to find character 3 +time to find character 4 = 4*time to find character

since the time to find a character is the same case as before but with CT2=Combinations = 26 ,then

t= 4*tc

E(t) = 4*E(tc) = 4*(CT2+1)/2 = 4*(26+1)/2 = 54 seconds

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