The maximum throughput for pure ALOHA is________per cent. A) 12.2 B) 18.4 C) 36.8 D) none of the above

The maximum throughput for pure ALOHA is________per cent.
A)
12.2
B)
18.4
C)
36.8
D)
none of the above
by

1 Answer

Answer :

18.4
by

Related questions

A pure ALOHA Network transmits 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is ______. a. 0.384 b. 0.184 c. 0.286 d. 0.58

Description : A pure ALOHA Network transmits 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is ______. a. 0.384 b. 0.184 c. 0.286 d. 0.58

Last Answer : b. 0.184

A pure ALOHA Network transmit 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is .............. (A) 0.384 (B) 0.184 (C) 0.286 (D) 0.586

Description : A pure ALOHA Network transmit 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is .............. (A) 0.384 (B) 0.184 (C) 0.286 (D) 0.586

Last Answer : (B) 0.184

For pure ALOHA, the maximum channel utilization is a. 100% b. 50% c. 36% d. 18%

Description : For pure ALOHA, the maximum channel utilization is a. 100% b. 50% c. 36% d. 18%

Last Answer : d. 18%

In__________, eachstation is forced to send only at the beginning of the time slot. A) pure ALOHA B) slotted ALOHA C) both (a) and (b) D) neither (a) nor (b)

Description : In__________, eachstation is forced to send only at the beginning of the time slot. A) pure ALOHA B) slotted ALOHA C) both (a) and (b) D) neither (a) nor (b)

Last Answer : slotted ALOHA

In pure ALOHA, the vulnerable time is ______the frame transmissiontime. A) the sameas B) two times C) three times D) none of the above

Description : In pure ALOHA, the vulnerable time is ______the frame transmissiontime. A) the sameas B) two times C) three times D) none of the above

Last Answer : two times

A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? a. 1 Mbps b. 2 Mbps c. 10 Mbps d. 12 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? a. 1 Mbps b. 2 Mbps c. 10 Mbps d. 12 Mbps

Last Answer : b. 2 Mbps

A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network? a. 2 Mbps b. 60 Mbps c. 120 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network? a. 2 Mbps b. 60 Mbps c. 120 Mbps

Last Answer : a. 2 Mbps

A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is __________ bytes/second. a. 2500 b. 2000 c. 1500 d. 500

Description : A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of ... is lost, the sender throughput is __________ bytes/second. a. 2500 b. 2000 c. 1500 d. 500

Last Answer : a. 2500

In IEEE 802.11, the access method used in the DCF sublayer is _________. A) ALOHA B) CSMA/CA C) CSMA/CD D) none of the above

Description : In IEEE 802.11, the access method used in the DCF sublayer is _________. A) ALOHA B) CSMA/CA C) CSMA/CD D) none of the above

Last Answer : CSMA/CA

Inslotted ALOHA, the vulnerable time is ______ the frame transmissiontime. A) the sameas B) two times C) three times D) none of the above

Description : Inslotted ALOHA, the vulnerable time is ______ the frame transmissiontime. A) the sameas B) two times C) three times D) none of the above

Last Answer : the sameas

A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be a. 4.25 b. 3.5 c. 350 d. 450

Description : A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be a. 4.25 b. 3.5 c. 350 d. 450

Last Answer : b. 3.5

The Aloha protocol is an OSI _______ protocol for LAN networks with broadcast topology A. layer 1 B. layer 4 C. layer 2 D. layer 3

Description : The Aloha protocol is an OSI _______ protocol for LAN networks with broadcast topology A. layer 1 B. layer 4 C. layer 2 D. layer 3

Last Answer : C. layer 2

With Slotted Aloha, a …………. sends out small clock tick packets to the outlying stations. A. distributed clock B. synchronized clock C. centralized clock D. Digital clock

Description : With Slotted Aloha, a …………. sends out small clock tick packets to the outlying stations. A. distributed clock B. synchronized clock C. centralized clock D. Digital clock

Last Answer : C. centralized clock

When propagationspeed is multiplied by propagation time, we get the ________. A) throughput B) wavelength of thesignal C) distortionfactor D) distancea signal or bit has traveled

Description : When propagationspeed is multiplied by propagation time, we get the ________. A) throughput B) wavelength of thesignal C) distortionfactor D) distancea signal or bit has traveled

Last Answer : distancea signal or bit has traveled

While transferring files between server and client, if the transmission rates along the path is 10Mbps, 20Mbps, 30Mbps, 40Mbps. The throughput is usually a. 20Mbps b. 10Mbps c. 40Mbps d. 50Mbps

Description : While transferring files between server and client, if the transmission rates along the path is 10Mbps, 20Mbps, 30Mbps, 40Mbps. The throughput is usually a. 20Mbps b. 10Mbps c. 40Mbps d. 50Mbps

Last Answer : b. 10Mbps

In a network, when the load is below the capacity of the network, the throughput ______________ A. increases sharply B. increases proportionally with the load C. declines sharply D. declines proportionally with the load

Description : In a network, when the load is below the capacity of the network, the throughput ______________ A. increases sharply B. increases proportionally with the load C. declines sharply D. declines proportionally with the load

Last Answer : B. increases proportionally with the load

In a network, after the load reaches the capacity,throughput _______. A. increases sharply B. increases proportionally with the load C. declines sharply D. declines proportionally with the load

Description : In a network, after the load reaches the capacity,throughput _______. A. increases sharply B. increases proportionally with the load C. declines sharply D. declines proportionally with the load

Last Answer : C. declines sharply

A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? (A) 1 Mbps (B) 2 Mbps (C) 10 Mbps (D) 12 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? (A) 1 Mbps (B) 2 Mbps (C) 10 Mbps (D) 12 Mbps

Last Answer : (B) 2 Mbps

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Description : 5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Last Answer : Each term in the series is the addition of successive prime numbers. Like Prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on. So first term is addition of (2 + 3). Ie 5, then second term is (3 + 5) ie ... +17)ie. 30, then (17+19)ie.36, then (19+23)ie., 42, then (23+29)i.e. 52. Answer: a)

In the sequence of numbers 8, 24, 12, X, 18, 54 the missing number X is : (A) 26 (B) 24 (C) 36 (D) 32

Description : In the sequence of numbers 8, 24, 12, X, 18, 54 the missing number X is : (A) 26 (B) 24 (C) 36 (D) 32

Last Answer : Answer: C In the sequence of numbers 8, 24, 12, X, 18, 54 8*3= 24, 12*3= 36, 18*3= 54, i.e multiple of 3.

Insert the missing number: 8 24 12 ? 18 54 (A) 26 (B) 24 (C) 36 (D) 32

Description : Insert the missing number: 8 24 12 ? 18 54 (A) 26 (B) 24 (C) 36 (D) 32

Last Answer : Answer: C Explanation: Numbers are alternatively multiplied by 3 and divided by 2. So, the next number = 3 x 12 = 36, 36 ÷ 2 = 18

A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ? (A) 2 Mbps (B) 60 Mbps (C) 120 Mbps (D) 10 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ? (A) 2 Mbps (B) 60 Mbps (C) 120 Mbps (D) 10 Mbps

Last Answer : (A) 2 Mbps Explanation: In data transmission, throughput is the amount of data moved successfully from one place to another in a given period of time, and typically measured in bits per second (bps), ... second (Mbps) or gigabits per second (Gbps). Here, Throughput = 15000 x 8000/60 = 2 Mbps 

Aloha and ATA Airlines cease operations; Are these signs that the US has hit a recession?

Description : Aloha and ATA Airlines cease operations; Are these signs that the US has hit a recession?

Last Answer : The powers at be don’t want to admit there is a recession, but yes along with other things these are signs the terrorist are getting what they wanted. To destroy the American economy.

Gunveen, Marshy and Aloha are three developers working on an exciting new app, and the launch day is just a day away. Gunveen creates -Technology

Description : Gunveen, Marshy and Aloha are three developers working on an exciting new app, and the launch day is just a day away. Gunveen creates -Technology

Last Answer : The team members are not working in a systematic way and they are not saving the versions of their work. Changes made in one part of the software can be incompatible with those made by another ... trusted until it's tested. So testing and development proceed together until a new version is ready.

LAN? a) CDMA b) CSMA/CA c) ALOHA d) CSMA/CD

Description : LAN? a) CDMA b) CSMA/CA c) ALOHA d) CSMA/CD

Last Answer : b) CSMA/CA

Which multiple access technique is used by IEEE 802.11 standard for wireless LAN? a. ALOHA b. CSMA/CA c. CDMA d. none of the mentioned

Description : Which multiple access technique is used by IEEE 802.11 standard for wireless LAN? a. ALOHA b. CSMA/CA c. CDMA d. none of the mentioned

Last Answer : b. CSMA/CA

You are measuring a voice channel at a -4 dB test point level, the meter reads -73 dBm (pure test tone) convert the reading in dBmCO. A. 16 B. 18 C. 22 D. 12

Description : You are measuring a voice channel at a -4 dB test point level, the meter reads -73 dBm (pure test tone) convert the reading in dBmCO. A. 16 B. 18 C. 22 D. 12

Last Answer : C. 22

what is 1/3 +1/12 + 1/18 + 1/36?

Description : what is 1/3 +1/12 + 1/18 + 1/36?

Last Answer : 18/36=1/2 because you convert the denominator which gives you 12/36+3/26+2/36+1/36=12+3+2+1/36= 18/36=1/2

In the arrangement shown below, two trucks A and B are held in position with spring compressed. When they are released and the spring has fallen away, A is found to be moving to the left at 2 m/s. The energy stored in the compressed spring was (a) 12 J (b) 24 J (c) 18 J (d) 36 J

Description : In the arrangement shown below, two trucks A and B are held in position with spring compressed. When they are released and the spring has fallen away, A is found to be moving to the left at 2 m/s. The energy stored in the compressed spring was (a) 12 J (b) 24 J (c) 18 J (d) 36 J

Last Answer : Ans:(b) Direction: In the following question, a statement of Assertion (A) is given followed by corresponding Reason (R), just below it. Read the statements carefully and mark the correct answer: (a) If both A and ... true but R is not the correct explanation of A. (c) If A is true but R is false

The time taken for a radioactive element to reduce to 50% of its original weight is __________ years, if its half life period is 12 years. (A) 24 (B) 18 (C) 6 (D) 36

Description : The time taken for a radioactive element to reduce to 50% of its original weight is __________ years, if its half life period is 12 years. (A) 24 (B) 18 (C) 6 (D) 36

Last Answer : (B) 18

There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field. Harish takes 12 minutes. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet ? (a) 36 minutes (b) 18 minutes (c) 6 minutes (d) They will not meet

Description : There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field. Harish takes 12 minutes. Suppose they both start at the same point and at the same time and go in the same ... will they meet ? (a) 36 minutes (b) 18 minutes (c) 6 minutes (d) They will not meet

Last Answer : (a) 36 minutes

Now it was Ravi‘s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12? a. 1 b. 5/36 c. 1/18 d. 0

Description : Now it was Ravi‘s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12? a. 1 b. 5/36 c. 1/18 d. 0

Last Answer : a. 1

The area of the circle that can be inscribed in a square of side 6 cm is (a) 36 π cm 2 (b) 18 π cm 2 (c) 12 π cm 2 (d) 9 π cm 2

Description : The area of the circle that can be inscribed in a square of side 6 cm is (a) 36 π cm 2 (b) 18 π cm 2 (c) 12 π cm 2 (d) 9 π cm 2

Last Answer : (d) 9 π cm 2

After cleaning a slow sand filter, the filtered water is not used for (A) 6 hours to 12 hours (B) 12 hours to 18 hours (C) 18 hours to 24 hours (D) 24 hours to 36 hours

Description : After cleaning a slow sand filter, the filtered water is not used for  (A) 6 hours to 12 hours  (B) 12 hours to 18 hours  (C) 18 hours to 24 hours  (D) 24 hours to 36 hours 

Last Answer : (D) 24 hours to 36 hours 

At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Description : At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Last Answer : ratio of their ages, 12 years hence = 6 : 5  Difference in ages = 1 ratio = 6 years wife’s age (12 years hence) = 5 ratio = 5 × 6 years = 30 years wife’s age at the time of marriage = 30 – 12 = 18 years Answer: a)

Amount of Alachlor 50% EC for spray ing 12 ha maize field (given the dose 1.5 kg ai/ha) a). 36 kg b). 18 kg c). 54 kg d).12 kg

Description : Amount of Alachlor 50% EC for spray ing 12 ha maize field (given the dose 1.5 kg ai/ha) a). 36 kg b). 18 kg c). 54 kg d).12 kg

Last Answer : a). 36 kg

The mean of the ages of father and his son is 27 years. After 18 years, father will be twice as old as his son. Their present ages are (A) 42, 12 (B) 40, 14 (C) 30, 24 (D) 36, 18

Description : The mean of the ages of father and his son is 27 years. After 18 years, father will be twice as old as his son. Their present ages are (A) 42, 12 (B) 40, 14 (C) 30, 24 (D) 36, 18

Last Answer : Answer: A

X and Y can do a job in 24 days, Y & Z can do the same job in 36 days. X,Y,Z together can finish it in 18 days. X and Z together will do it in. a) 24 b) 36 c) 12 d) 26

Description : X and Y can do a job in 24 days, Y & Z can do the same job in 36 days. X,Y,Z together can finish it in 18 days. X and Z together will do it in. a) 24 b) 36 c) 12 d) 26

Last Answer : A (X+Y+Z) s 1 day work =1/18 (X+Y)'s 1 day work = 1/24 (Y+Z)'s 1 day work = 1/36 (X+Z)'s 1 day work = 2[X+Y+Z]'s 1 day work - [(X+Y)'s 1 day work + (Y+Z)'s 1 day work]= 2 [1/18]- (1/24 + ... /24 + 1/36) = 1/9 -1/24 - 1/36 =(8-3-2)/72 =3/72 = 1/24 So X & Z together will do the work in 24 days

What is the maximum theoretical data rate of the above problem if one transponder is used for binary transmission? A. 36 Mbps B. 72 Mbps C.18 Mbps D. 144 Mbps

Description : What is the maximum theoretical data rate of the above problem if one transponder is used for binary transmission? A. 36 Mbps B. 72 Mbps C.18 Mbps D. 144 Mbps

Last Answer : B. 72 Mbps

Which was the world’s first minicomputer and when was it introduced? a. PDP-I, 1958 b. IBM System/36, 1960 c. PDP-II, 1961 d. VAX 11/780, 1962

Description : Which was the world’s first minicomputer and when was it introduced? a. PDP-I, 1958 b. IBM System/36, 1960 c. PDP-II, 1961 d. VAX 11/780, 1962

Last Answer : PDP-I, 1958

What is throughput ?

Description : What is throughput ?

Last Answer : In parallel processing, the maximum amount of output per unit time is called throughput.

Is the total useful information processed or communicated over a specific amount of time A. Baud rate B. Bit rate C. Information theory D. Throughput

Description : Is the total useful information processed or communicated over a specific amount of time A. Baud rate B. Bit rate C. Information theory D. Throughput

Last Answer : D. Throughput

Undercharging of coal in the by-product coke ovens results in (A) Decrease in the c.v. of coke oven gas (B) Increase in its throughput (C) Increase in the c.v. of coke oven gas

Description : Undercharging of coal in the by-product coke ovens results in (A) Decrease in the c.v. of coke oven gas (B) Increase in its throughput (C) Increase in the c.v. of coke oven gas

Last Answer : (A) Decrease in the c.v. of coke oven gas

What are the advantages of HDFS federation in Hadoop? a) Isolation b) Namespace scalability c) Improves throughput d) All of the above

Description : What are the advantages of HDFS federation in Hadoop? a) Isolation b) Namespace scalability c) Improves throughput d) All of the above

Last Answer : All of the above

The ____ of can assembly line to be I/t p: a. Clock period b. Pipelining c. Throughput d. Flow through

Description : The ____ of can assembly line to be I/t p: a. Clock period b. Pipelining c. Throughput d. Flow through

Last Answer : c. Throughput

Who is the determined by the time taken by the stages the requires the most processing time: a. Clock period b. Flow through c. Throughput d. None of these

Description : Who is the determined by the time taken by the stages the requires the most processing time: a. Clock period b. Flow through c. Throughput d. None of these

Last Answer : a. Clock period

Ti is the time taken for the ith stage and there are n stages in the: a. Throughput b. Assembly line c. Both A and B d. None of these

Description : Ti is the time taken for the ith stage and there are n stages in the: a. Throughput b. Assembly line c. Both A and B d. None of these

Last Answer : b. Assembly line

The time taken for all stages of the assembly line to become active is called the: a. Flow through time b. Clock period c. Throughput d. All of these

Description : The time taken for all stages of the assembly line to become active is called the: a. Flow through time b. Clock period c. Throughput d. All of these

Last Answer : a. Flow through time

How much water is lost through a commercial cooling tower system with a throughput of about 600 GPM?

Description : How much water is lost through a commercial cooling tower system with a throughput of about 600 GPM?

Last Answer : This question depends on many factors. It sounds like the tower is small. A rule of thumb suggests that the tower will see an evaporation loss of about 0.1% of the circulation flowrate for each ... and blow down. Blow down is simply a purge of tower water to prohibit the buildup of solids.