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almond37 [142]
3 years ago
10

Consider sending a 10000-byte datagram into a link that has an MTU of 4468 bytes. Suppose the original datagram is stamped with

the identification number 218. How many fragments are generated? What are the values in the various fields in the IP datagram(s) generated related to fragmentation?
Computers and Technology
1 answer:
AfilCa [17]3 years ago
5 0

Answer:

Number of fragments is 3

Explanation:

The maximum size of data field in each fragment = 4468 - 20(IP Header)

= 4448 bytes

Hence, the number of required fragment = (10000 - 20)/4448

= 3

Fragment 1

Id = 218

offset = 0

total length = 4468 bytes

flag = 1

Fragment 2

Id = 218

offset = 556

total length = 4468 bytes

flag = 1

Fragment 3

Id = 218

offset = 1112

total length = 1144 bytes

flag = 0

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What type of computer processing system is characterized by data that are assembled from more than one location and records that
ikadub [295]

Answer:

d) Online, real-time systems.

Explanation:

The options are:

a) Batch processing systems.

b) Personal computer systems.

c) Data compression systems.

d) Online, real-time systems.

And the correct option is D. Online, a real-time system. And this is because it is this which is characterized by the data which is assembled from more than one location as it is online, and various clients from various locations enter the data, and it's updated immediately, as it is a real-time system, which is updated in real-time. And hence D. is the correct option. The data compression system takes the data from one location at a time. The batch processing system is made up of different programs for input, output, and process. and hence is different from real-time, which requires continual all the three. And a PC is not made to a specific function, and it does different activities. Hence, its also not the right option here.

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4 years ago
HELPPPP In 5-6 sentences explain how technology has impacted engineers.
lorasvet [3.4K]

Answer:

Digitization is changing the playing field for engineers.

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3 0
3 years ago
Which type of operating system is usually used in personal computers?
jeyben [28]

Answer:

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4 0
3 years ago
Consider the following program that monitors two sensors. Here sensor1 and sensor2 denote
dusya [7]

Answer:

Check the explanation

Explanation:

a)

It is a technical impossible for the update in the value of sensor1 in ISR while the main function is still evaluating whether sensor1 is with a fault or not at the point of execution in the critical section block because in this block no other instructions are not executed until the completion of instructions in the critical section.

Also semaphores are used to execute instructions when they are get lock to execute. if we write the code in operating system in such a way that at any instant of time shared resources cannot be executed in parallel, then only it is not possible to update ISR value of sensor1 while main function checking for sensor1 faultiness.

-> It is possible that  to update ISR value of sensor1 while main function checking for sensor1 faultiness if operating system and compilers are coded for parallel execution even though shared resources.

for example:

int i=5;

main()

if (i==5)

{print "i is %d",i}

other()

i=7;

//if above program executes parallel boh main() and other() functions are parallel executed

then at first clock cycle in main method i=5 so goes to next instruction print.

but in the first clock cycle also in other method i value changes to 7.

so in the next clock cycle in main method result as " i is 7.

b)

if there is an occurrence of error related to only for faulty value of sensor 1 or sensor 2 then only it is possible for this code would report "Sensor1 faulty" or "Sensor2 faulty" when there is no error related to other issues.

Like power interrupt etc.

if a spurious error can cause not only sensor1 or sensor2 to be a faulty value but also interrupting whole program or suspending entire program etc may possible for this code would not report  "Sensor1 faulty" or "Sensor2 faulty".

c)

if we Assume the interrupt source for ISR() is timer-driven , then there are conditions could cause this code to never check whether sensors are faulty or not.

timer driven means set a clock for its execution how can we set the clock. if we set clock for ISR() method to stop the entire program or repeated continuously at setting up interrupts and enabling interrupts.

these two conditions 1)set clock to stop the program would not enter into checking sensors faultiness.

2)set the clock to spare entire time to execute  at setting up interrupts and enabling interrupts. repeated continuously.

It means never run ":while(1) {} " block instructions for checking faultiness of the sensors.

5 0
3 years ago
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