Answer:
Option a
Explanation:
The Critical Path Method is the arrangement of booked exercises that decides the term of the task. These planned exercises must be performed if the venture is to be viewed as a triumph.
Therefore, options b, c, d and e can't be true because:
b. Activities in the Critical Path Method has no or zero slack.
c. The duration of the critical path in CPM determined on the basis of the latest activity and the earliest initiation.
d. The CPM method schedules the activity of the longest duration.
Ok we will do it for you cause they are very mean
Answer:
In terms of efficient use of memory: Best-fit is the best (it still have a free memory space of 777KB and all process is completely assigned) followed by First-fit (which have free space of 777KB but available in smaller partition) and then worst-fit (which have free space of 1152KB but a process cannot be assigned). See the detail in the explanation section.
Explanation:
We have six free memory partition: 300KB (F1), 600KB (F2), 350KB (F3), 200KB (F4), 750KB (F5) and 125KB (F6) (in order).
Using First-fit
First-fit means you assign the first available memory that can fit a process to it.
- 115KB will fit into the first partition. So, F1 will have a remaining free space of 185KB (300 - 115).
- 500KB will fit into the second partition. So, F2 will have a remaining free space of 100KB (600 - 500)
- 358KB will fit into the fifth partition. So, F5 will have a remaining free space of 392KB (750 - 358)
- 200KB will fit into the third partition. So, F3 will have a remaining free space of 150KB (350 -200)
- 375KB will fit into the remaining partition of F5. So, F5 will a remaining free space of 17KB (392 - 375)
Using Best-fit
Best-fit means you assign the best memory available that can fit a process to the process.
- 115KB will best fit into the last partition (F6). So, F6 will now have a free remaining space of 10KB (125 - 115)
- 500KB will best fit into second partition. So, F2 will now have a free remaining space of 100KB (600 - 500)
- 358KB will best fit into the fifth partition. So, F5 will now have a free remaining space of 392KB (750 - 358)
- 200KB will best fit into the fourth partition and it will occupy the entire space with no remaining space (200 - 200 = 0)
- 375KB will best fit into the remaining space of the fifth partition. So, F5 will now have a free space of 17KB (392 - 375)
Using Worst-fit
Worst-fit means that you assign the largest available memory space to a process.
- 115KB will be fitted into the fifth partition. So, F5 will now have a free remaining space of 635KB (750 - 115)
- 500KB will be fitted also into the remaining space of the fifth partition. So, F5 will now have a free remaining space of 135KB (635 - 500)
- 358KB will be fitted into the second partition. So, F2 will now have a free remaining space of 242KB (600 - 358)
- 200KB will be fitted into the third partition. So, F3 will now have a free remaining space of 150KB (350 - 200)
- 375KB will not be assigned to any available memory space because none of the available space can contain the 375KB process.
Answer:
18, 13, 19
Explanation:
Number of computer programmers proficient only in Java = 45 - ( 1+1+6) = 37
Number of computer programmers proficient only in C++ = 30 - (6+1+5) = 18
Number of computer programmers proficient only in python = 20 - ( 1+1+5) = 13
Number of computer programmers are not proficient in any of these three languages = 100 - ( 37 + 18 + 13 + 1+ 1+ 5+ 6 ) = 100 - 81 = 19
Answer:
Explanation:
temporal locality can be defined as: when a particular memory is referenced or accessed several times within a specific period of time. In the question, i think the variable that exhibit temporal locality are I, J and 0(all the variable). This is because the variable J and 0 are accessed several times within the loop. I would not have been part of it, but in the A[I][J]=B[I][0]+A[J][I], the variable "I" is also accessed in the addition. this is why it is part of the temporal locality.
