Let assume are lettered A to E in that order. Thus, there
will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of
these potential lines has 4 possibilities. Therefore, the total number of
topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect
each and one of them.</span>
Answer: see description
Explanation:
first we accommodate the bit pattern in a matrix of 4x4 which is the minimum length checksum field, now with even parity two-dimensional scheme we need to complete this matrix by adding one row and one column by adding at the end of each row a 1 or a 0 to complete pairs of 1's:
we have
![\left[\begin{array}{cccc}1&1&1&0\\0&1&1&0\\1&0&0&1\\1&1&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%260%5C%5C0%261%261%260%5C%5C1%260%260%261%5C%5C1%261%260%261%5Cend%7Barray%7D%5Cright%5D)
so we complete with this, adding a row at the end which matches a pair number of 1's
![\left[\begin{array}{ccccc}1&1&1&0&1\\0&1&1&0&0\\1&0&0&1&0\\1&1&0&1&1\\1&1&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%261%261%260%261%5C%5C0%261%261%260%260%5C%5C1%260%260%261%260%5C%5C1%261%260%261%261%5C%5C1%261%260%260%260%5Cend%7Barray%7D%5Cright%5D)
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.
