1. The advantage of using a very large memory cell size is that the computer can store larger values in its memory cells. The disadvantage is that there would be fewer cells available.
2. This question asks how many bits are needed in the MAR with the following numbers
a. 1 million bytes = 20 bits
b. 10 million bytes = 24 bits
c. 100 million bytes = 27 bits
d. 1 billion bytes = 30 bits
3. 640 kb would be 655,360 memory cells. A 512MB would contain 536,870,912 memory cells. A 2GB would contain 2,097,152,000 memory cells
4. ROM is read only information. This information is put into ROM in the factory and normally not messed with. It contains information such as BIOS, and firmware information.
5. 1 MB of storage would be 1024 × 1024.
6. The …show more content…
The memory would have to be 2-dimensional.
8. There would have to be some sort of directions or conversion in order to translate the MDR to match what goes in the instruction. Without that there would be a fatal error because in the MDR would not be able to be read by something that is only looking for 32 bit instructions.
9. ((65*60)/780) *500 = 2500M floating points
10.
11. The main advantage of this way is that there is not as much emphasis is put on the memory, as the information is put into the hardware. The disadvantage is that the code would have to be changed based on the hardware and must be tested over and over again.
12. This asks some questions about disk storage
a. 1024 (characters per sector)*20 (sectors per track) *500 (tracks per surface)*2 (number of 2 surfaces) = 20,480,000 characters
b. Best = .42 Average = 12.59 Worst = …show more content…
Best case = .31 Average 11.44 worst 32.01
14. The best and fastest area to store the file would be on the first available sector as the access time would be .42 so that would mean being at one would be the fastest time. On average it takes close to 13 so that would also be a good spot to start. However, to be saving to the 60 track would be the safest as it is sure to be read in a timely fashion.
15. I believe the best would still be .42 and then the worst would be 10.25 and then the average would be 9.55. It would be a lot faster since it does not have to worry about 500 tracks per surface rather then just the 20 sectors per track
16. In order to read every sector on the whole disk it would be .42 (best case reading a sector) * 20 (number of sectors) *500 (number of tracks) *2 (number of side) = 8400
17. Someone would use sequential access storage devices when it is one file or one group of something that will be read in order from beginning to end. This could be used for backups as it can be moved somewhere, and then when restored will be read in order from start to finish.
18. 8929 instructions (rounded up)
19. This question asks about instruction register
a. 64
b. 256 kilobytes = 262,144
2. This question asks how many bits are needed in the MAR with the following numbers
a. 1 million bytes = 20 bits
b. 10 million bytes = 24 bits
c. 100 million bytes = 27 bits
d. 1 billion bytes = 30 bits
3. 640 kb would be 655,360 memory cells. A 512MB would contain 536,870,912 memory cells. A 2GB would contain 2,097,152,000 memory cells
4. ROM is read only information. This information is put into ROM in the factory and normally not messed with. It contains information such as BIOS, and firmware information.
5. 1 MB of storage would be 1024 × 1024.
6. The …show more content…
The memory would have to be 2-dimensional.
8. There would have to be some sort of directions or conversion in order to translate the MDR to match what goes in the instruction. Without that there would be a fatal error because in the MDR would not be able to be read by something that is only looking for 32 bit instructions.
9. ((65*60)/780) *500 = 2500M floating points
10.
11. The main advantage of this way is that there is not as much emphasis is put on the memory, as the information is put into the hardware. The disadvantage is that the code would have to be changed based on the hardware and must be tested over and over again.
12. This asks some questions about disk storage
a. 1024 (characters per sector)*20 (sectors per track) *500 (tracks per surface)*2 (number of 2 surfaces) = 20,480,000 characters
b. Best = .42 Average = 12.59 Worst = …show more content…
Best case = .31 Average 11.44 worst 32.01
14. The best and fastest area to store the file would be on the first available sector as the access time would be .42 so that would mean being at one would be the fastest time. On average it takes close to 13 so that would also be a good spot to start. However, to be saving to the 60 track would be the safest as it is sure to be read in a timely fashion.
15. I believe the best would still be .42 and then the worst would be 10.25 and then the average would be 9.55. It would be a lot faster since it does not have to worry about 500 tracks per surface rather then just the 20 sectors per track
16. In order to read every sector on the whole disk it would be .42 (best case reading a sector) * 20 (number of sectors) *500 (number of tracks) *2 (number of side) = 8400
17. Someone would use sequential access storage devices when it is one file or one group of something that will be read in order from beginning to end. This could be used for backups as it can be moved somewhere, and then when restored will be read in order from start to finish.
18. 8929 instructions (rounded up)
19. This question asks about instruction register
a. 64
b. 256 kilobytes = 262,144