10.4 (c).what are the minimum and maximum latency times for the disk? what is th
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10.4 (c).what are the minimum and maximum latency times for the disk? what is the average latency time for this disk?
answers needed for 8.3,8.4,8.7,10.10,10.3
known as CAV and CLV. The playing time of a CLV disk is approximately tha of a CAV disk, although the number of tracks, track width of the tracks on the di and amount of data per video frame is the same. Explain why this is so Ola-lashiohed twelve-inch laser twice A disk consists of two thousand concentric tracks. The disk is 5.2 inches in diameter The innermost track is located at a radius of 1/2 inch from the center. The outermo track is located 2 1/2 inches from the center. The density of the disk 1650 bytes per inch along the track. The transfer rate is specified as 256,000 bytes per 10.6 is specified as second. The disk is CLV. All blocks are of equal size. a. innermost track consists of ten blocks. How many bytes are contained in block? ow many blocks would the outermost track contain? he capacity of the disk is approximately equal to the capacity in bytes of th middle track times the number of tracks. What is the approximate capacity of the disk? What is the motor rotation speed when reading the innermost track? The outermost track? Why is the average seek time for a hard disk much shorter than for a CD-ROM or DVD-ROM? 10.7 There is a current proposal to cut the size of an individual bit in a DVD-ROM in half o as to increase the capacity of the disk. This would cut both the width of the track and the track length required per bit in half. If the current capacity of a DVD-ROM is approximately 4.7 GB, what would be the capacity of the new "high-density DVD-ROM? 10.8 Atypical published page consists ofapproximatelyforty lines at seventy-five characters per line. How many published pages of 16-bit Unicode text would fit on a typical 600 MB CD-ROM? How many published pages of text would fit on a netbook computer with an 80 GB flash memory? A high-quality photographic image requires 3 bytes per pixel to produce sixteen million shades of color. 10.9 10.10 a. How large a video memory is required to store a 640 x 480 image during display A 1600 × 900 image? A 1440 × 1080 image? A 2560 × 1440 image? How many 1920 × 1080 non-compressed color images will fit on 4 DVD-ROM? b. on 4.7 GB A 1024 × 768 image is displayed, noninterlaced at a rate of thirty frames per second. 10.11 If the image is stored with 64K-color resolution, which uses 2 bytes per how much memory is required to store the picture?Explanation / Answer
8.3)
a) With no stall for dependencies: R5 first gets 2 from ADD then overwritten by 24
With stall for dependencies: R5 first gets 24 from MULT then overwritten by 26
Because of the dependency, the second execution unit is idle for 16 clock cycles
and the first execution unit is idle for 1 clock cycle for writeback of R5
b) R5 gets values 24 then 26 then 12 as each of the 3 instructions finishes
The execution units finish all 3 instructions in 17 cycles as the 3rd instruction uses the second execution unit
10.10)
1) 640 x 480 = there are 307200 pixels in a 640x480 image... since one pixel requires 3 bytes (24 bit), 921600 bytes will be needed. (921600 bytes=900 kb)
2) 1600 x 900= there are 1440000 pixels in a 1600x900 image... since one pixel requires 3 bytes (24 bit), 4320000 bytes will be needed. (4320000 bytes=4218 kb)
2) 1440 x 1080 = there are 1555200 pixels in a 1440x1080 image... since one pixel requires 3 bytes (24 bit), 4665600 bytes will be needed. (4665600 bytes=4556 kb)
4) 1024 x 768= there are 786432 pixels in a 1024x768 image... since one pixel requires 3 bytes (24 bit), 2359296 bytes will be needed.
4.7 gigabytes is: 5046586572 bytes. a 4.7 gb dvd rom can store 2139 images in 1024x768 resolution.
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