Architecture, Systems 6e Ch. 5
Solutions-11
Chapter 5 Solutions
Vocabulary Exercises
1. Dynamic RAM requires frequent __________ to maintain its data content. refresh cycles
2.
The __________ rate is the speed at which data can be moved to or from a storage device over a communication channel. data transfer
3.
Three standard optical storage media that are written only during manufacture are called __________ and __________. CD-ROM, DVD-ROM, BD
4.
__________, __________, ___________, and __________ are competing standards for rewritable DVD discs. DVD-RW, DVD+RW, DVD+RW DL, and DVD-RW DL
5.
6.
The __________ of a hard disk drive generate or respond to a magnetic field. read/write heads Data stored on magnetic media for long periods of time might be lost because of _________ and __________. magnetic leakage, magnetic decay
7. 8.
A(n) __________ stores data in magnetically charged areas on a platter. hard disk drive The contents of most forms of RAM are __________, making them unsuitable for long-term data storage. volatile
9.
__________ and __________ are outdated technologies for nonvolatile primary storage. __________ and __________ are promising new technologies for implementing NVM. EPROM, ROM, magnetoresistive RAM (MRAM), phase-change memory (PCM)
10.
__________ is typically stated in milliseconds for secondary storage devices and nanoseconds for primary storage devices. Access time
11.
The three components that are summed to calculate average access time for a disk drive are __________, __________, and __________. head-to-head switching time, track-to-track seek time, rotational delay
12. 13.
In a magnetic disk drive, a read/write head is mounted on the end of a(n) __________. access arm The access method for RAM is __________ or __________ if words are considered the unit of data access. The access method is __________ if bits are considered the unit of data access.
Systems Architecture, 6e Ch. 5 random, direct, parallel
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14. 15.
Both DDS and AIT use __________ to record bits and tracks on a magnetic tape. helical scanning __________ and __________ are two current standards or formats for linear recording on magnetic tapes. LTO, SDLT A(n) __________ mimics the behavior and physical size of a magnetic disk drive but has no moving parts. solid-state drive
16.
17. 18.
A(n) __________ is a series of sectors stored along one concentric circle on a platter. track A magnetic disk drive's data transfer rate can be calculated by dividing 1 by the drive's access time and multiplying the result by the __________. data transfer unit
19.
__________, __________, and __________ are storage formats originally designed for music or video recording that have been applied to computer data storage. CD-DA, DVD, BD
20. 21. 22.
Tape drives are __________ devices. __________ are random or direct access devices. serial access, disk drives Average access time can usually be improved by __________ files stored on a disk. defragmenting Modern desktop and laptop computers generally use memory packaged on small standardized circuit boards called __________. SIMMs or DIMMs
23.
The __________ of a magnetic or optical storage medium is the ratio of bits stored to a unit of the medium's surface area. areal density (or recording density or bit density)
24. 25. 26.
For most disk drives, the unit of data access and transfer is a(n) __________ or __________. block, sector Software programs stored permanently in ROM are called __________. firmware Many open standards for cartridge tape storage have been defined by the __________. Quarter Inch Committee (QIC)
Review Questions
1. What factors limit the speed of an electrically based processing device? The most important factor is the device size and, therefore, the distance electricity must travel to enable the device to perform its function. Another factor is resistance of the internal conductive
Systems Architecture, 6e Ch. 5
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pathways, which is a function of the materials composing these pathways and the device's operating temperature. Packing density, the type of packaging, and the presence or absence of heat dissipation mechanisms influence operating temperature, which influences resistance and reliability. 2. What are the differences between static and dynamic RAM? The most important differences are speed and cost. Static RAM is 5 to 10 times faster and approximately 10 times more expensive than dynamic RAM. Static RAM is composed entirely of transistors, and dynamic RAM is composed of transistors and capacitors. Capacitors require frequent recharging, which slows access time.
3.
DRAM?
What improvements are offered by synchronous DRAM compared with ordinary Synchronous DRAM begins the next sequential read access on each clock cycle and overlaps processing stages of several read accesses. Ordinary DRAM accepts and processes a read request in its entirety before accepting and processing another read request. Synchronous DRAM is several times faster than ordinary DRAM when processing sequential read accesses.
4.
Why isn't flash RAM commonly used to implement primary storage? Flash RAM wears out after several hundred thousand read/write cycles. Therefore, it would have to be replaced frequently if used as primary storage in a general-purpose computer.
5.
Describe current and emerging nonvolatile RAM technologies. What potential advantages do the emerging technologies offer compared with current flash RAM technology? NVM technologies include ROM, EPROM, EEPROM, flash RAM, magnetoresistive RAM (MRAM), and phase-change memory (PCM). Flash RAM is currently the dominant NVM type. MRAM and PCM are emerging NVM types. The main advantage of MRAM is that it doesn't degrade with repeated writes, which gives it better longevity than conventional flash RAM. The main advantages of PCM are faster write times and better longevity than flash RAM. However, PCM has slower read times and lower storage density than flash RAM.
6. method?
Describe serial, random, and parallel access. What types of storage devices use each Serial access reads or writes data units in sequential order. Magnetic tape is the only widely used form of serial access storage. Random access can "jump" directly between two noncontiguous data units. All primary storage and disk storage devices use random access. Parallel access reads or writes portions of a data item in parallel on separate storage devices or media. RAM can be considered a parallel access device.
7.
How is data stored and retrieved on a magnetic mass storage device? For writing, electrical current is routed through a read/write head, which generates a magnetic charge at its tip. A bit area of a magnetic recording surface is placed in close proximity to the read/write head, and a charge is induced in the surface material. The direction of current flow through the read/write head determines the stored charge's polarity. For reading, a bit area of the recording surface is placed in close proximity to the read/write head, and the stored charge induces current to flow through the read/write head. A switch connected to the read/write head detects the current flow's direction to interpret it as a bit value.
8.
Describe the factors that contribute to a disk drive's average access time. Which of these factors is improved if spin rate is increased? Which is improved if areal density is increased? Disk average access time is a combination of rotational delay, track-to-track seek time (head movement), and head-to-head switching time. Rotational delay is reduced if spin rate is increased. If areal density is increased, TTT seek time is improved because tracks are narrower and packed together more tightly (more tracks per same unit of distance traveled by the read/write head).
Systems Architecture, 6e Ch. 5 9.
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What problems contribute to read/write errors on magnetic tapes? Are these problems also present with other magnetic storage media/devices? Magnetic decay and magnetic leakage affect tapes. Magnetic leakage is compounded by winding the tape on a reel. Loss of chargeable material can occur because of friction between the read/write head and the tape. The tension required to wind and rewind tape leads to stretching of the medium, which can weaken it and/or its chargeable coating. Stretching also alters the size of bit areas, thus distorting their data content. Magnetic disks are subject to data loss because of magnetic leakage and decay but not friction and stretching.
10.
What are the advantages and disadvantages of helical scanning compared with linear recording? Helical scanning can store more data in the same amount of tape than linear recording can, and linear recording can access data more quickly than helical scanning can. Linear recording doesn't require physical contact the between read/write head and the tape, which extends tape life and reliability compared with helical scanning.
11.
Why is the areal density of optical discs higher than the areal density of magnetic disks? What factors limit this areal density? A laser can be tightly focused to a narrow beam. Magnetic charge can't be focused easily, and an attempt to charge a very small area overwrites surrounding areas. Optical bit size is limited by the read laser's wavelength and the amount of reflected light required to activate the read mechanism.
12.
Describe the processes of reading from and writing to a phase-change optical disc. How does the performance and areal density of these discs compare with magnetic disks? A phase-change disc is coated with a material that's amorphous or crystalline. Each state has different reflective properties. A bit area in the amorphous state is changed to a crystalline state by heating it to a precise temperature with a laser. A bit area in the crystalline state is changed to the amorphous state by heating it to the material's melting point. Phase-change discs are read in the same fashion as other optical discs. Phase-change optical discs have slower access time than magnetic disks because more time is required to heat a bit with a laser than to alter its magnetic polarity. Phase-change discs also lose their capability to change state easily after repeated use and, therefore, have shorter lives than magnetic disk platters under continuous use.
13.
List and briefly describe the standards for recordable and rewritable CDs and DVDs. Are any of the standards clearly superior to their competitors? This answer is taken from Table 5-7:
Technology/for mat CD-ROM CD-R CD-RW DVD-ROM DVD+/-R
Writable? No One time only Yes No One time only
Description Adaptation of musical CD technology; 650 or 700 MB capacity. CD-ROM format with a dye reflective layer that can be written by a low-power laser. CD-ROM format with phase-change reflective layer; can be written up to 1000 times. Adaptation of DVD video technology; similar to CD-ROM but more advanced; 4.7 GB (single layer) or 8.5 GB (dual layer) capacity. DVD-ROM single- and dual-layer formats; similar to CD-R with improved performance and capacity. DVD-R and DVD+R are slightly different formats DVD-ROM single- and dual-layer formats with phase-change reflective layer. DVD-RW and DVD+RW are slightly different
DVD+/-RW
Yes
Systems Architecture, 6e Ch. 5 formats.
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Problems and Exercises
1.
Assuming a sector size of 512 bytes, the capacity is 5 platters 2 sides 1024 tracks 50 sectors 512 bytes, or 262,144,000 bytes (250 MB). Assuming the data is organized for maximum read efficiency, the read occurs cylinder by cylinder. This results in 1024 track-to-track seeks (1024 3 microseconds = 0.003072 seconds) and 10 head-to-head switches per track (1024 10 2 nanoseconds = 0.00002048 seconds). The number of disk rotations equals the number of tracks read (1024 10 = 10,240 tracks). The disk rotates at 10,000 rpm, so the time required to read the tracks (exclusive of seeks and switches) is 10,240 10,000, or 1.024 minutes (61.44 seconds). Because there's no rotational delay when reading sequentially, the total elapsed time to read the disk is 0.003072 + 0.00002048 + 61.44 = 61.44309248 seconds. Serial access time is the time required to read the second of two sequential sectors. If the sectors are on the same platter and track, it's only a function of rotation speed. Because there are 50 sectors per track, the sequential access time is 1/50 of a rotation, or 0.00012 seconds (60 seconds 10,000 50). An average serial access time should also account for the fraction of sequential reads that require a head-to head switch (1 every 50 sectors) and a track-to-track seek (1 every 500 sectors). The easiest way to compute this adjusted access time is to divide the time required to read the entire disk by the number of sectors (61.44309248 seconds 512,000 sectors = 0.00012000604 seconds). Each read requires one-half rotational delay, a seek over 512 tracks, switching through 5 heads, and 1/50 of a rotation to read the data. One-half rotational delay is 60 seconds 10,000 2, or 0.003 seconds. A seek over 512 tracks requires 512 0.000003, or 0.001536 seconds. Switching through 5 heads requires 5 0.000000002, or 0.00000001 seconds. Reading the data requires 0.00012 seconds (the unadjusted serial access time). The average access time is the sum of these numbers, or 0.00465601 seconds (4.65 milliseconds). 1 2,400,000,000 2 = approximately 0.2083 nanoseconds. Note that this answer allows no time for transmitting the access request. Processor cycle time is 0.4167 nanoseconds, so 12 CPU cycles (5 divided by 0.4167) are required to complete a fetch operation. Half of the next cycle (the fetch portion) is wasted because execution can't begin until 16.25 seconds have elapsed. Therefore, six wait states are incurred. 10 0.4167 = 24 CPU cycles to complete the fetch. Part of the next cycle is wasted waiting for the execution cycle to begin, so 24 wait states are incurred. Average access time 4 ns Data transfer unit size 64 bits Data transfer rate (1 .000000004 seconds) 64 bits = 16 Gbps (or 2 GBps) (1 .01 seconds) 512 bytes = 5,120 Bps (or 5.12 KBps) Note: This computation is a worst-case scenario. The data transfer rate is much higher for sequential access. (1 .005 seconds) 512
2.
3.
Storage device RAM
Optical disc
100 ms
512 bytes
Magnetic disk
5 ms
512 bytes
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bytes = 102,400 Bps (or 102.4 KBps) Note: This computation is a worst-case scenario. The data transfer rate is much higher for sequential access.
Research Problems
Project 1
Answers will vary based on the computers selected for the investigation, but students might note some of these general findings: Older computers might not be able to accept a 1 GB memory expansion. The cost of 1 GB of memory for most current desktops and laptops is similar. The cost of 1 GB of memory for most servers is higher because memory must often be added in groups of four.
Project 2
Students should find information at the IBM Web site (www.ibm.com). Holographic storage technology still appears to be in the experimental stages at IBM. The company has created limited prototypes for secondary storage capable of storing approximately 7 MB of data. Seeing products based on holographic technology in the near future is incredibly unlikely. Storage density and data transfer rate are much higher than with NVM, and costs are similar, so there's little incentive to commercialize the technology in its present state.
Project 3
LTO provides the highest capacity at the lowest cost, followed by SDLT and AIT (with variations based on standard version and manufacturer). The tape format with the fewest passes over read/write heads is likely to provide the best longterm reliability before developing defects. In practice, SDLT and AIT have a small advantage by this measure, but all media are roughly comparable. AIT and LTO have a small advantage in measures of mean time between failure because of frictionless reading and writing, but it's a less important selection factor than speed and capacity. LTO appears to hold the market advantage and will likely continue to because IBM and HP are its main backers, and as they represent the bulk of the mainframe market, users of their hardware are likely to follow their recommendations. Opinions vary on the changing role and usefulness of tape as a storage medium. However, tape remains a widely used medium in large data centers for archival storage. Students' conclusions on which tape format to choose will vary.
Project 4
Answers will vary based on students' opinions, but one comparison outcome (at the time of this writing) is as follows:
CD+R stores 700 MB and costs less than 25 each if bought in bulk. CD+RW also stores 700 MB but can be rewritten; costs about $1.25 each. DVD+R and DVD-R store 4.7 GB and cost about 25 each if bought in bulk. DVD+RW and DVD-RW also store 4.7 GB but cost about $1.25 each.
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BD-RE stores 25 GB and costs about $20 each in 2010, although this cost will probably drop quickly over the next year or two. The functional lifetime of all recordable formats is similar. All rewritable media are prone to data degradation if data is rewritten repeatedly over time.
Storing 20 GB of digital photographs would take 29 CD-Rs (at a cost of about $7.25) or 6 DVDRs (at a cost of about $3.75). DVD-R appears to be the preferred solution based on lower cost and more compact storage. High-capacity flash drives and memory cards are a viable alternative to recordable optical media, with costs in 2010 at around $4/GB and much faster access times.