unit1-2Review

1.Benchmarksa.With regard to computing, define the term benchmark.Benchmarking is the process of running standardized tests upon differing configurations todetermine the speed of components or software. Benchmarking: comparing disparate systems orcomponents via a standardized set of instructions or series of tasks.b.What tasks does benchmarking software perform in order to measure thesystem being tested?Testing your system performance, such as CPU, hard disk, CD/DVD ROM, and compare it withother systems.c.How can consumers use benchmarks to help them purchase a computer system?Consumers can run benchmarks on different machines. The comparison ismeasured by the time it takes to execute these instructions, the faster the better.d.How do benchmarks help identify slow points (bottlenecks) in the system?We can run many kinds of benchmarks, each kind of benchmark2.Binarya.Define bit.Each binary digit is called a bit.( Computer systems are made up of electricalcomponents that are either on, or off, repr esenting 1’s and 0’s, also called binary numbers)b.Define byte. Eight Bits equal to one bytec.Fill in the following chart, listing the prefixes used for the amountof bytes shown:d.Fill in the following chart, converting binary numbers to decimal anddecimal numbers to binary:e.Although in sales literature 1000 bytes and 1024 bytes are both commonlyreferred to as a kilobyte, in computing the only correct number of bytesin a kilobyte is1024. Explain why 1024 is correct.Since the data stored on computer are binary sequence as 1s and 0s, the capacity of memory is the exponent of 2, so a kilobyte is 1024 bytes.f.How can the difference in measuring bytes presented in part e impact aconsumer when they are purchasing a hard disk drive?For instance, a consumer purchased a 40G hard disk , but the actual capacity of it is only about 37G .The reason is that the capacity of memory is the exponent of 2ponent IdentificationChoose among the following components to label the image:Motherboard Power supplyBIOS ROM ChipCooling fan Expansion slotRAM chip Expansion cardDisk drives ChipsetIDE cable Ethernet cable PCI busbel A is _ Power supply ____.bel B is __ Cooling fan ___.bel C is __ Expansion slot ___.bel D is ___Expansion card __.bel E is __ Motherboard ___.bel F is ___ Disk drives __.bel G is _ IDE cable ____.Describe the functionality of each of the following components (in 2sentences or less):h.Microprocessor: Processes instructions stored in main memory. Sometimes, themicroprocessor is found underneath a cooling fan to prevent it from overheating.i.RAM: RAM (random access memory) is a temporary holding area for both data and instructions.It is also referred to as main memory. RAM stores data and instructions needed to executeprograms. The data in RAM is lost when the computer is turned off. In contrast to accessing dataserially, searching sequentially for the data to be retrieved, data in RAM can be accessed directlyvia its address.j.Bus: A bus is a pathway through which data is transferred from one part of a computer to another. It consists of the data bus and the address bus.k.Expansion card: Enables a computer to control peripheral devices such as the monitor and the microphonel.Disk drive: computer hardware that holds and spins a magnetic or optical disk and reads and writes information on itm.IDE cable: Transfers data from storage devices to the motherboardpressiona.What is compression? reduction of the space required for storage(of binarydata) by an algorithm which converts the data to a smaller number of bits whilepreserving the information content.b.Describe a benefit of compression. To reduce the file size.c.Describe dictionary-based compression.Replace common sequence of character with a single codeword,or symbol thatpoints to a dictionary of the original characters, or to the originaloccurrence of the word. 注意空格和#号d.Describe run-length encoding.A kind of compression algorithm which replaces sequences ("runs") of consecutive repeated characters (or other units of data) with a single character and the length of the run. This can either be applied to all input characters, including runs of length one, or a special character can be used to introduce a run-length encoded group. The longer and more frequent the runs are, the greater the compression that will be achieved. This technique is particularly useful for encoding black and white images where the data units would be single bit pixels.e.Describe a benefit of lossy compression. a file type that incorporates lossy compression. WINZIPputer Speeda.One Hz is one __cycle___ per __second___.b.Define IPS. Instructions per secondc.Answer the following questions on the relationship of Hz to IPS.i.Which is a better determiner of speed: MHz or IPS? MHzii.Does a high Hz equal a high IPS, and can they be inverselyproportional? Explain.High Hz unequal a high IPS, because Instructions can differ a great deal. Somecomplex instructions require a lot of cycles and take a comparatively long time to execute.Other instructions may be very simple and execute in very little time. For example, adivision operation takes longer than an addition operation.iii.How is it possible for a machine with a lower clock speed (Hz) to have a higher IPS than a machine with a higher clock speed?because Instructions can differ a great deal. Some complex instructions require a lot ofcycles and take a comparatively long time to execute. Other instructions may be very simpleand execute in very little time. For example, a division operation takes longer than anaddition operation.d.What is the relationship between the system clock and Hz?One Hz means one system clock cycle per second.6.Hexadecimala.Define hexadecimal notation.b.What advantages does hexadecimal have over binary?Because it will takes more digits to express a value in binary notation than in hexadecimalnotation, we have adopted hexadecimal notation as shorthand for binary so that we can express values more concisely.c.Which number system, binary or hexadecimal, is used when referencingmemory addresses and why? Hexadecimal,7.Imagesa.How are bitmapped images rendered?Bitmap is composed to a grid of dots, and the color of each dot is stored as a binary number.b.How are vectored images rendered?Vectored images consist of a set of instructions for re-creating a picture.Instead of storing the color value for each pixel, a vector graphic file contains the instructions that the computer needs to create the shape, size, position, and color for each object in an image.c.Which image type tends to be more photo-like in quality, bitmapped orvectored?Bitmap tends to be more photo-like in quality. Since the bitmap is composed of pixels, so is the photo.d.Which image type maintains a high image quality, regardless of how itis resized? Vectored images.e.List three image editors. CORELDRAW、PHOTOSHOP、ACDSEE328.I/Oa.Define I/O device. Stands for input/output device.b.Consider that a computer requires I/O devices to be able to interact withits environment. Explain why it is necessary for computers to have thesedevices.c.Give four examples of I/O devices.Digital camera, printer, speaker, keyboard.d.For each I/O device in your answer to part c, explain the device'sfunction from the perspective of a user.Users take pictures with digital camera. It is stored in digital form, and can be uploaded onto a compute r.The printer can print images or files. Keyboard is used to type data into computer. The speaker is the output device to enable the voice.9.Magnetic Mediaa.List two advantages of magnetic media over optical media.Magnetic disks, with the exception of floppy disks, can be written and read faster than optical disks. Most hard disk drives offer greater capacity than any currently available optical device. And usually the writable of magnetic disks is unlimited.b.Fill in the following chart matching the type of media with its storagecapacity. The types of media to choose from are Jaz disk, Zip disk, hardc.Consider that magnetic media is read by an electromagnet. How does theread/write head discern and transfer data from the magnetic media to thesystem?Before data is stored, the particles on the surface of the disk are scattered in random patterns. The disk drives red- write head magnetizes the particles, and orients them in either a positive or negative direction. These patterns of magnetized particles represent 0 and 1 bits.10.Moore's Lawa.Define Moore's Law.The observation made in 1965 by Gordon Moore, co-founder of Intel, that the number of transistors per square inch on integrated circuits had doubled every year since the integrated circuit was invented. Moore predicted that this trend would continue for the foreseeable future. In subsequent years, the pace slowed down a bit, but data density has doubled approximately every18 months, and this is the current definition of Moore's Law, which Moorehimself has blessed. Most experts, including Moore himself, expect Moore's Law to hold for at least another two decades.b.List two extrapolations of Moore's Law that are discussed in the coursematerials.Processing power (speed) doubles every 12-18 months. Storage capacity of RAM doubles every 12-18 months.c.How can Moore's Law and the extrapolations from Moore's Law assist aconsumer in purchasing a computer?User expect to pay for a computer today that is twice as powerful as the computer that is now 18 months old approximately the same amount of money paid for the old computer.d.Define Parkinson's Law of Data.Parkinson's law of data states that data expands to fill the space available.ing Parkinson's Law, what projection can be made regarding memory andhard disk usage?f.How can Parkinson's Law assist a consumer in purchasing software orcomponents for a computer?11.Optical Mediaa.List two advantages of optical media over magnetic media.Optical media are more durable. They are not ruined by dust or moisture, nor are they vulnerable to electrical damage (however, they can be damaged by physical damages such asscratches). Optical media's MTBF rating (average life expectancy) ranges between 30 and 300years, while magnetic media utilize magnetic properties that have a MTBF of about 3–7 years.Optical media are less expensive per MB than magnetic disks.b.Fill in the following chart. The types of media to choose from are CD-R,c. Consider that optical media is read by a laser. How does the laser discern and transfer data from the optical media to the system? In the optic readable CD-ROM, the data storage consists of millions of indentations burnt into the lacquer coated, light reflecting silver surface. The burnt dents reflect less light than the shiny surface. A weak laser beam is sent to the disk through a two-way mirror and the sensor registers the difference in light reflection from the burnt and shiny areas as zeros and ones.c.d.Consider that a DVD and a CD are the same size. List two reasons a DVDholds more information than a CD. First, DVDs have narrower tracks, so they cansqueeze more tracks onto the same size disk. Both CDs and DVDs are read using light from alaser. But, the CD laser is red, while the DVD laser is blue. Red light has a longer wavelengththan a blue light. The blue laser is thus able to produce a smaller beam, allowing it to focus onthe tinier tracks of the DVD. The second way that DVDs achieve increased capacity over CDs isby using multiple layers of tracks. The blue laser is not only narrower, but also more powerful.Its beams can penetrate the plastic and focus at different depths.12.Picture Qualitya.Explain resolution, with respect to a monitor.signifies the number of dots (pixels)on the entire screen.b.List two typical monitor resolutions. 800*600,1024*768c.Explain color depth, with respect to a monitor. The number of distinctcolors that can be represented by a piece of hardware or software. Colordepth is sometimes referred to as bit depth because it is directly relatedto the number of bits used for each pixel. A 24-bit video adapter, forexample, has a color depth of 2 to the 24th power (about 16.7 million)colors. One would say that its color depth is 24 bits.d.List two typical monitor color depths. 24bits,32bitse.Consider that the higher the resolution and the higher the color depth,the more system resources are required to display output on the monitor.e the resolutions from part b and the color depths from part dto calculate the amount of RAM required to display the image.1024*768, 32bits RAM = 1024*768*32/8= 3145728 Bytes ,about 3Mii.Explain the calculation you made in part e.i in a way that confirms the statement made at the beginning of part e—that "the higherthe resolution and the higher the color depth, the more systemresources are required to display output on the monitor."From the equation above we know ,the capacity of RAM needed todisplay depends on resolutions and the color depths13.Port IdentificationConsider the following devices: Mouse Printer Keyboard Modem Speaker Digital camera Ethernet jack MonitorFor the following questions, indicate which device(s) should be plugged in which port(s) A-H shown in the diagram below.a.What device(s) can be plugged into port A? mouse or keyboardi.What is the name of this port? PS/2 portsii.Is this a serial port or parallel port? a serial portb.What device(s) can be plugged into port B? Digital camerai.What is the name of this port? USBc.Is this a serial port or parallel port? PDA devices used to connect to DB-9ports before the advent of USB ports. Other devices that can be connected to a DB-9 port are the external modem and the barcode scanner, in general, older electronic devices.d.What device(s) can be plugged into port C? the external modem and thebarcode scanner, in general, older electronic devicesi.What is the name of this port? serial portii.Is this a serial port or parallel port? seriale.What device(s) can be plugged into port D? printer,i.What is the name of this port? parallel portii.Is this a serial port or parallel port? parallel portf.What device(s) can be plugged into port E? Speakerg.What device(s) can be plugged into port F? Monitorh.What device(s) can be plugged into port G? modemi.What device(s) can be plugged into port H? Ethernet jack14.RAMa.Define RAM. (random access memory) is a temporary holding area for both data andinstructions. It is also referred to as main memory.b.What is the unit of measurement typically used for the speed of RAM?Latency measures the speed of RAM. Latency is typically measured in nanoseconds (ns), billionth ofa second (10-9 sec). (The capacity of RAM is typically measured in megabytes (MB))c.What is SDRAM and how does it work?SDRAM (Synchronous Dynamic RAM) used in many personal computers. It is fast and relativelyinexpensive. It is synchronized to the clock so that data can be sent to the CPU at each tick of theclock, increasing the number of instructions the processor can execute within a given time.d.Data on hard disk drives is accessed indirectly via the file system.i.How is data in RAM accessed? data in RAM can be accessed directly via itsaddress.ii.What are the benefits of accessing RAM in this way?RAM derives its name from the fact that its contents can be accessed (i.e., read from and written to) in any order. (Perhaps it should have more descriptively been named nonsequential memory, as it is not accessed in a random manner.) This is in contrast to sequential memory devices, such as disks (hard, floppy and optical) and magnetic tapes, for which the data must be accessed to some extent in a fixed order. Thus, the times required for RAM to access any of its memory locations are virtually identical, in contrast to disks and tapes, which have delay times that vary according to the location on the media.RAM features much higher access speeds than hard disks and optical disks. That is, it takes far less time to read data from and write data to RAM than it does for discs. This is due to the fact that access is entirely electronic, i.e., there are no moving parts (e.g., magnetic heads and spindle motors) to slow things down. RAM access time is expressed in nanoseconds (millionths of a second), whereas hard disk access time is expressed in milliseconds (thousandths of a second).e.With regard to RAM, what is volatility?A characteristic of RAM is that it is volatile, in contrast to ROM (read-only memory), disks and tapes. This means that any data stored in RAM is retained only as long as the RAM chips are connected to a power supply. In the case of DRAM chips, moreover, not only must they be connected to a power supply but also they must be refreshed at frequent intervals (i.e., multiple times per second) by an electric current, hence the term dynamic in their name. The reason is that each bit of data is stored as a charge in a microscopic capacitor, and such charges quickly dissipate without refreshment. Thus, when the power supply is interrupted (e.g., the computer is turned off), the memory contents are lost. When a computer is turned on, the operating system and other files are once again loaded into RAM, usually from the hard disk.。