homework_1_ch2_20150303

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回家写作业英语说

Homework is a term commonly used in English to refer to assignments or tasks that students are given to complete outside of regular class hours.Here are some phrases and expressions related to homework in English:1.Assignment:This is another word for homework.For example,I have a lot of assignments to finish tonight.2.Do your homework:This is a direct command to complete the homework.For example, Dont forget to do your homework before you go out to play.3.Homework help:This refers to assistance given to someone with their homework.For example,I need some homework help with my math problems.4.Homework time:This is the time set aside for doing homework.For example,Its homework time,so please turn off the TV.5.Homework due:This indicates that the homework must be submitted by a certain date. For example,Your science homework is due on Friday.6.Homework project:A more extended piece of work that may take several days or weeks to complete.For example,Im working on a history homework project about the Civil War.7.Homework overload:When there is too much homework to handle.For example,I have a homework overload this week.8.Homework excuses:Reasons given for not having completed homework.For example,I didnt do my homework because my dog ate it.9.Homework club:A group or club where students can go to do their homework together, often with supervision or help available.For example,I go to the afterschool homework club to get help with my assignments.10.Homework stress:The anxiety or pressure felt from having a lot of homework.For example,All this homework is causing me a lot of stress.11.Homework planner:A tool or system used to organize and keep track of homework assignments.For example,I use a homework planner to make sure I dont miss any assignments.12.Homework buddy:A friend or classmate who helps with homework.For example, My homework buddy and I study together every evening.13.Homework pass:A permission slip or note that allows a student to be excused from homework for a particular day.For example,I got a homework pass from the teacher because I was sick.14.Homework folder:A designated place where students keep their homework materials. For example,Make sure to put all your homework in your folder.15.Homework policy:The rules or guidelines set by a school or teacher regarding homework.For example,Our school has a strict homework policy that must be followed.Remember,when speaking about homework in English,its important to be clear and specific about the type of work and the expectations involved.。

homework参考答案

第五章
、用户空间有个页面，每页，主存答：（）程序空间的大小为，因此逻辑地址的有效位
数是位。（）内存储空间的大小是，因此物理地址至少需
要位。
第五章
用户空间有个页面，每页号物理块号页，主存。
、答：
（）虚地址的二进制表示如下:
页号页内地址
其页号为，从表中可知该页对应的物理块号为，所以，将二进制表示中的页号换为块号，则物理地址用二进制表示为：
（）写出逻辑地址的格式。
（）进程的页表有多少项？每个页表项占多少位？
（）如果物理空间减少一半，则页表中页表项数怎样改变？每个页表项占多少位？
答：
（）位页号位页内偏移
15 11 10
0
页号页内地址
第五章
、逻辑地址空间页，每页，物理空间。
（）进程的页表有多少项？每个页表项占多少位？
答：实际存储页表时，每个页表项中不需要存储页号，只需存储块号即可。
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作业() 作业()
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第三章作业号到达运行 ()
完成要求内存磁带机打印机（）
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作业() 作业()
作业()
作业() 作业()
作业()
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第三章作业号到达运行 ()
第三章
（）采用短作业优先调度算法，则其调度顺序是、、、作业提交运行开始完成周转带权周转号时间时间时间时间时间时间
平均周转时间： () 平均带权周转时间：()

我的家庭作业英语作文

我的家庭作业英语作文My Homework。

As a student, doing homework is an essential part of my daily routine. It helps me to consolidate what I have learned in class and to develop my critical thinking skills. However, sometimes homework can be overwhelming, and I find myself struggling to keep up with the workload.To manage my homework effectively, I have developed a few strategies that work well for me. Firstly, I make a schedule of all the tasks that I need to complete, and I prioritize them according to their importance and deadline. This helps me to stay organized and focused, and to avoid procrastination.Secondly, I try to create a conducive environment for studying. I make sure that my study area is clean, quietand well-lit, and that I have all the necessary materials and resources at hand. This helps me to stay motivated andproductive, and to avoid distractions.Thirdly, I seek help when I need it. If I am struggling with a particular concept or assignment, I don't hesitate to ask my teachers or classmates for assistance. This not only helps me to understand the material better, but also fosters a sense of collaboration and teamwork.Finally, I try to take breaks and engage in activities that I enjoy, such as reading, listening to music or playing sports. This helps me to recharge my batteries and to avoid burnout, which can be detrimental to my academic performance.In conclusion, doing homework can be challenging, but with the right mindset and strategies, it can also be a rewarding and enriching experience. By staying organized, creating a conducive environment, seeking help when needed, and taking breaks, I am able to manage my homework effectively and achieve my academic goals.。

homework-1

Find the word or phrase from the list below that best matches the description in the following questions.1.1.1 Computer used to run large problems and usually accessed via a network ==> 3) servers1.1.2 10^15 or 2^50 bytes (i.e., 10 to the 15 or 2 to the 50) ==> 7) petabyte1.1.3 Computer composed of hundreds to thousands of processors and terabytesof memory ==> 5) supercomputers1.1.4 Today's science fiction application that probably will be available in nearfuture ==> 1) virtual worlds1.1.5 A kind of memory called random access memory==> 12) RAM1.1.6 Part of a computer called central processor unit ==> 13) CPU1.1.7 Thousands of processors forming a large cluster ==> 8) datacenters1.1.8 A microprocessor containing several processors in the same chip==> 10) multi-core processors1.1.9 Desktop computer without screen or keyboard usually accessed via a network==> 4) low-end servers1.1.10 Currently the largest class of computer (i.e., there are moreof these types of computers than any others) that runs one applicationor one set of related applications ==> 9) embedded computers1.1.11 Special language used to describe hardware components ==> 11) VHDL1.1.12 Personal computer delivering good performance to single users at lowcost ==> 2) desktop computers1.1.13 Program that translates statements in high-level language to assemblylanguage==> 15) compiler1.1.14 Program that translates symbolic instructions to binary instructions==> 21) assembler1.1.15 High-level language for business data processing ==> 25) cobol1.1.16 Binary language that the processor can understand==> 19) machine language1.1.17 Commands that the processors understand ==> 17) instruction1.1.18 High-level language for scientific computation ==> 26) fortran1.1.19 Symbolic representation of machine instructions ==> 18) assembly language1.1.20 Interface between user's program and hardware providing a variety ofservices and supervision functions ==> 14) operating system1.1.21 Software/programs developed by the users ==> 24) application software1.1.22 Binary digit (value 0 or 1) ==> 16) bit1.1.23 Software layer between the application software and the hardware thatincludes the operating system and the compilers ==> 23) system software1.1.24 High-level language used to write application and system software==> 20) C1.1.25 Portable language composed of words and algebraic expressions thatmust be translated into assembly language before run in a computer ==> 22) high-level language1.1.26 10^12 or 2^40 bytes==> 6) terabyteExercise 1.21.2.1 For a color display using 8 bits for each of the primary colors (red, green, blue) per pixel, what should be the minimum size in bytes of the frame buffer to store a frame?8 bits × 3 colors = 24 bits/pixel => 4 bytes/pixel.1280 × 800 pixels = 1,024,000 pixels.1,024,000 pixels × 4 bytes/pixel = 4,096,000 bytes (approxitly 4 Mbytes).1.2. 2 How many frames could it store, assuming the memory contains no other information?2 GB = 2000 MbytesNumber of frames = 2000 Mbytes/4 Mbytes = 500 frames1.2.3 If a 256 Kbytes file is sent through the Ethernet connection, how long it would take?1 gigabit network ==> 1 gigabit/per second = 125 Mbytes/second.File size: 256 Kbytes = 0.256 Mbytes.Time= Mbytes/network speed: 0.256/125 = 2.048 ms.Exercise 1.31.3.1 Which processor has the highest performance expressed in instructions per second?P2 has the highest performanceP1 (Performance = instructions/sec) = 2 × 10^9/1.5 = 1.33 × 10^9P2 (Performance = instructions/sec) = 1.5 × 10^9/1.0 = 1.5 × 10^9P3 (Performance = instructions/sec) = 3 × 10^9/2.5 = 1.2 × 10^91.3.2 If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions.Number of cycles = time × clock rateCycles (P1) = 10 × (2 × 10^9) = 20 × 10^9 sCycles (P2) = 10 × (1.5 × 10^9) = 15 × 10^9 sCycles (P3) = 10 × (3 × 10^9) = 30 × 10^9 sTime = (Number of instructions × CPI)/clock rate; Number of instructions = Number of cycles/CPI Instructions (P1) = 20 × 10^9/1.5 = 13.33 × 10^9Instructions (P2) = 15 × 10^9/1 = 15 × 10^9Instructions (P3) = 30 × 10^9/2.5 = 12 × 10^91.3.3 We are trying to reduce the time by 30% but this leads to an increase of 20% in the CPI. What clock rate should we have to get this time reduction?Time new = Time old × 0.7 = 7 sCPI new = CPI old × 1.2 => CPI (P1) = 1.8, CPI (P2) = 1.2, CPI (P3) = 3Time = Number of instructions × CPI/clock rateTime (P1) = 13.33 × 10^9 × 1.8/7 = 3.42 GHzTime (P2) = 15 × 10^9 × 1.2/7 = 2.57 GHzTime (P3) = 12 × 10^9 × 3/7 = 5.14 GHz1.3.4 Find the IPC (instructions per cycle) for each processor.IPC new = 1/CPI old = Number of instructions/ (time × clock rate)IPC (P1) = (20 x 10^9)/(7 x 3) = 1.42IPC (P2) = (30 x 10^9)/(10 x 2.5) = 2IPC (P3) = (90 x 10^9)/(9 x 4) = 3.331.3.5 Find the clock rate for P2 that reduces its execution time to that of P1.Time new/Time old = 7/10 = 0.7.Clock rate new = Clock rate old/0.7 = 1.5 GHz/0.7 = 2.14 GHz1.3.6 Find the number of instructions for P2 that reduces its execution time to that of P3.Time new/Time old = 9/10 = 0.9.Instructions new = Instructions old × 0.9 = (30 × 10^9)× 0.9 = 27 × 10^9Exercise 1.41.4.1 Given a program with 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C and 20% class D, which implementation is faster?P2 has a faster implementation timeClass A: 10^5 instr.Class B: 2 × 10^5 instr.Class C: 5 × 10^5 instr.Class D: 2 × 10^5 instr.Time = Number of instructions × CPI/clock rateP1: Time class A = 0.66 × 10^−4Time class B = 2.66 × 10^−4Time class C = 10 × 10^−4Time class D = 5.33 × 10^−4Total time P1 = 18.65 × 10^−4P2: Time class A = 10^−4Time class B = 2 × 10^−4Time class C = 5 × 10^−4Time class D = 3 × 10^−4Total time P2 = 11 × 10^−41.4.2 What is the global (i.e., average) CPI for each implementation?CPI = time × clock rate/ number of instructionsCPI (P1) = (18.65 × 10^−4) × (1.5 × 10^9)/10^6= 2.79CPI (P2) = (11 × 10^−4) × (2 × 10^9)/10^6 = 2.21.4.3 Find the clock cycles required in both cases.Clock cycles (P1) = (10^5 × 1) + ((2 × 10^5) × 2) + ((5 × 10^5) × 3) + ((2 × 10^5) × 4) = 28 × 10^5Clock cycles (P2) = (10^5 × 2) + ((2 × 10^5) × 2) + ((5 × 10^5) × 2) + ((2 × 10^5) × 3) = 22 × 10^51.4.4 Assuming that arithmetic instructions take 1 cycle, load and store 5 cycles and branch 2 cycles, what is the execution time of the program in a 2 GHz processor?(500 × 1) + (50 × 5) + (100 × 5) + (50 × 2) × (0.5 × 10^–9) = 675 ns1.4.5 Find the CPI for the program.CPI = time × clock rate/ Number of instructionsCPI = ((675 × 10^–9) × (2 × 10^9))/700 = 1.921.4.6 If the number of load instructions can be reduced by one-half, what is the speed-up and the CPI?Time = ((500 × 1) + (50 × 5) + (50 × 5) + (50 × 2)) × (0.5 × 10^–9) = 550 nsSpeed-up = 675 ns/550 ns = 1.22CPI = ((550 × 10^–9) × (2 × 10^9))/700 = 1.57Exercise 1.51.5.1 Assume that peak performance is defined as the fastest rate that a computer can execute any instruction sequence. What are the peak performances of P1 and P2 expressed in instructions per second?1.5.2 If the number of instructions executed in a certain program is divided equally among the classes of instructions except for class A, which occurs twice as often as each of the others. Which computer is faster? How much faster is it?1.5.3 If the number of instructions executed in a certain program is divided equally among the classes of instructions except for class E, which occurs twice as often as each of the others? Which computer is faster? How much faster is it?1.5.4 Assuming that computes take 1 cycle, loads and store instructions take 10 cycles, and branches take 3 cycles, find the execution time of the program on a 3 GHz MIPS processor.1.5.6 Assuming that computes take 1 cycle, loads and store instructions take 2 cycles, and branches take 3 cycles, what is the speed-up of a program if the number of compute instruction can be reduced byone-half?Exercise 1.61.6.1 For the same program, two different compilers are used. The table above shows the execution time of the compiled program. Find the average CPI for the program given that the processor has a clock cycle time of 1 nS.1.6.2 Assume the average CPI found in 1.6.1, but that the compiled program runs on two difference processors. If the execution times on the two processors are the same, how much faster is the clock of the processor running compiler A’s code versus the clock of the processor running compiler B’s code?1.6.3 A new compiler is developed that uses only 600 million instructions and has an average CPI of 1.1. What is the speed-up of using this new compiler versus using Compiler A or B on the original processorof 1.6.1?1.6.4 Assume that peak performance is defined as the fastest rate that a computer can execute any instruction sequence. What is the peak performance of P1 and P2 expressed in instructions per second?1.6.5 If the number of instructions executed in a certain program is divided equally among the classes of instructions in Problem2.36.4 except for class A, which occurs twice as often as each of the others, how much faster is P2 than P1?1.6.6 At what frequency does P2 have the same performance as P1 for the instruction mix given in 1.6.5? Exercise 1.71.7.1 What is the geometric mean of the ratios between consecutive generations for both clock rate and power? (The geometric mean is described in Section 1.7.)Geometric mean clock rate ratio = (1.28 × 1.56 × 2.64 × 3.03 × 10.00 × 1.80 × 0.74)1/7 = 2.15Geometric mean power ratio = (1.24 × 1.20 × 2.06 × 2.88 × 2.59 × 1.37 × 0.92)1/7 = 1.621.7.2 What is the largest relative change in clock rate and power between generations?Largest clock rate ratio = 2000 MHz/200 MHz = 10Largest power ratio = 29.1 W/10.1 W = 2.881.7.3 How much larger is the clock rate and power of the last generation with respect to the first generation?Clock rate: 2.667 × 10^9/12.5 × 106 = 212.8Power: 95 W/3.3 W = 28.781.7.4 Find the average capacitive loads, assuming a negligible static power consumption.Capacitive = Power/Voltage^2 × clock rate80286: C = 0.0105 × 10^−680386: C = 0.01025 × 10^−680486: C = 0.00784 × 10^−6Pentium: C = 0.00612 × 10^−6Pentium Pro: C = 0.0133 × 10^−6Pentium 4 Willamette: C = 0.0122 × 10^−6Pentium 4 Prescott: C = 0.00183 × 10^−6Core 2: C = 0.0294 × 10^−61.7.5 Find the largest relative change in voltage between generations.Pentium Pro/Pentium 4 Willamette = 3.3/1.75 = 1.781.7.6 Find the geon1etric mean of the voltage ratios in the generations since the Pentium.Pentium Pro / Pentium => 3.3/5 = 0.66Pentium 4 Willamette/ Pentium Pro => 1.75/3.3 = 0.53Pentium 4 Prescott / Pentium 4 Willamette => 1.25/1.75 = 0.71Core 2 / Pentium 4 Prescott => 1.1/1.25 = 0.88Geometric mean = 0.68Exercise 1.81.8.1 How much has the capacitive load been reduced between versions if the dynamic power has been reduced by 10%?Power1 = V^2× clock rate × Capacitive; Power2 = 0.9 Power1C2/C1 = 0.9 × 5^2 × 0.5 × 10^9/3.3^2 × 1 × 10^9 = 1.031.8.2 By how much has the dynan1ic power been reduced if the capacitive load does not change?Power2/Power1 = V22 × clock rate2/V12 × clock rate1Power2/Power1 = 0.87 => Reduction of 13%1.8.3 Assuming that the capacitive load of version 2 is 80% the capacitive load of version 1, find the voltage for version 2 if the dynamic power of version 2 is reduced by 40% from version 1.Power2 = V22 × 1 × 10^9 × 0.8 × C1 = 0.6 × Power1Power1 = 52 × 0.5 × 10^9 × C1V22 × 1 × 10^9 × 0.8 × C1 = 0.6 × 52 × 0.5 × 10^9 × C1V2 = ((0.6 × 52 × 0.5 × 10^9)/(1 × 10^9 × 0.8))1/2 = 3.06 V1.8.4 By what factor does the dynamic power scales?Power new = 1 × C old × V2old/(2−1/4)2× clock rate × 21/2 = Power oldAccording to this the power scales by 1.1.8.5 Find the scaling of the capacitance per unit area.1/2−1/2 = 21/21.8.6 Using data from Exercise 1.7, find the voltage and clock rate of the Core 2 processor for the next process generation.Voltage = 1.1 × 1/2−1/4 = 0.92 VClock rate = 2.667 × 21/2 = 3.771 GHzExercise 1.91.9.1 Find the percentage of the total dissipated power comprised by static power.1.9.2 If the total dissipated power is reduced by 10% while maintaining the static to total power rate of problem 1.9.1, how much should the voltage be reduced to maintain the same leakage current?1.9.3 Determine the ratio of static power to dynamic power for each technology.1.9.4 Determine the static power for each version at 0.8 V, assuming a static to dynamic power ratio of 0.6.Power st/Power dyn = 0.6 => Power st = 0.6 × Power dyna)Power st = 0.6 × 35 W = 21 Wb)Power st = 0.6 × 30 W = 18 W1.9.5 Determine the static power and dynamic power dissipation assuming the rates obtained in problem 1.9.1.L lk = Voltage / Power sta) I lk = 21/0.8 = 26.25 Ab) I lk = 18/0.8 = 22.5 AExercise 1.101.10.1 The table above shows the number of instructions required per processor to complete a program on a multiprocessor with 1, 2, 4, or 8 processors. What is the total number of instructions executed per processor? What is the aggregate number of instructions executed across all processors?Instructions per processor = (Arith Inst x Arith Cpi)+( load Inst x load Cpi)+( branch Inst x branch Cpi)program on I, 2, 4, and 8 processors. Assume that each processor has a 2 GHz clock frequency.1.10.3 If the CPI of arithmetic instructions was doubled, what would the impact be on the execution time of the program on 1,2,4, or 8 processors?1.10.4 Assuming a 3 GHz clock frequency, what is the execution time of the program using 1, 2, 4, or 8 cores.1.10.5 Assuming that the power consumption of a processor core can be described by the following equation:Power = (5.0n1A)/(MHz)* Voltage^2Where the operation voltage of the processor is described by the following equation:Voltage = 1/5*Frequency+0.4With the frequency measured in GHz. So, at 5 GHz, the voltage would be 1.4V. Find the power consumption ofthe program executing on 1, 2, 4, and 8 cores assuming that each core is operating at a 3GHz clock frequency. Likewise, find the power consumption of the program executing on I, 2, 4, or 8 cores assuming that each core is operating at 500 MHz.Exercise 1.111.11.1 Find the yield.Wafer area = π × (d/2)2a. Wafer area = π × 7.52 = 176.7 cm2b. Wafer area = π × 12.52 = 490.9 cm2Die area = wafer area/dies per wafera. Die area = 176.7/90 = 1.96 cm2b. Die area = 490.9/140 = 3.51 cm2Yield = 1/ (1 + (defect per area × die area)/2)2a. Yield = 0.97b. Yield = 0.921.11.2 Find the cost per die.Cost per die = cost per wafer/ (dies per wafer × yield)a. Cost per die = 0.12b. Cost per die = 0.161.11.3 If the number of dies per wafer is increased by 100/0 and the defects per area unit increases by 150/0, find the die area and yield.a. Dies per wafer = 1.1 × 90 = 99Defects per area = 1.15 × 0.018 = 0.021 defects/cm2Die area = wafer area/Dies per wafer = 176.7/99 = 1.78 cm2Yield = 0.97b. Dies per wafer = 1.1 × 140 = 154Defects per area = 1.15 × 0.024 = 0.028 defects/cm2Die area = wafer area/Dies per wafer = 490.9/154 = 3.19 cm2Yield = 0.931.11.4 Find the defects per area unit for each technology given a die area of 200 mm2Yield = 1/(1 + (defect per area × die area)/2)2Defect per area = (2/die area) (y−1/2− 1)Replacing values for T1 and T2 we getT1: defects per area = 0.00085 defects/mm2 = 0.085 defects/cm2T2: defects per area = 0.00060 defects/mm2 = 0.060 defects/cm2T3: defects per area = 0.00043 defects/mm2 = 0.043 defects/cm2T4: defects per area = 0.00026 defects/mm2 = 0.026 defects/cm2Exercise 1.121.12.1 Find the CPI if the clock cycle time is 0.333 ns.CPI = clock rate × CPU time/instr. countClock rate = 1/cycle time = 3 GHza. CPI (pearl) = 3 × 10^9 × 500/2118 × 10^9 = 0.7b. CPI (mcf) = 3 × 10^9 × 1200/336 × 10^9 = 10.71.12.2 Find the SPEC ratio.SPECratio = ref. time/execution time.a. SPECratio (pearl) = 9770/500 = 19.54b. SPECratio (mcf) = 9120/1200 = 7.61.12.3 For these two benchmarks, find the geometric mean.(19.54 × 7.6)1/2 = 12.191.12.4 Find the increase in CPU time if the number of instruction of the benchmark is increased by 100/0 without affecting the CPI.CPU time = number of instructions × CPI/clock rateIf CPI and clock rate do not change, then the CPU time will increase equal to the number ofinstructions, which would be by 10%.1.12.5 Find the increase in CPU time if the number of instruction of the benchmark is increased by 10% and the CPI is increased by 5%.CPU time (before) = number of instructions × CPI/clock rateCPU time (after) = 1.1 × number of instructions × 1.05 × CPI/clock rateCPU times (after)/CPU time (before) = 1.1 × 1.05 = 1.155.CPU time is increased by 15.5%1.12.6 Find the change in the SPECratio for the change described in 1.12.5.SPECratio = reference time/CPU timeSPECratio (after)/SPECratio (before) = CPU time (before)/CPU time (after) = 1/1.1555 = 0.86The SPECratio is decreased by 14%.Exercise 1.131.13.1 Find the new CPI.CPI = (CPU × clock rate)/Number of instr.a. CPI = 450 × 4 × 10^9/ (0.85 × 2118 × 10^9) = 0.99b. CPI = 1150 × 4 × 10^9/ (0.85 × 336 × 10^9) = 16.101.13.2 In general, these CPI values are larger than those obtained in previous exercises for the same benchmarks. This is due mainly to the clock rate used in both cases, 3 GHz and 4 GHz. Determine whether the increase in the CPI is similar to that of the clock rate. If they are dissimilar, why?Clock rate ratio = 4 GHz/3 GHz = 1.33.a. CPI at 4 GHz = 0.99, CPI at 3 GHz = 0.7, ratio = 1.41b. CPI at 4 GHz = 16.1, CPI at 3 GHz = 10.7, ratio = 1.50They are different because the CPU time has been reduced by a lower percentage, even though the number of instructions has been reduced by 15%,1.13.3 How much has the CPU time been reduced?a. 450/500 = 0.90CPU time reduction = 10%b. 1150/1200 = 0.958CPU time reduction = 4.2%1.13.4 If the execution time is reduced by an additional 10% without affecting the CPI and with a clock rate of 4 GHz, determine the number of instructions.Number of instructions = CPU × clock rate/CPI.a. Number of instructions = 820 × 0.9 × 4 × 10^9/0.96 = 3075 × 10^9b. Number of instructions = 580 × 0.9 × 4 × 10^9/2.94 = 710 × 10^91.13.5 Determine the clock rate required to give a further 10% reduction in CPU time while maintaining the nun1ber of instructions and CPI unchanged.Clock rate = Number of instructions × CPI/CPU time.Clock rate new = Number of instructions × CPI/0.9 × CPU time = 1/0.9 clock rate old = 3.33 GHz.1.13.6 Determine the clock rate if the CPI is reduced by 15% and the CPU time by 20% while the number of instructions is unchanged.Clock rate = Number of instructions × CPI/CPU time.Clock rate new = Number of instructions × 0.85 × CPI/0.80 CPU time = 0.85/0.80 clock rate old = 3.18 GHz. Exercise 1.141.14.1 One usual fallacy is to consider the computer with the largest clock rate as having the large perfonl1ance. Check if this is true for PI and P2.Number of instructions = 106Tcpu (P1) = 106 × 1.25/4 × 10^9 = 0.315 × 10–3 sTcpu (P2) = 106 × 0.75/3 × 10^9 = 0.25 × 10–3 sClock rate (P1) > clock rate (P2), but performance (P1) < performance (P2)1.14.2 Another fallacy is to consider that the processor executing the largest number of instruction will need a larger CPU time. Considering that processor PI is executing a sequence of 106 instructions and that the CPI of processors PI and P2 do not change, detenl1ine the number of instructions that P2 can execute in the same time that PI needs to execute 106 instructions.P1: 106 instructions, Tcpu (P1) = 0.315 × 10–3 sP2: Tcpu (P2) = N × 0.75/3 × 109N = 1.26 × 1061.14.3 A common fallacy is to use MIPS (millions of instructions per second) to con1pare the performance of two different processors, and consider that the processor with the largest MIPS has the largest perforn1ance. Check if this is true for PI and P2.MIPS = Clock rate × 10−6 /CPIMIPS (P1) = 4 × 10^9 × 10–6/1.25 = 3200MIPS (P2) = 3 × 10^9 × 10–6/0.75 = 4000MIPS (P1) < MIPS (P2), performance (P1) < performance (P2) in this caseAnother common performance figure is MFLOPS (millions of floating-point operations per second), defined asMFLOPS = Number of FP operations/(execution time x 10^6)1.14.4 Find the MFLOPS figures for the programs.a. FP op = 106 × 0.4 = 4 × 105, clock cylesfp = CPI × Number of FP instr. = 4 × 105Tfp = 4 × 105 × 0.33 × 10–9 = 1.32 × 10–4 then MFLOPS = 3.03 × 103b. FP op = 3 × 106 × 0.4 = 1.2 × 106, clock cylesfp = CPI × Number of FP instr. = 0.70 × 1.2 × 106Tfp = 0.84 × 106 × 0.33 × 10–9 = 2.77 × 10–4 then MFLOPS = 4.33 × 1031.14.5 Find the MIPS figures for the programs.5 CPU clock cycles = FP cycles + CPI (L/S) × Number of instr. (L/S) + CPI (Branch) ×Number of instr. (Branch)a. 5 × 105 L/S instr., 4 × 105 FP instr. and 105 Branch instr.CPU clock cycles = 4 × 105 + 0.75 × 5 × 105 + 1.5 × 105 = 9.25 × 105Tcpu = 9.25 × 105 × 0.33 × 10–9 = 3.05 × 10–4MIPS = 106/ (3.05 × 10–4 × 106) = 3.2 × 103b. 1.2 × 106 L/S instr., 1.2 × 106 FP instr. and 0.6 × 106 Branch instr.CPU clock cycles = 0.84 × 106 + 1.25 × 1.2 × 106 + 1.25 × 0.6 × 106 = 3.09 × 106Tcpu = 3.09 × 106 × 0.33 × 10–9 = 1.01 × 10–3MIPS = 3 × 106/ (1.01 × 10–3 × 106) = 2.97 × 1031.14.6 Find the performance for the programs and con1pare with MIPS and MFLOPS.a. Performance = 1/Tcpu = 3.2 × 103b. Performance = 1/Tcpu = 9.9 × 102The second program has the higher performance, but the ﬁrst program has the higher MIPSﬁgure.Exercise 1.151.15.1 By how much is the total time reduced if the time for FP operations is reduced by 20%?a. T fp = 35 × 0.8 = 28 s; T p1 = 28 + 85 + 50 + 30 = 193 s; Reduction of 3.5%b. T fp = 50 × 0.8 = 40 s; T p4 = 40 + 80 + 50 + 30 = 200 s; Reduction of 4.7%1.15.3 Can the total time be reduced by 20% by reducing only the time for branch instructions?a. T p1 = 200 × 0.8 = 160 s; T fp + T int + T l/s = 170 s. No the total time cannot be reduced by 20%b. T p4 = 210 × 0.8 = 168 s; T fp + T int + T l/s = 180 s. No the total time cannot be reduced by 20% Assume that each processor has a 2 GHz clock rate.1.15.4 By how much must we improve the CPI of FP instructions if we want the program to run two times faster?Clock cyles = CPI fp × Number of FP instr. + CPI int × Number of INT instr. + CPI l/s × Number of L/Sinstr. + CPIbranch × Number of branch instr.Tcpu = clock cycles/clock rate = clock cycles/2 × 109a. 1 processor=> clock cycles = 8192; T cpu = 4.096 sb. 8 processors=> clock cycles = 1024; T cpu = 0.512 sTo half the number of clock cycles by improving the CPI of FP instructions:CPI improved fp × Number of FP instr. + CPI int × Number of INT instr. + CPI l/s × Number of L/S instr. + CPI branch × Number of branch instr. = clock cycles/2CPI improved fp= (clock cycles/2 − (CPI int × Number of INT instr. + CPI l/s × Number of L/S instr. +CPI branch × Number of branch instr.))/Number of FP instr.a. 1 processor: CPI improved fp = (4096 – 7632)/560 => not possibleb. 8 processors: CPI improved fp = (512 – 944)/80 => not possible1.15.5 By how much must we improve the CPI of L/S instructions if we want the program to run two times faster?Using clock cycle from 1.15.4:To half the number of clock cycles improving the CPI of L/S instructions:CPI fp × Number of FP instr. + CP Iint × Number of INT instr. + CP Iimproved l/s × Number of L/S instr. +CPI branch × Number of branch instr. = clock cycles/2CPI improved l/s = (clock cycles/2 − (CPIfp × Number of FP instr. + CPI int × Number of INT instr. +CPI branch × Number of branch instr.))/Number of L/S instr.a. 1 processor: CPI improved l/s = (4096 – 3072)/1280 = 0.8b. 8 processors: CPI improved l/s = (512 – 384)/160 = 0.81.15.6 By how much is the execution time of the program improved if the CPI of INT and FP instruction is reduced by 40% and the CPI of L/S and branch is reduced by 30%?Clock cyles = CPIfp× Number of FP instr. + CPI int× Number of INT instr. + CPI l/s× Number of L/S instr. +CPI branch × Number of branch instr.T cpu = clock cycles/clock rate = clock cycles/2 × 109CPI int = 0.6 × 1 = 0.6; CPI fp = 0.6 × 1 = 0.6; CPI l/s = 0.7 × 4 = 2.8; CPI branch = 0.7 × 2 = 1.4a. 1 processor => T cpu (before improved) = 4.096 s; T cpu (after improved) = 2.739 sb. 8 processors =>T cpu (before improved) = 0.512 s; T cpu (after improved) = 0.342 s。

家庭作业英

家庭作业英
homework，英文单词，主要用作名词，译为“家庭作业，课外作业”。

短语搭配：
Setting homework 布置作业 ; 布置家庭作业
in homework 上交作业 ; 交作业 ; 在做作业时 ; 交功课
homework help 家庭作业助手 ; 家庭作业帮助 ; 作业辅导 ; 功课帮助Homework corner 家庭作业角 ; 功课角 ; 作业的角落
Homework Time 作业时间 ; 到写作业的时间了 ; 户外活动 ; 写作业时间Good homework 好好写作业 ; 擅长功课 ; 好好做作业
Set homework 布置作业 ; 布置家庭作业
Copying homework 抄作业 ; 抄袭家庭作业
math homework 数学作业
例句：
You do the homework only half.
你做的作业只有一半。

I have to do my homework.
我不得不做我的作业。

You do your homework.
你做你的家庭作业。

homework 英语作文

homework 英语作文Homework is an essential part of the learning process for students. It helps them to reinforce what they have learned in class and develop important skills such as time management and critical thinking. However, homework can also be a source of stress and anxiety for many students, especially when they are overwhelmed with a heavy workload.For many students, homework is a daily struggle. They spend hours each night completing assignments, often sacrificing sleep and leisure time in order to keep up with the demands of their classes. This can take a toll on their mental and physical well-being, leading to feelings of exhaustion and burnout. In extreme cases, excessive homework can even contribute to the development of anxiety and depression in students.In addition to the negative impact on students' health, excessive homework can also have a detrimental effect on their academic performance. When students are overwhelmedwith assignments, they may rush through their work orsimply give up, leading to lower quality work and poorer grades. This can create a vicious cycle, as poor grades can further demotivate students and make them even less likelyto complete their homework in the future.Furthermore, the burden of excessive homework can also strain the relationship between students and their families. Parents may feel pressured to help their children withtheir homework, adding to their own stress and reducing the quality time they can spend together as a family. This can create tension and conflict within the household, further exacerbating the negative effects of excessive homework.In order to address these issues, it is important for educators to carefully consider the amount and nature ofthe homework they assign to students. Homework should be purposeful and meaningful, designed to reinforce key concepts and skills rather than simply keeping students busy. Teachers should also be mindful of the workload they are placing on students and strive to strike a balance between academic rigor and student well-being.Additionally, schools and educators should work to provide support for students who are struggling with their homework. This could include offering tutoring services, creating study groups, or providing resources for time management and study skills. By providing students with the support they need, schools can help to alleviate the stress and anxiety that can come with excessive homework.In conclusion, while homework is an important part of the learning process, it is essential for educators to be mindful of the potential negative effects of excessive homework on students. By carefully considering the amount and nature of the homework they assign, and providing support for students who are struggling, schools can help to ensure that homework remains a valuable tool for learning without becoming a source of stress and anxietyfor students.。

Homework1(上交时间10月14号)

Homework 1(上交时间：10月14号)倒排索引
实现目标
通过Hadoop提供的API接口，对所给文件建立一个倒排索引，使能够根据单词对文件进行检索
实现要求
文件内容为英文，存储在本地文件中。

建立倒排索引的程序运行完时，Reduce过程的输出形式为。

<单词，<文件>:词频,…..>
<“mapreduce”, 0.txt:1, 2.txt:2>
Map 过程
●首先使用默认的TextInputFormat类对输入文件进行处理，得到文本中的偏移量及其类容。

●Map过程对输入的<key, value>进行分析，得到需要的信息，单词，文件名，词频。

由于<key, value>对只能有两个只，则需根据情况将其中的两个合并，例如将单词与文件名合并。

Combine 过程
●经过map方法后，Combine将key相同的value相加，得到一个单词在文件中的词频。

●由于具有相同单词的记录应该被同一个reduce处理，所以这里应该修改key为单词，value为词频与文件的组合
Reduce 过程
经过上述两个过程后，reduce过程只需将相同key值的value组合成所需的输出格式即可。

homework英语作文

homework英语作文Title: The Importance of Homework。

Introduction:Homework plays a crucial role in the educational system as it reinforces the concepts learned in class, develops essential skills, and prepares students for future challenges. In this essay, we will explore the significance of homework and how it contributes to a student's academic success.Body:1. Reinforcement of Concepts:Homework serves as a tool to reinforce the concepts taught in the classroom. It allows students to practicewhat they have learned and apply it to real-life situations. Through homework, students can solidify their understandingof various subjects, ensuring that the knowledge is retained for a longer duration.2. Skill Development:Homework aids in the development of essential skills such as time management, self-discipline, and independent learning. By completing assignments within a given timeframe, students learn to prioritize tasks, manage their time effectively, and work independently. These skills are invaluable and will benefit them not only academically but also in their future careers.3. Preparation for Exams:Homework prepares students for exams by providing them with an opportunity to revise and review the material covered in class. Regular practice through homework allows students to identify their weaknesses and seek help if needed. It also helps them become familiar with the exam format, reducing anxiety and increasing their chances of success.4. Enhancement of Critical Thinking:Homework assignments often require students to think critically and solve problems independently. By analyzing and interpreting information, students develop their critical thinking skills, which are essential for success in higher education and the workplace. Homework encourages students to think beyond the surface level, fostering creativity and innovation.5. Parental Involvement:Homework provides an opportunity for parents to be involved in their child's education. Parents can assist their children with their homework, offering guidance and support. This involvement not only strengthens the parent-child bond but also helps parents understand their child's strengths and weaknesses, enabling them to provide additional assistance if required.Conclusion:In conclusion, homework plays a vital role in astudent's academic journey. It reinforces concepts, develops essential skills, prepares students for exams, enhances critical thinking, and fosters parental involvement. While the debate on the effectiveness of homework continues, it is clear that when assigned appropriately and in moderation, homework can significantly contribute to a student's overall academic success. Thus, it is essential for educators, parents, and students to recognize and appreciate the importance of homework.。

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第一次作业的提交时间：
．．．．．
．．．．．．．．．，周一课间
．．．．．．．．．．．2015.3.23
关系模式如下：
⏹职工E（ename，eno，bdate，addr，salary，dno）
姓名工号出生日期家庭地址工资所在部门编号
⏹部门D（dname，dno，mgreno）
部门名称部门编号部门负责人的工号
⏹项目P（pname，pno，city，dno）
项目名称项目编号所在城市主管部门编号
⏹工作W（eno，pno，hours）
职工工号项目编号工作时间
⏹职工家属Depend（eno，name，sex）
职工工号家属的姓名家属的性别
请用关系代数表达式来表示下述数据查询操作。

1)检索职工Smith所参与的项目的编号和名称；
2)检索所有部门负责人的信息，结果返回部门名称、部门负责人的工号及姓名；
3)检索拥有两个或两个以上家属的职工的姓名；
4)检索不带家属的职工的姓名；
5)检索参加过‘p2’号项目的职工的工号；
6)检索只参加过‘p2’号项目的职工的姓名；
7)检索只参加过一个项目的职工的姓名；
8)检索参加了所有项目的职工的工号；
9)检索全体3号部门的职工都参加了的项目的编号和名称；
10)检索工资收入最高的职工的姓名；
11)查询每一个部门中工资收入最高的职工，结果返回部门编号以及该部门中工资收入
最高的职工的工号。