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双原子分子中两原子之间的作用力势可表示为 \[ V(r)=V_0[1 - \mathrm{e}^{-(r - R)/a}]^2 - V_0 \] $r$为两原子之间的相对距离，$R$与$a$为正的常数。$R > a$，且$\mathrm{e}^{R/a} \gg 1$。 当$r = R$时，$V(r)$取最小值$-V_0$，当$r \to \infty$时，$V(r) \to 0$，当$r \to 0$时，$V(r) \to \infty$。求轨道角动量$l = 0$的束缚定态能量。
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[
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Quantum Mechanics
|
Physics
|
某市某街道为南北走向，汽车高峰流量为2600辆/h，车速平均40km/h，每辆车排出碳氢化合物量为2.5×10⁻²g/s，汽车排放高度0.4m，试问某一阴天白天吹东风，风速3m/s，该街道西边400m处碳氢化合物浓度为多少？
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[
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Atmospheric Chemistry
|
Earth Science
|
对于现有对流层大气，如果欧美地区的氮氧化物排放增加1倍而一氧化碳排放增加50%，则北半球和南半球的对流层羟基自由基平均浓度分别会发生什么变化？
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
温度为 200℃、压力为 0.101 MPa 的含尘气体通过一旋风除尘器，筒体直径为 D，筒体长度 L = 2 D，锥体高度 H = 2 D，进口宽度 b = 1/5 D，进口高度 h = 3/5 D，尘粒密度为 2000 kg/m³。若旋风除尘器筒体直径为 0.65 m，进口气速为 21 m/s，试求： （1）气体处理量（标态） （2）气体通过旋风除尘器的压力损失 （3）尘粒的临界粒径
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[
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Atmospheric Chemistry
|
Earth Science
|
晚上空气温度为-8.8度，相对湿度为80%，如果在日出前气温为-14度，试求雾的液态水含量（克/米3）和雾形成时释放热量（焦耳/米3）。
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
据估计，某燃烧着的垃圾堆以3gs的速率排放氨氧化物。在风速为7m/s的阴天夜里，源的正下风方向3km处的平均浓度是多少?假设这个垃圾堆是一个无有效源高的地面点源。
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
一股含庚烷气流，流量为50m³/min，温度150°C，气压为1atm，庚烷浓度为50000ppm。需要冷却到多少温度，才能去除40%庚烷？若溶剂庚烷比重为0.75，市场价为7.0元/L，估算每天冷凝回收下来的庚烷金额。
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
对流层甲基氯仿受到羟基自由基的显著影响。在考虑全球对流层大气中氮氧化物的源总量和汇总量始终处于平衡状态的情况下，如果平衡状态A下的氮氧化物汇比平衡状态B下的氮氧化物汇多30%，则状态A下甲基氯仿的生命周期比状态B长还是短？
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
欲设计一个用于取样的旋风分离器，希望在入口气速为20m/s时，其空气动力学分割直径为1μm。1）估算该旋风分离器的筒体外径；2）估算通过该旋风分离器的气体流量。
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
燃料油的重量组成为：C86%，H14%。在干空气下燃烧，烟气分析结果（基于干烟气）为：O₂1.5%；CO 600×10⁻⁶（体积分数）。试计算燃烧过程的空气过剩系数。若实测烟尘浓度为24mg/m³，试问校正至过剩空气系数α=1.8时烟尘浓度是多少？
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[
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Atmospheric Chemistry
|
Earth Science
|
普通煤的成分分析结果（质量分数）如下：C: 65.7%; 灰分: 18.1%; S: 1.7%; H: 3.2%; 水分: 9.0%; O: 2.3%, 含N量不计。计算燃煤1kg所需要的理论空气量和SO₂在烟气中的浓度。
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
大气中的某种痕量气体对特定波长的可见光具有一定的弱吸收，卫星探测器可以利用这一特点探测该气体。假设该气体的体积混合比在对流层内均一，若地面以上2km处突然出现大量气溶胶，则卫星探测器对于对流层不同高度处该痕量气体的浓度变化的敏感性会发生什么变化？
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[
""
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[
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Atmospheric Chemistry
|
Earth Science
|
已知地面气压P = 1000 hPa，温度t = 10℃，露点ta=5℃ （1）请求出其比湿q，虚温tv，空气密度ρ（kg/m³） （2）若温度和比湿不随高度改变，求850 hPa等压面离地面的高度 z。 （3）若在850 – 1000 hPa的气柱中，比湿不随高度变化，求出底面积为1 m²的气柱中水汽总质量 mv (kg/m²)，如果其全部凝结成液体降落至地面，则层深有多厚（cm）？
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[
""
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[
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Atmospheric Chemistry
|
Earth Science
|
为了能使云中半径为10-6厘米和10-5厘米的纯水滴长大，云中的过饱和度Δf该多少？
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[
""
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[
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Atmospheric Chemistry
|
Earth Science
|
地面气压为760mm汞柱时，测量得到1.5~1.6μm波段内的太阳直接辐射S∆λ入下表所示：
天顶角分别为40°、50°、60°、70°时，对应的S∆λ（W·cm⁻²）分别为13.95、12.55、10.46、7.67。
请计算大气上界处S∆λ,0，光学厚度δ∆λ及透明系数P∆λ。
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
采用泥炭为滤料的生物过滤床处理苯乙烯废气，最大去除负荷在60-75 g/(m³·h)之间。最近的研究表明当负荷为40 g/(m³·h)，苯乙烯的去除率可以达到97%。为了满足某地区的臭气排放标准，利用生物过滤床处理造船厂排放出的苯乙烯废气。已知废气气量为10000m³/h（30°C，1atm），苯乙烯浓度50μL/L，要求处理效率至少90%以上。假设设计的床层深度至少1m，表面负荷不超过1.5m/min，则该过滤床的长、宽、高分别为多少？
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[
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[
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Atmospheric Chemistry
|
Earth Science
|
利用活性炭吸附处理脱脂生产中排放的废气，排气条件为294K，1.38×10⁵Pa，废气量25400m³/h。废气中含有体积分数为0.02的三氯乙烯，要求回收率99.5%。已知采用的活性炭的吸附容量为28kg三氯乙烯/100kg活性炭，活性炭的密度为577kg/m³，其操作周期为4h，加热和解析2h，各用1h，试确定活性炭的用量和吸附塔尺寸。
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[
""
] |
[
""
] |
Atmospheric Chemistry
|
Earth Science
|
假设现有地球大气中的氮气分子全部变成氩气分子，计算此时大气的干绝热递减率（单位：K/km）
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[
""
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[
""
] |
Atmospheric Chemistry
|
Earth Science
|
请列举3个在运用气块法评估大气不稳定性时忽略掉的物理过程。
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[
""
] |
[
""
] |
Atmospheric Chemistry
|
Earth Science
|
蜡烛在一个有少量空气的完全隔热箱体内燃烧。蜡烛燃烧的热值为3.48×10⁷J/kg，蜡烛燃烧速度为0.0454kg/h。每kg蜡烛消耗的空气量为50kg，空气初始温度15°C。计算排出箱体气体的温度。
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[
""
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[
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] |
Atmospheric Chemistry
|
Earth Science
|
在砂型中浇铸尺寸为 $300 \times 300 \times 20 \mathrm{~mm}$ 的纯铝板。设铸型的初始温度为 $20^{\circ} \mathrm{C}$ ，浇注后瞬间铸件－铸型界面温度立即升至纯铝熔点 $660^{\circ} \mathrm{C}$ ，且在铸件凝固期间保持不变。浇铸温度为 $670^{\circ} \mathrm{C}$ ，金属与铸型材料的热物性参数见下表： \begin{tabular}{|c|l|l|l|l|l|l|} \hline \multirow{2}{|c|}{{$材料~~~热物性$}} & 导热系数 $\lambda$ & 比热容$C$ & 密度 $\rho$ & 热扩散率 ${ }{\text {a }}$ & 结晶潜热 \\ \hline & $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$ & $\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})$ & $\mathrm{kg} / \mathrm{m}^3$ & $\mathrm{m}^2 / \mathrm{s}$ &$\mathrm{J} /\mathrm{kg}$ \\ \hline 纯铝 &$212$ & $1200$ & $2700$ & $6.5 \times 10^{-5}$ & $3.9 \times 10^{5}$ \\ \hline 砂型 & $0.739$ & $1840$ & $1600$ & $2.5 \times 10^{-7}$ & \\ \hline \end{tabular} 试求：（1）计算不同时刻铸件凝固层厚度$s$，并给出$0-120s$每隔$20s$的厚度值； （2）分别用＂平方根定律＂及＂折算厚度法则＂计算铸件的完全凝固时间，并分析差别。
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[
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[
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Material Synthesis and Processing
|
Materials Science
|
在平面挤压工艺中，金属坯料通过平行模具间隙发生塑性变形。已知入口厚度为 $H$ ，出口厚度为 $h$ ，满足 $H=2 h$ 。坯料入口水平速度为 $\dot{u}_0$ ，出口速度为 $\dot{u}_1=2 \dot{u}_0$ 。假设材料内部存在一菱形滑移区 $A B C D$ ，其顶点 $O$ 位于中心线上，各边与水平方向均成 $45^{\circ}$ 夹角（即滑移线方向为 $\pm 45^{\circ}$ ）。材料的最大剪应力为 $K$ 。 要求：运用上限法，推导该挤压过程中单位面积上的平均流动应力 $p$ 。（提示：需考虑滑移线速度不连续对剪切功率的贡献，并与外力功率平衡。）
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[
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Material Synthesis and Processing
|
Materials Science
|
塑料模成型零件的热处理应注意什么？试述用低碳合金渗碳钢制造的塑料模型腔件的制造工艺路线。
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[
""
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[
""
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Material Synthesis and Processing
|
Materials Science
|
为什么制造蜡模多采用糊状蜡料加压成形，而较少采用蜡液浇铸成形？为什么脱蜡时水温不应达到沸点？
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[
""
] |
[
""
] |
Material Synthesis and Processing
|
Materials Science
|
平行砧板间平面压缩变形，已知接触表面摩擦力 $\tau=\mu \sigma_y, \mu$ 为摩擦系数，最大剪应力为$K$，试用主应力法求 $\sigma_y$ 分布规律及平均流动应力$P$
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[
""
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[
""
] |
Material Synthesis and Processing
|
Materials Science
|
试简述不锈钢焊条药皮发红的原因？有什么解决措施？
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[
""
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[
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Material Synthesis and Processing
|
Materials Science
|
今有塑料制靠背椅（整体件），工作面要求光洁，无划痕，局部有细花皮纹，年产量 45 万件，试选用模具（型腔件）的钢材品种，并提出热处理工艺路线。
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[
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[
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Material Synthesis and Processing
|
Materials Science
|
请为下列工作条件下的塑料膜选用材料：
（1）形状简单，精度要求低，批量不大的塑料膜；
（2）高耐磨、高精度、型腔复杂的塑料膜；
（3）大型、复杂、产品批量大的塑料膜注塑膜；
（4）耐蚀、高精度塑料模具。
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[
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[
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Material Synthesis and Processing
|
Materials Science
|
Question: Industrial printer settings: 1) High-resolution mode requires cooling system active (temp <35°C); 2) Duplex printing disabled when paper weight >200gsm; 3) CMYK mode needs 4+ ink cartridges above 15%; 4) Maintenance cycle blocks all functions every 500 pages. Current job: 220gsm cardstock, 3 cartridges at 20% (C/M/Y), temp 32°C, page counter 499. What's possible?\nA. Single-side CMYK print\nB. High-res duplex\nC. Maintenance lockout\nD. Grayscale high-res
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[
""
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[
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Material Synthesis and Processing
|
Materials Science
|
在一套管式换热器中，用120℃的饱和水蒸气在环隙中冷凝放热，使内管中湍流流动的流量为3000kg/h的苯，从20℃加热到80℃。当流量增加到4500kg/h时，只能从20℃加热到76℃。试计算换热器的传热面积和流量为4500kg/h时的总传热系数K。 计算时，水蒸气冷凝的a值取用8000W/(m²·K)。可忽略管壁热阻及污垢热阻，并可当平壁处理。
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[
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[
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Chemical Engineering and Technology
|
Chemistry
|
现有 16 个处理器, 编号分别为 \( 0,1, \cdots, 15 \), 用一个 \( N = 16 \) 的互连网络互连。处理器 \( i \) 的输出通道连接互连网络的输入端 \( i \), 处理器 \( i \) 的输入通道连接互连网络的输出端 \( i \)。当该互连网络实现的互连函数分别为 (1) \( \mathrm{Cube}_{3} \) (2) \( \mathrm{PM} 2_{+3} \) (3) \( \mathrm{PM} 2_{-0} \) (4) \( \sigma \) (5) \( \sigma(\sigma) \) 时, 分别给出与第 13 号处理器所连接的处理器号。
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[
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Computer System Architecture
|
Computer Science
|
Read the following absdiff function written in .c and answer the questions. (3 pts)
/* C code starts here. */
long absdiff(long x, long y) {
long result;
if (x > y)
result = x-y;
else
result = y-x;
return result;
}
/* Assembly code starts here. */
absdiff:
cmpq %rsi, %rdi
jle .L4
movq %rdi, %rax
subq %rsi, %rax
.L4:
movq %rsi, %rax
subq %rdi, %rax
ret
subq [src], [dest] dest = dest - src
(1) X is stored in register: ________________; Y is stored in register: ________________; Return
value is stored in register: ________________. (Write the register name)
(2) Let x = 0x0000 0000, y = 0x0000 0000. Write the conditional flags after cmpq:
ZF = ________________; SF = ________________; OF = ________________;
(3) Let x = 0x0000 0001, y = 0x8000 0000. Write the conditional flags after cmpq:
ZF = ________________; SF = ________________; OF = ________________;
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[
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[
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Computer System Architecture
|
Computer Science
|
若干个等待访问磁盘者依次要访问的柱面为 38, 90, 55, 20, 65, 11, 70，假设每移动一个柱面需要 2ms 时间，移动臂当前位于 50 号柱面，磁头正向磁道号减少的方向移动，请按 FCFS, SSTF, SCAN 算法分别计算为完成上述访问总共花费的寻找时间。
|
[
""
] |
[
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] |
Computer System Architecture
|
Computer Science
|
我现在有三个任务，任务1，2，3，任务1输出数据到任务2和任务3，任务2接受任务1的输入，输出数据到任务3，任务3同时接受来自任务1和任务2的输入。所有数据传输都是FIFO模式，请问在什么情况下会发生死锁
|
[
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[
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Computer System Architecture
|
Computer Science
|
In this problem, we will look at the Blacklisting Memory Scheduler (BLISS) to reduce unfairness. There are two key aspects of BLISS that you need to know.
• When the memory controller services consecutive requests from a particular application, this application is blacklisted. We name this non-negative integer the Blacklisting Threshold.
• The blacklist is cleared periodically every 10000 cycles starting at t = 0.
To reduce unfairness, memory requests in BLISS are prioritized in the following order:
• Non-blacklisted applications’ requests
• Row buffer hit requests
• Older requests
The memory system for this problem consists of 2 channels with 2 banks each. Tables 1 and 2 show the memory request stream in the same bank for both applications at different times (t = 0 and t = 10). For both tables, a request on the left-hand side is older than a request on the right-hand side in the same table. The applications do not generate more requests than those shown in Tables 1 and 2. The memory requests are labeled with numbers that represent the row position of the data within the accessed bank. Assume the following for all questions:
• Arow buffer hit takes 100 cycles.
• Arow buffer miss (i.e., opening a row in a bank with a closed row buffer) takes 200 cycles.
• A row buffer conflict (i.e., closing the currently open row and opening another one) takes 250 cycles.
• All row buffers are closed at time t = 0
| Application A (Channel 0, Bank 0) | | | | | | | | |
|-----------------------------------|--------|--------|--------|--------|--------|--------|--------|--------|
| Application B (Channel 0, Bank 0) | Row 2 | Row 2 | Row 2 | Row 2 | Row 2 | Row 3 | Row 3 | Row 4 |
Table 1: Memory requests of the two applications at t = 0
| Application A (Channel 0, Bank 0) | Row 3 | Row 7 | Row 2 | Row 0 | Row 5 | Row 3 | Row 8 | Row 9 |
|-----------------------------------|--------|--------|--------|--------|--------|--------|--------|--------|
| Application B (Channel 0, Bank 0) | Row 2 | Row 2 | Row 2 | Row 2 | Row 2 | Row 3 | Row 3 | Row 4 |
Table 2: Memory requests of the two applications at t = 10. Note that none of the Application B’s existing requests are serviced yet.
If we use the BLISS scheduler, for what value(s) of η (the Blacklisting Threshold) will
the slowdowns of both applications be equal to those obtained with FR-FCFS?
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[
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[
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Computer System Architecture
|
Computer Science
|
请求分页管理系统中，假设某进程的页表内容如下表所示:
|页号 |页框(Page Frame)号 |有效位(存在位) |
|--|--|--|
|0 |101H| 1|
| 1 |---- |0|
| 2 |254H| 1|
|3|----|0|
页面大小为 4KB，一次内存的访问时间是 100ns，一次快表(TLB)的访问时间是 10ns，处
理一次缺页的平均时间 108 ns(已含更新 TLB 和页表的时间)，进程的驻留集大小固定为 2，采用最近最少使用置换算法(LRU)和局部淘汰策略。假设 (1) TLB 初始为空; (2) 地址转换
时先访问 TLB，若 TLB 未命中，再访问页表(忽略访问页表之后的 TLB 更新时间); (3) 有
效位为 0 表示页面不在内存，产生缺页中断，缺页中断处理后，返回到产生缺页中断的指 令处重新执行。设有虚地址访问序列 2347H、1565H、08B4H、15B0H，请问: 1) 依次访问上述四个虚地址，各需多少时间?给出计算过程。
2) 基于上述访问序列，虚地址 1565H和08B4H的物理地址是多少?请说明理由
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[
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Computer System Architecture
|
Computer Science
|
Assume an in-order processor that employs Runahead execution, with the following specifications:
- The processor enters Runahead mode when there is a cache miss.
- There is no penalty for entering and leaving the Runahead mode.
- There is a 64KB data cache. The cache block size is 64 bytes.
- Assume that the instructions are fetched from a separate dedicated memory that has zero access latency, so an instruction fetch never stalls the pipeline.
- The cache is 4-way set associative and uses the LRU replacement policy.
- A memory request that hits in the cache is serviced instantaneously.
- A cache miss is serviced from the main memory after $ X $ cycles.
- A cache block for the corresponding fetch is allocated _immediately_ when a cache miss happens.
- The cache replacement policy does _not_ evict the cache block that triggered entry into Runahead mode until after the Runahead mode is exited.
- The victim for cache eviction is picked at the same time a cache miss occurs, i.e., during cache block allocation.
- ALU instructions and Branch instructions take one cycle.
- Assume that the pipeline _never stalls for reasons other than data cache misses_. Assume that the conditional branches are always correctly predicted and the data dependencies do not cause stalls (except for data cache misses).
Consider the following program. Each element of Array A is one byte.
```
for(int i=0;i<100;i++){ \\ 2 ALU instructions and 1 branch instruction
int m = A[i*16*1024]+1; \\ 1 memory instruction followed by 1 ALU instruction
... \\ 26 ALU instructions
}
```
After running this program using the processor specified above, you find that there are 66 data cache hits. What are all the possible values of the cache miss latency X? You can specify all possible values of X as an inequality.
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
In this problem, we will look at the Blacklisting Memory Scheduler (BLISS) to reduce unfairness. There are two key aspects of BLISS that you need to know.
• When the memory controller services consecutive requests from a particular application, this application is blacklisted. We name this non-negative integer the Blacklisting Threshold.
• The blacklist is cleared periodically every 10000 cycles starting at t = 0.
To reduce unfairness, memory requests in BLISS are prioritized in the following order:
• Non-blacklisted applications’ requests
• Row buffer hit requests
• Older requests
The memory system for this problem consists of 2 channels with 2 banks each. Tables 1 and 2 show the memory request stream in the same bank for both applications at different times (t = 0 and t = 10). For both tables, a request on the left-hand side is older than a request on the right-hand side in the same table. The applications do not generate more requests than those shown in Tables 1 and 2. The memory requests are labeled with numbers that represent the row position of the data within the accessed bank. Assume the following for all questions:
• Arow buffer hit takes 100 cycles.
• Arow buffer miss (i.e., opening a row in a bank with a closed row buffer) takes 200 cycles.
• A row buffer conflict (i.e., closing the currently open row and opening another one) takes 250 cycles.
• All row buffers are closed at time t = 0
| Application A (Channel 0, Bank 0) | | | | | | | | |
|-----------------------------------|--------|--------|--------|--------|--------|--------|--------|--------|
| Application B (Channel 0, Bank 0) | Row 2 | Row 2 | Row 2 | Row 2 | Row 2 | Row 3 | Row 3 | Row 4 |
Table 1: Memory requests of the two applications at t = 0
| Application A (Channel 0, Bank 0) | Row 3 | Row 7 | Row 2 | Row 0 | Row 5 | Row 3 | Row 8 | Row 9 |
|-----------------------------------|--------|--------|--------|--------|--------|--------|--------|--------|
| Application B (Channel 0, Bank 0) | Row 2 | Row 2 | Row 2 | Row 2 | Row 2 | Row 3 | Row 3 | Row 4 |
Table 2: Memory requests of the two applications at t = 10. Note that none of the Application B’s existing requests are serviced yet.
Compute the slowdown of each application using the FR-FCFS scheduling policy after both threads ran to completion. We define:
slowdown = memory latency of the application when run together with other applications / memory latency of the application when run alone
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
Question 1. Choosing a right branch predictor to your baseline processor can speedup the program execution. Now consider two branch predictors: BP1 and BP2. Branches are 10% of all instructions.
Assume normal CPI is 1, but the branch mispredict penalty is 2 extra stall cycles. Evaluate the program execution time with each branch predictors and answer the questions.
• BP1: 10% misprediction rate
On average, one miss prediction will introduce the cycle time by 15%
• BP2: 12% misprediction rate
On average, one miss prediction will introduce the cycle time by 14%
1. Which predictor would you choose?
2. If branches are instead 30% of all instructions, which predictor would you choose?
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
设\(x(t)\)的傅里叶变换为\(X(j\omega)\)，\(f(t)=\frac{d^{2}x(t)}{dt^{2}}\)。 (1) 假定\(X(j\omega)= \begin{cases}1, & |\omega| < 1 \\ 0, & |\omega| > 1\end{cases}\) ，计算\(\int_{-\infty}^{+\infty}|f(t)|^{2}dt\)； (2) 求\(f(j\frac{\omega}{4})\)的傅里叶反变换。
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
You would like to understand the configuration of the DRAM subsystem of a computer using reverse engineering techniques. Your current knowledge of the particular DRAM subsystem is limited to the following information:
● The physical memory address is 16 bits.
● The DRAM subsystem consists of a single channel and 4 banks.
● The DRAM is byte-addressable.
● The most-significant 2 bits of the physical memory address determine the bank.
● The DRAM command bus operates at 500 MHz frequency.
● The memory controller issues commands to the DRAM in such a way that no command for servicing a later request is issued before issuing a READ command for the current request, which is the oldest request in the request buffer. For example, if there are requests A and B in the request buffer, where A is the older request and the two requests are to different banks, the memory controller does not issue an ACTIVATE command to the bank that B is going to access before issuing a READ command to the bank that A is accessing.
You realize that you can observe the memory requests that are waiting to be serviced in the request buffer. At a particular point of time, you take the snapshot of the request buffer and you observe the following requests in the request buffer.
Requests in the request buffer (in descending order of request age, where the oldest request is on the top):
Read 0x4C80
Read 0x0140
Read 0x4ECO
Read 0x8000
Read 0xF000
Read 0x803F
Read 0x4E80
At the same time you take the snapshot of the request buffer, you start probing the DRAM command bus. You observe the DRAM command type and the cycle (relative to the first command) at which the command is seen on the DRAM command bus. The following are the DRAM commands you observe on the DRAM bus while the requests above are serviced.
Cycle 0 -- PRECHARGE
Cycle 6 -- ACTIVATE
Cycle 10 -- READ
Cycle 11 -- READ
Cycle 21 -- PRECHARGE
Cycle 27 -- ACTIVATE
Cycle 31 -- READ
Cycle 32 -- ACTIVATE
Cycle 36 -- READ
Cycle 37 -- READ
Cycle 38 -- READ
Cycle 42 -- PRECHARGE
Cycle 48 -- ACTIVATE
Cycle 52 -- READ
What is the status of the banks prior to the execution of any of the the above requests? In other words, which rows from which banks were open immediately prior to issuing the DRAM commands listed above? Fill in the table below indicating whether a bank has an open row, and if there is an open row, specify its address. If there is not enough information to infer the open row address, write unknown.
| | Open or Closed? | Open Row Address |
|---|---|---|
| Bank 0 | | |
| Bank 1 | | |
| Bank 2 | | |
| Bank 3 | | |
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
Suppose we have a system with 32 cores that share a physical second-level cache. Assume each coreis running a single single-threaded application, and all 32 cores are concurrently running applicationsAssume that the page size of the architecture is &KB, the block size of the cache is 128 bytes. andthe cache uses LRU replacement. We would like to ensure each application gets a dedicated space inthis shared cache without any interference from other cores, We would like to enforce this using theOS-based page coloring mechanism to partition the cache, as we discussed in lecture. Recall that withpage coloring, the operating system ensures, using virtual memory mechanisms, that the applications donot contend for the same space in the cache.
Assume you would like to design a 32MB shared cache such that the operating system has the abilityto ensure that the cache is partitioned such that no two applications interfere for cache space.
What is the maximum associativity of the 32MB cache such that this is possible?
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
Assume an in-order processor that employs Runahead execution, with the following specifications:
- The processor enters Runahead mode when there is a cache miss.
- There is no penalty for entering and leaving the Runahead mode.
- There is a 64KB data cache. The cache block size is 64 bytes.
- Assume that the instructions are fetched from a separate dedicated memory that has zero access latency, so an instruction fetch never stalls the pipeline.
- The cache is 4-way set associative and uses the LRU replacement policy.
- A memory request that hits in the cache is serviced instantaneously.
- A cache miss is serviced from the main memory after $ X $ cycles.
- A cache block for the corresponding fetch is allocated _immediately_ when a cache miss happens.
- The cache replacement policy does _not_ evict the cache block that triggered entry into Runahead mode until after the Runahead mode is exited.
- The victim for cache eviction is picked at the same time a cache miss occurs, i.e., during cache block allocation.
- ALU instructions and Branch instructions take one cycle.
- Assume that the pipeline _never stalls for reasons other than data cache misses_. Assume that the conditional branches are always correctly predicted and the data dependencies do not cause stalls (except for data cache misses).
Consider the following program. Each element of Array A is one byte.
```
for(int i=0;i<100;i++){ \\ 2 ALU instructions and 1 branch instruction
int m = A[i*16*1024]+1; \\ 1 memory instruction followed by 1 ALU instruction
... \\ 26 ALU instructions
}
```
Is it possible that every memory access in the program misses in the cache? If so, what are all possible values of X that will make all memory accesses in the program miss in the cache? If not, why?
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
A memory system is composed of eight banks, and each bank contains 32K rows. The row size is 8KB.
Every DRAM row refresh is initiated by a command from the memory controller, and it refreshes a single row. Each refresh command keeps the command bus busy for 5~ns.
We define command bus utilization as a fraction of the total time during which the command bus is busy due to refresh.
The retention time of each row depends on the temperature (T). The rows have different retention times, as shown in the following Table 1:
Table 1: Retention time
Retention Time || Number of rows
(128-T) ms, 0^{\circ} C\leq T\leq 128^{\circ} || C2^{8} rows
2*(128-T) ms, 0^{\circ} C\leq T\leq 128^{\circ} || C2^{16} rows
4*(128-T) ms, 0^{\circ} C\leq T\leq 128^{\circ} || Call other rows
8*(128-T) ms, 0^{\circ} C\leq T\leq 128^{\circ} || C2^{8} rows
Assume that the memory controller implements an approximate mechanism to reduce refresh rate using Bloom filters, as we discussed in class. For this question we assume the retention times in Table 1 with a constant temperature of 64^{\circ} C.
One Bloom filter is used to represent the set of all rows that require a 64~ms refresh rate.
The memory controller's refresh logic is modified so that on every potential refresh of a row (every 64~ms), the refresh logic probes the Bloom filter. If the Bloom filter probe results in a "hit" for the row address, then the row is refreshed. Any row that does not hit in the Bloom filter is refreshed at the default rate of once per 128~ms.
The memory controller performs 2107384 refreshes in total across the channel over a time interval of 1.024 seconds. What is the false positive rate of the Bloom filter?Hint: 2107384=2^{3}*\left(2^{18}+2^{11}-2^{10}+2^{9}-2^{8}-1\right)
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
在有400个关键码的20阶B树查找，最大需要和________个关键码进行比较。
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
给定 $A= 2.6125 \times 10^1$, $B= 4.150390625 \times 10^{-1}$。其中A, B都是浮点表达格式（包含1bit 符号位，5bit指数位，10bit尾数位置）。通过浮点计算的对阶，尾数加减，规格化，舍入（舍入操作如IEEE754标准保留3位extra bits, 即guard-bit, round-bit, sticky-bit,并采用就近舍入round-to-nearest），检查溢出这五个步骤，计算 A + B的结果，最后给出最终结果并以$x.y \times 10^n$的形式表达（x<10）。
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
find a way to implement AND gate using only IEEE-754 floating-point subtraction. You need to define True and False as 2 fp numbers, then give a procedure that does AND gate calculation. You can use constant fp number in the procedure. {True, False} should be closed under the procedure, in other words, result given by the procedure must be either True or False, for example you can't assume result not equal to True nor False is cast to F
|
[
""
] |
[
""
] |
Computer System Architecture
|
Computer Science
|
在植物基因组中，现在流行用ATAC-seq来识别和鉴定顺式调控元件，这是根据染色质的科技型原理。问这种技术来识别调控元件需要注意哪些地方？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何快速鉴定两个近缘基因组的结构变异 ？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
土壤根际微生物的生物量氮与季节变化有如下关联：
\begin{tabular}{cc}
\hline 月份 & 生物量氮 $\left(10^{-4} \mathrm{mg} \cdot 100 \mathrm{~g}^{-1}\right)$ \\
\hline 5 & 6.57 \\
6 & 7.44 \\
7 & 8.72 \\
8 & 10.68 \\
9 & 11.55 \\
10 & 9.15 \\
11 & 5.87 \\
12 & 4.42 \\
\hline
\end{tabular}
生物量氮与月份之间存在怎样的回归关系？求出回归方程。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
针对鳞翅目昆虫的棉古毒蛾、春尺蠖以及弘复峻尺蛾，请问是否可以使用翅膀相关基因使用分子钟模型推断这两个物种分歧时间吗？请给出判断并说明原因：
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
用根瘤材料进行DNA测序时，因为其中存在细菌，经常掺入细菌DNA污染，如何避免和解决？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
据说罗布麻有降血压的功能。为了检验服药后的血压值是否与服药前有关，随机抽取 10 名受试者，测其服药前、后的收缩压如下表：
\begin{tabular}{ccccccccccc}
\hline 受试者 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline 服药前 $/ \mathrm{mmHg}$ & 137 & 147 & 161 & 127 & 130 & 134 & 135 & 158 & 147 & 142 \\
服药后 $/ \mathrm{mmHg}$ & 143 & 138 & 146 & 127 & 120 & 119 & 122 & 172 & 134 & 127 \\
\hline
\end{tabular}
问服药后的血压值是否与服药前的血压值有关？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
给定一个二代测序文库，比如就说RNA-seq测序，数据拿到以后，与参考基因组进行序列比对，如果唯一比对率10%，而多比对率是80%，可能是什么原因？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何构建全基因组导入系（CSSLs）群体？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何鉴定基因组中的着丝粒区域？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
竞争性等位基因特异性聚合酶链式反应（KASP）是在群体定位中具体的使用方法是什么？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
利用结构相似性画进化树的优势？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
心脏的冠状窦口直径 $(d)$ 与冠状窦瓣宽 $(w)$ 和窦瓣高 $(h)$ 存在一定关联，下面测量了从新生儿到儿童末期的 6 个年龄组的窦口直径、窦瓣宽和窦瓣高，结果见下表 ：
\begin{tabular}{lcccccr}
\hline & \multicolumn{6}{c}{ 组 } \\
\cline { 2 - 7 } & I & II & III & IV & V & VI \\
\hline 窦口直径 $/ \mathrm{mm}$ & 3.19 & 4.43 & 4.96 & 5.81 & 6.30 & 7.98 \\
窦瓣宽 $/ \mathrm{mm}$ & 4.64 & 6.42 & 7.32 & 7.68 & 8.99 & 10.30 \\
窦瓣高 $/ \mathrm{mm}$ & 1.68 & 3.93 & 4.08 & 4.41 & 4.94 & 5.02 \\
\hline
\end{tabular}
分别计算窦瓣宽和窦瓣高与突口直径间的相关系数，并检验相关系数的显著性。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
对表型进行全基因组关联定位前，需要对数据进行哪些预处理？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何判断一个基因是否是由转座活动产生的？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
现在研究5mC DNA甲基化的手段有whole genome bisulfite sequencing二代测序技术和Nanopore以及Pacbio三代测序技术，这两个技术分别利用bisulfite处理DNA序列使甲基化保护的Cytosine不会转换，而没有甲基化保护的cytosine会变成U从而识别甲基化了的位点。而三代测序由于没有PCR过程，是实时测序，可以检测到DNA序列上的修饰。目前已经很多工作利用这两种技术开展DNA甲基化研究。但是两种技术由于测序原理以及测序技术差异等原因，在一些应用场景上还是会有不同，你知道都有哪些？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何区分同源多倍体和异源多倍体？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
新疆维吾尔族和哈萨克族男生各100名，他们的立定跳远平均成绩与年龄之间的关系如下表所示：
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline 年龄／a & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline 维吾尔族／cm & 124.51 & 132.65 & 138.59 & 143.39 & 151.74 & 160.91 \\
\hline 哈萨克族／cm & 135.80 & 146.52 & 153.34 & 162.88 & 171.10 & 174.29 \\
\hline 年龄／a & 13 & 14 & 15 & 16 & 17 & 18 \\
\hline 维吾尔族／cm & 169.31 & 184.22 & 195.57 & 200.51 & 207.84 & 217.24 \\
\hline 哈萨克族／cm & 185.88 & 190.24 & 211.21 & 228.63 & 235.07 & 233.65 \\
\hline
\end{tabular}
分别计算两个民族的成绩与年龄之间的相关系数，并检验两个相关系数的显著性。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，社鼠头骨若干特征的度量值与年龄存在相关性，下表列出了 40 只社鼠的鉴定年龄（a）和头骨 8 个特征的度量值（ mm ）：
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline 序号 & 鉴定年龄 Y & $X_1$ & $X_2$ & $X_3$ & $X_4$ & $X_5$ & $X_6$ & $X_7$ & $X_8$ \\
\hline 1 & 3 & 34.60 & 33.62 & 31.26 & 16.10 & 5.44 & 8.74 & 6.12 & 6.74 \\
\hline 2 & 3 & 34.50 & 33.44 & 31.68 & 15.92 & 4.82 & 9.00 & 5.82 & 6.48 \\
\hline 3 & 4 & 37.36 & 36.36 & 34.28 & 17.46 & 5.48 & 9.96 & 6.08 & 6.72 \\
\hline 4 & 4 & 36.94 & 35.80 & 34.10 & 17.14 & 5.28 & 9.80 & 5.46 & 6.62 \\
\hline 5 & 5 & 38.00 & 37.72 & 35.74 & 17.46 & 5.14 & 9.92 & 5.84 & 6.68 \\
\hline 6 & 5 & 38.30 & 37.44 & 35.64 & 17.08 & 5.14 & 10.26 & 5.72 & 6.90 \\
\hline 7 & 5 & 39.72 & 39.18 & 36.72 & 17.84 & 5.60 & 10.50 & 5.76 & 6.62 \\
\hline 8 & 1 & 27.34 & 26.42 & 23.50 & 13.46 & 4.70 & 7.59 & 4.50 & 5.12 \\
\hline 9 & 4 & 36.78 & 36.36 & 34.52 & 16.48 & 5.36 & 9.44 & 5.96 & 6.78 \\
\hline 10 & 4 & 37.12 & 36.12 & 34.24 & 16.44 & 5.14 & 9.52 & 5.90 & 6.38 \\
\hline 11 & 3 & 34.78 & 33.56 & 31.40 & 15.46 & 5.14 & 8.42 & 5.68 & 5.88 \\
\hline 12 & 2 & 31.38 & 30.86 & 28.56 & 14.54 & 5.08 & 7.82 & 5.78 & 6.00 \\
\hline 13 & 4 & 36.50 & 35.72 & 33.48 & 16.42 & 5.06 & 8.90 & 5.44 & 6.40 \\
\hline 14 & 2 & 33.80 & 32.92 & 30.70 & 16.88 & 5.08 & 8.24 & 5.66 & 6.00 \\
\hline 15 & 2 & 32.28 & 31.14 & 28.50 & 15.38 & 4.88 & 7.68 & 5.60 & 5.38 \\
\hline 16 & 4 & 37.88 & 37.06 & 34.54 & 16.60 & 5.66 & 9.92 & 5.52 & 6.84 \\
\hline 17 & 2 & 32.74 & 31.82 & 29.58 & 15.30 & 5.14 & 8.00 & 6.00 & 5.08 \\
\hline 18 & 1 & 30.00 & 28.56 & 26.18 & 13.92 & 4.98 & 7.12 & 5.10 & 5.12 \\
\hline 19 & 2 & 33.22 & 32.10 & 29.62 & 15.58 & 4.96 & 8.00 & 5.56 & 5.66 \\
\hline 20 & 4 & 37.08 & 36.90 & 33.78 & 17.38 & 5.72 & 9.60 & 6.04 & 6.68 \\
\hline 21 & 3 & 35.32 & 34.32 & 32.18 & 15.70 & 5.00 & 8.88 & 6.02 & 6.46 \\
\hline 22 & 2 & 32.66 & 31.08 & 28.92 & 15.34 & 4.76 & 7.80 & 5.72 & 5.42 \\
\hline 23 & 2 & 32.64 & 31.50 & 29.46 & 14.64 & 5.08 & 7.40 & 5.74 & 5.20 \\
\hline 24 & 2 & 32.68 & 31.50 & 29.18 & 14.94 & 4.76 & 7.86 & 5.82 & 5.68 \\
\hline 25 & 1 & 30.94 & 30.20 & 27.70 & 14.36 & 5.22 & 7.22 & 5.70 & 4.92 \\
\hline 26 & 4 & 36.84 & 35.96 & 34.04 & 17.02 & 5.36 & 9.08 & 6.16 & 6.00 \\
\hline 27 & 5 & 37.58 & 36.88 & 34.44 & 16.72 & 5.46 & 10.00 & 5.60 & 6.36 \\
\hline 28 & 5 & 37.88 & 37.06 & 34.54 & 16.60 & 5.66 & 9.92 & 5.52 & 6.84 \\
\hline 29 & 3 & 34.28 & 33.34 & 31.30 & 16.64 & 5.18 & 9.22 & 5.58 & 6.46 \\
\hline 30 & 3 & 35.80 & 35.00 & 32.70 & 16.64 & 5.82 & 10.00 & 5.68 & 6.00 \\
\hline 31 & 3 & 34.12 & 33.10 & 31.14 & 15.68 & 5.46 & 9.32 & 5.62 & 6.00 \\
\hline 32 & 3 & 34.22 & 33.26 & 31.60 & 16.00 & 5.22 & 9.12 & 5.56 & 6.28 \\
\hline 33 & 4 & 37.54 & 36.80 & 34.62 & 16.44 & 5.24 & 10.00 & 5.74 & 6.70 \\
\hline 34 & 3 & 33.94 & 33.38 & 31.36 & 16.84 & 5.08 & 8.72 & 5.70 & 6.24 \\
\hline 35 & 3 & 34.00 & 33.02 & 30.54 & 15.56 & 5.12 & 8.86 & 5.96 & 6.42 \\
\hline 36 & 2 & 31.54 & 30.46 & 28.04 & 15.20 & 4.92 & 7.78 & 5.46 & 5.68 \\
\hline 37 & 5 & 38.10 & 37.62 & 34.86 & 17.44 & 5.72 & 10.16 & 6.14 & 7.16 \\
\hline 38 & 2 & 30.50 & 30.00 & 27.92 & 14.84 & 5.00 & 7.12 & 5.70 & 5.30 \\
\hline 39 & 2 & 32.26 & 30.82 & 28.62 & 15.30 & 4.94 & 7.82 & 5.50 & 5.46 \\
\hline 40 & 4 & 37.38 & 36.20 & 34.22 & 16.90 & 5.30 & 9.44 & 5.54 & 6.42 \\
\hline
\end{tabular}
注：$X_1$ ：颅全长。 $X_2$ ：颅基长。 $X_3$ ：基底长。 $X_4$ ：颧宽。 $X_5$ ：眶间宽。 $X_6$ ：齿隙长。$X_7$ ：上裂齿长．$X_8$ ：门齿孔长。
在计算完多元回归方程，复相关系数后，给出用逐步回归方法选出的包含 3 个自变量的回归方程。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
4到10月龄胎儿的肝重与肝的 Ca 含量存在以下关系：
\begin{tabular}{cccccccc}
\hline 肝 重 $/ \mathrm{g}$ & 6.48 & 13.02 & 24.17 & 44.86 & 58.39 & 75.58 & 86.47 \\
\hline Ca 含量 $/\left(\mu \mathrm{g} \cdot \mathrm{g}^{-1} \mp\right.$ 重 $)$ & 1271.0 & 1440.9 & 1016.6 & 663.7 & 516.3 & 535.9 & 492.5 \\
\hline
\end{tabular}
求钙含量在肝重上的回归方程并检验回归的显著性。
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[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
GWAS的定位方法都有哪些算法，适合哪种表型？
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[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
近缘基因组之间如何提升基因注释质量？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何证明基因组组装中的一段序列是准确的
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
用不同浓度的草甘膦异丙胺盐（除草剂）溶液处理后，中华大蟾蜍心电图的三项指标平均值如下表：
\begin{tabular}{|l|l|l|l|}
\hline 浓 度 ／（ $\mathrm{mL} \cdot \mathrm{L}^{-1}$ ） & P 波 ／mV & R 波 ／mV & $\mathrm{P}-\mathrm{R}$ 间期 ／ms \\
\hline 0 & 0.160 & 1.319 & 0.182 \\
\hline 0.82 & 0.147 & 0.965 & 0.156 \\
\hline 1.23 & 0.118 & 0.725 & 0.196 \\
\hline 1.64 & 0.104 & 0.804 & 0.223 \\
\hline 2.05 & 0.117 & 0.683 & 0.230 \\
\hline 2.46 & 0.102 & 0.797 & 0.255 \\
\hline 2.87 & 0.095 & 0.651 & 0.258 \\
\hline
\end{tabular}
分别计算 P 波， R 波及 $\mathrm{P}-\mathrm{R}$ 间期对浓度的回归方程，并检验回归系数的显著性。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
青菜对 ${ }^{14} \mathrm{CO}_2$ 的富集系数（CF 值）如下：
\begin{tabular}{ccc}
\hline 时间／d & 菜 心 & 叶 子 \\
\hline 6 & 24.6 & 13.8 \\
12 & 53.4 & 30.9 \\
18 & 82.0 & 41.9 \\
24 & 100.1 & 63.2 \\
36 & 114.1 & 96.8 \\
48 & 156.4 & 135.6 \\
\hline
\end{tabular}
以时间为自变量，菜心和叶子分别为因变量，计算回归方程，并比较两者回归系数的差异显著性。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何通过生物信息学手段预测互作基因？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
目前传统序列比对算法的不足在哪里？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
这个是我面试博士后常问的问题，1. 给定一个转录组或者其他二代测序的测序文库，问生成的reads，在进行序列比对的时候，唯一比对率偏低，比如10%左右，问是什么原因？如何检测？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
我们往往用转录组测序来估计基因的表达情况，那现在随着蛋白质质谱技术的进步和价格下降呢，很多人开始直接选择蛋白质组来估计基因表达量。如果我有一个相同材料的RNA-seq和蛋白质组，但是我发现个别基因转录水平很高，但是在蛋白质组里却表达很低，可能是什么原因引起的，如何验证？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
在一项罕见遗传病的研究中，研究人员怀疑疾病相关基因中存在结构变异（Structural Variations, SVs），这些变异可能影响基因功能并导致疾病表型。请设计一项实验以推测基因组中与疾病潜在相关的SVs。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
如何鉴定两个基因组间的同源基因
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，社鼠头骨若干特征的度量值与年龄存在相关性，下表列出了 40 只社鼠的鉴定年龄（a）和头骨 8 个特征的度量值（ mm ）：
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline 序号 & 鉴定年龄 Y & $X_1$ & $X_2$ & $X_3$ & $X_4$ & $X_5$ & $X_6$ & $X_7$ & $X_8$ \\
\hline 1 & 3 & 34.60 & 33.62 & 31.26 & 16.10 & 5.44 & 8.74 & 6.12 & 6.74 \\
\hline 2 & 3 & 34.50 & 33.44 & 31.68 & 15.92 & 4.82 & 9.00 & 5.82 & 6.48 \\
\hline 3 & 4 & 37.36 & 36.36 & 34.28 & 17.46 & 5.48 & 9.96 & 6.08 & 6.72 \\
\hline 4 & 4 & 36.94 & 35.80 & 34.10 & 17.14 & 5.28 & 9.80 & 5.46 & 6.62 \\
\hline 5 & 5 & 38.00 & 37.72 & 35.74 & 17.46 & 5.14 & 9.92 & 5.84 & 6.68 \\
\hline 6 & 5 & 38.30 & 37.44 & 35.64 & 17.08 & 5.14 & 10.26 & 5.72 & 6.90 \\
\hline 7 & 5 & 39.72 & 39.18 & 36.72 & 17.84 & 5.60 & 10.50 & 5.76 & 6.62 \\
\hline 8 & 1 & 27.34 & 26.42 & 23.50 & 13.46 & 4.70 & 7.59 & 4.50 & 5.12 \\
\hline 9 & 4 & 36.78 & 36.36 & 34.52 & 16.48 & 5.36 & 9.44 & 5.96 & 6.78 \\
\hline 10 & 4 & 37.12 & 36.12 & 34.24 & 16.44 & 5.14 & 9.52 & 5.90 & 6.38 \\
\hline 11 & 3 & 34.78 & 33.56 & 31.40 & 15.46 & 5.14 & 8.42 & 5.68 & 5.88 \\
\hline 12 & 2 & 31.38 & 30.86 & 28.56 & 14.54 & 5.08 & 7.82 & 5.78 & 6.00 \\
\hline 13 & 4 & 36.50 & 35.72 & 33.48 & 16.42 & 5.06 & 8.90 & 5.44 & 6.40 \\
\hline 14 & 2 & 33.80 & 32.92 & 30.70 & 16.88 & 5.08 & 8.24 & 5.66 & 6.00 \\
\hline 15 & 2 & 32.28 & 31.14 & 28.50 & 15.38 & 4.88 & 7.68 & 5.60 & 5.38 \\
\hline 16 & 4 & 37.88 & 37.06 & 34.54 & 16.60 & 5.66 & 9.92 & 5.52 & 6.84 \\
\hline 17 & 2 & 32.74 & 31.82 & 29.58 & 15.30 & 5.14 & 8.00 & 6.00 & 5.08 \\
\hline 18 & 1 & 30.00 & 28.56 & 26.18 & 13.92 & 4.98 & 7.12 & 5.10 & 5.12 \\
\hline 19 & 2 & 33.22 & 32.10 & 29.62 & 15.58 & 4.96 & 8.00 & 5.56 & 5.66 \\
\hline 20 & 4 & 37.08 & 36.90 & 33.78 & 17.38 & 5.72 & 9.60 & 6.04 & 6.68 \\
\hline 21 & 3 & 35.32 & 34.32 & 32.18 & 15.70 & 5.00 & 8.88 & 6.02 & 6.46 \\
\hline 22 & 2 & 32.66 & 31.08 & 28.92 & 15.34 & 4.76 & 7.80 & 5.72 & 5.42 \\
\hline 23 & 2 & 32.64 & 31.50 & 29.46 & 14.64 & 5.08 & 7.40 & 5.74 & 5.20 \\
\hline 24 & 2 & 32.68 & 31.50 & 29.18 & 14.94 & 4.76 & 7.86 & 5.82 & 5.68 \\
\hline 25 & 1 & 30.94 & 30.20 & 27.70 & 14.36 & 5.22 & 7.22 & 5.70 & 4.92 \\
\hline 26 & 4 & 36.84 & 35.96 & 34.04 & 17.02 & 5.36 & 9.08 & 6.16 & 6.00 \\
\hline 27 & 5 & 37.58 & 36.88 & 34.44 & 16.72 & 5.46 & 10.00 & 5.60 & 6.36 \\
\hline 28 & 5 & 37.88 & 37.06 & 34.54 & 16.60 & 5.66 & 9.92 & 5.52 & 6.84 \\
\hline 29 & 3 & 34.28 & 33.34 & 31.30 & 16.64 & 5.18 & 9.22 & 5.58 & 6.46 \\
\hline 30 & 3 & 35.80 & 35.00 & 32.70 & 16.64 & 5.82 & 10.00 & 5.68 & 6.00 \\
\hline 31 & 3 & 34.12 & 33.10 & 31.14 & 15.68 & 5.46 & 9.32 & 5.62 & 6.00 \\
\hline 32 & 3 & 34.22 & 33.26 & 31.60 & 16.00 & 5.22 & 9.12 & 5.56 & 6.28 \\
\hline 33 & 4 & 37.54 & 36.80 & 34.62 & 16.44 & 5.24 & 10.00 & 5.74 & 6.70 \\
\hline 34 & 3 & 33.94 & 33.38 & 31.36 & 16.84 & 5.08 & 8.72 & 5.70 & 6.24 \\
\hline 35 & 3 & 34.00 & 33.02 & 30.54 & 15.56 & 5.12 & 8.86 & 5.96 & 6.42 \\
\hline 36 & 2 & 31.54 & 30.46 & 28.04 & 15.20 & 4.92 & 7.78 & 5.46 & 5.68 \\
\hline 37 & 5 & 38.10 & 37.62 & 34.86 & 17.44 & 5.72 & 10.16 & 6.14 & 7.16 \\
\hline 38 & 2 & 30.50 & 30.00 & 27.92 & 14.84 & 5.00 & 7.12 & 5.70 & 5.30 \\
\hline 39 & 2 & 32.26 & 30.82 & 28.62 & 15.30 & 4.94 & 7.82 & 5.50 & 5.46 \\
\hline 40 & 4 & 37.38 & 36.20 & 34.22 & 16.90 & 5.30 & 9.44 & 5.54 & 6.42 \\
\hline
\end{tabular}
注：$X_1$ ：颅全长。 $X_2$ ：颅基长。 $X_3$ ：基底长。 $X_4$ ：颧宽。 $X_5$ ：眶间宽。 $X_6$ ：齿隙长。$X_7$ ：上裂齿长．$X_8$ ：门齿孔长。
计算多元回归方程，复相关系数。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
问一个简单的，HiC测序技术，最开始是用来研究染色体空间互作的技术。那么为什么现在越来越多的人使用HiC来辅助组装基因组呢？这个技术辅助组装基因组可能会引起什么样的错误？或者需要注意的地方？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
判断一个非常重要的蛋白质编码基因在另一个物种中是否存在直系同源基因，你如何做？要求是结果一定要准确。假设给定的基因A，问你在拟南芥中是否存在直系同源基因，存在几个？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
请参考生物信息学的方式（代码）得到解答，
马鹿下臼齿咀嚼面宽度与年龄之间存在以下关系：
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 序号 & 年龄 ／a & \multicolumn{18}{|c|}{下臼齿咀嚼面宽度／mm} \\
\hline 1 & 2.5 & 8.65 & 8.90 & 8.30 & 8.80 & & & & & & & & & & & & & & & \\
\hline 2 & 3.5 & 9.60 & 8.35 & 8.30 & 8.40 & 7.80 & 8.40 & 8.70 & 9.40 & 7.50 & 7.90 & 8.90 \\
\hline 3 & 4.5 & 10.13 & 9.10 & 8.65 & 10.17 & 10.00 & 9.80 & 10.90 & 9.72 & 9.92 & 9.82 & 10.00 & 10.00 & 10.14 & 10.15 & 10.12 & 8.80 & 10.15 \\
\hline 4 & 5.5 & 10.75 & 10.68 & 11.68 & 10.30 & 10.22 & 10.00 & 11.90 & 11.85 & 11.90 & 11.85 \\
\hline 5 & 6.5 & 11.30 & 11.00 & 12.70 & 11.30 & 11.48 & 11.87 & 10.20 & 10.82 & 11.52 & 11.60 & 10.25 \\
\hline 6 & 7.5 & 10.40 & 11.00 & 12.50 & 13.50 & 9.98 & & & & \\
\hline 7 & 8.5 & 12.16 & 12.80 & 11.88 & 11.10 & 11.48 & 11.40 & 12.10 & 10.15 & \\
\hline 8 & 9.5 & 12.72 & 11.68 & 12.80 & 11.35 & 13.33 & & & & \\
\hline 9 & 13.5 & 12.20 & & & & & & & & \\
\hline 10 & 17.5 & 14.03 & & & & & & & & \\
\hline
\end{tabular}
以年龄为自变量，咀嚼面宽度为因变量，计算回归方程。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
全基因组关联分析（Genome-Wide Association Study, GWAS）是一种用于识别与特定性状或疾病相关的遗传变异（通常是单核苷酸多态性，SNP）的研究方法。GWAS通常需要大样本量（数千至百万级个体）以提高统计效力，使用基因芯片或测序技术获取全基因组范围的SNP数据。然而，在GWAS研究时会出现膨胀系数（Genomic Inflation Factor, λ）偏高，请分析可能的原因以及解决方法。
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
近缘物种基因组大小差异较大可能是什么原因造成的，该如何验证？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
在缺少基因组注释文件的情况下如何快速鉴定近缘物种的系统发育关系？
|
[
""
] |
[
""
] |
Genetics and Bioinformatics
|
Biology
|
惯性参考系 $S$ 中有一个半径为 $R$ 的固定圆环，一个静质量为 $m$ 的小球初始时刻静止在圆环上某一点，然后对小球施加大小恒定为 $F$ 的切向外力，同时环壁也会对小球反应弹力。真空中的光速为 $c$ 。本题须考虑狭义相对论效应。假设每圈外力做功恰好等于小球静能。 （1）求小球绕行一圈的过程中弹力大小对轨道弧长的平均值； （2）若把圆环换为通径 $2 p$ ，离心率 $e$ 的椭圆环，关于外力做功量的假设仍然成立。求小球绕行一圈的过程中弹力大小对曲率半径线扫过面积 $S$ 的平均值。
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[
""
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[
""
] |
Classical Mechanics
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Physics
|
一个均匀带电的金属球，半径为R带电量为Q，如果把这个金属球切为两部分，切面距离球中心的最小距离是H，请问这两部分之间的排斥力有多大
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[
""
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[
""
] |
Classical Mechanics
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Physics
|
质量为1kg的小物块以5m/s的初速度滑上一块原来静止在水平面上的木板，木板质量为4kg，木块与木板之间的动摩擦因数为0.2。经过2s以后，物块恰好滑出木板，在这一过程中木板的位移为0.5m，求木板和地面之间的动摩擦因数。
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[
""
] |
[
""
] |
Classical Mechanics
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Physics
|
不可伸长轻细绳吊着质量为\(m_0\)的摆球（视为质点）在竖直向下的匀强电场\(E\)的作用下绕平衡位置小幅摆动，不计摆球重力，球电量为\(q > 0\)，摆球初始振幅为\(\theta_0\)，求以下两种不同情况下的末振幅： （1）摆球质量随时间缓慢减小（不断有质量从摆球上无相对速度地分离），由\(m_0\)变为\(m_1\)，过程中摆球电量\(q\)保持不变； （2）空间中有弥散的、静止的小水珠，摆球运动过程中吸收撞击到球上的水珠，其质量随时间缓慢增大，由\(m_0\)变为\(m_2\) ，过程中摆球电量\(q\)保持不变。
|
[
""
] |
[
""
] |
Classical Mechanics
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Physics
|
太阳系中均匀分布的尘埃会对行星施加一额外的引力，其表达式为：\[ \mathbf F = -mC\mathbf r \]，其中 \( m \) 为行星质量，\( C \) 是与引力常数和尘埃密度成正比的常数，\( \mathbf r \) 为太阳指向行星的径矢（两者均视为质点）。该附加力远小于太阳与行星之间的直接引力。行星在此复合力场中作半径为 \( r_0 \) 的圆周运动，计算该圆周运动轨道受微小扰动后的径向振动频率$\omega$（ \( \tau_0 = 2\pi r_0^{3/2}\sqrt{m/k} \) 为无微扰势时的圆轨道周期）。
|
[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
在\( \frac{1}{r} \) 势场下的双曲运动中，偏心近点角类比量 \( F \) 的定义为： \[ r = a(e \cosh F - 1), \] 其中 \( a(1 - e) \) 是近心点距离。请求出类似于Kepler's 方程的表达式，将时间 \( t \)（从近心点开始计时）表示为 \( F \) 的函数（能量为$E$）。
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[
""
] |
[
""
] |
Classical Mechanics
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Physics
|
一个点电荷质量为m，带电量为q，被从一个很大的固定金属板上方d高度处释放，板上有极小的孔洞可供电荷来回穿过。只考虑静电力，不考虑能量损耗，求电荷在金属板两侧来回运动的周期是多少？
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[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
一个无阻尼的质点$m$，受到稳定的方波激励，方波周期为$T$，在前半个周期受力$-\omega_n^2(1+\mu)mx$,其中$x$是质点偏离平衡位置的位移；在后半个周期受力$-\omega_n^2(1-\mu)mx$。问何时质点的振动会不断被放大
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[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
一桶水底部有一个半径为r的圆形孔，孔被一个质量为m半径为R的球给堵住了，水桶里有高度为h的水，水将球压在孔洞上，球的一部分也从洞下方露出来。假设水始终能没过整个球，请问水的高度要满足什么条件球会从孔洞处浮起？
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[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
有一根水平放置的圆柱杆子，截面半径为$R$，在它上面搭了一个角钢，也就是两个长边相连并短边相互垂直的薄长条，各自的短边宽度均等于圆柱的截面直径$2R$，总质量$2m$。放置时角铁长边与杆子长边一个方向，同时一个长条为水平，一个长条为竖直，求杆子和角铁之间的静摩擦系数要满足什么情况才能让它不掉下来
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[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
有一根线质量为$\rho$的重绳子，两端由恒力F拉着伸展开来。此时通过某种操作让这个绳子在一端形成了一个小环，半径为R，已知这个环可以像横波一样传播，让它沿着绳子前进。推导单个绳环携带的动能。
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[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
求在实验室参考系中两个分裂后质点飞出方向之间夹角 \theta 的取值范围（粒子初始是运动的，速度为V。粒子分裂后在质心系中粒子速度大小分别是v_{10}和v_{20}。最终结果应该用这三个参数表示）
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[
""
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[
""
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Classical Mechanics
|
Physics
|
有一个倾角为$\alpha$的传送带，传送带由一系列间距为d的水平圆柱滚筒组成，每个质量为m，半径为r，表面粗糙但并非总是和货物无滑动。求一个质量为M的长木板在这个传送带上滑下来的最终速度
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[
""
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[
""
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Classical Mechanics
|
Physics
|
一根均匀的质量为m长度为l的棍子，其两端被两个滑块水平支撑着， 此时缓慢移动两个滑块，使他们最终在棍子的原质心处汇合，搭在上面的棍子会受摩擦力作用滑动。如果静摩擦系数为μ1，动摩擦系数为μ2，动摩擦系数小于等于静摩擦系数，请注意不考虑两端同时发生动摩擦的状况。在此过程中滑块总共做了多少功？
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[
""
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[
""
] |
Classical Mechanics
|
Physics
|
计算由排斥中心力 \( f = kr^{-3} \) 产生的散射现象的微分散射截面$\sigma(\theta) d\Omega$。假设 \( \theta/\pi \) 的比值为\( x \) ，能量为\( E \)。
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[
""
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[
""
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Classical Mechanics
|
Physics
|
一根长度为L质量为M的薄纸片被盘绕成半径为R的一卷，纸卷一端被系在地上，并以初速度v_0开始水平滚动，忽略弹性和摩擦，求这个纸卷完全展开需要多少时间
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[
""
] |
[
""
] |
Classical Mechanics
|
Physics
|
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