hp exams questions

本文提供HP认证考试HP0-538的备考资料,包括完整的题库和实践考试,覆盖了从基础知识到高级技巧的所有方面。文章还包含了关于内存调优、磁盘控制器缓存配置及系统性能测量的选择题示例。
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1.Which statement is false?
 
A: Lack of memory can exhibit itself as a disk bottleneck
B: If the sum of operating system memory, Oracle8i memory, and non-Oracle application memory exceeds physical RAM, memory paging will occur.
C: You should tune memory after you tune disk subsystem
D: You should tune the application and SQL before making final determination of the physical memory in your system.
Correct Answers:  C   
2.How would you configure the disk controller cache for the highest performance in an I/O request-intensive Oracle environment (for example, an OLTP system)?
A: Disable caching for the redo logs, and 25%read cache/75% write-back cache for the database
B: Read-ahead caching for the redo log and the database
C: Write-back caching for the redo logs and the database
D: Write-back caching for the redo logs, and 75% read cache/25% write-back cache for the database
Correct Answers:  C
HP0-094 
3.A system is capable of processing 2500 transactions per minute. By adding another processor, the transactions-per-minute rate increases to 4200. What is this type of measurement called?
A: Processing latency
B: Processor affinity
C: Processor scale-up
D: Processor speed-up
Correct Answers:  C
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03-13
在数字逻辑设计中,有限状态机(FSM)是描述时序电路行为的重要工具。根据提供的参考资料[^1],Moore 型状态机的输出仅取决于当前状态,并且该类型的状态机会通过不同的状态来反映不同的输入序列历史。 在实际设计中,状态数目的优化是一个关键问题。一种常见的优化策略是将某些检测逻辑集中在特定状态中完成,从而减少整体状态数目[^2]。例如,在一个基于序列检测的任务中,可以将对特定输入序列(如 `101`)的检测集中到某个状态中进行,这样虽然减少了状态数量,但可能增加了次态转移逻辑的复杂度。 针对 2014 年某考试中第 3 题关于 FSM 的题目,尽管没有直接提供具体的考题内容,但可以根据常规的 FSM 设计原则推测其大致结构。通常这类题目要求设计一个能够响应特定输入序列的状态机,涉及状态图绘制、状态编码、次态逻辑推导以及输出逻辑实现等步骤。 以下是一个简化的 Verilog 实现示例,用于说明如何构建一个 Moore 型状态机: ```verilog module fsm_example ( input clk, input reset, input x, output reg out ); // 状态定义 typedef enum logic [2:0] { A, B, C, D, E, F } state_t; state_t current_state, next_state; // 状态寄存器 always_ff @(posedge clk or posedge reset) begin if (reset) current_state <= A; else current_state <= next_state; end // 次态逻辑 always_comb begin case (current_state) A: next_state = x ? B : A; B: next_state = x ? C : A; C: next_state = x ? C : D; // 连续三个周期为 101 的部分逻辑 D: next_state = x ? E : F; E: next_state = F; F: next_state = A; default: next_state = A; endcase end // 输出逻辑(Moore 型) always_ff @(posedge clk or posedge reset) begin if (reset) out <= 1'b0; else out <= (current_state == D || current_state == E); end endmodule ``` 此代码片段展示了状态从 `A` 到 `F` 的迁移过程,其中状态 `C` 和 `D` 被用来检测特定输入序列并决定下一步状态跳转。
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