分 Suppose the hard disk with SCAN scheduling Alg has a single platter (盘面) as illustrated below , and the rotation speed for the platter is 6000 rounds per minute . The disk has only 200 tracks , and each track is composed of 100 sectors . The time cost for the disk arm to navigate between neighboring track is about 1ms. Suppose the disk arm currently hangs over the 100th track , and moves centrifugally towards lower - order tracks on the outside . The pending disk block requests are positioned on the tracks with numbers 50, 180,90,30,120, respectively , and form a queue . For each block in this queue , we do not know what are its exact sector numbers , and we can only assume that the corresponding sector randomly distributes on each track as shown below . Then , The expected time to finish the retrieval of the five sectors is1] ms . If SSTF is used , the expected time is [] ms . 0 磁道 磁头运动方向 .

时间: 2024-02-14 19:05:05 浏览: 100
根据题目所给的信息,我们可以画出磁道与扇区的示意图: ``` | Track 1 | | Sector 1 | Sector 2 | |---------------------| | Track 2 | | Sector 1 | Sector 2 | |---------------------| | ... | |---------------------| | Track 200 | | Sector 1 | Sector 2 | ``` 其中,每个磁道有 100 个扇区,总共有 200 个磁道。 根据题目所描述的情况,我们可以计算出使用 SCAN 调度算法时,磁头移动的距离为: - 从 100 磁道移动到 50 磁道,需要移动 50 个磁道,耗时 50ms; - 从 50 磁道移动到 30 磁道,需要移动 20 个磁道,耗时 20ms; - 从 30 磁道移动到 90 磁道,需要移动 60 个磁道,耗时 60ms; - 从 90 磁道移动到 120 磁道,需要移动 30 个磁道,耗时 30ms; - 从 120 磁道移动到 180 磁道,需要移动 60 个磁道,耗时 60ms。 因此,总共需要移动 220 个磁道,耗时为 220ms。 对于 SSTF 调度算法,我们可以按照距离当前磁头位置最近的磁道顺序来访问请求的磁道。因此,访问的顺序应该是 90、120、50、30、180。这样,总共需要移动的磁道数为 130,耗时为 130ms。 因此,答案为: - 使用 SCAN 调度算法时,耗时为 220ms; - 使用 SSTF 调度算法时,耗时为 130ms。
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用C++编写程序,实现以下问题2、题目ID Codes(POJ1146) Time Limit: 1000MS Memory Limit: 10000K 描述: It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.) An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set. For example, suppose it is decided that a code will contain exactly 3 occurrences of a', 2 of b' and 1 of c', then three of the allowable 60 codes under these conditions are: abaabc abaacb ababac These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order. Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message No Successor' if the given code is the last in the sequence for that set of characters. 输入: Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #. 输出: Output will consist of one line for each code read containing the successor code or the words 'No Successor'. 样例输入 abaacb cbbaa # 样例输出 ababac No Successor

3)A digital clock consists of a screen to display the time and a dial for setting in turn the year, month, day, hour and minute. Twisting the dial to the left reduces by one the value being changed but twisting it to the right increases it by one. Pushing the dial alters which value is being adjusted. At first, it is the year but after the dial is pushed once, it is the month, then after the dial is pushed again, it is the day and so on. Imagine the clock is represented by a class with attributes year, month, day etc. The following is what the code for a method rotateDialLeft() might look like. public void rotateDialLeft() { if (mode == YEAR_MODE) { year--; } else if (mode == MONTH_MODE) { month--; } else if (mode == DAY_MODE) { day--; } else if (mode == HOUR_MODE) { hour--; } else if (mode == MINUTE_MODE) { minute--; } } The code for rotateDialRight() is similar. Apply the Open-Closed Principle to explain why the above code is unsatisfactory from the design viewpoint, considering the possibility of future change to the code, giving an example of such a change. 5)Give the code required for the classes introduced in question 3), focusing on the code for a method selectState() which changes the value that is being adjusted from years to months. Make it clear in which classes the code is to be found. Assume the existence of other methods that are needed such as getMonthSetUpState(). 8)Suppose that in a multiplayer role-playing game, a class Client has a dependency to an interface Fighter with public methods attack(), defend() and escape(). The game designer now wishes for Client to use a class Wizard with three different but equivalent public methods castDestructionSpell(), shield() and portal(). Explain how it is possible to do this using an appropriate design pattern.

(a) Consider the case of a European Vanilla Call option which is path independent. Examine the convergence of the Monte Carlo Method using the programme given in ‘MC Call.m’. How does the error vary with the number of paths nP aths? The current time is t = 0 and the Expiry date of the option is t = T = 0.5. Suppose that the current value of the underlying asset is S(t = 0) = 100 and the Exercise price is E = 100, with a risk free interest rate of r = 0.04 and a volatility of σ = 0.5. (b) Now repeat part (a) above but assume that the volatility is σ = 0.05. Does the change in the volatility σ influence the convergence of the Monte Carlo Method? (c) Now repeat part (a) but instead of taking one big step from t = 0 to t = T divide the interval into nSteps discrete time steps by using the programme given in ‘MC Call Small Steps.m’. Confirm that for path independent options, the value of nP aths determines the rate of convergence and that the value of nSteps can be set to 1. (d) Now let us consider path dependent options. The programme given in ‘MC Call Small Steps.m’ is the obvious starting point here. We assume that the current time is t = 0 and the expiry date of the option is t = T = 0.5. The current value of the underlying asset is S(t = 0) = 100 and the risk free interest rate is r = 0.05 and the volatility is σ = 0.3. (i) Use the Monte Carlo Method to estimate the value of an Arithematic Average Asian Strike Call option with Payoff given by max(S(T) − S, ¯ 0). (ii) Use the Monte Carlo Method to estimate the value of an Up and Out Call option with Exercise Price E = 100 and a barrier X = 150. (iii) Comment on the the rate of convergence for part (i) and (ii) above with respect to the parameters nP aths and nP aths使用matlab编程

用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

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