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 .

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磁盘调度算法(SCAN和SSTF)

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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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(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

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