Operating System Concepts - Assignment Example

 Operating Systems Assignment 1. (20 points) If the operating system has a user-level thread library what, if any, issues must be addressed to support concurrent programming? Provide a detailed answer. A user-level thread library provides support for thread creation, scheduling and management with no support from the kernel. Supporting concurrent programming is paramount in a user level thread library. Yet, to do concurrent programming is not easy as various operations interleave among threads making the execution of program un-deterministic. To support concurrent computation, at user level thread, we must address the following issues thus: 1. Scheduling of threads: Usually it's done by default scheduling which user-level thread library does. Usually four approaches are used by multi processors Load Sharing, Gang Scheduling, dedicated processor assignment and Load Sharing. For concurrent programming I support Gang Scheduling as a set of related threads are scheduled to run on the same time. 2. Methods and Object Synchronization (mutual exclusion): One process should be able to use only one resource at a time. The request by other process should be delayed till the resource has been released. 3. Deadlock prevention: Deadlock can be prevented by using different schemes or by avoiding the necessary conditions (mutual exclusion, hold and wait, no preemption and circular wait (Stalling, 274)) which hold true for it. The different schemes are: requesting all resources at once Preemption Resource ordering 2. (20 points) Given that the first three necessary conditions for a deadlock are in place, provide detailed comments on the feasibility of the following strategy and the potential outcomes based upon its application. All processes are given unique priorities. When more than one process is waiting for a resource and the resource becomes available, allocate the resource to the waiting process with the highest priority. The first three conditions for the deadlock are: Mutual exclusion: It holds for non shareable resources. Shareable resources do not need mutual exclusion but all resources cannot be shareable, hence the condition holds. Hold and Wait: To make sure that hold-and-wait condition never occurs, we make sure that a process which requests resource does not hold any. No Preemption: There cannot be preemption of resource that has already been allocated. Now, in the given scenario all processes have their unique priority. Let us assume there are two resources R1 and R2 and process P1 has higher property than P2. Consider P1 holds R1 and P2 holds R2 and P2 is requesting for R1. Now, if P1 requests for R2 it will not get the resource even if it has the higher priority, because resources are non preemptive. So deadlock is still possible in this condition. 3. (20 points) Given a four level hierarchical storage system consisting of: cache, primary storage, secondary storage, and tertiary storage. Assume the following: programs may be executed on any of the four levels; each level consists of the same amount of real storage and the range of addresses on each level is identical. The speed at which programs are run is grouped from slowest (tertiary storage) to fastest (cache), where each layer is 10 times faster than the previous lower layer. There exists one CPU in this system, which may run one program at a time. Should information be allowed to move from any level to any level or should transfers only occur from adjacent levels? Explain in detail. If we allow transfer from any level to any level then it requires direct interconnection between each level. This will be too complex and so is not a recommended approach. Another way to transfer data is centralized approach. Means every level must be connected to one central level (for e.g. primary storage). So in this data can be transferred from any level to any level by routing through the central level. This will require less interconnection. The last approach is transfer of data b/w adjacent levels. But due to serial connection there will be more delay in the transfer. 4. (20 points) Gopher Gallery consists of a shopping mall and a cart ride that covers the 150 acre habitat. There are m visitors and n single-person vehicles. Visitors stroll around the mall at their leisure, then line up for the cart ride. When a cart is available, it allows the single passenger to climb aboard and drives around the habitat for a random amount of time. If the n carts are all taken, then a future rider waits; if a vehicle is available but no one is waiting, then the vehicle waits. The solution to this problem must synchronize visitor tasks and vehicle tasks using semaphores. Below is a potential solution. Correct any issues with this code, if any exist. Explain your position in detail. Visitor() { vehicleAvailable.wait(); vehicleTaken.signal(); vehicleFilled.wait(); visitorReleased.wait(); } Vehicle() { while(True) { vehicleAvailable.signal(); vehicleTaken.wait(); vehicleFilled.signal(); Drive through habitat for some arbitrary amount of time; vehicleReleased.signal(); } } First, we should analyse the vehicle part. The semaphore first transmits the signal of availability. When it's been taken the semaphore value is decreased due to wait condition. The signal operation will again increment the value on release. Now, the visitor will be receiving a signal if the cart is available. The value will be increased (for other visitors) when it's taken and the wait condition will again decrease semaphore value on release. The given solution is correct. But there is an unnecessary semaphore “vhicleFilled”. This semaphore can be removed from the code without affecting the results. 5. (20 points) Given the following statements, justify which are true and which are false: a. Timesharing systems normally use non-preemptive CPU scheduling. (False) CPU scheduling decisions may take place when a process switches from running to ready state or switches from waiting to ready which is preemptive scheduling. b. Response times are less predictable in preemptive vs. nonpreemptive systems. (False) It's more a function of scheduling algorithm. c. Shortest remaining time first always has a lower response time than Shortest Job First.( True) If we assume that SJF is non preemptive d. Since Shortest Job First gives preference to short jobs, it is useful in timesharing systems. ( False) it may cause starvation. Works Cited Silbershatz, G. G. (2004). Operating System Concepts. Wiley. Stalling, W. (2001). Operating System. Prentice Hall. Read More