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Question:
a) What do you understand by Active directory domain in Windows Server 2008? Support your answer with a diagram.
b) Distinguish between active directory trees and forests
c) Describe an Active Directory Forest with a single tree using a diagram.
d) Describe an Active Directory Forest with multiple trees using a diagram.
e) Describe briefly the three models which are used in designing an organisational unit (OU)? Support your answer with diagrams
Explain the Novell NetWare NetWare doesn't really have the concept of processes in the architecture, as the most closely associated element in the NetWare environment to
Synchronization In communication among processes messages passed may be blocking or nonblocking - also called as synchronous and asynchronous. Blocking send : The sending pr
What are the different types of semaphore? Principally the semaphore is able to be classified into two on the basis of their values a) Counting semaphore: In counting s
Explain the Worker Model The Worker Model of client-server application architecture provides a very good understanding of threads and their power to the developer. This exercis
Explain why SSTF scheduling tends to favor middle cylinders over the innermost and outermost cylinders. The center of the disk is the location having the fewest average distanc
when demand is 24000 units/year, production rate is 48000 units/year, setup cost is rs 200 per setup, carring cost is rs 20 per units/year, and economic batch quantity is 692.8203
ADVANTAGES AND INCONVENIENCES OF ULT Advantages: Thread switching doesn't involve the kernel - no mode switching Scheduling is able to be application specific -
How free-space is managed using bit vector implementation? The free-space list is implemented as a bit map or bit vector. Each block is shown by 1 bit. If the block is free, th
Q. Dynamically loadable kernel modules give elasticity when drivers are added to a system however do they have disadvantages too? Under what situations would a kernel be compiled
Worst Fit Algorithm Here we obtain the largest space available for the smallest. Therefore after that process entered there will be much more space remaining. In this space we
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