Solution manual william stalling

Data and computer communications william stallings 8th edition pdf.

1.11 Define Ci = Average cost per bit, memory level i Si = Size of memory level i Ti = Time to access a word in memory level i Hi = Probability that a word is in memory i and in no hher-level memory Bi = Time to transfer a block of data from memory level (i 1) to memory level i Let cache be memory level 1; main memory, memory level 2; and so on, for a total of N levels of memory.

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The address portion of the IR (940) is loaded into the MAR. The I/O transfer rate is therefore 25,000 words/second. The number of machine cycles available for DMA control is 106(0.05 × 5 0.95 × 2) = 2.15 × 106 If we assume that the DMA module can use all of these cycles, and nore any setup or status-checking time, then this value is the maximum I/O transfer rate. A reference to the first instruction is immediately followed by a reference to the second. The ten accesses to a[i] within the inner for loop which occur within a short interval of time.

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Unformatted text preview: INSTRUCTORS MANUAL OPERATING SYSTEMS: INTERNALS AND DESN PRINCIPLES FOURTH EDITION WILLIAM STALLINGS Copyrht 2000: William Stalling TABLE OF CONTENTS PART ONE: SOLUTIONS MANUAL ...............................................................................1 Chapter 1: Computer System Overview ......................................................................2 Chapter 2: Operating System Overview ......................................................................6 Chapter 3: Process Description and Control ...............................................................7 Chapter 4: Threads, SMP, and Microkernels.............................................................12 Chapter 5: Concurrency: Mutual Exclusion and Synchronization.........................15 Chapter 6: Concurrency: Deadlock and Starvation..................................................26 Chapter 7: Memory Management ...............................................................................34 Chapter 8: Virtual Memory..........................................................................................38 Part One SOLUTIONS MANUAL This manual contains solutions to all of the problems in Operating Systems, Fourth Edition. We begin with the result from probability theory that: N Expected Value of x = ∑ i Pr[x = 1] i =1 We can write: N Ts = ∑T i H i i =1 We need to realize that if a word is in M1 (cache), it is read immediately.

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If you spot an error in a solution or in the wording of a problem, I would greatly appreciate it if you would forward the information via email to me at [email protected] The value in location 300 (which is the instruction with the value 1940 in hexadecimal) is loaded into the MBR, and the PC is incremented. If it is in M2 but not M1, then a block of data is transferred from M2 to M1 and then read. From Equation 1.1 : 1.1 × T1 = T1 (1 – H)T2 (0.1)(100) = (1 – H)(1200) -5- H = 1190/1200 1.13 There are three cases to consider: Location of referenced word In cache Not in cache, but in main memory Not in cache or main memory Probability 0.9 (0.1)(0.6) = 0.06 Total time for access in ns 20 60 20 = 80 (0.1)(0.4) = 0.04 12ms 60 20 = 12000080 So the average access time would be: Avg = (0.9)(20) (0.06)(80) (0.04)(12000080) = 480026 ns 1.14 Yes, if the stack is only used to hold the return address.


Solution manual william stalling:

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