It is sometimes said that tape is a sequential-access medium, whereas magnetic d
ID: 3542016 • Letter: I
Question
It is sometimes said that tape is a sequential-access medium, whereas magnetic disk is a random-access medium. In fact, the suitability of a storage device for random access depends on the transfer size. The term streaming transfer rate denotes the data rate for a transfer that is underway, excluding the effect of access latency. By contrast, the effective transfer rate is the ratio of total bytes per total seconds, including overhead time such as the access latency.
Suppose that, in a computer, the level 2 cache has an access latency of 8 nanoseconds and a streaming transfer rate of 800 megabytes per second, the main memory has an access latency of 60 nanoseconds and a streaming transfer rate of 80 megabytes per second, the magnetic disk has an access latency of 15 millisecond and a streaming transfer rate of 5 megabytes per second, and a tape drive has an access latency of 60 seconds and a streaming transfer rate of 2 megabytes per seconds.
a. Random access causes the effective transfer rate of a device to decrease, because no data are transferred during the access time. For the disk described, what is the effective transfer rate if an average access is followed by a streaming transfer of 512 bytes, 8 kilobytes, 1 megabyte, and 16 megabytes?
b. The utilization of a device is the the ratio of effective transfer rate to streaming transfer rate. Calculate the utilization of the disk drive for random access that performs transfers in each of the four sizes given in part a.
c. Suppose that a utilization of 25% (or higher) is considered acceptable. Using the performance figures given, compute the smallest transfer size for disk that gives acceptable utilization.
d. Complete the following sentence: A disk is a random-access device for transfers larger than bytes, and is a sequential-access device for smaller transfers.
e. Compute theminimum transfer sizes that give acceptable utilization for cache,memory, and tape.
f. When is a tape a random-access device, and when is it a sequential-access device?
Explanation / Answer
a) The effective transfer rate, for 512 is calculated as:
Latency time=15 ms
Streaming transfer = 5 MB/s
Transfer size,(N) = 512 B
Effective Transfer Rate(ETT) = ((N/Streaming Transfer) + latency).
= (512B/5MB per second) + 15 ms = 15.0097ms.
Therefore, effective transfer Rate = 512B/15.0097ms= 33.12 KB/sec.
Using the above formula, calculate Transfer time for 8KB,1MB and 16 MB.
Effective Transfer Rate for 8KB = .47MB/sec.
Effective Transfer Rate for 1MB = 4.65MB/sec.
Effective Transfer Rate for 16MB = 4.98MB/sec.
b) Utilization of the device = ETT/STT
Use the above given formula, then calculate
For 512 B = 33.12KB/sec / 5MB / sec = .0064 = .64%
For 8KB = 9.4%.
For 1MB = 93%.
For 16MB = 99.6%.
c) Calculate 25% of ETT/STT, solving for transfer size N.
STT = 5MB, so1.25MB/S = ETT.
1.25MB/S * ((N/5) + .015) = N.
.25X + .01875 = N.
N = .025MB.
d) For smaller transfers, there will be sequential access. Whereas, for larger transfer than K bytes, it will be random access.
Note : K > disk block size
e) Calculate minimum transfer size,
For acceptable utilization of cache memory:
STT = 800MB,
ETT = 200,
Latency = 8 * 109.
200 (NMB/800 + 8 * 109) =NMB.
.25NMB + 1600 * 109= NMB.
N = 2.24 bytes.
Calculate for memory:
STT = 80MB,
ETT = 20, L = 60 * 109.
20 (NMB/80 + 60 * 109) =NMB.
.25NMB + 1200 * 109= NMB.
X = 1.68 bytes.
Calculate for tape:
STT = 2MB, ETT = .5, L = 60s.
.5 (NMB/2 + 60) = NMB.
.25NMB + 30 =NMB.
N=40MB.
f) Tape can be used for both the purposes like random access device as well as sequential access device. If tape is used to restore backup, then it is used as sequential access.
In any other case, if tape is used to access some records, then in that case the tape is used as random.
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