Q. It is occasionally said that tape is a sequential-access medium whereas magnetic disk is a random-access medium. In fact the correctness of a storage device for random access depends on the transfer size. The term streaming transfer rate indicates the data rate for a transfer that is underway excluding the effect of access latency. Through contrast the effective transfer rate is the ratio of total bytes per total seconds including overhead time such as the access latency.

Presume that in a computer the level-2 cache has an access latency of 8 nanoseconds as well as a streaming transfer rate of 800 megabytes per second the major 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 as well as a streaming transfer rate of 5 megabytes per second as well as a tape drive has an access latency of 60 seconds and a streaming transfer rate of 2 megabytes per seconds.

a. Random access results 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 and 8 kilobytes and 1 megabyte and 16 megabytes?

b. The utilization of a device is the ratio of effective transfer rate to streaming transfer rate. Compute the utilization of the disk drive for random access that perform transfers in each of the four sizes given in part a.

c. Suppose that a use of 25 percent (or higher) is considered acceptable. Utilizing 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 transport larger than bytes as well as is a sequential access device for smaller transfers.

e. Calculate the minimum transfer sizes that give acceptable utilization for cache and memory and tape.

f. When is a tape a random-access device, and when is it a sequential access device?

Answer:

a. For 512 bytes the efficient transfer rate is calculated as follows. ETR = transfer size/transfer time. If X is transfer size after that transfer time is ((X/STR) + latency). Transfer time is 15ms + (512B/5MB per second) = 15.0097ms. Effective transfer rate is consequently 512B/15.0097ms = 33.12 KB/sec.

ETR for 8KB = .47MB/sec.

ETR for 1MB = 4.65MB/sec.

ETR for 16MB = 4.98MB/sec.

b. Utilization of the device for 512B = 33.12 KB/sec / 5MB/sec = .0064 = .64

For 8KB = 9.4%.

For 1MB = 93%.

For 16MB = 99.6%.

c. Compute .25 = ETR/STR solving for transfer size X.

STR = 5MB, so 1.25MB/S = ETR.

1.25MB/S * ((X/5) + .015) = X.

.25X + .01875 = X.

X = .025MB.

d. A disk is the random-access device for transfers larger than Kbytes (where K > disk block size) as well as is a sequential-access device for smaller transfers.

e. Compute minimum transfer size for acceptable utilization of cache memory:

STR = 800MB, ETR = 200, latency = 8 * 10-9.

200 (XMB/800 + 8 X 10-9) = XMB.

.25XMB + 1600 * 10-9 = XMB.

X = 2.24 bytes.

Compute for memory:

STR = 80MB, ETR = 20, L = 60 * 10-9.

20 (XMB/80 + 60 * 10-9) = XMB.

.25XMB + 1200 * 10-9 = XMB.

X = 1.68 bytes.

compute for tape:

STR = 2MB, ETR = .5, L = 60s.

.5 (XMB/2 + 60) = XMB.

.25XMB + 30 = XMB.

X = 40MB.

f. It relies upon how it is being used. Presume we are using the tape to restore a backup. In this example the tape acts as a sequential-access device where we are sequentially reading the contents of the tape. As another instance presumes we are using the tape to access a variety of records stored on the tape. In this example access to the tape is arbitrary and therefore considered random.