CS501 Assignment # 5 Solution

Question No 1
What percentage of time will a 10 MIPS processor spend in the busy wait loop of an 65 - character line printer when it takes 1 msec to print a character and a total of 950 instructions need to be executed to print an 85 character line? Assume that 3 instructions are executed in
the polling loop?


Here we are given that 950 instructions are required to 85 character line, but we are required to print
65 characters on a line therefore, number of instructions required to print 65 character line will be calculated as under
instructions required for printing 65 character line  950
65 85
 instructions required for printing 65 character line  950  65  726.47 or 726 instructions
85



Out of 726 instructions (65 3)  195

are required for polling

Remaining instructions = 726 195  531

For a 10 MIPS processor, the execution of remaining instructions takes

531
10 106


 5.31105 sec  0.0531ms



Printing of 65 character line takes 65 × 1 = 65 ms



 65ms  0.0531ms  64.9469ms

is spent on polling loop

before the 65 characters can be printed


This makes 64.9469 100  99.91% of total time
65
Question No 2
A 4 K byte block of data is read from a disk drive. What is the overall data transmission rate if
the disk drive has a latency of 9 ms, and a burst bandwidth of 1 MB per second?

Time required for execution = 4K/(1MB/s) + 9ms = 4ms + 9ms = 13ms

Above calculation can be done simply as follows
4K = 4 × 103 bytes
1MB/s = 1 × 106 bytes/sec

Overall transmission rate will be 4K/13ms = 0.30MB/s

Above calculation can be done simply as follows
4K = 4 × 103 bytes
13ms = 13 × 10-3 =0.031 sec