Answer to Question #365 in Assembler for sharjeel

Question

What percentage of time will a 20 MIPS processor spend in the busy wait loop of 65-character line printer when it takes 3 m-sec to print a character and a total of 457 instructions need to be executed to print 65 character lines? Assume that 4 instructions are executed in the polling loop?

Answer 1

The processor executes 20 million instructions per second, and we know that printing a single character takes 3 ms. Thus, in order to print all the characters one must take 195 ms. To do this we need to perform 457 instructions, which in total is 0.002285 ms, hence the delay in following the instructions in the processor will make no changes in the total time of the printer. Response: 195 ms.