The cycles are on the right of the code 5. Given the following three loop struct

Question

The cycles are on the right of the code

5. Given the following three loop structures discussed in class METHOD 1: METHOD 2: METHOD 3: org$800 jsr bgmac(4) stop org$800 jsr bgmac(4) stop org $800 jsr bgmac(4) stop #XA #YA #0,MA #0,MA+2 bgmac lax #XA ;(2) | bgmac lax ; (2) | bgmac1dx #XA Idy movw movw clrb #YA #0,MA #0,MA+2 Idy movw movw ldab Idy movw movw ldab #YA #0,MA #0,MA+2 ; (5) ; (5) ; (5) ; (5) ; (5) ; (5) : (1) loop dbeg b,exit: (3/3) loop emacs MA (13) +1 loop cmpb #N :(13) beq exit (3/1) emacs MA emacs MA leax 2, x leay 2,y bra loo leax 2,x leay 2,y dbne b,loop; (3/3) (13) ; (3) exit rts incb bra ; (1) exit rts loop equ ? XA rmb*N YA rmb*N MA rmb 4 exit rts ; (5) N equ ? equ ? rmb 2*N XArmb 2*N YA rmb 2*N MArmb 4 XA MA rmb 4 Write a formula for the total execution time (in machine cycles) of each of the programs, above, as a function of N (for N2. Note that the cycle count for each instruction is given; for conditional branches, these are given as cycles for the branch taken/not taken. Then, using the showcycles command in the CW Command window, determine the number of cycles consumed by each method for N=255 (set a breakpoint at the stop instruction) Method 1 Method 2 Method 3

Explanation / Answer

Formula for total execution time is:

Method 1: 25*N+26

Method 2: 23*N+27

Method 3: 20*N+24

Number of cycles consumed when N=255:

Method 1: 25*255+26 = 6401

Method 2: 23*255+27 = 5892

Method 3: 20*255+24 = 5124

Explaination:

There are two types of instructions in each method: sequential(outside loop block) and looping(inside loop block). All the sequential instructions are taking same time in all the methods i.e. 4[jsr]+2[ldx]+2[ldy]+5[movw]+5[movw]+1[clrb/ldab]+5[exit] = 24 machine cycles.

Total Execution Time = (num_times_loop_executed * total_time_taken_looping_instr )+ time_consumed_exiting_loop + time_taken_seq_instr

Method1:

num_times_loop_executed = N (clrb initializes the register b with 0 and loop increments b by value 1 every time its executed. Loop exits when it's value equals N. Hence loop is executed from value 0,1,.....,N-1.)

total_time_taken_looping_instr = 1+3+13+2+2+1+3=25

time_consumed_exiting_loop = 1[cmpb] + 1[beq] = 2 (When value stored in regesiter is N then loop will exit and only these two instructiona will be executed)

Hence,

Total Execution Time = N * 25 + 2 + 24 = 25 * N +26

Method 2:

num_times_loop_executed = N (ldab initailizes register with N+1 and when dbeq is executed, it first decrements the value by 1 then checks branching condition. Thus loop is being executed for the values N+1,N.....,2. When this value becomes 1, dbeq will decrement it to 0 & loop will exit.)

total_time_taken_looping_instr = 3+13+2+2+3 = 23

time_consumed_exiting_loop = 3 (When value stored in registar reaches 0 then loop will exit and only dbeq instruction will be executed)

Hence,

Total Execution Time = N * 23 + 3 + 24 = 23 * N +27

Method 3:

num_times_loop_executed = N (ldab initailizes register with N , loop is being executed for the values N,N-1.....,1).

total_time_taken_looping_instr = 13 + 2 +2 +3 = 20

time_consumed_exiting_loop = 0 (When value stored in registar reaches 1, dbeq will decrement it to 0 and exit immediately, hence no further execution will take place

Hence,

Total Execution Time = N * 20 + 0 + 24 = 20 * N +24

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