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###################6S############## ##5####6##*##6D####4V########6#####6#####6#####6#####6D####6#####6#####6#####6### ##6#####6#####Engineering Computing 600.105 - 1998Lecture 6 Control Flow and Loops6.1 Concept of control flow Programming is not rigidly following one statement to the next, from the starting to the end, without jumping order, but it often needs to do repetitive calculations (iterations) at certain stage of the programming, then it continues from where it was interrupted. The decision when to carry out these repetitive task (iterations) temporarily is a decision making process, which is implemented by one of the four control-flow structures, to be discussed in this session. These four control structures are: for loops, while loops, if-else-end structures and switch-case structures. 6.2 for loops (MATLAB Version 5 User's Guide, 1997) A for loop allow a group of commands to be repeated over a fixed, predetermined number of times between the for and end statements. After the completion of the loop, the control then return to normal execution order, ie. after the end of the loop. General form of a for loop: for x = predetermined numbers then perform a group of commands ... end Examples:1). to calculate 9 sine values from 0 to 180 degrees @22.5 for n=1:9 % for n from 1 to 9 x(n)=sin(((n-1)*22.5*pi/180); % 0 to 8 @22.5, x is an array end But, it is more efficient in run time if for loop is replaced by an equivalent array approach, and also with preallocated array before the for loop, to maximise execution speed: (WHY?) x=zeros(1,10); % preallocated memory to x n=0:8 % for 0 to 8 directly x=sin(((n1)*22.5*pi/180); % the statement is to be repeated 9 times2). use different input format for a for loop data=[1 3 5 7; 9 13 17 21]; for n=data x=n(1)-n(2) % find the difference of two specified numbers end % answer: 8, 10, 12, 143). double loop, or nested loop f for n=1:5 f for m=5:-1:1 A(n,m)=n^2+m^2; % create a x by m matrix end disp(n) end disp(A) % ? * Remember: reassigning the loop variable "n" within the loop using the same variable, such as n = 5, is not allowed. 6.3 while loops A while loop executes a group of statements over an indefinite number of times, as long as the conditional expression between the while and end statements are True. General form: while conditional expression is True a group of commands . . . end Examples: num=0; EPS=1; while (1+EPS)>1 EPS=EPS/2; num=num+1; end % final answer: EPS=2*EPS num=53, EPS=2.2204e-16* Conditional statements give a scalar value, but array is also valid. Another way to jump out pf an infinite loop, by using a break command: EPS=1; for n=1:1000 % n over indefinite times EPS=EPS/2; if(1+EPS)<=1 % condition to break off from the loop EPS=EPS*2 break % command for breaking off loop end end6.4 if-else-end constructions A simply if-else-end loop is not different to the for and while loops, for example CD-disk = 10; cost=CD_disk*4.99; if CD_disk>5 % discount for large purchase cost=0.7*cost; % 30% discount end But, it becomes useful if there are two alternatives or more: if expression1 commands evaluated if expression1 is True elseif expression2 commands evaluated if expression2 is True elseif... . . else commands evaluated if NO other expression is True end * Only of the cases is executed, and the last else statement may or may not appear.6.5 switch-case constructions Similar to if-elseif-else-end loop, if repeated conditional evaluations are required against a common argument, the switch-case constructions can be used: switch expression (or argument) case {test_expression1} commands1 ... case {test_expression2} commands2 ... ... ... otherwise last_commands end * where the "expression" is either in scalar or a character string It is tested against the test_expression1
in the first case; If the first comparison is not true, then the second case is tested, and so on ... If any of the test_expression is satisfied, the rest are skipped. Example: (to convert x=1.84 metres into centimetres) x=1.84; units='m'; % given 1.84 metres switch units case {'inch','in'} y=x*2.54; case {'feet','ft'} y=x*12*2.54; case {'metre",'m'} % matching case with the input units y=x*100; % answer: y=184 case {'centimetre",'cm'} y=x; case {'millimetre",'mm'} y=x/100; otherwise disp(['Unknown units: ',units]) y=nan; end6.6 A amortisation example (MATLAB Version 5 User's Guide, 1997) John Citizen wants to purchase a car, for which he borrow $15,000 at annual interest rate of 8.5% over 3 years. After making equal monthly payments, the interest is calculated based on the remaining principal after each payment. Produce a script M-file and print an amortisation schedule, showing payment number, balance, interest and principal paid at each monthly payment. Assume the monthly payment P on a loan of A dollars, having a monthly interest rate of R, paid off in M months, is given by # % amort.m script file (MATLAB v.5 User's Guide) A = 10,000; % loan M= 3*12; % number of months aR=8.9; % annual interest rate in % R=(aR/100)/12; % monthly rate in % P=A*(R*(1+R)^M /((1+R)^M -1)); % monthly payment % % preallocate storage B=zeros(M,1); % storage for monthly remaining balance Ip=B; % storage for interest payment Pr=B; % storage for principal paid % for m=1:M if m==1 Ip(m)=R*A; % interest for first month else Ip(m)=R*B(m-1); % interest for m >= 2 end % Pr(m)=P-Ip(m); % Principal paid if m==1 B(m)=A-Pr(m); % remaining balance at first month else B(m)=B(m-1)-Pr(m); % remaining balance end end % disp(['Amount = ' num2str(A)]) disp(['Interest Rate = ' num2str(R)]) disp(['No. of Months = ' int2str(M)]) disp(['Payment = ' num2str(P)]) disp( ' ') disp(' AMORTISATION SCHEDULE') disp(' Payment No. Balance Interest Principal Paid') disp([(1:M)' B Ip Pr]) format short g OUTPUT running the amort.m fileAmount = 10000Interest Rate = 8.9No. of Months = 36Payment = ---.-AMORTISATION SCHEDULE Payment No. Balance Interest Principal Paid 1 1.0 2 2.0 3 3.0 4 4.0 5 5.0 . . . 3 30.0 . 3 35.0 36.0 -0.00 6.##u#######K#######'#v###=#########'#v#-#a#6#####&## #######,#####Times####
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