MegaTech Insurance Co. has just bought out Old Line Insurance from Stodgy Corp. They have put the 10,000 policies from Old Line into their computer system, and today they will begin accepting new policies. Everything is computerized, but just to make sure, they will follow Old Line's long-time practice and have Old Foskins double-check all the policies.
The policies are stored alphabetically in the computer. Each morning a technician prints out the next 100 policies in sequence for Old Foskins to check by paper and pencil examination. Each evening 200 new policies are merged into the computer in alphabetic order.
How long will it take Old Foskins to reach the end of the list?
We'll assume the new clients being added will have positions in the alphabet that generally match the distribution in the existing set of clients, so that some of the new clients will be behind where Old Foskins is in his checking. We'll assume that's approximately proportional to the fraction of the alphabetic list he's already covered to the total to be done.
behind=0; ahead=10000; ct=0;
while ahead>0
behind=behind+100;
ahead=ahead-100;
tot=behind+ahead;
newbehind=round(200*behind/tot); % considering fraction added
newahead=ahead+200-newbehind; % the whole new ahead, including old
behind=behind+newbehind;
ahead=newahead;
ct=ct+1;
end
ct
finds that after 317 days, Foskins will have reached the end of the data base, having examined all the old policies and some of the new.
In that time he will have examined 31,700 policies.
But of course there are now 10,000 + 200*317 = 73,400 policies altogether, so there's still a considerable backlog.
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Posted by Charlie
on 2023-12-14 14:13:49 |
most likely answer
