Word Trans, 4th ed., Ch. 8, pg 140 Question #9

[futjim]

Dear all,

I read the answer to this question but I am still confused. Does someone have an easier to solve this problem?

Q: A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10am and the last showing of each program ends at 4 pm. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?

I'm not sure how to organize this data. Can anyone help?

Thanks!

[tim]

Okay, this one is very non-traditional and requires you to step out of your comfort zone and experiment with some possibilities rather than plug in a pre-fabricated algorithm. In this one, your first observation should be that even if you could choose the start times at will the process will take at least 160 minutes (movies of 15+30+40+45 and breaks of 10+10+10)..

After a little experimenting you'll notice that three of the four movies (the 15, 30, and 45) always start and end on a quarter-hour (:00, :15, :30, or :45 after the hour). Well, if you have a movie ending on a quarter-hour and another starting on a quarter-hour, your 10-minute break will be forced to extend to 15 minutes anytime two of those movies are adjacent, which means 5 minutes of wasted time..

Thus we want to minimize the number of times that two of those movies (the 15, 30, or 45) are adjacent. This is accomplished by placing the 40 in between two of them. Even if we do that though we cannot avoid 5 minutes of wasted time somewhere along the line, which means it will be impossible to push our time lower than 165 minutes. At this point, it should be pretty easy to play around with the values and the constraint that the 40 is in between two other movies to come up with a 165. Since we've already ascertained that the answer cannot be lower than 165, once we have our 165 we know it's the answer..

[aditisoni1987]

Okay, this one is very non-traditional and requires you to step out of your comfort zone and experiment with some possibilities rather than plug in a pre-fabricated algorithm. In this one, your first observation should be that even if you could choose the start times at will the process will take at least 160 minutes (movies of 15+30+40+45 and breaks of 10+10+10)..

After a little experimenting you'll notice that three of the four movies (the 15, 30, and 45) always start and end on a quarter-hour (:00, :15, :30, or :45 after the hour). Well, if you have a movie ending on a quarter-hour and another starting on a quarter-hour, your 10-minute break will be forced to extend to 15 minutes anytime two of those movies are adjacent, which means 5 minutes of wasted time..

Thus we want to minimize the number of times that two of those movies (the 15, 30, or 45) are adjacent. This is accomplished by placing the 40 in between two of them. Even if we do that though we cannot avoid 5 minutes of wasted time somewhere along the line, which means it will be impossible to push our time lower than 165 minutes. At this point, it should be pretty easy to play around with the values and the constraint that the 40 is in between two other movies to come up with a 165. Since we've already ascertained that the answer cannot be lower than 165, once we have our 165 we know it's the answer..

Tim can you please explain how did you arrive at 5 mins of wasted time.

[tim]

i already did: "if you have a movie ending on a quarter-hour and another starting on a quarter-hour, your 10-minute break will be forced to extend to 15 minutes anytime two of those movies are adjacent, which means 5 minutes of wasted time."

was there something about this explanation that did not make sense to you?