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Hi all,

I was trying to analyze the following problem but couldnt get any idea of how the total amount of terabytes can be calculated. Could you please help me in understanding the problem. And let me know the better way to do the (dynamic) program.

Problem:

Suppose n=4 and the values of x and s are given by the following table.

Day1 Day2 Day3 Day4

x 10 1 7 7

s 8 4 2 1

The best solution would be to reboot on day2 only; this way, you process 8 terabytes

on day1, then 0 on day2, then 7 on day3, then 4 on day4, for a total of 19.

Note that if you didn't reboot at all, you'd process 8+1+2+1 =12; and other rebooting strategies

give you less than 19 as well.

Give an efficient algorithm that takes values for x1,x2,x3..........xn and

s1,s2,s3........sn and returns the total number of terabytes processed by an optimal solution.

Thanks & Regards,

Venkat.

I was trying to analyze the following problem but couldnt get any idea of how the total amount of terabytes can be calculated. Could you please help me in understanding the problem. And let me know the better way to do the (dynamic) program.

Problem:

Suppose n=4 and the values of x and s are given by the following table.

Day1 Day2 Day3 Day4

x 10 1 7 7

s 8 4 2 1

The best solution would be to reboot on day2 only; this way, you process 8 terabytes

on day1, then 0 on day2, then 7 on day3, then 4 on day4, for a total of 19.

Note that if you didn't reboot at all, you'd process 8+1+2+1 =12; and other rebooting strategies

give you less than 19 as well.

Give an efficient algorithm that takes values for x1,x2,x3..........xn and

s1,s2,s3........sn and returns the total number of terabytes processed by an optimal solution.

Thanks & Regards,

Venkat.

Problem Statement:

To run a high performance computing system capable of

processing several terabytes of data per day .For each of n days ,you are

being presented with a quantity of data; on day i ,you’re presented with xi

terabytes .For each terabyte you process ,you receive a fixed revenue, but any

unprocessed data becomes unavailable(i.e expires) at the end of the day and

thus you can not work on it in subsequent days.

You also can not always process everything each day because you are

constrained by the capabilities of your computing system one-of-a kind software that, while very sophisticated ,is not totally reliable, and so the amount of data you can process goes down with each day that passes since the most recent reboot of the system. On the first day after a reboot, you can process s1 terabytes, on the second day after the reboot ,you can process s2 terabytes ,and so on, up to sn; it is assumed that s1>s2>s3>….>sn>0.Note that on day i you can only process up to xi terabytes, even if the corresponding sj is larger(note the applicable subscripts can be different values).To get the system back to peak performance, you can choose to reboot it; but on any day that you choose to reboot the system, you can not process any data.

• the xi values(i.e x1,x2,….xn) represent the amount of data available(in terabytes) for processing on day i.

• the si values(i.e. s1,s2,…..sn) represent the amount of data(in terabytes) that your system can process on the ith day after a reboot.

Program to print:

• The maximum total amount of data that the system could process( in terabytes).

• The day numbers on which a reboot should occur. Note that there may well be more than one such day number and days start numbering at 1(not 0).

Thanks & regards,

Venkat.

Don't think that is true. Look at the following table:

x 10 1 2 8

s 8 4 2 1

Obviously it is best to reboot at day 2 *and* at day 3 to catch the 8 capacity at day 4.

I don't have a good algorithm either but you could consider to have a loop where you check whether a reboot is good or not. For that check you recursively need to check the next day and maybe some more. If I understand the statement "note the applicable subscripts can be different values" correctly, the number of indices of s is (much) less than that of x. If so, the number of recursive checks is limited by the number of indices of s.

Regards, Alex

Somehow this is not obvious to me. If you reboot on days 2 and 3, your output is 8 + 0 + 0 + 8 = 16. If you reboot on day 3 only, your output is 8 + 1 + 0 + 8 = 17.

You are right. That means two reboots on a row never is an option.

That reduces the number of possible solutions to 2^(n-1) - (n-1) what - unfortunately - means that checking all possibilities isn't an efficient method for large n.

Regards, Alex

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So for every day, the reboot decision is a trade-off. If you reboot, you have no output that day, but you restore maximum capacity for the next day. If you don't reboot, you can use that day's capacity up to that day's x, but the capacity will be lower the next day.

For n days, there are 2 to the nth power possible reboot schedules. The simplest algorithm would be just to calculate the output totals for all of them and pick the one with the highest total. This will not be practical unless n is fairly small, so you will probably have to design a cleverer algorithm.

Some elementary reasoning can reduce the solution space to be searched. For example, there is no point in rebooting on the last day, since that would just throw away capacity with no benefit. Similarly, there is no point in rebooting two days in a row.

I don't have a good algorithm for you, but I hope this helps you understand the problem.