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Math Help - How long for a run of papers

  1. #1
    Junior Member froodles01's Avatar
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    How long for a run of papers

    Hi
    An applied maths problem to do with printing an amount of newspapers.


    I have a requirement to pubish 350,000 copies of a newspaper.
    2 presses with a capacity of 35,000 copies per hour can start at 22:00
    This takes 5 hours to complete, but is too long.
    So a second press with capacity of 20,000 copies per hour can start at 23:30

    What is the quickest time, assuming no hold ups, that this can be done?

    Really I need the formula to work this out, not the straight answer, so that if any variable changes I can just alter the variable (either print capacity or time that presses are available).

    I know that optimum end time is 02:15
    print capacity from 22:00 to 23:30 is 70,000
    print capacity from 23:30 onwards is 90,000

    105,000 copies are printed in first 1.5 hours
    time of remaining run = remaining copies to print / print capacity
    time of remaining run = (350,000 - 105,000) / 90,000
    = 2.72 hours (approx 2 3/4 hours)

    so end time is 23:30 + 2:45 = 02:15

    so what can I use as a formula, please?


    Thank you
    Last edited by froodles01; July 26th 2013 at 09:28 AM.
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  2. #2
    Junior Member froodles01's Avatar
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    Re: How long for a run of papers

    Ok done it.

    press capacity 1 = x = 30,000 copies per hour
    press capacity 2 = y = 35,000 copies per hour
    press capacity 3 = z = 20,000 copies per hour

    presses x and y start at 22.00
    press z starts at 23.30

    planned quantity = Q = 350,000

    OK
    2 presses work for t hours
    1 press works (t-1.5) hours

    (capacity 1) t + (capacity 2) t + (capacity 3) (t-1.5) = 350,000

    xt + yt + z(t-1.5) = Q
    xt + yt + zt - z(1.5) = Q
    xt + yt + zt = Q + z(1.5)

    t = (Q + z(1.5)) / xt + yt + z

    In real terms with real numbers then
    The two presses work for t hours. The third one for ( t - 1.5 ) hrs

    70000 t + 20000 ( t - 1.5 ) = 350000

    70000 t + 20000 t - 30000 = 350000

    90000 t = 380000

    ......... t = 380000/ 90000

    ......... t = 4.2 = 4hrs 13 min

    The machines started at 22 hrs, plus 4 hrs 13 min . It ends at 2 hrs 13 min

    It ends at 2 hrs 13 min

    Ta dah!
    Not all my own work, sorry.
    Thanks from HallsofIvy
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    Re: How long for a run of papers

    Quote Originally Posted by froodles01 View Post
    Hi
    An applied maths problem to do with printing an amount of newspapers.


    I have a requirement to pubish 350,000 copies of a newspaper.
    2 presses with a capacity of 35,000 copies per hour can start at 22:00
    This takes 5 hours to complete, but is too long.
    So a second press with capacity of 20,000 copies per hour can start at 23:30
    You mean a third press don't you?
    Suppose the run goes T hours after 22:00. Then the first two presses will each have done a total of 35000T copies and the two together will have done 2(35000T)= 70000T copies.
    Since the third press started 1 and half hours later, it will have run for x- 1.5 hours and will have completed 20000(x- 1.5)= 20000x- 30000 copies.
    The three presses, together will have done 70000T+ 20000T- 30000= 90000T- 30000 copies and we want that to be equal to 350000 copies:
    Solve the equation 90000T- 30000= 350000.

    90000T= 380000 so T= 380000/90000= 4.22222= 4 and 2/9 hours.
    (I get 4 hours, 13 minutes, and 20 seconds so the run would be finished at 2:13:20 AM.)


    What is the quickest time, assuming no hold ups, that this can be done?

    Really I need the formula to work this out, not the straight answer, so that if any variable changes I can just alter the variable (either print capacity or time that presses are available).
    Unfortunately, mathematics does not often consist of just plugging numbers into formulas. Hopefully you can understand my and ***'s reasoning to see how the calculations would be modified in other circumstances.

    I know that optimum end time is 02:15
    Then, unfortunately, you know wrong.

    print capacity from 22:00 to 23:30 is 70,000
    70000 copies per hour, don't forget that part.

    print capacity from 23:30 onwards is 90,000

    105,000 copies are printed in first 1.5 hours
    time of remaining run = remaining copies to print / print capacity
    time of remaining run = (350,000 - 105,000) / 90,000
    = 2.72 hours (approx 2 3/4 hours)
    Were you told to round to the nearest quarter hour? You didn't tell us any ofthat.

    so end time is 23:30 + 2:45 = 02:15

    so what can I use as a formula, please?


    Thank you
    The only 'formula' involved is precisely the one you used to find all of those numbers.
    Last edited by HallsofIvy; July 27th 2013 at 06:54 PM.
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