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Logistics Term Paper IV


A#2, Part 4 Purchasing

Your phone:   Ring, ring.
[Your name speaking, how can I help you?]
Hi [your name].  This is Tom Dollar.  I am down in purchasing and we have run into a snag.  Can you run down and see if you can give us a hand?
So what is this all about?
            Upon arrival you get a one-minute update.  Tim Apple is “Mr. fixture model L901”.  He functions as the product manager.
            Before Tim had gone away on a two-week vacation, Tim had worked out an optimal purchase quantity for a part for the L901 fixture.
            Purchasing was trying to process the order.  But a problem had surfaced.  Someone had spilled coffee on the worksheet that Tim had given as backup for the order.  Most of the information was still legible.  EXCEPT the quantity used each day could not be deciphered.
            Purchasing could process the order but would like to know the rate that Tim had used just as a “sanity check”. 
            This is almost as good as CSI, yes?  You get to do something akin to a forensic analysis.
            You can make out from Tim’s coffee-stained notes:
                        Cost to produce the L901 = $540.00
                        Q* = 70
                        Holding cost $1.50 per day
                        Set-up cost for the order $105.00
            Everyone agrees that these are the figures.  And that they are accurate (e.g., ordering cost is $105). 

            #4-1 (100 points) What is the daily use rate that Tim had used?
            #4-2 (50) Briefly explain

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Question: prove by induction 2^2 + 4^2 + 6^2 + ... + (2n)^2 = (2n)(2n+1)(2n+2)/6 ANSWER we will use induction on n base case : n=1 we have, 2^2 = 2*3*4/6 = 4 which is true inductive hypothesis let it be true for n = k i.e.,  2^2 + 4^2 + ... + (2k)^2 =   [(2k)(2k+1)(2k+2)]/6 inductive case let n = k+1 then we have 2^2 + 4^2 + .... + (2k)^2 + (2(k+1))^2 =   [(2k)(2k+1)(2k+2)]/6 + (2k+2)^2 =(2k+2)*[(2k)(2k+1)/6 + (2k+2)] =(2k+2)*[ (4k^2+2k)/6 + (12k + 12)/6 ] =(2k+2)*[ (4k^2+14k+12)/6 ] = =(2k+2)*[(2k)(2k+1)/6 + (2k+2)] =(2k+2)*[ (4k^2+2k)/6 + (12k + 12)/6 ] =(2k+2)*[ (4k^2+14k+12)/6 ] = (2k+2)*[ (4k^2 + 8k + 6k + 12)/6 ] = (2k+2)*[ (4k(k + 2) +6(k+2))/6 ] = (2k+2)*[ (4k+6)(k+2)/6 ] =  (2k+2)*[ 2 (2k+3)(k+2)/6  ] =   (2k+2)*[  (2k+3)*2*(k+2)/6  ] =   (2k+2)*[  (2k+3)(2k+4)/6  ] = [(2*(k+1))(2*(k+1)+1)(2*(k+1)+2)]/6 replacing k+1 by m, we get replacing k+1 by m, we get [(2*m)(2*m+1)(2*m+2)]/6 this completes our proof b...

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