Summing Appropriate Cells If Last Entry Of A Unique Id Is Encountered?

Jul 1, 2014

ID = 1: The B Number is the same for entries of ID =1. Thus where ID = 1 and Last is indicated, Cost = 10.

ID = 2: The B Number changes four times. In cell G20 (Last of all ID =2) Cost = 20+30+40+30 = 120. You are basically summing at each instance the B Number changes.

ID = 3: Cell G24 = 100 +30 (two instances of B Number changing)

ID = 4: B Number is always the same. Cost = 50

ID = 5: B number changes 3 times, Cost = 50+120+140 = 310

In essence, the idea is that if the ID matches for all rows of particular client, the total cost = individual cost. However, if the ID changes multiple times for a single client, the total cost becomes the sum of changed costs, but not the entire column, just summing at each instance it changes. And this summed cost must be entered whenever we see the last instance of a unique ID (This is indicated using Column F).

The problem that I'm running into is that currently I am doing this manually with my actual data set (has nearly 200,000 rows). What excel function or VBA code could I write to automate this entire process?

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# Of Unique Dates Per Unique List Entry

Oct 22, 2009

Each product is represented by a serial number (column A).
The can be sorted on column A from smallest to largest prior to calculating results if that helps.

The repair list contains 1 entry per spare part used, so the same serial number may occur several times.

Furthermore, a product may have been repaired on several instances - so the serial numbers can span several dates (column B).

The solution i am looking for should return the number of unique repair dates per serial number. That way i can see, how many times each product has been repaired. Results can be displayed in an individual column.

Sample list:
Serial........Repair date
207742052008-09-04
207755082008-12-17
207755212008-12-31
207755212009-01-22
207755212009-01-22
207755212009-01-22
207755212009-02-13
207755212009-07-24
207755362009-05-20................................

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Jan 3, 2008

I've found part of my answer from searching for a previous thread and altering to suit my needs.
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[URL].....

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