FANDOM


20 January 2014
Item Base date Base price Price on adjustment date Comments
Christmas cracker 31 December 2008 688,800,000 2,147,483,632 Unchanged
Blue partyhat 340,100,000 2,147,483,647
Green partyhat 114,800,000 1,490,829,327
Purple partyhat 83,100,000 1,279,369,134
Red partyhat 129,400,000 1,696,893,747
White partyhat 183,300,000 2,146,833,194
Yellow partyhat 96,300,000 1,335,202,311
Pumpkin 5,300,000 174,290,712
Easter egg 4,300,000 70,272,905
Blue h'ween mask 12,800,000 132,769,255
Green h'ween mask 10,600,000 110,890,723
Red h'ween mask 17,300,000 185,134,793
Santa hat 14,800,000 135,913,306
Disk of returning 4,700,000 223,384,911
Half full wine jug 31,100,000 333,709,743
Fish mask 29 September 2012 4,555,462 944,822
Christmas tree hat 20 January 2013 2,295,576 12,593,302
Crown of Seasons 23 July 2013 8,307,542 3,926,892
Black Santa hat 222,967,707 Added item

Calculations

From the old divisor obtained from the templates:

$ {div}_{\text{old}} = 15.2129 $


We need to calculate a new divisor:

$ {div}_{\text{new}} = {div}_{\text{old}} \times \frac{\sum \left ( \frac{p}{q} \right )_{\text{new}}}{\sum \left ( \frac{p}{q} \right )_{\text{old}}} $


To calculate the new divisor, we need to find:

$ \begin{align} \sum \left ( \frac{p}{q} \right )_{\text{old}} & = \text{sum of ratios prior to change} \\ & = \frac{2,147,483,557}{688,800,000} + \frac{2,147,483,647}{340,100,000} + \frac{1,490,829,327}{114,800,000} + \dots + \frac{12,593,302}{2,295,576} + \frac{3,926,892}{8,307,542} \\ & = 230.87579845 \text{ (up to 8 d.p.)} \end{align} $


And also:

$ \begin{align} \sum \left ( \frac{p}{q} \right )_{\text{new}} & = \text{sum of ratios prior to change} - \text{sum of removed ratios} + \text{sum of added ratios} \\ & = \sum \left ( \frac{p}{q} \right )_{\text{old}} - \text{sum of removed ratios} + \text{number of added items} \\ & = 230.87579845 - 0 + 1 \\ & = 231.87579845 \text{ (up to 8 d.p.)} \end{align} $


Thus, the new divisor is:

$ \begin{align} {div}_{\text{new}} & = {div}_{\text{old}} \times \frac{\sum \left ( \frac{p}{q} \right )_{\text{new}}}{\sum \left ( \frac{p}{q} \right )_{\text{old}}} \\ & = 15.2129 \times \frac{231.87579845}{230.87579845} \\ & = 15.2787 \text{ (4 d.p.)} \end{align} $
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