FANDOM


29 September 2012
Item Base date Base price Price on adjustment date Comments
Christmas cracker 31 December 2008 688,800,000 2,146,908,554 Unchanged
Blue partyhat 340,100,000 2,147,483,647
Green partyhat 114,800,000 1,124,859,082
Purple partyhat 83,100,000 895,254,988
Red partyhat 129,400,000 1,232,179,013
White partyhat 183,300,000 1,652,939,303
Yellow partyhat 96,300,000 948,567,763
Pumpkin 5,300,000 165,935,075
Easter egg 4,300,000 56,837,419
Blue h'ween mask 12,800,000 112,172,382
Green h'ween mask 10,600,000 90,439,249
Red h'ween mask 17,300,000 149,074,067
Santa hat 14,800,000 116,884,816
Disk of returning 4,700,000 171,076,782
Half full wine jug 31,100,000 261,000,577
Fish mask 4,555,462 Added item

Calculations

From the old divisor obtained from the templates:

$ {div}_{\text{old}} = 15.0000 $


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{2,146,908,554}{688,800,000} + \frac{2,147,483,647}{340,100,000} + \frac{1,124,859,082}{114,800,000} + \dots + \frac{261,000,577}{31,100,000} \\ & = 181.52107486 \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} \\ & = 181.52107486 - 0 + 1 \\ & = 182.52107486 \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}}} \\ & = 15.0000 \times \frac{182.52107486}{181.52107486} \\ & = 15.0826 \text{ (4 d.p.)} \end{align} $