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20 January 2013
Item Base date Base price Price on adjustment date Comments
Christmas cracker 31 December 2008 688,800,000 2,147,476,523 Unchanged
Blue partyhat 340,100,000 2,147,483,647
Green partyhat 114,800,000 1,439,919,758
Purple partyhat 83,100,000 1,194,593,128
Red partyhat 129,400,000 1,588,813,583
White partyhat 183,300,000 2,115,586,339
Yellow partyhat 96,300,000 1,288,308,900
Pumpkin 5,300,000 153,543,323
Easter egg 4,300,000 73,851,614
Blue h'ween mask 12,800,000 130,080,270
Green h'ween mask 10,600,000 107,141,857
Red h'ween mask 17,300,000 168,034,752
Santa hat 14,800,000 130,038,816
Disk of returning 4,700,000 193,631,082
Half full wine jug 31,100,000 282,814,298
Fish mask 29 September 2012 4,555,462 1,778,504
Christmas tree hat 2,295,576 Added item

Calculations

From the old divisor obtained from the templates:

${div}_{\text{old}} = 15.0826$

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{1,439,919,758}{114,800,000} + \dots + \frac{1,778,504}{4,555,462} \\ & = 209.14534690 \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} \\ & = 209.14534690 - 0 + 1 \\ & = 210.14534690 \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}}} \\ & = 15.0826 \times \frac{210.14534690}{209.14534690} \\ & = 15.1547 \text{ (4 d.p.)} \end{align}
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