FANDOM


19 June 2014
Item Base date Base price Price on adjustment date Comments
Christmas cracker 31 December 2008 688,800,000 2,147,483,644 Unchanged
Blue partyhat 340,100,000 2,147,483,647
Green partyhat 114,800,000 1,512,978,741
Purple partyhat 83,100,000 1,242,748,792
Red partyhat 129,400,000 1,706,390,538
White partyhat 183,300,000 2,147,481,758
Yellow partyhat 96,300,000 1,325,097,236
Pumpkin 5,300,000 157,608,053
Easter egg 4,300,000 62,666,643
Blue h'ween mask 12,800,000 121,249,470
Green h'ween mask 10,600,000 98,258,298
Red h'ween mask 17,300,000 170,663,885
Santa hat 14,800,000 125,533,632
Disk of returning 4,700,000 211,466,440
Half full wine jug 31,100,000 348,603,157
Fish mask 29 September 2012 4,555,462 669,574
Christmas tree hat 20 January 2013 2,295,576 12,559,556
Crown of Seasons 23 July 2013 8,307,542 4,555,009
Black Santa hat 20 January 2013 222,967,707 410,924,805
Cloak of Seasons 7,614,236 Added item

Calculations

From the old divisor obtained from the templates:

{div}_{\text{old}} = 15.2787


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{2,147,483,644}{688,800,000} + \frac{2,147,483,647}{340,100,000} + \frac{1,512,978,741}{114,800,000} + \dots + \frac{4,555,009}{8,307,542} + \frac{410,924,805}{222,967,707} \\

    & = 221.84047459 \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} \\

    & = 221.84047459 - 0 + 1 \\

    & = 222.84047459 \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}}} \\

    & = 15.2787 \times \frac{222.84047459}{221.84047459} \\

    & = 15.3476 \text{ (4 d.p.)}

  \end{align}

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