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We first CLAIM: KARTHIKCS VERSION

Theoretically we can establish a bijection between all concepts offered as incentives in super markets to

their concept of discount.

Let us define, Discount (D) = summation of i from 1 to n (over all items) of Ni(CPiSPi)

Discount percent (d) = D X 100 / summation of i from 1 to n (over all items) of Ni(CPi)

Profit (P) = summation of i from 1 to n (over all items) of Ni(SPiCPi) = summation of i from 1 to n (over

all items) of Ni(CPiCPi)D

Where Ni is the quantity of the ith item (in the super market),

CPi is the cost price (the price at which the item is bought by the super market) of the ith item,

CPi is the intermediate cost/selling price (the maximum retail price offered by the super market over which

equivalent discount would be given) of the ith item,

SPi is the final selling price (the final price at which the customer buys the item) of the ith item.

The following are values considered fixed: P, D, d, CP, CP.

We also have from a realistic point of view, the obvious inequality:

CP < SP < CP

Since we have CP and CP fixed, SP can be squeezed anywhere in betweenand this freedom to vary SP is

the basis of this concept to be developed!!...

We also define valued customers here. These are customers who have purchased at least Rs 5000/-, and

only after this would be given this status. To separate them from the normal customers we always use a

factor of 3/2. As far as implementation goes, we can make every purchase bill personalized with name,

for all normal customers. If the customer produces all required Rupees five thousand plus summed up bills,

under one name, then he is eligible to be a valued customer, with an unique id and password.

If a customer buys a certain set of items at final amount A.

(i) for valued customers

x + (xrt0/100)> dA/100

x + (xrt0/100) = (A/100)*(d + 0.67d)

(ii) for normal customers

x + (xrt0/100)< dA/100

x + (xrt0/100) = (A/100)*(ddd)

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Calculation of d :

We need d to be in range of [0,3]. At d=0, d=0. Now we want d + 0.67dto be less than 100, for all

range of d. So this implies at d=100, d=0. We also want d are maximum at d=25. This is chosen on

careful observation, thatafter d=25, the super market cant afford to offer a lot of extra discount, so d

has to start decreasing... Since we want a polynomial expression ofd in terms of d, and looking that wehave 3 conditions, we can go for a fourth degree equation with the constant part zero.

d = a x + b x2+ c x3

We figure out that a = 4/15, b = - 13/1875, c = 2/46875

Also now since most super markets work on weekly attraction, for normal customers we can keep t0 = 8

days (not 7 days as customers are indifferent between Saturdays and Sundays). For our valued customers

we use our golden factor 3/2. So for the valued customers t0 = 12 days.

Now we need a coordinating factor function between x and (xrt0)/100.

(i) x = xrt0/100 (as in the initial money at use would be doubled after t0)

This implies r =12.5%

(ii) 3/2 x = xrt0/100 (as in the initial money would be grown by 150%)

This implies r =12.5%

We can summarize conclusions and give these results:

(i) For normal customers, for amount A, given d for a super market, we have fixed r = 12.5%, t0 = 8 days.

We can also calculate d from d. We thus get x, and then calculate their growth.

(ii) For valued customers, for amount A, given d for a super market, we have fixed r = 12.5%, t0= 12 days.

We can also calculate d from d. We thus get x, and thencalculate their growth.

Example: 1. If the super market equivalent of d = 22%, and Dr Rajagoura purchases for Amount Rs

2400/-, calculate his propositions:

For normal customers, x = Rs 227, and it grows everyday at rate of 12.5%, for 8 days. At the end of 8 days

the equivalent discount amount would be Rs 454/-

For valued customers, x = Rs 230, and it grows everyday at rate of 12.5%, for 12 days. At the end of 12

days the equivalent discount amount would be Rs 575/-