A newsvendor boy wants to make a decision on how many papers to buy at the begin

Question

A newsvendor boy wants to make a decision on how many papers to buy at the beginning of the day in order to maximize his sale by the end of the day. Consider these notations and assumptions:

1- The selling price of newspaper is P

2- The purchasing price of newspaper is C

3- One unsold unit of the newspaper will be sold by the price V at the end of the day

4- The shortage cost of one unit of the newspaper is S

5- The probability density function of the daily demand is f(d) (d is the demand)

6- The cumulative distribution of the daily demand is F(d)

How many newspapers should the newsvendor boy buy at the beginning of the day in order to maximize his profit by the end of the day. Explain your solution and logic behind it.

Explanation / Answer

Newsvendor's profit is represented by the following equation

Profit = Selling price * least of the quantity stocked and Demand - Cost * quantity stocked

Profit = P * min (q, d) - q * C + max (0, q - d) * V , which takes the following forms

Profit = P * d - q * C + (q - d) * V , if q > d

Profit = P * q - q * C , if q <= d

In order to maximise the profit function ,

F(q) = (P-C) / (P - V),

In the above equation, P-C represent the opportuity cost of lost sales or shortage cost of one unit, P-C = S

Therefore optimal quantity, that the newsvendor should buy = F-1 ( S / (P-V)) or F-1 ( S / (S+C-V))

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