Question

A bakery sells rolls in units of a dozen. The demand X (in 1000 units) for rolls has a gamma distribution with parameters α = 3, θ = 0.5, where θ is in units of days per 1000 units of rolls. It costs $ 2 to make a unit that sells for $ 5 on the first day when the rolls are fresh. Any leftover units are sold on the second day for $ 1. How many units should be made to maximize the expected value of the profit?

Answer 1

Answer:

The value  is  $E(X) = \ 1.7067$

Step-by-step explanation:

From the question we are told that

   The  parameters  are  α = 3, θ = 0.5

    The cost of making a unit on the first day  is  c = $2

    The selling price of a  unit on the first day is  s = $5

    The selling price of a leftover unit on the second day is  v  = $ 1

Generally the profit of a unit on the first day is

        $p_1 = 5 - 2$

           $p_1 = \ 3$

The profit of a unit on the second day is

       $p_2 = 1 - 2$

=>     $p_2 = - \ 1$

Generally the probability of making profit greater than $ 1 is mathematically represented as

    $P(X > 1 ) = Gamma (X ,\alpha , \theta)$

=>   $P(X > 1 ) = Gamma (1 ,3 , 0.5)$

Now from the gamma distribution table  we have that

    $P(X > 1 ) = 0.67668$

Generally the probability of making profit less than or  equal to  $ 1 is mathematically represented as

       $P(X \le 1 ) = 1 - P(X > 1 )$

=>     $P(X \le 1 ) = 1 - 0.67668$

=>     $P(X \le 1 ) = 0.32332$    

So  the probability of making  $3  is    $P(X > 1 ) = 0.67668$

and  the probability of making  -$1  is   $P(X \le 1 ) = 0.32332$  

Generally the value of profit per day is mathematically represented as

      $E(X) = 3 * P(X > 1 ) + (-1 * P(X \le 1 ) )$

=>     $E(X) = 3 * 0.67668 + (-1 * 0.32332 )$

=>     $E(X) = \ 1.7067$