Hamptonshire Express Case Solution

1. How many Hamptonshire Express newspapers should Sheen stock?

The underlying problem in the Hamptonshire Express case is that the demand for newspapers is erratic. To avoid product waste and financial losses, the sheen would have had to throw away all of her unsellable books. Even if Armentrout offered to purchase back the Express, Sheen was unsure whether she would accept. Armentrout's Express was completely out of stock. Because of this, buyers were forced to choose between buying the private line or not buying anything at all from the newsstand.

By varying the stocking quantity, the optimal stocking quantity was calculated to determine the maximum expected profit per day that could be made. The complete set of calculations is provided below.

Expected Demand = 500 = M. Standard deviation = s = 100. Marginal cost of production = 0.2 per copy and selling price per unit is 1.2. Any unsold copies are discarded. Hence the cost of understocking = Cu = 1.2 and Cost of Overstocking is Co = 0.2. The service level corresponding to optimal ordering is given by Cu / (Cu + Co) i.e., = 1.2/(1.2 + 0.2) = 85%. i.e. the optimal order quantity will be equal to the mean demand plus k times the standard deviation where k is the value of the curve as per the normal table. Looking up the normal table we get k = 1.07.

Hence quantity = 500 + 100*1.07 = 607 units.

Consequently, Sheen should stock approximately 607 units of newspapers. n. According to the following formula, this will result in the following amount of profit:

At this order quantity the Sales = 500 * 1.2 = $600. Cost = 607*0.2 + 10 + 30 = Cost of 607 units + Cost per day of leasing + Cost per day of rental = 161. Hence profit = Sales - Cost = 600 - 161 = $439.

To generate nearly $439 in profit, 607 units of newspapers must be purchased and stored.

2. How many hours should Sheen invest?

Information provided: -

1. Rental cost for space = $ 10/day

2. Printing cost for newspaper = $ 0.2 /copy

3. Newsstand rental cost = $ 30/day

4. The selling price of a newspaper =$ 1/copy

5. Daily demand for the newspaper is normally distributed with a mean of 500 and a standard deviation of 100. (This means that demand varies from 400 to 600)

6. Demand is the function of time that the editor invest.

7. The editor records the time each day as per the given diagram. ( i,e at t=0 , D=500 , t=1 ,D= 550 , t =4 ,D =600)

8. Demand Function is given by D= 500 + 2 SQ ROOT(h)

9. Opportunity cost for working at another plane = $ 10 /Hr

Let us calculate the profit of the editor at D= 400,500,600

Profit = Total revenue - Total cost

When D=400,

Profit = (1*400 - (10+30+ 0.2*400) = (Price of 1 copy * total demand)- ((Rental cost of space + rental cost of newsstand) + Printing cost/copy * demand /day)

= $ 80

Similarly When D=5

Profit = (1*500 - (10+30 +0.2 *500) = 360

When D = 600

Profit = 440

From given Relation :

D=500 + 50 SR(h) h = ((D-500)/50)2

From above relation

At D= 400; h= 4 but this is not valid from given graph

At D=500; h= 0, but working hrs can not be zero