# Hamptonshire Express

Pages: 8 (2153 words) Published: March 30, 2013
Arjun R. Sabhaya
Production 529
Hamptonshire Express
October 16, 2012

PROBLEM #1
A. The simulated function given in the Excel spreadsheet “Hamptonshire Express: Problem_#1” allows the user to find the optimal quantity of newspapers to be stocked at the newly formed Hamptonshire Express Daily Newspaper. Anna Sheen estimated the daily demand of newspapers to be on a normal standard distribution; stating that daily demand will have a mean of 500 newspapers per day with a standard deviation of 100 newspapers per day. Using the function provided, the optimal stocking quantity, which maximizes expected profit, is determined to be approximately 584 newspapers. If 584 newspapers were to be ordered, Hamptonshire Express will net an expected profit of \$331.436 per day with an expected fill rate of 98%. Any inventory ordered above 584 will produce a loss of profit due to stocking inventory over expected demand causing an imbalance between the gains and losses due to the respective overage and underage costs. The table below outlines the optimal amount of daily expected profit. Profits rise until the 584 newspaper mark; any potential increase in quantity stocked will decrease daily expected profit for every newspaper ordered above 584.

Stocking Quantity| Daily Expected Profit|
575 newspapers| 331.323|
576 newspapers| 331.347|
577 newspapers| 331.369|
578 newspapers| 331.387|
579 newspapers| 331.403|
580 newspapers| 331.415|
581 newspapers| 331.425|
582 newspapers| 331.431|
583 newspapers| 331.435|
584 newspapers| 331.436|
585 newspapers| 331.435|
586 newspapers| 331.430|
587 newspapers| 331.423|
588 newspapers| 331.413|
589 newspapers| 331.400|
590 newspapers| 331.385|
Calculations: Cr=cu/cu+co where Cr= critical ratio. Cu=1-0.2=.8 Co=0.2 Therefore, Cr= .8/.8+.2=.8 which is equal to .84 (z value) on the standard normal distribution function table. To find the optimal stocking quantity that maximizes expected profit, we will use mean and standard deviation in formula shown: Q=mean +z*(SD): 500+.84*100=584. B. Using the Newsvendor Formula given, Q=μ+θ-1(Cu/Cu+Co )σ, where: Q=optimal quantity

μ= mean of expected demand
σ= standard distribution of expected demand
Θ-1= the inverse of the standard normal distribution function Cu= underage costs (Sale price – cost), or (\$1.00-\$0.20)
Co= overage costs (cost – salvage value), or (\$0.20 - \$0),

We can estimate the optimal stock quantity. The values allow us to input the z statistic, and the overage/underage costs in the given equation to derive the equation and answer shown below. The output of the Newsvendor equation, while different than the excel function (due to rounding error), is consistent with the optimal stocking quantity found by the Excel model.

Q= 500+ (.8601)-1*(.80/.80 + .20) * 100 = 593.01244
PROBLEM #2
A. The given simulation model in the Excel spreadsheet “Hamptonshire Express: Problem_#2” allows the user to find the optimal number of hours per day to be invested into creating the profile section to maximize expected profits for the Hamptonshire Express Daily Newspaper. Looking at the table below, Anna can spend 4 hours, where (H=4), Hours Spent (H)| Optimal Expected Profit|

2.00| \$367.91|
2.25| \$368.84|
2.50| \$369.58|
2.75| \$370.17|
3.00| \$370.61|
3.25| \$370.94|
3.50| \$371.16|
3.75| \$371.29|
4.00| \$371.33|
4.25| \$371.29|
4.50| \$371.18|
4.75| \$371.01|
5.00| \$370.77|

B. Anna Sheen’s choice of effort is at the point where the marginal cost of spending the extra time to develop the profile section = marginal benefit of spending the extra time to develop the profile section; or the point where profit is maximized. If she spends the extra time to develop the profile section past where marginal cost = marginal benefit, she won’t be able to create enough demand for her newspaper, but if she spends less time than...