I’m an organizational economist focused on applying data science to provide decision makers robust understanding of economic behavior, performance, and potential policy impacts across an organization.

I grew up in southeastern Minnesota and have also lived in Colorado, Iowa, North Dakota, and now Ohio.

• B.S. in Kinesiology (North Dakota State University, Fargo, ND)
• M.S. in Cost Analytics (Air Force Institute of Technology, Dayton, OH)
• Ph.D. in Logistics (Air Force Institute of Technology, Dayton, OH)

Current:

• AFIT: Assistant Professor of Logistics and Supply Chain Management
• UC: Adjunct Assistant Professor of Business Analytics

My day-to-day work includes:

• performing empirical econometric research
• writing data analysis software
• teaching people about business analytics and R programming
• writing books
• creating digital projects.

• Programming with R since 2013
• Diverse knowledge:
  • Data mining
  • Longitudinal cluster analysis
  • Multivariate regression
  • Multi-level models (hierarchical regression)
  • Data envelopment analysis
  • Text and semantic analysis
  • Flexdashboard & Shiny apps

I have also written a book on Data Wrangling with R

1. Compute \(100(1 + \frac{0.05}{12})^{24}\)

100 * (1 + (0.05 / 12))^24
## [1] 110.4941

2. What is the remainder when 3333 is divided by 222?

3333 %% 222
## [1] 3

3. Investigate the behavior of \((1 + \frac{1}{n})^n\) for large, integer values in n.

n <- c(1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000)

(1 + (1 / n))^n
##  [1] 2.000000 2.593742 2.704814 2.716924 2.718146 2.718268 2.718280
##  [8] 2.718282 2.718282 2.718282

4. The Economic Order Quantity (EOQ) gives the optimal order quantity as \(Q = \sqrt{\frac{2DK}{h}}\) where D is the annual demand, K is the fixed cost per order, and h is the annual holding cost per item.

Create and set the variables \(D = 1000\), \(K = 5\), and \(h = 0.25\) and compute the associated value of Q.

D <- 1000
K <- 5
h <- 0.25

Q <- sqrt((2 * D * K) / h)
Q
## [1] 200

5. For an initial principal amount P and a nominal annual interest rate r that is compounded n times per year over a span of t years, the final value of a certificate of deposit is \(F = P(1 + \frac{r}{n})^{nt}\).

Create and set the variables \(P = 100\), \(r = 0.08\), \(n = 12\), and \(t = 3\) and compute the associated value of F.

P <- 100
r <- 0.08
n <- 12
t <- 3

F <- P * (1 + (r / n))^(n * t)
F
## [1] 127.0237