1 1 Wrap Up John H. Vande Vate Spring, 2012

2 2 Review of Topics Variability Basics –The big idea Forecasting –The 5 Laws of Forecasting –Actual/Forecast Distribution Inventory basics – –Continuous Review, Periodic Review –Safety Stock and Safety Lead time News Vendor Model Sport Obermeyer Case Revenue Management Make-to-Stock/Make-to-Order

3 3 Projects Halstead/Vertex –The world is messy –Different channels –Developing a supply chain strategy Ziqitza –Competing objectives in location –Comparing apples and apples –The process, not the model or the answer Tiffany’s –Trading off transport and risk –Translation to practical policies GPRE –Freight Optimization – are the savings real?

4 4 Last Year’s Exam Our company currently provides inventory to 2 major customers via a dedicated VMI hub for each one. We are exploring the bene¯ts of combining the two inventories into a single shared inventory for the two customers, i.e., maintaining a single VMI inventory for the two customers. We replenish each VMI inventory on a weekly basis with an order lead time of 3 weeks. The operations run 52 weeks per year. We are currently maintain a safety lead time of 7 weeks for each supplier in order to ensure the 95% cycle service level we have committed to Our cumulative distribution of Actual to Forecast ratios is: A/F ratio Cumulative % % % % % % % % % % % % % We are responsible for transportation to the customer and delivering from the single combined VMI hub to the customers averages $0.02 per unit higher transportation costs than delivering from the two dedicated VMI hubs. Each unit of the product costs $100 and we work with a 25% annual inventory carrying cost rate.

5 5 Savings per unit? On the basis of inventory savings and added transportation costs only, what's the per unit difference in cost between operating a shared VMI hub versus operating two dedicated VMI hubs? Today, we maintain a safety lead time of 7 weeks and 4 weeks of planned inventory to cover the one week period between orders (T) and the three week average lead time (L). So our order up to level for both customers is 11 weeks. With the combined VMI hub, the A/F ratios suggest we need an order up to level of 2.5 times the forecast over the planned 4 weeks (T +L) to ensure a cycle service level of 95%. So, with a combined VMI hub, the order up to level would be 10 weeks of demand. Let's assume the total combined average weekly demand of our two customers is D. The combined VMI hub saves us one week of inventory, which has an annual carrying cost of 25%*$100*D = $25D. On the other hand, the combined VMI hub requires we spend $0.02 per unit in extra transportation cost, which on an annual basis amounts to $0.02*52*D = $1.04D since there are 52 weeks in the year. So, we clearly come out ahead financially (ignoring any fixed costs and any savings from only having to staff a single facility, etc.) The marginal savings appear to be $0.46 = $23.96/52 per unit.

6 6 Challenges? The main challenge will be the customers' concern about who gets the last items in safety inventory in those rare (about twice per year) occasions when we do run out of inventory in the combined VMI hub. Those will be exactly the times when product is scarce or demand is high (occasions that represent great opportunities to win market share and improve margin because demand is higher than supply). Beyond the obvious operational concerns of having the VMI hub farther away, customers may also be concerned about the amount of information the competitor can glean about their sales by seeing the inventories in the combined VMI hub and knowing his own sales.

7 7 Obermeyer Product Line Wholesale Price Min. Demand Expected Demand Max. Demand Stephanie $ ,000 2,000 Seduced $ ,000 4,000 5,000 Anita $ ,000 3,000 5,000 Here, Min. Demand, Expected Demand and Max. Demand give us our estimated distribution of actual demand in the very simple way: the probability density rises linearly from 0 at m = Min.Demand to a = Expected Demand. It then falls linearly from that point to M = Max. Demand. 0If x < m f(x) =p(x - m) If m < x < a p(M - x)If a < x < M 0 If x > M where p = 1/(a - m) 2 We assume that demand is symmetrically distributed about the expected level, i.e., a - m = M - a The Cumulative Density Function is 0 If x < m p (x - m) 2 /2If m < x < a F(x) =0.5 + p *M(x - a) - p(x 2 - a 2 )/2 If a < x < M 1 If x > M

8 8 Newsvendor Assume that when Obermeyer sells these products in the primary markets, its profit margin is 25% of the wholesale price. So Obermeyer's cost for these items is 75% of the wholesale price. When Obermeyer cannot sell these items in the primary markets, it must salvage them in the secondary markets for only 65% of the wholesale price. Assume too, for this part of the question, that we are considering the problem AFTER the Las Vegas show. Obermeyer did not order any of these product lines before the show and the distributions of demand are based on information Obermeyer received at the Las Vegas show. Finally, assume capacity is unlimited. How many of each of these three product lines should Obermeyer make? The first thing to recognize is that this is AFTER the Las Vegas show. Obermeyer has just about as much information as it's going to get. So, this is a question of maximizing expected profits: a straightforward newsvendor application. The second thing is to compute P, the probability demand is less than the quantity we should purchase of a given product. That should be pretty standard now: P = 25/35 = The last thing is to figure out how much we need to purchase of each item to achieve the value P = Since P > 0.5, the quantity we should purchase is clearly greater than a in each case and so we just need to solve: p *M(x - a) - p(x 2 - a 2 )/2 = 5/7 or p*M(x - a) - p(x 2 - a 2 )/2 = 3/14 Simplifying, this reduces to p*Mx - px 2 = 3/14+ p*M * a - pa 2 /2 i.e., which gives us approximately 1,244 for Stephanie, 4,244 for Seduced and 3,488 for Anita.

9 9 With Capacity Now suppose Obermeyer only has the capacity to make only 5,000 units in total for these three product lines. How many of each of each product line should Obermeyer make? Since this is AFTER the Las Vegas show, we want to solve: max Expected Profit s.t. Total Volume Purchased ≤ 5,000 One natural way to approach this is to use the approach we did for the problem before the Las Vegas show. Here, we aren’t interested in Marginal Expected Return on Invested Capital. Instead, let’s consider: Maximize Expected Profit – λ Total Volume Purchased Total Volume Purchased = ΣQ i Setting the partial derivatives to 0, we get Marginal Expected Profit – λ = 0 So as before, we are sorting the items by Marginal Expected Profit and taking the top ones. Adjust λ till we find the top 5,000 items

10 With Capacity For λ fixed, how to solve Maximize  Expected Profit(Q i ) - λ  Q i s.t. Q i  0 Remember it is separable (separate decision for each item) Exactly the same thinking as the News Vendor Last item: –Reward: Profit*Probability Demand exceeds Q –Risk: (Cost – Salvage)* Probability Demand falls below Q –λ  λ is like a tax or interest on the investment that adds λ to the cost. We pay it whether the item sells or not. –If it sells, get the original profit – λ –If it doesn’t sell, get (Salvage – Cost – λ)

11 Solving for Q i Last item: –Reward: (Revenue – Cost – λ)*Prob. Demand exceeds Q (Revenue – Cost – λ)*(1-P) –Risk: (Cost + λ – Salvage) * Prob. Demand falls below Q (Cost + λ – Salvage) * P –As though Cost increased by λ, the “Tax” or “Interest” we pay for capacity –P = Profit/(Revenue - Salvage) – λ/(Revenue - Salvage) = 5/7 – λ/0.35

12 With Capacity Scale λ to represent λ/0.35 We want to solve: p*M(x - a) - p (x 2 - a 2 )/2 = 5/7 – λ which has the solution: We can further simplify this by replacing with a simpler parameter, call it λ’. Now, substituting the values of a and M for the three products and adding up the results to get the total quantity purchased, we see that the problem reduces to finding the value of λ’ that satisfies: 12,000 – 4000 λ’ = 5000 Now, substituting this value for λ’, we see that we should purchase 250 units of Stephanie, 3,250 units of Seduced and 1500 units of Anita.

13 Oops But that adjusted critical ratio is < 0.5 –Adjusted Critical Ratio = 5/7 – λ/0.35 –How do we know that? So, we should be solving p (x - m) 2 /2 = 5/7 – λ, i.e., Now find λ, so that we get 5,000 in total The answer – exactly the same.

14 The Exam 4 – 5 Questions To keep things fair n  3 questions about projects You choose n-2 questions to answer You can’t choose the question about your project. Got it?