This post outlines the generalized solution to Multi-Echelon Inventory Decisions by finding the lower, upper and average bound for an optimal solution. It then outlines a genetic algorithm approach to finding the optimal solution. The generalized solution procedure is applied to the Multi-Echelon Inventory Decisions at Jefferson Plumbing Supplies To Store or Not to Store? Case Study

# Introduction:

Alex, a NASCAR enthusiast, is the Inventory Manager of Jefferson Plumbing Supplies, and he faces an interesting quandary. He has been notified by one of his suppliers that there is an upcoming increase to their minimum order quantity. The following case study summary presents a description, a system model, and a summary and conclusion for the problem Alex faces.

# Problem:

Jefferson Plumbing has know demand of 100 units annually, from loyal customers, of a specialty faucet that is ordered from a long time supplier. Alex received a letter from the supplier informing him that they would be increasing their minimum order quantity. The new minimum order quantity is triple what would normally be ordered. Due to their storefront’s city location, inventory holding costs for the faucet are expensive. Jefferson Plumbing Supplies has a warehouse with a cheaper annual holding cost, and Alex would like to determine the optimum costs previous to, and after the policy change.

## Options:

1. convince the company to maintain relations to drop the minimum order amount

2. not stock the item -> lost revenue and customers

3. order and store in the store

4. order and store in the storage facility (multi-echelon system)

5. find alternative storage

Given that option 1, 2, and 5 are not feasible there is option 3 (the base case) and option 4.

## Variables

Holding cost h [$ unit/time]

Stockout Penalty p [$/unit/time ]

Fixed Cost k [$/order ]

Purchase cost c [$/ unit ]

Demand λ [units/time ]

Order quantity Q [units ]

Optimal order quantity Q* [units]

Re-order interval u [ time]

Average Annual Cost C(Q) [$/year]

Optimal order quantity C* =C(Q*) [units]\

## Assumptions:

Assumption given in case study:\

- Continuous deterministic demand of λ = 100 units/year

2. The cost of not having inventory is very high (lost of profit and lost of customers and future profit)

3. Minimum order quantity \(Q_{min} = 130\) units, previous order quantity was about 130/3 units 4. No capacity limits 5. Must always have positive inventory (no backorders) based on assumption 2

Assumptions taken to relax problem:\

- No lead time

2. Purchase cost c is ignored in optimization as it remains constant

3. The holding cost includes the cost of held capital (no interest rate)

## Givens:

Purchase Fixed ording cost \(k_p = \$100/order\)

Warehouse ordering cost \(k_w = \$50/order\)

Store holding cost \(h_s = \$10/unit/year\)

Warehouse hoding cost \(h_w = \$1/unit/year\)

# System model:

## Base Case: Deterministic Economic Order Quantity Model

In the base case, the system can be modeled as a simple single order system where the quantity is stored at the retail store.

Manufacturer -> Retailer

Using a basic economic order quantity (EOQ) model and the assumptions previously made, the yearly cost of the base case would be:

\[C(Q) = \frac{k_p \lambda}{Q} + \frac{h_s Q}{2}\]Given the first order condition:

\[\frac{\partial C} {\partial Q} = - \frac{k_p \lambda}{Q^2} + \frac{h_s}{2} = 0\]Given this is a local minimum, the optimal quantity would be:

\[Q^* = \sqrt{\frac{2 k_p \lambda}{h_s}} = 44.72135955 = 45 units\]With the optimal annual cost would be:

\[C^* = C(Q^*) = \sqrt{2 \lambda k_p h_s} \cong \$447.21\]Since the minimum order quantity \(Q_{min} \ge 130\) as stated in the problem and the actual optimal annual cost would be:

\[C(Q_{min}) = \frac{k_p \lambda}{Q_{min}} + \frac{h_s Q_{min}}{2} \cong \$726\]The annual cost is convex optimum, therefore the smallest minimum order quantity would be optimal in this model.

## Two stage deterministic multi-echelon serial system

An alternative system model would be two stage: after an order is placed from the manufacturer it would be stored at a lower cost in the warehouse and be shipped to the retailer. In shematic form:

Manufacturer -> Warehouse -> Retailer

Each stage functions like an EOQ system and determines the optimal order quantity and period. Solution will have a zero inventory ordering (ordering will occur when inventory is zero) assuming no lead time.

Since the minimum order quantity is greater than yearly demand, therefore there will only be a maximum of 1 order within a year from the manufacturer. The initial assumption will be made that it is cheaper to store inventory at the warehouse and order parts when required. This assumption will be later verified by comparing the total cost to the base case.

We are going to introduce the reorder interval \(\textbf{u} = Q /
\lambda\), and can be thought of as a vector for each stage (j) of the
ordering process. In our case, \(u_w\) is the ordering period made **at**
the warehouse **from the store** and \(u_p\) is the ordering period **at
the production factory** from the warehouse.

The averaged annual cost function (to be minimized) of the most ideal function would be the following:

\[C (\textbf{u}) = \sum_{j} \left(\frac{k_j}{u_j}+ \frac{h_j \lambda u_j}{2}\right) = \frac{k_w}{u_w}+ \frac{(h_s -h_w)\lambda u_w}{2} + \frac{k_p}{u_p}+ \frac{h_w \lambda u_p}{2}\]Such that:

\[u_j = \theta_j u_{j+1},\] \[u_j \ge 0,\] \[\theta_j \in \mathbb{Z}^+ = \{1, 2, 3, ...\}\]There is also the added constraint due to the minimum quantity order for $u_p$

\[u_p \ge \frac{Q_{min}}{\lambda} = \frac{13}{10}\]Solving this optimization problem for the optimum reorder interval \(\textbf{u*}\) leads to a non-convex mixed integer non-linear programming. Since the constrained solution is non-convex, it does not have a guaranteed optimal solution except in limit. For this reason, a generalized approach to finding the lower, upper and mean average annual cost must be used.

## Generalized Solution:

The generalized solution to an non-convex, mixed interger non-linear programming optimization problem is to get an upper bound by solving a relaxed problem (non-interger constraint) which represents the worst cast scenario.

The relaxed optimization problem is the following:

\[C (\textbf{u}) = \sum_{j} \left(\frac{k_j}{u_j}+ \frac{h_j \lambda u_j}{2}\right) = \frac{k_w}{u_w}+ \frac{h_s \lambda u_w}{2} + \frac{k_p}{u_p}+ \frac{h_w \lambda u_p}{2}\]Such that:

\(u_j \ge u_{j+1}\) \(u_j \ge 0\) \(u_p \ge \frac{Q_{min}}{\lambda} = \frac{13}{10}\)

This can be solved using multi-variate calculus or [numerically]

Using calculus, the partial derivative of the cost function with respect to \(u_p\) can be taken as:

\[\frac{\partial}{\partial u_p}(\frac{k_w}{u_w}+ \frac{h_s \lambda u_w}{2} + \frac{k_p}{u_p}+\frac{h_w \lambda u_p}{2})= \frac{\lambda h_w}{2}- \frac{k_p}{u_p^2}\]Now solving for $u_p$

\[u_p^* = ±\frac{\sqrt{\frac{2k_p}{h_w}} }{\sqrt{\lambda}} = \sqrt{2} \ge \frac{13}{10}\]Similarly, the partial derivative of the cost function with respect to \(u_w\) can be taken as:

\[\frac{\partial}{\partial u_w}(\frac{k_w}{u_w}+ \frac{h_s \lambda u_w}{2} + \frac{k_p}{u_p}+ \frac{h_w \lambda u_p}{2})= \frac{h_s \lambda}{2}-\frac{k_w}{u_w^2}\]Now solving for $u_w$

\[u_w^* = ± \frac{\sqrt{2 k_w}}{\sqrt{h_s \lambda}} = \frac{\sqrt{10}}{10}\]The upper bound of the overall cost function can now be calculated as:

\[C_* = C(\textbf{u*}) \ge 100(\sqrt{2} + \sqrt{10}) \cong 457.649122254\]In this case, about every 1.414 years the warehouse would order:

\[Q_p^* = u_p \lambda = 100 \sqrt{2} \cong 141.421 = 142 units\]The retail store would order:

\[Q_w^* = u_w \lambda = 100 \sqrt{10} \cong 31.522 = 32 units\]This would result in the overall cost to be:

\[C(\textbf{Q}) = \frac{129979}{284} \cong \$457.67\]Using the same approach using the un-relaxed cost function (and assuming incorrectly an optimal solution), the optimal order quantities becomes:

\[Q_p^* = u_p \lambda = 100 \sqrt{2} \cong 141.421 = 142 units\]The retail store would order:

\[Q_w^* = u_w \lambda = 100/3 \cong 33.33 = 34 units\]This results in the following approximate ideal average annual cost \(C(\textbf{Q})_{ideal} \cong\) $441.81 This lower bound would be better estimated by evaluating the cost of storing the remaining inventory in the warehouse. For example, after the first order cycle of 142 parts from the supplier and first three shipments to the store, there will be 14 parts remaining. At a cost of $1 per year per item, this also adds a $14 cost over the three cycles. In the next cycle (order the minimum quantity of 130 units), there will be 16 left over. This continues by units of 2 (over 30 cycles) until 30 units and then finally there will be no remain parts and the cycle starts again. This means over 30 order cycles from the store to the warehouse, we would face a mean unit period cost of 198 units * $1 /30 or $6.6 per order cycle. This means the actual lower bound of the solution is around 448.41. The difference between the upper and ideal lower bound of the solution is around $16.19 or the price of a beer and a Speedway Big Hog Chedder Dog at a Nascar event and the approximate average annual cost would sit closer to $448.41. Since the optimal order quantity from the warehouse is \(Q_w* = 34\) units, \(\theta_{theoretical}^* = Q_w^* / Q_p^* \cong 4.176470588\) and \(\theta_{actual}^* = 4\). Since the difference between \(\theta_{theoretical}^*\) and \(\theta_{actual}^*\) is small, the actual solution would be much closer to the lower bound.

# Constrained Optimization Solution

To improve upon the generalized approach and attempt to find the true optimal solution, the optimization problem can be solved using a non-linear programming tool such as Goal Seek in Excel, as demonstrated in this spreadsheet. Alternatively, the lower bound of the optimal solution can be estimated numerically

or using calculus. Then $\theta_j$ can be calculated and rounded to find the order quantities (as shown in the generalized solution).

The results from both methods are in agreement and the result in $Q_w^* = 34$, $Q_p^*= 136$ and an overall average annual cost of about $441.59 which is $5.62 dollars cheaper than before the policy change.

# Summary and Conclusion

## Optimum Cost before policy change

The optimum annual cost before the policy change in the problem is $447.21/year

## Optimum Cost Ordering and Storing to Retail Store Alone

Given the assumptions in this problem, using the optimum solution of a basic EOQ model results in an optimum ordering 130 units (when stock reaches zero), results in an actual optimal annual cost of around $726

This means that this would cost the store about $277.79 more if they choose to order, ship and store to the retail store.

## Optimum Cost Using a Warehouse

Given the assumptions in this problem, the approximate optimum solution of the two stage deterministic multi-echelon serial system provides an annual cost of around $447.21. This means that the policy change would cost the store at most about $5.61 less if they order 144 parts to the warehouse and 136 parts from the warehouse to the retail store. Alex Halek wwould be able to rest at ease and possibly use those savings towards the purchase of a delicious Speedway Big Hog Cheddar Dog at next year’s Spring Cup.