Optimization of Order Volume
Inaccuracy of Wilson’s model.
If your goal is to optimize order volume, use of Wilson’s model is not the proper tool.
Specialists in inventory management are familiar with the graph below (Figure one) that shows how Wilson’s model works. It is used to calculate dependence of unit cost on the value of the delivery (to calculate order volume). Use of it as an interpretive tool is provided I Figure 1.:
Figure 1 – Dependence of unit cost on the value of delivery
line graph = unit costs for the storage order
hyperbola graph = the unit costs of placing the order
optimum graph = overall unit costs
q * = Point Wilson (Wilson), where storage costs are caused by the delay in the placement of orders.
Obviously, for some portion of the order, there are optimal conditions for purchase. The graph clearly shows these by referencing an assumed economic order quantity (EOQ). However, the purpose of any business is not to minimize costs, but to maximize profit. Use of this model presumes that large profits are optimal; however, large profits do not correspond to low cost.
For proof we can take an example from the Wilson’s book. The table below (Table 1) shows the dependence of average inventory levels and costs based on a number of factors:
— The number of orders per year;
— Order size;
— The cost of placing orders, in USD;
— Costs for the maintenance of the average stock, in USD, and
— Total cost (also in USD).
The minimum cost is seen on line two. However, optimal results from an economic point of view would be achieved by the regime on line three. Therefore, it is necessary to obtain information about cost for the second and third rows. For simplicity, the rows will be referred to as options, so we are looking at options two and three.
Unfortunately, information about trade margin is missing here, so it is not possible to determine exact actual profit. We will fill this gap in our hypothesis as follows. First, we will presume that the profit margin on sales was $150 US. Now we can calculate the net annual profit for each purchase option:
We will use the following formula:
δC = (∆P–ZT)∙Q∙N ,
The pieces of the formula are defined as follows:
δC – annual net profit on the money used in the purchase of the analyzed goods, USD;ΔP – trading margin per unit of goods, USD;
ZT – the overall cost per unit of goods, USD;
Q – the quantity of goods in one order, units;
N – the amount procured during the year.
The annual net profit is therefore calculated as follows:
δC, v2 = (150-100) ∙ 500 ∙ 2 = 50000 (USD), and
δC, v3 = (150-108) ∙ 333 ∙ 3 = 41958 (USD).
It would seem that the conclusion is obvious, but further analysis is in order. In option three, there was less capital expenditure. Instead of 500 units being purchased, only 333 were. That is a difference of 167 units. Thus, in option three, the business owner has more “free money” available to launch into circulation in the future – on this or other purchases – that will result in ongoing profits.
Here is how it works:
δCf, v3 = (QZ, v2 – QZ, v3) ∙ P ∙ δCN,
δCf, v3 – annual net profit of the “free money”
QZ, v2, QZ, v3= size of the order for options two and three, in units
P – the purchase price in USD;
δCN =average annual yield of the enterprise (also in USD)
Unfortunately we do not have information about the profitability of the enterprise in the above model, and will have to again make an assumption. Because the company operates using the Wilson model, we will take into account the appropriate statutory δCN, which is that freed money can be considered under option two:
δCf, v3 = (500-333) ∙ (150-100) ∙ 2 = 16700 (USD).
Gross profit for option 3 consists of two parts:
δCt, v3 = δC, v3 + δCf, v3.
Now, we get comparable value profits for (δCt). In option two the profit is $50,000; in option three it is $58,000. Option three allows for a profit that is 16% higher than that in option two.
When calculating option two via the formula that Wilson recommended (two purchases a year of 500 units sold) it appears that the profit is indeed larger. However, our more accurate calculations show that this is not the case. No business should miss the prospect of profit growth in the 15-to 20 percent range.
There is little effort required to come to the best solution. It is merely necessary to learn how to count EOQ. Therefore, as shown, Wilson’s model does not provide the desired results.
Vladimir Chemeris (Researcher ID: J-1265-2014)
Authors affiliation: Ekaterinburg, Russia. Corresponding author: Vladimir Chemeris, firstname.lastname@example.org
Annotation: Minimum costs do not match the most highly effective size of order. Minimum costs and point Wilson do not correspond to maximum income.
Wholesale trade; purchase of goods; stocks; turnover; Wilson’s model.