Holding Cost Reduction in the EOQ Model
Dr. Peter J. Billington, Colorado State University – Pueblo, Pueblo, CO
ABSTRACT
The introduction of a capital expense to reduce the setup cost has expanded the classic EOQ model into numerous new research insights. The original intent of that research was to reduce the order quantity to better fit the JIT lean manufacturing model of setup reduction. In this paper, the EOQ model is further studied with a reduction in the per unit holding cost to determine if total cost can be reduced, even though the order quantity is increased. Results show that the total cost can be reduced under specific situations. This new model is combined with previous research on setup cost reduction to show that further total cost reduction is possible.
INTRODUCTION
The classic economic order quantity model (EOQ) is often studied as a way to analyze the trade-off between setup and holding cost to minimize total annual cost of holding inventory and setup (or ordering). With the publication of two seminal articles (Porteus, 1985 and Billington, 1987), the EOQ model was expanded to include a capital expense to reduce setup costs. This new model was in line with and helped explain the practice of setup reduction, necessary for JIT systems to work effectively by allowing a more economical, smaller order quantity. The resulting series of research on setup reduction includes articles by Spence and Porteus (1987), Paknejad and Nasri (1988), Nasri et al (1990), Kim et al (1992), Hong et al (1992), Hong et al (1993), Hong et al (1996), among others.
The remaining component of the EOQ that has not been studied is the holding cost, which includes the cost of funds invested in inventory, the storage facility, handling inventory, insurance, taxes, obsolescence, spoilage, deterioration and theft (Heitger, 1992). The ability to reduce the holding cost per unit is limited due to the fixed nature of many of these holding cost components. Internal rates of return are set, and insurance and taxes are out of the control of decision makers. However, the cost of handling inventory could be reduced through automation, and the cost of obsolescence and spoilage can be reduced through capital expenditure.
Consider the following examples. Without refrigeration and freezers, the spoilage rate of produce and frozen goods is high and very fast. If a supermarket did not have freezers and refrigeration, produce would spoil and frozen goods would thaw. The result would be a very limited amount of produce, and consumers would be forced to buy on a daily basis. In fact, in the early part of the 1900’s in large cities, it was not uncommon for the “vegetable man” and the “ice man” to daily cruise the residential streets selling their produce and ice. Shoppers would buy just enough to be consumed before spoilage of the produce. The advent of electric refrigerators for the home allowed consumers to purchase larger quantities, often a week’s worth. Supermarkets install refrigeration units and freezers to allow a larger quantity of items to be stocked. The result is that the capital expenditure in freezers and refrigerators allowed the spoilage cost to drop dramatically, and the order quantity to increase, perhaps with a reduction in total cost.
In the following model, Q is the quantity to order, S is the cost per order (or setup), D is the annual demand, and H is the per-unit cost of holding one unit of inventory for one year. The classic EOQ formula,
(1)
shows that if H were decreased, then the Q value would increase, just as shown to have occurred in the previous examples. The total cost would decrease since the minimum total cost
(2)
However, the total cost in (2) does not include any capital cost to reduce H. The full holding cost reduction model will introduce this capital expense.
Although some detractors will suggest that the EOQ has limited real applicability, this should not stop further analysis of the model to understand the trade-offs inherent in many types of inventory situations. The analysis in this paper should provide insights and stimulate further discussions of holding cost reduction.
HOLDING COST REDUCTION MODEL
The EOQ model as shown in (1) and (2) will be expanded to consider an annual capital cost to buy a certain amount of reduction in H. The classic EOQ model is formulated with the assumptions that demand is stationary and deterministic over an infinite horizon, costs are known, and backlogs are not allowed. The objective is to minimize total annual cost of holding and setup. Total cost is expanded here to include an annual capital cost to reduce the per-unit holding cost, H. K will designate the annual cost to reduce the original H value to a lower value. The full formula now is
(3)
where H = f(K). If the first derivative of TC is taken with respect to Q, set to zero and solved, the result is the classic EOQ equation (1). If the first derivative of (3) is taken with respect to K, set to zero and solved, the result is the optimal condition
(4)
Assume that H is a declining exponential function of K as shown in Figure 1. If no capital is spent, then H = U. H will decrease exponentially with increasing K toward a technological lower limit L, as shown in Figure 1 and represented by the formula
(5)
Figure 1
Exponential Holding Cost Function
The value of b equals the natural log of 2 divided by the amount of K that is required to reduce the holding cost from U to halfway between U and L.
The first derivative of (5) with respect to K is
(6)
which satisfies the first order condition for optimality. Substitute (6) into (4) and use (1) to find the optimal H*. After solving, the resulting optimal conditions are:
let (7)
then
(8)
and
(9)
With H restricted to be less than U, the following limiting condition is found.
(10)
If then we would not invest and K*=0. Only if will we invest K* to reduce total cost.
Example 1
Consider the following example modified from Billington (1987).
D=3600S=1000U=20L=15
First, calculate from (10)
Figure 2 below shows the total cost lines for three different b values. When b = 0.0001, which is b < B, the total cost curves starts at $12,000 for no investment and continues to rise. The result is not to invest since the lowest total cost occurs at K = 0. For b=B, the line is straight across and again the result would not to invest. When b = 0.004, which is b > B, the line starts at $12,000 and forms a parabolic shape that reaches a minimum at K*.
Figure 2
Total Cost as a Function of Capital Expense to Reduce H
Let’s assume that management found that a cost of about $173 would reduce H to 17.5, half the way to the lower limit of 15. Using the result that b is equal to the natural log of 2 divided by this K value, then b = ln 2/173 = 0.004 which is greater than B of 0.000666. Continuing:
(7)
(8)
(9)
(1)
(3)
Table 1 below illustrates the results. Under the original problem formulation with no reduction in holding cost, column 1a, the optimal Q is 600 with a total cost of $12,000. In column 1b, If b<B, then the decision should be to not invest. Let’s say we did invest $477.97. At the b = 0.0001 value, this would reduce the holding cost to 19.77, but the total cost would increase to $12,408. The proper conclusion would be to not invest as indicated by b<B. With b = 0.004, then b>B, the calculations done above are placed in column 1c. Q increases to 676 but the total cost, with the optimal capital expense of $477.97, is reduced to $11,123.19.
Table 1 - Comparison of Decisions, Example 1
1aOriginal Problem
EOQ / 1b
b<B
but invested / 1c
b>B
b / 0.0001 / 0.004
H* / 20 / 19.77 / 15.739
K* / 0 / $477.97 / $477.97
Q* / 600 / 603.5 / 676.36
TC* / $12,000 / $12,408 / $11,123.19
Conclusion / Do not invest / Invest in H reduction
SETUP AND HOLDING COST COMBINED NAIVE MODEL
It is likely that managers would make decisions to reduce both the setup cost and the holding cost at the same time. A combined model will illustrate the complexities of this. Each model separately could likely find a different optimal Q. Combined, would each result in an optimal solution, or would another Q value result in lower total cost?
The Billington (1987) paper introduced the holding cost reduction model with an exponential setup cost reduction. The following are the results for an example presented in that paper. Notation has been changed so as not to conflict with the model presented earlier in this paper. Assume
u = upper level of setup cost
l = lowest technological level of setup cost
k = capital expense for setup cost reduction
(11)
and the following setup function similar to the holding cost function.
(12)
The optimal conditions:
let (13)
then
(14)
and
(15)
With S restricted to be less than u, the following limiting condition is found.
(16)
If then we would not invest and k*=0. Only if will we invest k* to reduce total cost.
Example 2
Consider the following example using the same parameters as in example 1 with the addition of u and l for the setup cost parameters.
D=3600S=1000U=20L=15u = 1000l = 100
Table 2 shows the results for the example in Billington (1987) and from example 1 in this paper, each solved individually without regard to the other. The original problem solved as a basic EOQ is shown in column 2a. In column 2b, with c = 0.0004, the setup cost reduction model applies, the formulas (11) to (15) and (1) above are solved and placed in the column. Total cost does decrease.
In column 2c, with b = 0.004, the holding cost reduction model applies as shown above in example 1, the formulas (7) to (10), (1) and (3) are solved and placed in the column. Total cost does decrease.
We now naively assume that we should invest $3257 from column 2b to reduce the setup cost and invest $477.97 from column 2c to reduce the holding cost, both at the same time. In column 2d, we have taken the reduced setup cost and reduced holding cost and found the resulting new Q and total cost. The total cost does indeed go down. However, is this the best result, or can we do better?
Table 2 - Setup and Holding Cost Reductions Combined
2aOriginal Problem
EOQ / 2b
c>C
S reduced only / 2c
b>B
H reduced only / 2d
Combined naïve, both S and H reduced
b / 0.004
c / 0.0004
H* / 20 / 20 / 15.739 / 15.739
S* / 1000 / 344.6 / 1000 / 344.6
K* / 0 / $477.97 / $477.97
k* / 0 / $3257 / $3257
Q* / 600 / 352.2 / 676.36 / 397.0
TC* / $12,000 / $10,301.20 / $11,123.19 / $9984.00
SETUP AND HOLDING COST COMBINED OPTIMAL MODEL
The combined optimal model will be
(17)
(18)
(5)
with the assumption that sufficient funds are available to fully fund any level of k and K. Taking first derivatives of (17) results in the following optimal conditions, in addition to the same Q result (1).
(4)
and
(19)
If we take the first derivatives of (5) and (18), the result will be
(6)
(20)
Combine (6) and (4), and (19) and (20) with (1), and let
(21)
then the optimal Q is
(22)
and
(23)
(24)
with capital costs
(15)
(9)
Solving this new model with the previous data results in a better optimal condition, as shown in Table 3. Columns 3a, 3b, 3c, and 3d are the results in columns 2a to 2d in Table 2 above copied here. In column 3e are the results using the combined optimal model from formulas (21) to (24), (9), (15) and (17) above. The total cost does indeed decrease. The interesting result is that, compared to the naïve approach in column 3d, reducing the capital expenses with larger holding and setup costs results in a larger Q but lower total costs.
Table 3 - Setup and Holding Cost Reductions Combined
3aOriginal Problem
EOQ / 3b
c>C
S reduced only / 3c
b>B
H reduced only / 3d
Combined naïve, both S and H reduced / 3e
Combined Optimal Model
b / 0.004 / 0.004
c / 0.0004 / 0.0004
H* / 20 / 20 / 15.739 / 15.739 / 16.2
S* / 1000 / 344.6 / 1000 / 344.6 / 388.55
K* / 0 / 477.97 / 477.97 / 356.1
k* / 0 / 3257 / 3257 / 2843.7
Q* / 600 / 352.2 / 676.36 / 397.0 / 415.52
TC* / $12,000 / $10,301.20 / $11,123.19 / $9984.00 / $9932.64
IMPLEMENTATION ISSUES
The use of the combined optimal model can be complicated by the values of c and b in the setup and holding cost functions. It may be possible that even if the value of c suggests that the setup cost should be reduced, the combined model may result in reduced setup cost but with a higher total cost. The same can be said about the b value in the holding cost reduction model. The following examples illustrate these possibilities.
Example 3
Consider the following example using the same values as in examples 1 and 2, but with different c and b values.
D=3600S=1000U=20L=15u = 1000l = 100
Using formulas (10) and (16) B = 0.000666 and C = 0.000185 as found earlier. In this example b = 0.0009 and c = 0.0004, both of which are greater than their respective B and C values. In this example, the combined optimal model does not give the lowest total cost, rather, the setup reduction model alone would be the best solution.
Table 4 – Example 3
4aOriginal Problem
EOQ / 4b
c>C
S reduced only / 4c
b>B
H reduced only / 4d
Combined naïve, both S and H reduced / 4e
Combined Optimal Model
b / 0.0009 / 0.0009
c / 0.0004 / 0.0004
H* / 20 / 20 / 18.57 / 18.57 / 20
S* / 1000 / 344.6 / 1000 / 344.6 / 329.48
K* / 0 / 374.7 / 374.7 / 0
k* / 0 / 3257.02 / 3257.02 / 3416.5
Q* / 600 / 352.21 / 622.7 / 365.53 / 330.44
TC* / $12,000 / $10,301.24 / $11,937.35 / $10,419.21 / $10,310.38
Note that in the combined optimal model, the optimal H would be greater than upper limit of 20 using the formulas provided. In that case, the optimal H must be the maximum value possible, or 20. Even though the c and b values are greater than the C and B limits suggesting the optimal model is the best, in this case the setup reduction model, column 4 b, gives the best solution.
The conclusion is that all models must be calculated in every situation to insure that the lowest cost solution is found.
CONCLUSION
The results of this research extend the analysis of the classic EOQ model to include a reduction in the per-unit holding cost. Under certain situations dependent on the values of the model parameters, it may be possible to reduce total cost by reducing the holding cost and increasing the order quantity. In combination with the setup cost, under certain conditions based on the model parameters, it may be possible to reduce the total cost even further by reducing both the setup cost and the per-unit holding cost.
An important result for managers is to recognize that if spending some money on reduction of the setup or holding cost does not mean that more money can reduce the total cost even more. Since the resulting total cost curve is parabolic as shown in Figure 1, there is an optimal amount to spend, beyond which the total cost will increase.
Another important result is that a naïve combination may not give the lowest total cost. A combination with slightly higher setup and holding costs may result in lower total costs. Also, the introduction of reduced holding cost actually increases the optimal quantity. This is a counter-intuitive result from the conventional wisdom in JIT and lean manufacturing that lower quantities are the result of lower cost. For the setup cost, this is true; but for the holding cost, lower holding cost results in larger quantities, with lower total cost.
REFERENCES
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