Introduction

Facing strong competition from both overseas steel industries and domestic mini-mills which take advantage of newer technology and lower costs, integrated steel companies exist within an economic environment suffused with strong competition, shrinking markets and lagging long-term profitability. This highly competitive global steel market makes the old topic, cost reduction, generalized widely. Since carrying inventories for various reasons can cost the steel companies anywhere from 20 to 40% of their value a year, inventory management, which has been called and quite fairly so, one of the classics in operation management literature during the last few decades, has recently been the focus of attention. Since raw material inventories account for the largest part of the inventories in the iron and steel industry, minimizing them in a quantitative manner is of vital economic significance. Thus, improving the management level of raw material inventory is clearly of extreme importance.

The raw material inventory problem studied in this paper is to determine the fixed order size and fixed interval of the replenishment process for each material based on the minimization of the total cost attributed to raw material inventories. This inventory problem arises from the production of BaoSteel. BaoSteel is not only the largest and most advanced iron and steel enterprise in China, but also one of the most profitable steel enterprises in the world enjoying international competence. Its annual production consists of over 20 million tonnes of steel, and its auto sheet accounts for more than 60% of the domestic market share. Therefore, each year it must consume large quantities of raw materials, for example, about 17 960 000 tonnes of iron ore per year. Obviously, the replenishment of raw materials in Baosteel is a matter of great significance.

After an investigation that dealt with the issue of managing raw material inventories in Baosteel has been made, we find it convenient to describe the entire process of raw material inventory management in Baosteel using Figure 1. When the demand plan of raw materials submitted by the production department is received, the raw material control centre, which is responsible for raw material inventory management, will set corresponding inventory policies to provide guiding principles for the purchasing department. The main purpose of our research is to help the raw material control centre determine the optimal order interval and inventory level for each material.

Figure 1.

The institution of the replenishment plan for raw materials.

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However, generating an efficient raw material inventory plan for a large steel company is not an easy task for several reasons. First, the raw material inventory problems derived from iron and steel production are new and important research topics, which are seldom considered in previous researches. Although similar problems have been studied and solved, the effort required to find better plans for problems with different backgrounds may vary considerably. Therefore, they still deserve studying carefully. Second, the problem is characterized by large production demands, high inventory costs, no stockout, limited inventory capacities, finite storage time for some kinds, etc. These features that are often disregarded add extra difficulties to the inventory problem under our consideration. Moreover, in order to ensure the continuity of production, large inventories usually have to be maintained to compensate for the unexpected demand fluctuations as well as variability in the replenishment process. In view of these facts, how to best balance raw material inventory and production demands under capacity constraints in the iron and steel industry becomes a difficult task.

For the convenience of research, a logical grouping concept is introduced in this paper to divide raw materials according to their own properties, which is different from physical grouping that concerns actual storage positions. This grouping principle aims to centralize the scattered inventory capacities of stock yards. It is very helpful to determine the total capacity of these stock yards for each raw material group which consists of raw materials with common properties. To understand it better, the schematic diagrams about the comprehensive stock yard of Baosteel are given in Figures 2, 3 and 4. This comprehensive stock yard, including stock yard phase I and II and stock yard phase III, is in charge of the centralized management and handling of raw materials, aiming at smooth supplying to the complex. According to the principle of logical grouping, one imaginary auxiliary raw material yard is defined to include all the real auxiliary raw material yards in the Baosteel comprehensive stock yard. The inventory capacity of this imaginary yard is determined to be the total capacities of the real yards included in it. The materials stored in those real yards compose the 'Auxiliary Raw Material Group'. Since out-of-stock (OOS) will bring enormous losses to Baosteel, it is not allowable here.

Figure 2.

Schematic diagram of the comprehensive stock yard.

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Figure 3.

Schematic diagram of stock yard phase I and II.

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Figure 4.

Schematic diagram of stock yard phase III.

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In the remaining part of the paper, we first review the related works as well as the current state of research concerning inventory. A novel mathematical programming model is then formulated to solve the inventory problem. The solution methodology and the computational results are presented in the following part. The final section concludes our study.

Literature review

Various inventories exist at practically every stage of the production process as raw materials, work-in-process semi-finished or finished goods. During the last few decades, inventory control has received considerable attention in literature. Therefore, a large collection of results is available. In spite of that, few of the previous research, however, dealt with the raw material inventory problem derived from iron and steel production. Bowman (1956) and Manne (1958) solved lot sizing problems by linear programming, respectively. Wagner and Whitin (1958) proposed the WW-model for the uncapacitated lot sizing problem and gave a corresponding polynomial-time algorithm. Newhart et al (1993) quantified the effects of inventory required for locating parts of the supply chain in different geographic areas by using a two-phased approach. Chiang and Gutierrez (1996) analysed a periodic review inventory system in which there are two modes of re-supply, namely a regular mode and an emergency mode, within the framework of an order-up-to-R inventory control policy. They simultaneously proposed procedures to determine the two policy parameters, that is, the order-up-to level and the indifference inventory level. Hoshino (1996) proposed two theoretical criteria allowing for selection from fixed-size ordering and fixed-interval ordering policies and the choice between the pull-type ordering and the push-type ordering policies. Ishii and Imori (1996) proposed an effective production ordering system for two-item, two-stage, capacity-constraint production and inventory systems, which reduced fluctuations in the total workload and inventory levels. Maia and Qassim (1999) presented an analytical solution for an optimized model that determined when it was preferable to incur inventory or opportunity cost. Ganeshan (1999) proposed a near-optimal (s,Q)-type inventory policy for a production/distribution network with multiple suppliers replenishing a central warehouse, which in turn distributed to a large number of retailers. Nechval and Nechval (1999) investigated the effect of estimation risk on the simplest of inventory problems. Matheus and Gelders (2000) considered an inventory problem subject to a probabilistic non-unit sized demand pattern and proposed an exact and an approximate reorder point calculation method for the (R,Q) inventory policy. Relph and Barrar (2003) argued that excess was important because there was evidence that even in well- managed businesses a significant proportion of the inventory existed in excess at any given time.

Problem description and formulation

The EOQ model is a classical approach to determining order and production lot sizes as well as setting order intervals for a single product with an unconstrained capacity. Countless modifications have been developed in order to adapt the model to various real-world situations. The approach has two major advantages: one is its simplicity and the other is the flatness of the function around the curve's minimum. Our raw material inventory management is a multi-item problem requiring considerations of inventory capacities, demand rates, set up costs as well as inventory costs. In general, this problem can be viewed as an economic lot size problem. The classical economic lot size model (ELSM) making simple tradeoffs between ordering and inventory costs is applicable to unconstrained cases. To make it more realistic, some capacity restrictions must be imposed. Inspired by the idea of ELSM, in this paper, we establish a constrained optimization model to solve the inventory management problem arising from Baosteel taking into account all the features summarized in Table 1. The objective is to determine the optimal order interval and inventory level for each raw material.

Table 1 - Features of the raw material inventory problem in Baosteel.

Full table

Assumptions

The problem characteristics described above form the basis for developing the mathematical models. The modelling assumptions are outlined below.

1. The distribution functions of demands and order lead times are given.
2. All the raw materials are bought in tonnes.
3. The raw materials are consumed in large quantities.
4. The demand for each raw material is relatively stable.
5. The decision horizon is suggested to be one year by the planners of Baosteel.
6. The time horizon is divided into T time units.
7. OOS is not allowable.

Safety stock and safety lead time

To absorb the random fluctuations in demands and transportation times, safety stock and safety lead time are introduced in this section. In the following part, we will give the estimation methods for them.

Estimation of safety stock

A method for determining the safety stock of each raw material is proposed in this part. Here material i is used as an example. Other materials could be treated in the same way. If the distribution function, mathematical expectation and variance of the demand for material i are given as F_i(s)=P(Ss), _i1 and _i1², respectively, the safety stock of material i can be determined as follows.

1. Calculate s_i1, which is deduced from P(S_i1+s_i1)=_i1. Let _i1 be a small positive real number which is predetermined by the planners of Baosteel.
2. Calculate s_i2, the minimum safety stock level specified by these planners.
3. Calculate s_i3=_i1t_i2, where t_i2 is the shortest time needed for purchasing material i from the nearest supplier.
4. The safety stock of raw material i,s_0i, is defined as s_0i=max{s_i1,s_i2,s_i3}.

Estimation of safety lead time

Since using safety stock to absorb the random fluctuations in transportation times will lead to overstock, in this section, we introduce the concept of safety lead time and give the estimation method for it. Safety lead time is the additional time, beyond the expected lead time, needed to guarantee receipt of the materials ordered. The method used to determine the safety lead time of material i is presented on condition that the distribution function, mathematical expectation and variance of its order lead time are given as F_i(t)=P(Tt), _i2 and _i2², respectively. The safety lead time of material i can be determined as follows.

1. Calculate t_i1, which is deduced from P(T_i2+t_i1)=_i1. Let _i1 be a small positive real number which is predetermined by the planners of Baosteel.
2. Calculate t_i2, the minimum lead time needed for Baosteel to prepare for the replenishment.
3. Calculate t_i3, the shortest time needed for purchasing the material from others.
4. The safety lead time of raw material i, t_0i, is defined as t_0i=max{t_i1,t_i2+t_i3}.

Notations

Parameters

T

decision horizon

N

set of raw materials stored in Baosteel comprehensive stock yard, index i

D_i

demand rate of raw material i, iN

h_i

inventory holding cost per unit of material i, that is, the capital occupied by keeping one unit of material i in stock, iN

K_i

fixed set-up cost incurred every time Baosteel places an order of material i, iN

vl_i

storage life, that is, the longest storage time, of material i, iN

I_k

raw material group k (according to logical grouping), 1kn; I₁I₂I_n=N;I_iI_j=,i,j=1,...,n,ij

V_k

the largest inventory capacity of Baosteel comprehensive stock yard for the raw materials included in I_k, 1kn.

Decision variables

Q_i

fixed maximum inventory level of raw material i, iN

t_i

fixed order interval of raw material i, iN

m_i

the number of the orders placed in the decision horizon for raw material i, iN

The decision model

In this section, the decision model for determining the optimal order interval and inventory level of each raw material is formulated. The objective is to minimize the manufacturer's total cost attributed to raw materials inventories. The details of the model are presented below.

Subject to

The two terms in objective function (1) represent setup cost and inventory holding cost, respectively. Constraints (2) reveal the relationship among m_i, t_i and T. Constraints (3) indicate that the inventory level of each material is determined by its safety stock and the consumption during its order interval. Constraints (4) are the inventory capacity restrictions. The maximum inventory level, Q_i, is forced to be not greater than the consumption of material i during its longest storage time, vl_i, by (5). Constraints (6)–(8) define the value ranges of the variables. The relationship between Q_i and t_i can be seen clearly from Figure 5.

Figure 5.

Inventory level of material i as a function of time.

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Solution methodology

Owing to the integer decision variables included, the inventory problem described above is difficult to solve directly. As we know, Lagrangian relaxation is one of the most widely used and efficient techniques for performing constrained optimization. The central idea of Lagrangian relaxation is to formulate a decomposable relaxed problem which is easier to solve than the original one. Based on the ideas of Luh et al (1990), Diaby et al (1992) and Tang et al (2002), we use Lagrangian relaxation as the main part of our solution methodology. For a minimization problem, the solution to the Lagrangian dual problem provides a lower bound, while each feasible solution provides an upper bound.

Lagrangian relaxation

The problem formulation presented previously shows that only constraints (4) result in the coupling of different raw materials. By relaxing these constraints, the original problem can be decomposed into smaller and easier subproblems, each for one material. For that reason, we form the following relaxed problem by introducing constraints (4) to the objective function through Lagrangian multipliers {uk}: LR

Subject to constraints (2), (3), (5), (6), (7), (8).

Let {{m_i^*},{Q_i^*},{t_i^*}} be an optimal solution of LR for a given set of multipliers. The dual problem of LR is to maximize the dual function C_D(u): LD

Subject to constraints (2), (3), (5), (6), (7), (8) and

Here u is a vector of non-negative Lagrangian multipliers with elements {u_k}. Since constraints (2), (3) and (5), (6), (7) and (8) are independent of each other, the relaxed problem can be decomposed into the following material-level subproblems: LR_i

Subject to constraints (2), (3), (5), (6), (7), (8).

Here (i) denotes the raw material group that material i belongs to.

Ordinal introduction of constraints algorithm for subproblems (LR_i)

Based on constraints (2) and (3), m_i and Q_i can be represented using t_i:

Constraints (5) can then be expressed as t_ivl_i-s_0i/D_i. This can be combined with constraints (6) into the following constraints.

Note that when t_i>0, constraints (8) are automatically satisfied. With these manipulations, problem (LR_i) becomes

Subject to

The objective function takes the form of

where

Obviously, a₁,a₂ and b are all positive. It is easy to see, by checking its first- and second-order derivatives, that y(t) is a strictly convex function. A graph of function y for a₁=100, a₂=70 and b₂=29 is shown in Figure 6 as an example.

Figure 6.

Plot of y as a strictly convex function of t at a₁=100, a₂=70 and b₂=29.

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Therefore, the subproblem is to minimize a strictly convex function of one variable in a fixed interval with additional integer constraints, which can be solved easily using the ordinal introduction of constraints algorithm. The steps of the algorithm to determine t_i are detailed below.

Step 1:

Let y'(t)=0, calculate the optimal solution of unconstrained problem (13) and denote it as t₁^*.

Step 2:

If 0<t₁^*min{vl_i-s_0i/D_i,T}, set t₂^*=t₁^*; otherwise set t₂^*=min{vl_i-s_0i/D_i,T}. Here t₂^* stands for the optimal solution of constrained problem (13) and (14).

Step 3:

If T/t₂^* is an integer, let t^*=t₂^*; otherwise, set c₁=T/t₂^*,t₁=T/c₁, t₂=T/(c₁+1). Calculate y(t₁) and y(t₂). If y(t₁)<y(t₂), set t^*=t₁; otherwise set t^*=t₂. t^* obtained here is the optimal solution of constrained problem (13)–(15).

Construction of a feasible solution to the original problem

The solution to the relaxed problem is generally infeasible for the original problem because the inventory capacity constraints (4) might be violated. To construct a feasible solution, a heuristic approach is proposed. The detailed algorithmic steps of this heuristics are described as follows.

Step 1:

Search for the first raw material group k, 1kn, which violates the corresponding inventory capacity constraint. If there exists no such group, set flag=1 and go to Step 6; otherwise calculate sum=_iIk(Q_i-s_0i), sum1=_iIkQ_i.

Step 2:

Calculate sub=sum1-V_k. If sub>sum, this heuristics is of no effect for this problem, set flag=0 and go to Step 6.

Step 3:

Create a material sequence for I_k by indexing the materials in the ascending order of K_i-0.5D_ih_i(T/m_i-T/(m_i+1)),iI_k.

Step 4:

Take the first unselected material i from the above sequence. Set a=Q_i,m_i=m_i+1, update t_i=T/m_i and Q_i=s_0i+D_it_i, then calculate sub=sub-(a-Q_i).

Step 5:

If sub>0, go to Step 3; otherwise, go back to Step 1.

Step 6:

Stop. If flag=1, {(m_i,Q_i,t_i)|iN} obtained above defines a feasible replenishment plan satisfying all the constraints. If flag=0, the manufacturer must reduce the safety stocks of the materials included in I_k or increase the value of V_k.

Updating Lagrangian multipliers

It is well known that subgradient algorithm is commonly used to solve Lagrangian dual problems. In order to solve the dual problem with respect to (10), the algorithm is adopted to update the vector of Lagrangian multipliers, u, by

where t^m is the step size at the mth iteration and ^m(u_k^m)=_iIk Q_i-V_k,1kn is the subgradient component related to the kth stock yard. The step size t^m is given by

where C^U is the best objective value found so far and C^L is equal to the value of C_D(u) at the mth iteration. The algorithm terminates, if one of the following conditions is satisfied.

1. (C^U-C^L)/C^L<, where >0 is a very small number;
2. m> the maximum iteration number given by the user.

Improvement method

When 10 consecutive iterations fail to improve the current optimal solution, a three-phase improvement method will be activated. Let {(m_1i,Q_1i,t_1i)|iN}, {(m_i,Q_i,t_i)|iN}, up_bound and lo_bound denote the relaxed solution, the current best feasible solution, upper bound and lower bound, respectively. The detailed description of this method is outlined below.

Phase I:

Step 1:

Take the solution obtained from solving the relaxed problem as the initial solution, that is, set m_i⁽¹⁾=m_1i,Q_i⁽¹⁾=Q_1i,t_i⁽¹⁾=t_1i, for each iN.

Step 2:

Search for the first raw material group k, 1kn, which violates the corresponding inventory capacity constraint. If there exists no such group, set flag=1 and go to Step 7; otherwise calculate sum=_iIk(Q_i⁽¹⁾-s_0i), sum1=_iIkQ_i⁽¹⁾.

Step 3:

Calculate sub=sum1-V_k. If sub>sum, let flag=0 and go to Step 10.

Step 4:

Search for the first unselected material lI_k which satisfies the conditions: Q_l⁽¹⁾-s_0lD_l (T/m_l⁽¹⁾-T/(m_l⁽¹⁾+1)) and m_l⁽¹⁾+1<T. If there exists no such material, set flag=0 and go to Step 10.

Step 5:

Set m_l⁽¹⁾=m_l⁽¹⁾+1, t_l⁽¹⁾=T/m_l⁽¹⁾, Q_l⁽¹⁾=s_0,l+D_lt_l⁽¹⁾, sub=sub-Q_l, where Q_l represents the decrement of Q_l⁽¹⁾.

Step 6:

If sub>0 and material l still satisfies the condition stated in Step 4, go back to Step 5; If sub>0 but material l cannot meet the condition, go back to Step 4; If sub0, go back to Step 2.

Step 7:

Calculate the objective value C₁=_iNK_im_i⁽¹⁾+_iN 1/2h_i(Q_i⁽¹⁾+s_0i), if C₁<up_bound, go to Step 10; otherwise set flag=0.

Step 8:

Form set S composed of all the raw material groups that violate inventory capacity constraints, based on the relaxed solution.

Step 9:

Search for raw material group kS in which there exist unused exchanges. If there is no such group, let flag=0; otherwise change the order of the materials in I_k and go back to Step 1.

Step 10:

Stop. If flag=1, update up_bound=C₁, let {(m_i⁽¹⁾,Q_i⁽¹⁾,t_i⁽¹⁾)|iN}{(m_i,Q_i,t_i)|iN}; Otherwise this heuristics is not effective for this problem.

Phase II:

Step 1:

Take the current best feasible solution as the initial solution, that is, set m_i⁽²⁾=m_i,Q_i⁽²⁾=Q_i,t_i⁽²⁾=t_i, for each iN.

Step 2:

Search for the first unselected material lN which satisfies the conditions: T/(m_l⁽²⁾-1)min{vl_i-(s_0l/D_l),T} and m_l⁽²⁾2; if there exist no such materials, stop.

Step 3:

Search for the raw material group k that l belongs to. If we can obtain a better feasible solution only by decreasing the value of m_l⁽²⁾ to m_l^(2'), let C₂ denote the corresponding objective value, update the up_bound=C₂, let {(m_i^(2'),Q_i^(2'),t_i^(2')|iN}{(m_i,Q_i,t_i)|iN} and then go back to Step 1.

Step 4:

Search for the material r that satisfies the conditions: rI_k and rl. If we can get a better feasible solution by both decreasing the value of m_l⁽²⁾ to m_l^(2'') and increasing the value of m_r⁽²⁾ to m_r^(2''), let C₂ denote the corresponding objective value, update up_bound=C₂, let {(m_i^(2''),Q_i^(2''),t_i^(2'')|iN}{(m_i,Q_i,t_i)|iN}. Finally go to Step 1.

Phase III:

Step 1:

Take the current best feasible solution as the initial solution, that is, set m_i⁽³⁾=m_i,Q_i⁽³⁾=Q_i,t_i⁽³⁾=t_i, for each iN.

Step 2:

Search for the first unselected material lN; if there exists no such material, stop.

Step 3:

Search for the raw material group k that material l belongs to.

Step 4:

Search for the material r that satisfies the conditions: rI_k and rl. If we can get a better feasible solution by both increasing the value of m_l⁽³⁾ to m_l^(3') and decreasing the value of m_r⁽³⁾ to m_r^(3'), let C₃ denote the corresponding objective value, update up_bound=C₃, let {(m_i^(3'),Q_i^(3'),t_i^(3'))|iN}{(m_i,Q_i,t_i)|iN}. Finally go to Step 1.

Computational results

Generation of problem instances

To study the performance of the algorithms described above, two types of experiments are performed across a range of test problems. In the first group of the tests, the number of raw material groups is fixed, and in the second group, the number of raw materials is fixed. To generate representative problem instances, we analyse the actual production data from Baosteel. A day is taken as the basic time unit. The decision horizon is set to be one year. The number of raw material groups is chosen from n{2,3,5,8,10}, while the number of raw materials changes from 4 to 200. For each combination of the number of materials and the number of raw material groups, 10 instances are randomly generated, therefore resulting in a total of 1030 test problems used in this research. The parameters of raw materials are generated based on the actual production data from Baosteel. Let G denote the set composed of raw materials whose practical data have been obtained. In what follows, material j is used as an example to illustrate the generation of those parameters.

1. One material i is chosen from G randomly for material j.
2. The demand rate of material j is randomly generated from [0.8, 3] times of D_i.
3. The inventory holding cost per unit of material j is generated uniformly from [0.5*h_i,1.5*h_i].
4. The safety stock and the fixed set-up cost of material j is randomly generated from [0.5, 1.5] times of s_0i and K_i, respectively.
5. To be closer to the practical production, the storage life of material j is randomly generated from a uniform distribution [60,365].
6. The largest inventory capacity of Baosteel comprehensive stock yard for the materials included in each raw material group is randomly generated from [1.05,1.1] times of their total safety stocks.

Testing results

The algorithms have been implemented using Visual C++ and the experiments are carried out on a Pentium-IV 2.4 GHz PC. Because Lagrangian relaxation cannot guarantee optimal solutions, the relative dualilty gap (C^UB-C^LB)/C^LB was used as a measure of solution optimality, where C^UB is the upper bound to the original problem and C^LB is the lower bound. The maximum iteration number 4000 is imposed on the stopping criterion. As mentioned earlier, for each problem structure we randomly generate and solve 10 problem instances. All the values shown in the tables below represent the average of performance measures and running times for these 10 instances of the corresponding problem structure. The results of these experiments are shown in Tables 2 and 3. It can be observed that the average duality gap obtained from the Lagrangian relaxation algorithm is 0.090648% in about 25 s of computation time. From the computational results, we also have the following observations.

1. When the number of raw material groups is fixed, the duality gap decreases but the computation time increases as the number of raw materials increases.
2. When the number of raw materials is fixed, the duality gap and the computational time increase as the number of raw material groups augments.

Table 2 - Average duality gap (%) of our method.

Full table

Table 3 - Average running time (s) of our method.

Full table

Conclusions

Since the steel industry is resource intensive, effective inventory management in this industry assumes even greater importance. This research studied the raw material inventory problem arising from Baosteel and formulated it as a mixed integer programming model. A Lagrangian relaxation algorithm was developed to solve it. In this algorithm, a 'separatable' relaxed problem was constructed by relaxing inventory capacity constraints. The subproblems derived from this relaxed problem were solved optimally by the ordinal introduction of constraints algorithm, while the Lagrangian multipliers were iteratively updated along a subgradient direction. Numerical results indicated that our algorithms generated high-quality solutions in a short computation time. Although this study is only a theoretical research, it provides the scientific foundation for practical applications. Future efforts may be focused on the following lines of research.

1. Modifying the model to describe the inventory problems in other industries.
2. Extending the model to solve the multi-echelon inventory problems in the iron and steel industry.
3. Studying inventory control problems from the perspective of supply chain management, such as VMI.

References

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Acknowledgements

We would like to thank the Production Manufacturing Center in Baosteel for providing a lot of production information and data. We also thank the anonymous referees for the valuable suggestions. This research is partly supported by National Natural Science Foundation for Distinguished Young Scholars of China (Grant No. 70425003), National Natural Science Foundation of China (Grant No. 60274049 and Grant No. 60674084).