Optimization on Production-Inventory Problem with Multistage and Varying Demand

Abstract

This paper addresses production-inventory problem for the manufacturer by explicitly taking into account multistage and varying demand. A nonlinear hybrid integer constrained optimization is modeled to minimize the total cost including setup cost and holding cost in the planning horizon. A genetic algorithm is developed for the problem. A series of computational experiments with different sizes is used to demonstrate the efficiency and universality of the genetic algorithm in terms of the running time and solution quality. At last the combination of crossover probability and mutation probability is tested for all problems and a law is found for large size.

1. Introduction

Production-inventory control plays a vital role in the management of manufacturing enterprises. During production process we often hope that the volume of a given product is just enough to satisfy customers’ demand without overextending the production line and manufacturing too many. The redundant inventory will cut down our net profit. On the other hand, lacking supply will also make a heavy loss when a big order is placed. So there must be an optimal production quantity.

Many researchers are interested in this problem and it has been investigated from various perspectives. Goyal and Giri [1] develop two different models for the production-inventory problem, in which the demand, production, and deterioration rates of a product are assumed to vary with time over an infinite planning horizon. Shortages of a cycle are allowed to be backlogged partially. Huang [2] proposes an optimal integrated vendor-buyer inventory policy for flawed items in a just-in-time manufacturing environment. The production process is assumed to deteriorate during processing and produces a certain number of defective items. The objective is to minimize the total joint annual costs incurred by the vendor and the buyer. Hill [3] considers the problem of a vendor supplying a product to a buyer with the product manufactured in batches at a finite rate. And the structure of the globally-optimal solution is derived. Gayon et al. [4] formulate the joint production-control and inventory-allocation problem as a Markov decision process and characterize the structure of the optimal policy. They draw a conclusion that the optimal inventory-allocation policy is a state-dependent multilevel rationing policy, with the rationing level for each class nondecreasing in the number of announced orders. An adaptive control approach with a feedback is applied to track the inventory levels toward their goal levels by Alshamrani [5, 6].

Considering uncertainty, Doğru et al. [7] develop a stochastic program to allow preferential component allocation for minimizing total inventory cost in assemble-to-order inventory systems with identical component lead times. Zijm and Houtum [8] compare the optimal base stock policy under stochasticcircumstance with an MRP systemin terms of cost effectiveness given a predefined target customer service level. And they also take stochasticlead times into consideration in amultistage production-to-ordersystem. According to setting safety lead times, it leads to similar decomposition structures as those derived formultistage inventory systems. Kleywegt et al. [9] formulate the inventory routing problem as a Markov decision process. Gupta and Wang [10] investigate credit terms in inventory problem with random demand. They prove that the structure of the optimal inventory policy is not affected by credit terms under a discrete time of the retailer’s operations, although the value of the optimal policy parameter is.

Ouyang et al. [11] address the same problem based on a deterministic model by fuzzifying the rate of interest charges, the rate of interest earned, and the deterioration rate into the triangular fuzzy number. Hsieh [12] proposes two fuzzy inventory models with fuzzy parameters for crisp order quantity, or for fuzzy order quantity under decision maker’s preference. He demonstrates that the optimal order quantity or the optimal fuzzy order quantity of the two models are the real numbers. The optimal solutions can be specified to meet classical production inventory models under a given condition. References [13–16] also focus on the fuzzy optimal inventory problem.

Chao et al. [17] research operational decisions in dynamic inventory management are correlated with and constrained by financial flows of the firm. A general framework for incorporating financial states of an organization in multiperiod inventory models with lost sales is proposed. Hopp and Xu [18] propose a static approximation of dynamic demand substitution behavior based on a fluid network model and a service-inventory mapping. The interdependent inventory/service, price, and product assortment decisions in noncompetitive and competitive scenarios with demand substitution are analyzed. Olsen and Parker [19] investigate consumer behavior of potentially leaving the firm’s market when he encounters an inventory stockout at a retailer under the stochastic demand distribution in a time-dynamic context. A single firm and a duopoly case are modeled, respectively.

Many researches of perishable inventory problem are investigated. Broekmeulen and Donselaar [20] take into account of the age of the inventory for perishable products at retailers. The new policy they conclude leads to substantial cost reductions compared with a base policy that does not take into account the age of inventories. Duan et al. [21] propose two kinds of inventory models, with and without backlogging, respectively, for perishable items. At last they draw a conclusion that the proposed model is generalization of present ones. Gumasta et al. [22] consider the transportation of perishable goods from distributer to retailer. They simultaneously maximize the revenue and minimize transportation and inventory cost so as to maximize the net profit. Other related researches under stochasticcircumstance can be referenced in [23, 24]. And Markov model for an inventory system with perishable items can be referenced in [25–27] and so on.

Some researchers are interested indiscussing backorder problem in inventory. Yao and Chiang [28] take the storing cost 𝑎, backorder cost 𝑏, cost of placing an order 𝑐, total demand 𝑟, order quantity 𝑞, and shortage quantity as the triangular fuzzy numbers and use the signed distance method to defuzzify. Gürler et al. [29] consider bayesian updating of demand and backorder distributions in a partial backorder newsvendor model. In order to model the relationship between risks and risk propagation in supply chain, Shin et al. [30] apply bayesian belief network to develop alternative backorder replenishment plan to minimize the total replenishment cost and expected risk cost. Hu et al. [31] and Shahand Soni [32] assume production elapsed time and demand to be a fuzzy random variable with backorder, respectively. We can also refer to [33–38] about backorder aspect.

The method for assembly line balancing problem is very similar to the inventory problem. Sotskov et al. [39, 40] minimize the number of stations 𝑚 for processing 𝑛 partially ordered operations within the given cycle time. They derive necessary and sufficient conditions when optimality of the line balance 𝑏  is stable with respect to sufficiently small variations of the operationtimes. Özcan and Toklu [41–43] and Purnomo et al. [44] research two-sided assembly lines balancing problem. Kellegöz and Toklu [45] address a branch and bound algorithm to solve the problem. Compared with the existing algorithm, the proposed algorithm outperforms it in terms of both CPU times and quality of feasible solutions found.

The traditional deterministic inventory models are based on the assumption that the demand rate is constant and order cycle is inflexible. As we all know the total cost is the integral of the production-inventory function because the production time is the same and the maximum inventory in every stage is also equal. But actually there is more than one customer. Demand cycles of these customers are different and the demand rate of every customer is also varying. This is multistage and varying demand production-inventory problem without stockout. In the planning horizon, which is the sum of all stages, what we face to decide is also when and how many quantity to produce in order to minimize the total cost subject to meeting every stage demand. Under the circumstance the traditional unconstrained optimization method is not suitable for the issue although it is still linear. Because according to every demand and its cycle there are different maximum inventory in every stage under different production schedules which means production frequency and production time. We have to define the maximum inventory in every stage for the sake of obtaining the objective function. And this definition relation is only determined by the constraint. So constrained optimization method will be applied to solve the problem. Motivated by the above case we model a hybrid integer programming to characterize the problem and solve it by using genetic algorithm.

The remainder of the paper is organized as follows. Section 2 is dedicated to the description of the modeling approach. Section 3 presents a genetic algorithm to solve the problem. Section 4 reports our computational results on test instance. Finally, conclusions are drawn in Section 5.

2. Production-Inventory Model

2.1. Problem Description

In multistage production-inventory system, every stage extent is versatile, demand rate is also different and production rate is a constant and is more than demand rate in every stage. The production-inventory status is drawn in Figure 1 according to the production rate and demand rate in every stage. The planning horizon is made up of five stages. 𝑇_𝑖 denotes the extent of stage 𝑖. There are twice production in Figure 1. The production mode is as shown in Figure 1(a). The first production begins from Stage 1 and ends in Stage 2 lasting 𝑡₁ time long. And at time 𝑡₁ the inventory reaches maximum. The maximum inventory in the first production is consumed in the end of Stage 3. We call it a production cycle from Stage 1 to Stage 3. Stage 3 is called the end of one production cycle. The second production cycle includes Stage 4 and Stage 5 lasting 𝑡₂ time long. Figure 1(b) depicts another case of twice production. The first production cycle is from Stage 1 to Stage 4 and the second production cycle is in Stage 5 alone.

(a)   Twice production

(b) Another two times production

Figure 1 

Production-inventory status.

In fact there are four production schemes under the circumstance of twice production. Except the above two schemes mentioned, one is that the first production is in Stage 1 alone and the second production starts from Stage 2 to Stage 5. The other is that the first production begins from Stage 1 to Stage 3 and the second production starts from Stage 4 to Stage 5.

One extreme case is five times production in planning horizon, that is, production occurs in every stage and the quantity as inventory in every stage only satisfies the current stage demand. The other extreme case is only once production in planning horizon. The best production times and quantity are specified by the total cost which bears setup cost of production and carrying cost held in inventory. And we will find the best production scheme under all production times.

2.2. Model

2.2.1. Denotations Definition

First we define some denotations in our model to express the above conception.  Parameters are defined as follows. 𝑃: Production rate. 𝑛: The extent of stages. 𝑚: Production times, 𝑚=1,2,…,𝑛. 𝑇_𝑖: Demand extent of stage 𝑖,𝑖=1,2,…,𝑛. 𝑅_𝑖: Demand rate in stage 𝑖,𝑖=1,2,…,𝑛. 𝐾: Setup cost of production once. ℎ: Holding cost per unit volume per unit time held in inventory. 𝑇: Planning horizon. Obviously 𝑇 equals to the sum of 𝑇_𝑖, that is, 𝑇=∑^𝑛_𝑖=1𝑇_𝑖. Variables are defined as follows: 𝑥_𝑖=1,iftheproductionbeginsfromstage𝑖,0,otherwise,𝑖=1,2,…,𝑛. (2.1) 𝑡_𝑖𝑘: The production time of one production cycle which begins from stage 𝑖 and ends in stage 𝑘, 1≤𝑖≤𝑘≤𝑛𝑣_𝑖𝑘=1,if𝑡_𝑖𝑘>0,0,otherwise,1≤𝑖≤𝑘≤𝑛. (2.2) 𝑄_𝑖: The inventory level at the beginning of stage 𝑖,𝑖=1,2,…,𝑛 𝑄_𝑖𝑗𝑘: The maximum inventory level at time 𝑡_𝑖𝑘 in stage 𝑗, 1≤𝑖≤𝑗≤𝑘≤𝑛. The variable maximum inventory 𝑄_𝑖𝑗𝑘 is positive if and only if 𝑡_𝑖𝑘 belongs to the interval (∑^𝑗−1_𝑙=𝑖𝑇_𝑙,∑^𝑗_𝑙=𝑖𝑇_𝑙)𝑤_𝑖𝑗𝑘=1,if𝑄_𝑖𝑗𝑘>0,0,otherwise,1≤𝑖≤𝑗≤𝑘≤𝑛. (2.3)

2.2.2. Constraints

(1)Because the stockout is not permitted, the total quantity should be equal to the sum of demands in the planning horizon. And the quantity in every production in a certain stage interval should satisfy the demand in the same stage interval: 𝑃^𝑛𝑛_{𝑖=1𝑘≥𝑖}𝑡_𝑖𝑘=^𝑛_𝑖=1𝑅_𝑖𝑇_𝑖,𝑃𝑡_𝑖𝑘=𝑣^𝑘_{𝑖𝑘𝑙=𝑖}𝑅_𝑙𝑇_𝑙,1≤𝑖≤𝑘≤𝑛. (2.4)(2.5)(2)Production times in the planning horizon are calculated by the following formula: ^𝑛_𝑖=1𝑥_𝑖=𝑚 (2.6)(3)Production time should be less than demand time: 𝑡_𝑖𝑘≤^𝑘_𝑙=𝑖𝑇_𝑙,1≤𝑖≤𝑘≤𝑛. (2.7)(4)The logic variables should satisfy the following: 𝑣_𝑖𝑘=𝑥_𝑖1−Φ^𝑘_𝑙=𝑖+1𝑥_𝑙𝑥_𝑘+1,1≤𝑖≤𝑘≤𝑛. (2.8) The binary function Φ(𝑎) is defined as Φ(𝑎)=1,if𝑎>0,0,otherwise. (2.9) Formula (2.8) means 𝑣_𝑖𝑘=1 if and only if 𝑥_𝑖=1, 𝑥_𝑘+1=1 and every 𝑥_𝑙=0,𝑙=𝑖,…,𝑘 are met simultaneously. (5)The inventory level 𝑄_𝑖 at the beginning of stage 𝑖 is derived by the following: 𝑄_𝑖=max_{𝑙≤𝑗≤𝑘}𝑤_𝑖𝑗𝑘Ψ𝑗−𝑖^𝑖−1_𝑠=𝑙𝑃−𝑅_𝑠𝑇_𝑠+Ψ𝑖−𝑗+1^𝑘_𝑠=𝑖𝑅_𝑠𝑇_𝑠, (2.10) where 𝑖≥2,1≤𝑙≤𝑖≤𝑘≤𝑛 and Ψ(𝑎)=1,if𝑎≥0,0,otherwise. (2.11) It is the fact that a new production begins in stage 𝑖 when 𝑄_𝑖 becomes zero. In other case 𝑄_𝑖 is always more than zero. When the production time lasts 𝑡_𝑖𝑘 time long in a production cycle, the first item Ψ(𝑗−𝑖)∑^𝑖−1_𝑠=𝑙(𝑃−𝑅_𝑠)𝑇_𝑠 in formula (2.10) indicates the inventory at the beginning of stage 𝑗 which is before stage 𝑖. And the second item Ψ(𝑖−(𝑗+1))∑^𝑘_𝑠=𝑖𝑅_𝑠𝑇_𝑠 in formula (2.10) indicates the inventory at the beginning of stage 𝑗 which is after stage 𝑖. For different 𝑗 and 𝑘, the variable 𝑤_𝑖𝑗𝑘 may be zero. Only when 𝑤_𝑖𝑗𝑘 equals to one, 𝑄_𝑖 is the right inventory at the beginning of stage 𝑖. So we solve it by maximization.(6)The relationship between the logic variables 𝑤_𝑖𝑗𝑘 and 𝑣_𝑖𝑘 and continuous variable 𝑡_𝑖𝑘 is characterized as follows: 𝑤_𝑖𝑗𝑘=𝑣_𝑖𝑘Φ𝑡_𝑖𝑘−^𝑗−1_𝑙=𝑖𝑇_𝑙Ψ^𝑗_𝑙=𝑖𝑇_𝑙−𝑡_𝑖𝑘,1≤𝑖≤𝑗≤𝑘≤𝑛. (2.12)(7)The maximum inventory level 𝑄_𝑖𝑗𝑘 is deduced according to the following: 𝑄_𝑖𝑗𝑘=𝑤_𝑖𝑗𝑘𝑄_𝑗+𝑡_𝑖𝑘−^𝑗−1_𝑙=𝑖𝑇_𝑙𝑃−𝑅_𝑗,1≤𝑖≤𝑗≤𝑘≤𝑛, (2.13)(8)when 𝑤_𝑖𝑗𝑘=1 and production time is 𝑡_𝑖𝑘, the item (𝑡_𝑖𝑘−∑^𝑗−1_𝑙=𝑖𝑇_𝑙)(𝑃−𝑅_𝑗) in formula (2.13) means the new inventory level during stage 𝑗. So 𝑄_𝑖𝑗𝑘 is the sum of the inventory 𝑄_𝑗 at the beginning of stage 𝑗 and (𝑡_𝑖𝑘−∑^𝑗−1_𝑙=𝑖𝑇_𝑙)(𝑃−𝑅_𝑗). At last the variables are defined as: 𝑥_𝑖,𝑡_𝑖𝑘,𝑄_𝑖,𝑄_𝑖𝑗𝑘≥0,𝑣_𝑖𝑘,𝑤_𝑖𝑗𝑘∈{0,1},1≤𝑖≤𝑗≤𝑘≤𝑛. (2.14) The initial conditions are 𝑡₁=0 and 𝑆₁=𝑆_𝑛+1=0.

2.2.3. Objective

The sum cost holds setup cost of production and carrying cost held in inventory. So the objective is to minimize the sum cost.

When producing 𝑚 times in the planning horizon the total setup cost is 𝑚𝐾 where 𝐾 is setup cost producing once. In every production cycle we suppose that production time is 𝑡_𝑖𝑘 and 𝑤_𝑖𝑗𝑘=1,1≤𝑖≤𝑗≤𝑘≤𝑛. Then we employ formula (2.4) to formula (2.13) to get the values of 𝑄_𝑙 and 𝑄_𝑖𝑗𝑘,1≤𝑖≤𝑗≤𝑘≤𝑛.

𝑄_𝑙 and 𝑄_𝑙+1 are the inventory levels at the beginning of stage 𝑙 and stage 𝑙+1, respectively. Because the inventory figures from stage 𝑖 to stage 𝑗−1 are all trapezoids. The area of a trapezoid is calculated by (𝑄_𝑙+𝑄_𝑙+1)𝑇/2. In fact the inventory figure of stage 𝑖 is a triangle for the inventory level 𝑄_𝑖 is zero such as inventory figure of Stage 1 in Figure 1(b). We may take a triangle as a degenerate trapezoid in which one edge length is zero. So we use uniform area formula. From stage 𝑖 to stage 𝑗−1 when 𝑗 is more than 𝑖, the holding cost is ℎ2^𝑗−1_𝑙=𝑖𝑄_𝑙+𝑄_𝑙+1𝑇_𝑙, (2.15) where ℎ is holding cost per unit volume per unit time held in inventory.

In stage 𝑗 where production stops the inventory figure is made up of two trapezoids such as the inventory figure of the Stage 3 in Figure 1(b). So the holding cost in stage 𝑗 is ℎ2𝑄_𝑗+𝑄_𝑖𝑗𝑘𝑡_𝑖𝑘−^𝑗−1_𝑙=𝑖𝑇_𝑙+𝑄_𝑖𝑗𝑘+𝑄_𝑗+1^𝑗_𝑙=𝑖𝑇_𝑙−𝑡_𝑖𝑘. (2.16)

The case from stage 𝑗+1 to stage 𝑘 is similar to stage 𝑖 to stage 𝑗−1. Therefore, the holding cost from stage 𝑗+1 to stage 𝑘 is ℎ2^𝑘_𝑙=𝑗+1𝑄_𝑙+𝑄_𝑙+1𝑇_𝑙. (2.17)

So given production times 𝑚, we could solve the minimal average total cost under every possible production mode in the planning horizon 𝑇 as 1min𝑇ℎ𝑚𝐾+2^𝑛_{𝑖=1𝑗≥𝑖𝑘≥𝑗}𝑤_𝑖𝑗𝑘^𝑗−1_𝑙=𝑖𝑄_𝑙+𝑄_𝑙+1𝑇_𝑙+𝑄_𝑗+𝑄_𝑖𝑗𝑘𝑡_𝑖𝑘−^𝑗−1_𝑙=𝑖𝑇_𝑙+𝑄_𝑖𝑗𝑘+𝑄_𝑗+1^𝑗_𝑙=𝑖𝑇_𝑙−𝑡_𝑖𝑘+^𝑘_𝑙=𝑗+1𝑄_𝑙+𝑄_𝑙+1𝑇_𝑙. (2.18)

For every possible production times 𝑚, we minimize the average total cost in the planning horizon. So the objective is min_{1≤𝑚≤𝑛}1𝐹=min𝑇ℎ𝑚𝐾+2^𝑛_{𝑖=1𝑗≥𝑖𝑘≥𝑗}𝑤_𝑖𝑗𝑘^𝑗−1_𝑙=𝑖𝑄_𝑙+𝑄_𝑙+1𝑇_𝑙+𝑄_𝑗+𝑄_𝑖𝑗𝑘𝑡_𝑖𝑘−^𝑗−1_𝑙=𝑖𝑇_𝑙+𝑄_𝑖𝑗𝑘+𝑄_𝑗+1^𝑗_𝑙=𝑖𝑇_𝑙−𝑡_𝑖𝑘+^𝑘_𝑙=𝑗+1𝑄_𝑙+𝑄_𝑙+1𝑇_𝑙. (2.19)

3. Solution Procedure

We solve the solution by employing Genetic Algorithm (GA). GA is a stochastic search method that works on a population of the solutions simultaneously and searches large and complicated fields based on the mechanics of natural genetics and evolutionary principles. In addition, it is particularly suitable for optimization problems with an objective function subject to numerous constraints. GA has demonstrated considerable success in these optimization problems and received more and more attentions during the past decades.

3.1. Chromosome Encoding and Decoding

We encode each chromosome as an 𝑚 string of integers whose components are the production cycles. Suppose that there are 𝑛 stages in the planning horizon. We generate 𝑚 integers 𝑎_𝑖,𝑖=1,2,…,𝑚 randomly to satisfy 𝑎₁+𝑎₂+⋯+𝑎_𝑚=𝑛. For example, let 𝑛=10 and 𝑚=4. A chromosome is may be (2 5 1 2). It means there is total 4 times production. The first production cycle is from Stage 1 to Stage 2. The second is from Stage 3 to Stage 7. The third is in Stage 8 alone. And the last production cycle is from Stage 9 to Stage 10.

The decoding procedure is as follows. When a chromosome is determined, we could decode this chromosome to get 𝑣_𝑖𝑘 first. Then according to formula (2.5) we could obtain 𝑡_𝑖𝑘. By formula (2.10) and formula (2.13) we would find 𝑤_𝑖𝑗𝑘, 𝑄_𝑖 and 𝑄_𝑖𝑗𝑘 finally.

3.2. Crossover

Crossover generates offspring by operating parent chromosomes. Let parameter 𝑃_𝑐 be the crossover probability. A chromosome will be selected as a parent when 𝑟<𝑃_𝑐, where 𝑟 is a random number generated from the interval [0,1]. We apply single parent crossover operator. For a selected parent chromosome with 𝑚 integers, 𝑛₁ and 𝑛₂ are stochastically generated from the interval [1,𝑚] and different. We exchange the value of location 𝑛₁ and the value of location 𝑛₂ in the parent chromosome. Figure 2 shows an example. The stochastically generated number is 2 and 3. An offspring (2 1 5 2) is obtained after exchanging 5 and 1. The crossover operator aims to change production scheme under a fixed production times.

Figure 2 

Example of single parent crossover operating.

3.3. Mutation

Mutation modifies a chromosome to form an offspring. Let parameter 𝑃_𝑚 be the mutation probability. A chromosome will be selected as a parent when 𝑟<𝑃_𝑚, where 𝑟 is a random number generated from the interval [0,1]. We randomly choose a location 𝑝 in a selected chromosome. Then generate an integer stochastically, 𝑞, from the interval [1,𝑛]. Then we compare the value of location 𝑝 in the chromosome and 𝑞. For the offspring (2 1 5 2), 𝑝 is 3 and 𝑞 is 6. Obviously 5 is less than 6. So we sum the values of right position 𝑝 and substitute the new sum, 7, for origin value, 5. The case of mutation is depicted in Figure 3(a). If 𝑞 is less than 5, let the sum of two numbers which are both more than 0 equals to 5. For example, 𝑞=2, which is less than 5, we find two numbers 𝑎₁ and 𝑎₂ randomly satisfying 𝑎₁+𝑎₂=5. Then 𝑎₁ and 𝑎₂ replace 5. The case of mutation is drawn in Figure 3(b) where 𝑎₁=4 and 𝑎₂=1. The mutation operator aims to change production times by increasing or decreasing it.

(a)   Case 1: q is more than the value of position p

(b) Case 2: q is less than the value of position p

Figure 3 

Example of mutation operating.

3.4. Evaluation Function and Selection

Evaluation function is to evaluate the quality of a chromosome. We define evaluation value of a chromosome as the objective function of the corresponding solution.

The selection strategy means how to choose the chromosomes in the current population will create offspring for the next generation. We take the roulette wheel as selection mechanism, in which each chromosome is assigned a slice of a circular roulette wheel and the size of the slice is proportional to the chromosome’s evaluation value.

4. Computational Results

In this section we perform a series of computational experiments to evaluate the GA proposed in Section 3 in different size problems. The experiments have been coded in C programming and implemented on an Athlon 3.10 GHz PC, with 3.12 GB of RAM, running on Windows XP.

In order to test the efficiency and universality of the GA, we generate four groups of data randomly in which stages are 10, 20, 50, and 100, named problems 1, 2, 3, and 4, respectively. Tables 1, 2, 3 and 4 show the exact values of demand extent and demand rate of every problem. The production rate is set as 320 in these problems. The setup cost of production once and holding cost per unit volume per unit time are $2 and $10000, respectively. The parameters in our GA are chosen as follows. The population size PS=30, crossover probability 𝑃_𝑐=0.6, and mutation probability 𝑃_𝑚=0.5.

Table 1 

Demand data in Problem 1.

Table 2 

Demand data in Problem 2.

Table 3 

Demand data in Problem 3.

Table 4 

Demand data in Problem 4.

We execute the GA in 30 independent runs on each problem. The evaluation value is recorded for every trial in every problem and shown in Figure 4. Table 5 displays the computational results including number of generations, the best found evaluation value, average evaluation value, standard deviation, and average running time for the 30 trials in the four problems. Because the scale of every problem is different, the smaller the problem is the fewer the number of generations is. The results reveal that the proposed GA could effectively solve the problem in different sizes.

Table 5 

Computational results statistics.

Figure 4 

Computational results.

Because the parameters in the proposed GA, crossover probability and mutation probability, play a crucial role of the result, we also test the best combination of the two parameters for all problems. Both crossover probability and mutation probability vary from 0.1 to 0.9 when the step is 0.1. So there are 81 combinations in every problem. The average evaluation value is obtained from 30 independent trials under all combinations. Figure 5 displays the comparison results. The best combination is shown as Table 6. It is obvious that the average evaluation value in Table 6 is much better than that in Table 5 for the same problem.

Table 6 

The best combination of (𝑃_𝑐,𝑃_𝑚).

(a) For Problem 1

(b) For Problem 2

(c) For Problem 3

(d) For Problem 4

Figure 5 

Comparison results with combination of 𝑃𝑐 and 𝑃𝑚.

For Problem 1 the best combination parameters are (0.4, 0.5) and (0.9, 0.8). This indicates that Problem 1 is not sensitive to the two parameters. Owing to the small size of Problem 1, if the number of generations increases we could find better solution than the former under any combination parameters. For example, when the number of generations reaches 1000, the average evaluation value is 1419.8273 from the 30 trials under the worst combination (0.1, 0.2) whose average evaluation value is 1482.9856 and number of generations is 100 in Figure 5(a). At the same time the average running time is only one second.

We find a law in Figures 5(c) and 5(d) that when given a value of crossover probability, the average evaluation value almost increases with the increase of mutation probability. And we test other experiments generated randomly where stage is from 60 to 90 (in which step is 10) under all combination parameters. The results reflect the same law. However, it is not the same case in Figures 5(a) and 5(b). This demonstrates that when the problem size is large the mutation probability should be small. So we could only find the best crossover probability. The best combination in Problems 3 and 4 is the same in Table 6. But it is just by chance because in the latter experiments we do not find the same best combination.

Table 7 reports the partial best solutions and evaluation values found during search under the best combination parameters for the four problems. The stages in a bracket in Table 7 indicate that they are in one continuous production cycle. And the stage which does not appear in Table 7 for every problem means a production cycle. For example, the first production cycle is from Stage 1 to Stage 2 in Problem 2. The second and the third production cycle are in Stage 3 and Stage 4, respectively. And the last production cycle starts from Stage 17 to Stage 20. So there are total twelve production cycles in Problem 2.

Table 7 

The partial best solutions and evaluation values.

The most cases are that one production cycle includes two stages in Table 7. When one production cycle includes two stages, the sum cost is obviously smaller if the demand rate in the first stage is more than that in the second stage. So we can conclude from Table 7 that all demands rate of two successional stages in one production cycle satisfy the above condition. There is similar regular pattern for demands rate of three and four successional stages in one production cycle except a case in Problem 2 from Stage 14 to Stage 16. For demand rate in Stage 1 is almost equivalent to that in Stage 2 in Problem 3, we consider that it does not violate the rule. The demand rate in Stage 14 in Problem 2 is much less than those in the next two stages, but the three stages are still included in one production cycle. The main reason is that the demand extent in Stage 14 is very short (only 3). Among the four problems, the maximum number of stages included in a continuous cycle are four, whereas the last three problems it occurs only once. And demand rate in the first stage is greater than the sum of those in the latter three stages in Problem 2 and 4.

By increasing the number of stages and demands rate, the total cost must increase. But the average cost may not always be so. Although the stage number in Problem 2 is twice of it in Problem 1 and the demand rate in Problem 2 is also more than that in Problem 1, the average cost of Problem 2 is less than that of Problem 1. This suggests that the best production mode (production times and volume of production) could reduce the average cost. So the total cost must cut down.

5. Conclusion

In this paper it is found that the efficient and effective production-inventory strategy could reduce unnecessary funding, inventory cost, and production cycle. Furthermore, the production-inventory problem with multistage and varying demands, in which every stage extent is different, is investigated. Moreover, we generate a hybrid integer model to optimize the production-inventory process and a GA to find a better solution. Computational results illustrate that the proposed GA yields a high-quality solution relatively fast. According to the tests of combination parameters we find the law between the average evaluation value and mutation probability under the large size.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (Grant nos. 60870008 and 61164003) and the Lanzhou Jiaotong University Young Scientific Research Fund Project (2011020).