Integrated versus hierarchical approach to aggregate production planning and master production scheduling

Abstract

The hierarchical planning concept is commonly used for production planning. Dividing the planning process into subprocesses which are solved separately in the order of the hierarchy decreases the complexity and fits the common organizational structure. However, interaction between planning levels is crucial to avoid infeasibility and inconsistency of plans. Furthermore, optimizing subproblems often leads to suboptimal results for the overall problem. The alternative, a monolithic model integrating all planning levels, has been rejected in the literature because of several reasons. In this study, we show that some of them do not hold for an integrated production planning model combining the planning tasks usually attributed to aggregate production planning and master production scheduling. Therefore, we develop a hierarchical and an integrated model considering both levels, aggregate production planning and master production scheduling. Computational tests show that it is possible to solve the integrated model and that it outperforms the hierarchical approach for all instances. Moreover, an indication is given why and when integration is beneficial.

Appendices

Appendix 1: Relationships among parameters

For our test instances, we assumed the following relationships among the parameters.

Time related parameters:

• The (expected) number of setups for product family j in $T^\mathrm{MPS}$ periods. Note that this approximation only holds for small TBO. For instance, it is very unlikely that for $T^\mathrm{MPS}=12$ and $\mathrm{TBO}=8,$ one or two setups are sufficient. This will be considered in (46).

  $$\begin{aligned} T^\mathrm{MPS}/\mathrm{TBO}_j \end{aligned}$$

  (42)

• The total capacity in $T^\mathrm{MPS}$ periods: capacity per week, summed over all weeks, multiplied by the maximum number of production lines and parameter $M^1$ stating the percentage usage of these lines. Note that a target utilization of 80 % was aspired. Therefore, the capacity $\mathrm{Cap}_{t_w}$ was multiplied by 0.8.

  $$\begin{aligned} \mathrm{TotalCap}^\mathrm{MPS} = \sum _{t_w=1}^{T^\mathrm{MPS}} 0.8\cdot \mathrm {Cap}_{t_w} \cdot \sum _n L_n^{\mathrm{max}} \cdot M^1. \end{aligned}$$

  (43)

• Total setup time S for all products: $M^2$ indicates the proportion of the total capacity which is used for setup activities.

  $$\begin{aligned} S = M^2 \cdot \mathrm{TotalCap}^\mathrm{MPS}. \end{aligned}$$

  (44)

• Disaggregate the total setup time into setup times for every product family j according to $M^3_j$ stating the proportion of j. Divide by the maximum number of setups in $T^\mathrm{MPS}$ periods. Moreover, the setup times should be limited such that $\sum _j s_j \le \mathrm{TotalCap}^\mathrm{MPS}/T^\mathrm{MPS}$ holds. Hence, we take the minimum value.

  $$\begin{aligned} s_j'&= S \cdot M_j^3 \cdot \left( \frac{T^\mathrm{MPS}}{\mathrm{TBO}_j}\right) ^{-1} = \frac{S \cdot M_j^3 \cdot \mathrm{TBO}_j}{T^\mathrm{MPS}}\end{aligned}$$

  (45)
  $$\begin{aligned} s_j&= \min \left\{ s_j', \frac{\mathrm{TotalCap}^\mathrm{MPS}\cdot M_j^3}{T^\mathrm{MPS}} \right\} . \end{aligned}$$

  (46)

• Production time product type k: The equality is just an estimation. We multiply the average production time of that product type with a buffer for setup operations. For a more accurate production time $a_k,$ the MPS could be solved in advance.

  $$\begin{aligned} a_k = \frac{\sum _{j \in \Gamma (k)} a_j^\mathrm{MPS}}{|\Gamma (k)|} \cdot (1-M^2)^{-1}. \end{aligned}$$

  (47)

• Maximum production amount for all production lines (without overtime) in $T^\mathrm{MPS}$ periods: the total capacity is reduced by setup operations. The remaining time can be used for production. Dividing by the average production time provides the maximum production amount.

  $$\begin{aligned} \mathrm{MaxProd} = \frac{(1-M^2) \cdot \mathrm{TotalCap}^\mathrm{MPS}}{\sum _j a_j^\mathrm{MPS}/J}. \end{aligned}$$

  (48)

• The maximum production amounts for product family j are calculated by disaggregating MaxProd. $\mathrm{MaxProd}_j$ is used to determine the demand (see Sect. 6.2).

  $$\begin{aligned} \mathrm{MaxProd}_j = \mathrm{MaxProd} \cdot M_j^3. \end{aligned}$$

  (49)

• Maximum overtime:

  $$\begin{aligned} O^{\mathrm{max}} = M^4 \cdot \mathrm{Cap}_{t_w}. \end{aligned}$$

  (50)

Cost-related parameters:

We can distinguish between variable and fixed cost in terms of the production amount. Fixed costs are $C_n^L$ and $C^O$, whereas $C_n^P$, $C^U$, $C^B$, and $C_k^I$ and $C_j^I$ are variable and depend on the production amount. Setup costs $C_j^S$ do not depend directly on the production amount, but the calculation is typically based on the inventory holding cost $C_j^I$, $\mathrm{TBO}_j$, and demand. We have to consider these two kinds of cost when we create the relationships between several costs. For one cost parameter, the values have to be set in advance. We chose the production cost $C_n^P$.

• Cost (per month) for running one production line of type n. We multiply the maximum production amount with production cost for one unit. How similar/close these cost are defines $M^5$. A value close to 1 might be reasonable.

  $$\begin{aligned} C_n^L = M^5 \cdot \frac{\mathrm{Cap}_{t_w}}{\sum _j a_j^\mathrm{MPS}/J} \cdot C_n^P. \end{aligned}$$

  (51)

• Overtime cost (per time unit): calculate the average cost for running a production line 1 h by dividing by the monthly capacity (in time units). The results are cost per time unit. $M^6$ represent extra cost which have to be paid additionally to the regular cost. Variable cost, e.g., energy cost, are already considered in $C_n^P$, so $M^6$ has only to define a percentage for the overtime premium of workers.

  $$\begin{aligned} C^O = M^6 \cdot \frac{\sum _n C_n^L/N}{\mathrm{Cap}_{t_m}}. \end{aligned}$$

  (52)

• Cost for unmet demand: we want to avoid losing sales, so we have to assign a sufficiently high value. We chose the average cost for running a production line for one month.

  $$\begin{aligned} C^U = \sum _n C_n^L/N \end{aligned}$$

  (53)

• Cost for backorders depend on the production cost for one unit. The relationship is given by factor $M^7$.

  $$\begin{aligned} C^B = M^7 \cdot \sum _n C_n^P/N. \end{aligned}$$

  (54)

• Inventory holding cost for one unit of family $j \in \phi (n)$ is related to the average production cost for one unit on machine n.

  $$\begin{aligned} C_j^I = M^8 \cdot \sum _n C_n^P/N. \end{aligned}$$

  (55)

• Inventory holding cost for product type k is similar (or equal) to the average cost for product families j multiplied with number of weeks per month (4):

  $$\begin{aligned} C_k^I = \frac{\sum _{j \in \Gamma (k)} C_j^I}{|\Gamma (k)|}\cdot 4. \end{aligned}$$

  (56)

• Setup cost for family j is determined as usual:

  $$\begin{aligned} C^S_j = \frac{C_j^I\cdot 1/T^\mathrm{MPS} \sum _{t_w} D_{jt_w} \mathrm{TBO}_j^2}{2}. \end{aligned}$$

  (57)

• Cost for opening/closing a production line of type n depends on the cost for running the line for one period $t_m$:

  $$\begin{aligned} C_n^{L+} \cdot M^{9} = C_n^{L} = C_n^{L-} \cdot M^{10}. \end{aligned}$$

  (58)

We propose the following values for $M^r$, $r=1,\ldots ,10$:

• $M^1=0.75$, i.e. 75 % of all machines are open on average.

• $M^2 \in \{0.1,0.3\}$, choice depends on instances.

• $M_j^3 = 1/J$, $\forall j$, i.e. Products have identical setup times.

• $M^4 = 0.5$. Assume 6 working days with two shifts. Allow one additional shift every day.

• $M^5=1$, i.e. running a production line equals cost for producing the maximum amount of one month on that line.

• $M^6=0.5$, i.e. we have to pay 50 % more, compared to regular production time, when we use overtime.

• $M^7=1.25$, i.e. backorder cost are 25 % higher than the production cost.

• $M^8=0.1$, i.e. production cost are ten times higher than inventory holding cost.

• $M^{9}=1.5$, i.e. opening a production line leads to costs 1.5 times the cost for running the line. Cost drivers: new workers have to be hired, lines have to installed.

• $M^{10}=0.5$, i.e. shutting down a production line costs 50  % of the costs for running that line for one month. Cost drivers: layoff cost, cleansing cost.

Remark 8.1

To avoid producing all products in period $t_w=1,$ we set the demand $D_{j1}=0$, $\forall j=1,\ldots ,J$. That means, initial production can take place in period $t_w=1$ or $t_w=2$.

Appendix 2: Including forecast errors

We follow the assumptions of Clark (2005) in his experimental design:

• Forecast accuracy increases by decreasing distance (called lead-time) to the forecast period.

• How the forecast is made is not relevant.

The forecast demand values are calculated as follows (for simplification, the explanation is product independent).

• First, we have to determine/set the true demand $V_0$.

• Then, we calculate the so called base value depending on the planning horizon length T: $V_T = \max \{0,V_0(1+T \alpha r)\}$, with $\alpha \ge 0$ as a degree of error and r as standardized normally distributed random variable.

• To ensure the convergence to the true demand $V_0$, we determine values $V_t$, $t=T,T-1,\ldots ,1,0$, by interpolation: $V_t = V_0 + (t/T)(V_T-V_0)$.

• Based on that values we calculate the forecast demands $F_t$ for every period t by using a formula similar to the first one: $F_t = \max \{0,V_t(1+t \alpha r)\}$, $t=T,T-1,\ldots ,1,0$.

In that way, we can test different degrees of forecast accuracy as Clark (2005) did. However, note that $\alpha$ does not represent the realized forecast error. Some computational tests have shown that we get a good estimation of the average forecast error by dividing the target error by ten. For instance, for a forecast which differs on average 30 % from the true demand, $\alpha$ equals 0.03. Naturally because of the random number r, we might get forecasts over and under the average.

Appendix 3: Result tables

See Table 18.

Table 18 Results for one iteration in case of low forecast error