Production Planning and Control: From Strategy to Scheduling
**Item 14: Planning and Production Control**
Operations Planning: Medium and Short Term
The process of planning and controlling production should follow a hierarchical approach in which vertical integration is achieved between the strategic and tactical objectives of the production system. It should also establish its horizontal relationship with other functional areas of the company. There are different hierarchical levels when making operations planning, depending on the horizon, but it is conditioned by the economic activity sector in which we operate. We can distinguish four levels:
Strategic Planning
Covers a time horizon greater than the medium. At this level, the installed capacity is determined. These become obstacles for lower levels.
Aggregate Planning
With a medium-term horizon of between 6 and 18 months, it determines the necessary labor, inventory levels, and supply contracts.
Master Production Schedule
Starting from the restrictions set forth in the aggregate plan, it determines the amount to make for each article and the period within which such production should be started.
Short-Term Production Planning and Control
We proceed to designate the planned production at each work center, controlling the proper execution of the delivery.
**Aggregate Production Planning**
Its main function is to determine the combination of the production rate, labor, and inventory levels to minimize costs and meet the predicted demand. The aggregate planning process begins with the estimation of future needs of end products from orders already placed by customers or via the demand forecast made by the marketing department, based on information obtained in market investigations. Demand is added, it does not refer to each product individually, but groups them into families of products. From this estimation, resource requirements, materials, labor, etc., are earmarked.
The company would wish to establish a stable aggregate plan with a similar production rate in all periods. However, the demand is generally not stable; it can present major ups and downs. Therefore, an aggregate plan is well prepared to adapt production to achieve the oscillations of the demand or try to reduce the intensity of these oscillations, as well as achieve efficient utilization of the company’s productive capacity.
**Master Production Schedule**
For a time horizon of several weeks, it indicates the quantity of each item we produce to meet market needs. It is a detailed plan that establishes the specific amount and exact dates of manufacture of final products to liaise between aggregate planning and very short-term planning. Its objective is to determine the production schedule for each type of product in a manner that respects the established deadlines and the existing capacity constraints, trying to exploit the installed capacity efficiently.
Most authors agree that the planning horizon can be variable depending on product type, volume of production, and lead time. This can range from hours to a few weeks and months, with weekly reviews generally.
**Production Planning in the Very Short Term**
This production plan has a planning horizon of less than 3 months and includes a series of activities concerning the planning of production schedules, assigning work to individual posts, the organization of material deliveries, and completed orders. This planning is known as workshop management and is responsible for scheduling, monitoring, and evaluating production operations in the very short term to achieve compliance with the master schedule with the available capacity and the most efficient way possible. Among its major functions are:
- Evaluation and control of manufacturing orders
- Define priorities between orders or work to be done, sorting them by work centers and assigning each one of them
- Trace the appraisement of current orders
- Control the conduct of operations
- Control the ability of each workplace
- Provide feedback to system planning and capacity control
**Linear Model of Production Scheduling**
One of the most widespread quantitative tools for short-term production programs is linear programming. Linear programming studies problems concerning the allocation of limited resources among competitive activities so that it is optimal. For linear programming, a mathematical model that describes the problem being treated is used.
The term “programming” has the sense of planning, the formulation of a plan to be carried out optimally, and the adjective “linear” indicates that all functional relations must be of a linear mathematical character. The objective of linear programming is to know which products must be obtained and in what amounts to obtain maximum yield without exceeding limited resources.
The formulation of a linear programming problem algebraically means planning the model, for which it will be necessary to determine the following issues:
- The decision variables of the model, i.e., those that indicate the decision that has been taken.
- The limitations or restrictions.
- The objective function so that we distinguish a mathematical expression of the objective we want to deploy.
Formulation of the Problem
Objective function Max (Min) Z i xi + c2 x2 + an xn
Restrictions a1 * X1 + a2 * x2 + an * xn >->-= b1
A2 * X2 + A2 * X2 + An * Xn> – <-b2
Other restrictions characteristic of the type of variables X1, X2 … …> o = ó <
Decision variables (unknowns) X1 (i = 1 … 2 …. N)
Resources available (data) b1, b2 … bn ..
Technological coefficients Aij + ej (i = 1 … 2 … n)
Usually, the solution of linear programming problems is obtained through one of the most efficient algorithms that exist in operations research, the simplex algorithm. It is an interactive search procedure for the optimal solution. The linear programming problem was developed by George Dantzig in 1947 and is one of the most known and used algorithms because of the simplicity and effectiveness of successful techniques in general. Linear programming, in particular, is due largely to the technological breakthrough of the second half of the twentieth century. Today, the resolution of linear programming problems is solved using some of the proposals also available in computer spreadsheets incorporating the routines for the treatment of linear planning problems, such as the Solver function in Microsoft Excel.
Example Company: Gepetto, SL
Each doll: Net Profit of €3, 2 hours of finishing work, 1 hour of carpentry work.
Each train: Net Profit of €2, 1 hour of finishing work, 1 hour of carpentry work.
Every week, you can have all the material you need, only 100 hours of completion, only 80 carpentry hours. Also, the demand for trains can be anything without limit. The demand for dolls is at most 40. Gepetto wants to maximize profits. How many dolls and trains should he make?
X = number of dolls produced a week.
Y = number of trains produced each week.
Objective function: To maximize profit and minimize cost.
Inequality Constraints: These limit the possible values of decision variables. In this exercise, the restrictions are given by the time of finishing, carpentry, and demand for dolls.
Signs or non-negativity: X = greater than or equal to or Y = greater than or equal to or.
The objective is for Gepetto to choose values of X and Y to maximize 3x + 2y. The variable Z will be used to demonstrate the value of the objective function. The objective function is Gepetto MAXZ = 3x + 2y.
Restrictions: When X and Y increase, the objective function increases, but not indefinitely, because Gepetto’s values for X and Y are limited by the following restrictions:
- Constraint 1: No more than 100 hours of finishing can be used.
- Constraint 2: No more than 80 hours of carpentry can be used.
- Constraint 3: No more than 40 dolls should be manufactured.
These three constraints can be expressed with the following inequalities:
- Constraint 1: 2x + y greater than or equal to 100
- Constraint 2: x + y less than or equal to 80
- Restriction 3: x less than or equal to 40
We also have the sign X equal to or greater than or equal to or less than or
TRAIN DOLL
Benefit 3 2
FINISH 2 1 < 100
CARPENTRY 1 1 < 80
DEMAND < 40
Max2 = 3x + 2y (objective function)
2x + y < 100 finish
X + y < 80 woodwork
X <= 40 (demand dolls)
X> o = y = 20
Graphical Solution of a Linear Programming Model
To solve graphically, we must first identify the feasible region (the set of all points that meet all restrictions). This is the region specified by the system of inequalities that form the constraints.
In an optimal solution to a maximization problem, the optimal solution is a point in the feasible region in which the objective function has a maximum value. For a minimization problem, it is where the objective function has a minimum value. Most linear programming problems have only one optimal solution, but some problems have no optimal solution, and some linear programming problems have an infinite number of solutions. One can show that the optimal solution is always at the frontier of the feasible region at a vertex if the solution is unique or a continuous segment between two vertices if there are infinite solutions.