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Chapter 2
Introduction to modeling: Linear
Programming
Our approach to address and solve logistics decision problems is the following:
- to formulate a suitable mathematical model to represent the decision problem;
- to implement and solve the model.
See Figure 2.1 for a graphical representation of the considered problem solving approach. Precisely, we move within Management Science, and specifically Operations Research, field of study which uses computer science, mathematics and statistics to solve decision problems.
Today, electronic spreadsheets provide a simple and useful way for business people to implement a model and analyze decision alternatives, although many other, more sophisticated and powerful solvers exist: spreadsheet models (i.e. models implemented via a spreadsheet) will be used hereafter.
Identify problem
Model formulation
Implementation & Analysis
Test Results
Implement Solution
Unsatisfactory Results
Figure 2.1: The problem solving approach
14 2.1. Optimization problems
2.1 Optimization problems
The addressed problems consist in deciding how to use the limited resources available in an efficient way. Typically, this must be accomplished by maximizing profits or minimizing costs: i.e. the addressed problems are optimization problems.
Mathematical Programming (MP) is the area of Operations Research aiming at mod- eling and solving optimization problems. Its applications include, among the others, manufacturing, financial planning and logistics.
In any case, an optimization problem involves:
- decisions to be taken (e.g. how much of each product should be produced, shipped, etc.);
- restrictions, or constraints, to be placed on the alternatives available to the deci- sion maker (e.g. limited amount of raw materials and labor when producing);
- the goal, or objective, to be considered by the decision maker when deciding (e.g. to choose the mix of products that maximizes profits).
2.1.1 Expressing optimization problems via mathematical models
How can we mathematically represent decisions, constraints and objective?
- decisions are represented by decision variables: x 1 , x 2 ,... x (^) n ;
- constraints are formulated by (in)equalities:
f (x 1 , x 2 ,... x (^) n ) b or f (x 1 , x 2 ,... x (^) n ) b or f (x 1 , x 2 ,... x (^) n ) = b
- the objective is modelled by an objective function to be maximized or minimized: max f (x 1 , x 2 ,... x (^) n ) or min f (x 1 , x 2 ,... x (^) n ).
Therefore, the general mathematical model for an optimization problem is:
max 8 / min f 0 (x 1 , x 2 ,... x (^) n ) subject to
<
:
f 1 (x 1 , x 2 ,... x (^) n ) b (^1) .. . f (^) k (x 1 , x 2 ,... x (^) n ) b (^) k .. . f (^) m (x 1 , x 2 ,... x (^) n ) = b (^) m
16 2.2. Linear Programming
resources and the operating requisites, so as to maximize the profit during the next production cycle?
In order to express the problem via an LP model, we follow these steps:
- identify the decision variables: what are the fundamental decisions that must be made to solve the problem? - x 1 : number of Aqua-Spa hot tubs to produce; - x 2 : number of Hydro-Lux hot tubs to produce;
- state the constraints as linear combinations of the decision variables:
- only 200 pumps are available, and each hot tub requires one pump:
x 1 + x 2 200;
- only 1,566 labor hours are available, and each Aqua-Spa requires 9 labor hours while each Hydro-Lux requires 6 labor hours:
9 x 1 + 6x 2 1 ,566;
- only 288 feet of tubing is available, and each Aqua-Spa requires 12 feet while each Hydro-Lux requires 16 feet:
12 x 1 + 16x 2 2 ,880;
- state the objective function as a linear combination of the decision variables: the manager has a profit of 350 on each Acqua-Spa he/she sells, and of 300 on each Hydro-Lux he/she sells; therefore the total profit, to be maximized, is
max 350x 1 + 300x 2 ;
- nonnegativity constraints: we cannot produce a negative number of hot tubs:
x 1 0 , x 2 0.
The overall LP model is therefore:
max 350x 1 + 300x (^2) x 1 + x 2 200 9 x 1 + 6 x 2 1 , 566 12 x 1 + 16 x 2 2 , 880 x 1 0 x 2 0
Chapter 2. Introduction to modeling: Linear Programming 17
2.2.2 Solving LP problems: an intuitive approach
How can we compute the best (i.e. optimal) solution? In the case of only two decision variables, we can use a graphical approach (see G. Bigi, A. Frangioni, G. Gallo, and M. Scutellà ((2014)) for more rigorous mathematical approaches, based on the LP Duality Theory):
- Plot the constraints and identify the LP feasible region; this is done by plotting the “boundary lines” of the constraints and identifying the points (x 1 , x 2 ) which satisfy all the constraints.
For example, the boundary of the first constraint in the Blue Ridge Hot Tubs model is the straight line defined by x 1 + x 2 = 200:
x (^1)
x (^2)
(0, 200) (^) x 1 + x 2 = 200
The shaded area is the region satisfying the first constraint (x 1 + x 2 200 ) and x 1 0 , x 2 0. However, feasible points must also satisfy the other constraints. So, plot the boundary line for the second constraint, i.e. 9 x 1 + 6x 2 = 1, 566 :
x (^1)
x (^2)
(0, 261) (^9) x 1 + 6x 2 = 1, 566
Since we have to satisfy both constraints:
Chapter 2. Introduction to modeling: Linear Programming 19
2.2.3 Other possible outcomes in solving LPs
Other possible outcomes in LP solving are:
- Alternate optimal solutions; e.g. if we want to maximize x 1 (the dashed line in the figure represents a corresponding level curve) over the feasible region depicted below:
x (^1)
x (^2)
alternate optimal solutions
- Unbounded optimal solutions; e.g. if we want to maximize x 1 +x 2 over the feasible region depicted below:
x (^1)
x (^2)
In this case, the objective function value can be made infinitely large (for a max- imization problem). In practice, this usually indicates that there is something wrong with the problem formulation.
- Infeasibility; for example:
max x 1 + x (^2) x 1 + x 2 150 x 1 + x 2 200 x 1 0 x 2 0
This may be due to an error in the problem formulation; some constraints have to be eliminated or loosened to get feasible solutions.
Finally, observe that redundant constraints may be present. A constraint is redundant if it has no role in determining the feasible region, therefore we can remove it from the model.
20 2.3. Solving LP models via spreadsheets
x (^1)
x (^2) redundant
References C. Ragsdale (2004): Chapter 2
2.3 Solving LP models via spreadsheets
Solving LP problems is an easy task by using spreadsheet packages, for example Excel: just formulate the problem correctly and implement it in an accurate way. The same concepts and techniques apply to spreadsheet packages other than Excel. Furthermore, specialized, more powerful optimization packages do exist: Cplex, Lindo, Coin...
In order to implement the Blue Ridge Hot Tubs model as a spreadsheet model, we refer the interested reader to C. Ragsdale (2004): Sections 3.3 to 3.6.
References C. Ragsdale (2004): Chapter 3
2.3.1 A production and inventory planning problem
“Upton Corporation” produces air compressors. The manager wants to plan its pro- duction and inventory levels for the next 6 months. Table 2.1 summarizes the monthly production costs, demands and production capacities that are expected over the next 6 months.
1 2 3 4 5 6 unit production cost 240 250 265 285 280 260 units demanded 1,000 4,500 6,000 5,500 3,500 4, maximum production 4,000 3,500 4,000 4,500 4,000 3,
Table 2.1: Production costs, demands and capacities in the next six months for Upton Corporation
The following constraints must be satisfied:
- size of the warehouse: a maximum of 6,000 units can be held in inventory at the end of each month;
- safety stock: at least 1,500 units must be held in inventory at the end of each month, to meet unexpected demand;
22 2.3. Solving LP models via spreadsheets
- the inventory levels at the end of each month must lay between 1,500 and 6,000:
1 , 500 b 2 6 , 000 1 , 500 b 3 6 , 000 1 , 500 b 4 6 , 000 1 , 500 b 5 6 , 000 1 , 500 b 6 6 , 000 1 , 500 b 7 6 ,000;
- the relationship among production variables, inventory variables and monthly demands can be expressed as follows:
b 2 = b 1 + p 1 1 , 000 b 3 = b 2 + p 2 4 , 500 b 4 = b 3 + p 3 6 , 000 b 5 = b 4 + p 4 5 , 500 b 6 = b 5 + p 5 3 , 500 b 7 = b 6 + p 6 4 ,000;
- the initial inventory level is 2,750:
b 1 = 2, 750.
The goal is to minimize the total cost. There are two kinds of cost: production costs and inventory costs. The total production cost is obtained simply by multiplying the unit production costs by the number of units produced:
240 p 1 + 250p 2 + 265p 3 + 285p 4 + 280p 5 + 260p 6.
Inventory costs in each month are estimated as the 1.5 % of the unit production cost multiplied by the average number of units in inventory in that month:
b 1 + b (^2) 2
b 2 + b (^3) 2
b 3 + b (^4) 2
b 4 + b (^5) 2
b 5 + b (^6) 2
b 6 + b (^7) 2
Putting all together, the overall LP model is the one presented in Model 2.1. In order to implement the Upton Corporation model as a spreadsheet model, we refer the interested reader to C. Ragsdale (2004): Section 3.12.
Chapter 2. Introduction to modeling: Linear Programming 23
min
production cost z}|{ 240 p 1 + 250p 2 + 265p 3 + 285p 4 + 280p 5 + 260p 6 +
- 3.6(b 1 + b 2 )/2 + 3.75(b 2 + b 3 )/2 + 3.98(b 3 + b 4 )/2 +
- 4.28(b 4 + b 5 )/2 + 4.20(b 5 + b 6 )/2 + 3|{z}. 9 inventory cost (1.5 % of 260)
(b 6 + b 7 )/ 2
2 , 000 p 1 4 , 000 1 , 750 p 2 3 , 500 constraints on 2 , 000 p 3 4 , 000 monthly production 2 , 250 p 4 4 , 500 levels 2 , 000 p 5 4 , 000 1 , 750 p 6 3 , 500 1 , 500 b 2 6 , 000 1 , 500 b 3 6 , 000 constraints on 1 , 500 b 4 6 , 000 monthly final 1 , 500 b 5 6 , 000 inventory levels 1 , 500 b 6 6 , 000 1 , 500 b 7 6 , 000 b 2 = b 1 + p 1 1 , 000 b 3 = b 2 + p 2 4 , 500 relationship b 4 = b 3 + p 3 6 , 000 between initial b 5 = b 4 + p 4 5 , 500 and final inventory b 6 = b 5 + p 5 3 , 500 levels + demand b 7 = b 6 + p 6 4 , 000 satisfaction b 1 = 2, 750
Model 2.1: The Upton Corporation problem
