Lecture 4: Linear Programming

Lecture 4: Linear Programming#

Note

The first module of this course focuses on developing objective function, defining decision variables, and formulating constraints for linear programming problems. We will explore associated solution methods such as the graphical approach and the simplex algorithm, along with decision-making principles for interpreting results, conducting sensitivity analysis, and understanding duality.

Definition#

Linear Programming involves a linear objective function and a set of linear constraints, i.e., the objective function and constraints are expressed as linear combination of decision variables. Thus, a general linear optimisation problem can be expressed as,

Objective:

\[ \min_{\mathbf{x}} \ f(\mathbf{x}) = c_1x_1 + c_2x_2 + ... + c_nx_n \]

Subject to:

\[\begin{split} \begin{aligned} a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n & \geq b_1 \\ a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n & \geq b_2 \\ ... \\ a_{m1}x_1 + a_{m2}x_2 + ... + a_{mn}x_n & \geq b_m \\ x_i & \geq 0 \ \forall \ i \in [1,n] \end{aligned} \end{split}\]

Here, \(f(\mathbf{x})\) is the objective function, \(x_i\) is a decision variable, \(a_{i1}x_1 + a_{i2}x_2 + ... + a_{in}x_n \geq b_i\) is a technical constraint, and \(x_i \geq 0\) is a domain constriant.

Note, an inequality constraint with \(a_{i1}x_1 + a_{i2}x_2 + ... + a_{in}x_n \leq 0\) can be re-written in the standard form as \(- a_{i1}x_1 - a_{i2}x_2 - ... - a_{in}x_n \geq 0\), while an equality constraint \(a_{i1}x_1 + a_{i2}x_2 + ... + a_{in}x_n = 0\) can be re-written in the standarad form as \(a_{i1}x_1 + a_{i2}x_2 + ... + a_{in}x_n \geq 0\); \(- a_{i1}x_1 - a_{i2}x_2 - ... - a_{in}x_n \geq 0\).

Assumptions#

  1. Proportionality - The contribution of each decision variable to the objective function and constraints is poroportional to the value of the variable.

  2. Additivity - The contribution of each decision variable to the objective function and constraints is independent of the values of other decision variables.

  3. Divisibility - The decision variables are allowed to take fractional values.

  4. Certainty - All the parameters are known with certainty.

Example #1#

Consider a textile firm operating in Kochi. This company needs to ship 100 tons of textile goods from Kanchipuram and can rent two types of trucks - \(\text{T}_1\) and \(\text{T}_2\). Each truck of type \(\text{T}_1\) can carry 10 tons of goods and costs ₹5000 per trip, while each truck of type \(\text{T}_2\) can carry 20 tons and costs ₹8000 per trip. Considering the managerial capacity of the firm (warehouse capacity, staff numbers, etc.), the management committee has imposed an upper limit on truck rental of 20 trucks in total, with a maximum of 12 any individual type of truck. Thus, all things considered, how many \(\text{T}_1\) and \(\text{T}_2\) type trucks should the company deploy so as to minimize the total costs?

Objective:

\[ \min_{x_1, x_2} \ z = 5000x_1 + 8000x_2 \]

Subject to:

\[\begin{split} \begin{aligned} 10x_1 + 20x_2 & \geq 100 \\ x_1 + x_2 & \leq 20 \\ x_1 & \leq 12 \\ x_2 & \leq 12 \\ x_1, x_2 & \in \mathbb{Z}_+ \end{aligned} \end{split}\]

Example #2#

Consider the two routes, \(\text{R}_1\) and \(\text{R}_2\), connecting Chennai Central Railway Station to Madras International Meenambakkam Airport. Here, \(\text{R}_1\) is a major arterial route with a travel time of 30 minutes and capacity of 1000 vehicles per hour. On the other hand, \(\text{R}_2\) is a collector route with a travel time of 45 minutes and capacity of 750 vehicles per hour. Assuming that Chennai Unified Metropolitan Transport Authority (CUMTA) has approporiate traffic control systems to route traffic through either of these routes, how should CUMTA allocate peak traffic of 1200 vehicles per hour to each route in order to minimise the total vehicle hours traveled (VHT)?

Objective:

\[ \min_{x_1, x_2} \ z = 30x_1 + 45x_2 \]

Subject to:

\[\begin{split} \begin{aligned} x_1 + x_2 & \geq 1200 \\ x_1 & \leq 1000 \\ x_2 & \leq 750 \\ x_1, x_2 & \in \mathbb{Z}_+ \end{aligned} \end{split}\]

Example #3#

A transportation firm in Chennai aims to manage commuter flow between Anna Nagar and Guindy by deploying two types of bus services, \(\text{B}_1\) (standard) and \(\text{B}_2\) (express), with the goal of minimising total costs. Specifically, bus service \(\text{B}_1\) is a standard 2-hour service priced at ₹50 per ticket with a capacity for 50 passengers, while bus service \(\text{B}_2\) is an express 1.5-hour service priced at ₹100 per ticket with a capacity for 30 passengers. Note, the purchase cost of the bus for standard service is ₹25,00,000 while that for the express service is ₹30,00,000. On the other hand, the operational costs amount to ₹1500 for the standard and ₹1000 for the express service. Given a planning horizon of 10 years, 330 working days in a year, and a discount rate of 3%, how many buses should the transportation firm deploy in order to minimise the total cost while ensuring a mimum service level with at least 5 standard service buses and a maximum total commuter-hours of 25. Assume ticket prices and operational costs to grow at the discount rate.

  • Revenue: \(\text{R} = (50 * 50)x_1 + (100 * 30)x_2\)

  • Operating Cost: \(\text{OC} = 1500x_1 + 1000x_2\)

  • Fixed Cost (amortised): \(\text{FC} = (2500000x_1 + 3000000x_2) \times (0.03 / (1 - 1.03^{-10})) / 330\)

    Note, for cashflow \(c_i\) in year \(i\), the net present value is given by \(c_{NPV} = \sum_{i=1}^y c_i/(1+r)^i\). However, if the cashflow is fixed every year such that \(c_i = c_o \ \forall \ i\), then, \(c_{NPV} = c_o(1 - (1+r)^{-y})/r\). Consequently, an initial investment \(c\) can be amortised as, \(c = c_A(1 - (1+r)^{-y})/r\), rendering, \(c_A = cr/(1 - (1+r)^{-y})\)

Objective: \(z = \text{R - (OC + FC)}\)

\[ \max_{x_1, x_2} \ z = 111.89x_1 + 934.27x_2 \]

Subject to:

\[\begin{split} \begin{aligned} 2x_1 + 1.5x_2 & \leq 25 \\ x_1 & \geq 5 \\ x_1, x_2 & \in \mathbb{Z}_+ \end{aligned} \end{split}\]

Note

Notice, in all the example problems we are given a defined problem (step 1 of the five step process) with relevant data (step 2) and thus we formulate the problem (step 3). In the next few lectures, we shall develop and employ appropriate optimisation techniques (step 4) before moving onwards to inferring and implementing the results (step 5).

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