Linear Programming

287. Linear Programming

287.1. Mathematical Programs

287.1.1. Formulation

$\begin{aligned} min & 𝑓 (𝑥_{1}, 𝑥_{2}, \dots, 𝑥_{𝑛}) & (objective function) \\ 𝑠 . 𝑡 . & 𝑔_{𝑖} (𝑥_{1}, 𝑥_{2}, \dots, 𝑥_{𝑛}) \leq 𝑏_{𝑖} & \forall 𝑖 \in 1, \dots, 𝑚 & (constraints) \\ 𝑥_{𝑗} \in ℝ & \forall 𝑗 \in 1, \dots, 𝑛 & (decision variables) \end{aligned}$

$𝑚$ : Number of constraints
$𝑛$ : Number of variables
$𝑥_{1}, 𝑥_{2}, \dots, 𝑥_{𝑛}$ : decision variables
$𝑓 : ℝ^{𝑛} \to ℝ$
$𝑔_{𝑖} : ℝ^{𝑛} \to ℝ$

\begin{aligned} min & \sum_{𝑗 = 1}^{𝑛} 𝑐_{𝑗} 𝑥_{𝑗} \\ 𝑠 . 𝑡 . & \sum_{𝑗 = 1}^{𝑛} 𝐴_{𝑖 𝑗} 𝑥_{𝑗} \leq 𝑏_{𝑖} \forall 𝑖 = 1, \dots, 𝑚 \end{aligned}

$𝐴_{𝑖 𝑗}$ : constraint coefficients
$𝑏_{𝑖}$ : right-hand-side values (RHS)
$𝑐_{𝑗}$ : objective coefficients

Vector

\begin{aligned} min & 𝑐^{𝑇} 𝑥 \\ 𝑠 . 𝑡 . & 𝑎_{𝑖}^{𝑇} 𝑥 \leq 𝑏_{𝑖} \forall 𝑖 = 1, \dots, 𝑚 \end{aligned}

$𝑎_{𝑖} \in ℝ^{𝑛}$
$𝑏_{𝑖} \in ℝ$
$𝑐 \in ℝ^{𝑛}$
$𝑥 \in ℝ^{𝑛}$

Matrix

\begin{aligned} min & 𝑐^{𝑇} 𝑥 \\ 𝑠 . 𝑡 . & 𝐴 𝑥 \leq 𝑏 \end{aligned}

$𝐴 \in 𝑅^{𝑚 \times 𝑛}$
$𝑏 \in ℝ^{𝑚}$

𝐴 = [\begin{matrix} 𝑎_{11} & 𝑎_{12} & \dots & 𝑎_{1 𝑛} \\ 𝑎_{21} & 𝑎_{22} & \dots & 𝑎_{2 𝑛} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 𝑎_{𝑚 1} & 𝑎_{𝑚 2} & \dots & 𝑎_{𝑚 𝑛} \end{matrix}]

𝑥 = [\begin{matrix} 𝑥_{1} \\ 𝑥_{2} \\ ⋮ \\ 𝑥_{𝑛} \end{matrix}] 𝑐 = [\begin{matrix} 𝑐_{1} \\ 𝑐_{2} \\ ⋮ \\ 𝑐_{𝑛} \end{matrix}] 𝑏 = [\begin{matrix} 𝑏_{1} \\ 𝑏_{2} \\ ⋮ \\ 𝑏_{𝑛} \end{matrix}]

𝐴 𝑥 = [\begin{matrix} 𝑎_{11} & 𝑎_{12} & \dots & 𝑎_{1 𝑛} \\ 𝑎_{21} & 𝑎_{22} & \dots & 𝑎_{2 𝑛} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 𝑎_{𝑚 1} & 𝑎_{𝑚 2} & \dots & 𝑎_{𝑚 𝑛} \end{matrix}] [\begin{matrix} 𝑥_{1} \\ 𝑥_{2} \\ ⋮ \\ 𝑥_{𝑛} \end{matrix}] = [\begin{matrix} 𝑎_{11} 𝑥_{1} & 𝑎_{12} 𝑥_{2} & \dots & 𝑎_{1 𝑛} 𝑥_{𝑛} \\ 𝑎_{21} 𝑥_{1} & 𝑎_{22} 𝑥_{2} & \dots & 𝑎_{2 𝑛} 𝑥_{𝑛} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 𝑎_{𝑚 1} 𝑥_{1} & 𝑎_{𝑚 2} 𝑥_{2} & \dots & 𝑎_{𝑚 𝑛} 𝑥_{𝑛} \end{matrix}] \leq [\begin{matrix} 𝑏_{1} \\ 𝑏_{2} \\ ⋮ \\ 𝑏_{𝑛} \end{matrix}]

𝑐^{𝑇} 𝑥 = [𝑐_{1}, 𝑐_{2}, \dots, 𝑐_{𝑛}] [\begin{matrix} 𝑥_{1} \\ 𝑥_{2} \\ ⋮ \\ 𝑥_{𝑛} \end{matrix}] = \sum_{𝑗 = 1}^{𝑛} 𝑐_{𝑗} 𝑥_{𝑗}

287.1.2. Transformation

Min and Max

max 𝑓 (𝑥) \Leftrightarrow min - 𝑓 (𝑥)

$=$ and $\geq$

𝑔_{𝑖} (𝑥) \geq 𝑏_{𝑖} \Leftrightarrow - 𝑔_{𝑖} (𝑥) \leq - 𝑏_{𝑖}

𝑔_{𝑖} (𝑥) = 𝑏_{𝑖} \Leftrightarrow 𝑔_{𝑖} (𝑥) \geq 𝑏_{𝑖} \land 𝑔_{𝑖} (𝑥) \leq 𝑏_{𝑖}

E.g.,

\begin{aligned} max & 𝑥_{1} & - & 𝑥_{2} \\ 𝑠 . 𝑡 . & - 2 𝑥_{1} & + & 𝑥_{2} & \geq & - 3 \\ 𝑥_{1} & + & 4 𝑥_{2} & = & 5 \end{aligned}

\begin{matrix} \Leftrightarrow \end{matrix}

\begin{aligned} min & - 𝑥_{1} & + & 𝑥_{2} \\ 𝑠 . 𝑡 . & 2 𝑥_{1} & - & 𝑥_{2} & \leq & 3 \\ 𝑥_{1} & + & 4 𝑥_{2} & \leq & 5 \\ - 𝑥_{1} & + & 4 𝑥_{2} & \leq & - 5 \end{aligned}

287.1.3. Constraints

Sign Constraints

𝑥_{𝑖} \geq 0 or 𝑥_{𝑖} \leq 0

Functional Constraints

\sum_{𝑖 = 1}^{𝑛} 𝑐_{𝑖} 𝑥_{𝑖} \leq 𝑏_{𝑖}

287.1.4. Feasible Solutions

Feasible solution: Satisfies all constants
Infeasable solution: violates at least one constraint

287.1.5. Feasible Region and Optimal Solution

Feasible Region: Set of feasible solutions
Optimal Solution: Attains largest (maximization) or smallest (minimization) objective value

287.1.6. Binding Constraint

Let $𝑔 (\cdot) \leq 𝑏$ be an inequality constraint and $| 𝑥 |$ be a solution. $𝑔 (\cdot) \leq 𝑏$ is binding at $| 𝑥 |$ if $𝑔 (| 𝑥 |) = 𝑏$

Example

\begin{aligned} 2 𝑥_{1} & + 𝑥_{2} \leq 100 is 𝐛𝐢𝐧𝐝𝐢𝐧𝐠 (𝐚𝐜𝐭𝐢𝐯𝐞) at the point & (𝑥_{1}, 𝑥_{2}) = (30, 40) \\ 2 𝑥_{1} & + 𝑥_{2} \leq 100 is 𝐧𝐨𝐧-𝐛𝐢𝐧𝐝𝐢𝐧𝐠 (𝐢𝐧𝐚𝐜𝐭𝐢𝐯𝐞) at the point & (𝑥_{1}, 𝑥_{2}) = (20, 20) \end{aligned}

287.1.7. Strict and Weak Inequalities

Strict: The two sides cannot be equal (e.g., $𝑥_{1} + 𝑥_{2} > 5$ )
Weak: The two sides can be equal (e.g., $𝑥_{1} + 𝑥_{2} \geq 5$ )

287.1.8. Graphical Approach

Example

\begin{aligned} max & 40 & 𝑥_{1} & + & 30 & 𝑥_{2} \\ 𝑠 . 𝑡 . & 2 & 𝑥_{1} & + & 𝑥_{2} & \leq & 100 \\ 𝑥_{1} & + & 2 & 𝑥_{2} & \leq & 80 \end{aligned}

287.1.9. Types of LP

287.1.10. LP Formulation

Example

Production

We produce desks and tables

Producing a desk requires 3 units of wood, 1 hour of labor and 50 minutes of machine time
Producing a table requires 5 units of wood, 2 hours of labor and 20 minutes of machine time

For each day, we have

200 workers that each work 8 hours
50 machines that each run for 16 hours
A supply of 3600 units of wood

Desks and tables are sold $700 and $900 per unit, respectively

Everything that is produced is sold

1. Define Variables

$𝑥_{1}$ : number of desks produced per day

$𝑥_{2}$ : number of tables produced per day

2. Formulate Objective Function

max 700 𝑥_{1} + 900 𝑥_{2}

3. Formulate Constraints

Resource	Consumption per		Total supply
Resource	Desk	Table	Total supply
Wood	3 units	5 units	3600 units
Labor	1 hours	2 hours	200 workers $\times$ 8 hours = 1600 hours
Machine time	50 minutes	20 minutes	50 machines $\times$ 16 hours = 800 hours

\begin{aligned} 3 𝑥_{1} & + & 5 𝑥_{2} & \leq & 3600 \\ 𝑥_{1} & + & 2 𝑥_{2} & \leq & 1600 \\ 50 𝑥_{1} & + & 20 𝑥_{2} & \leq & 48000 \end{aligned}

4. Complete Formulation

\begin{aligned} max & 700 & 𝑥_{1} & + & 900 & 𝑥_{2} \\ 𝑠 . 𝑡 . & 3 & 𝑥_{1} & + & 5 & 𝑥_{2} & \leq & 3600 & (wood) \\ 𝑥_{1} & + & 2 & 𝑥_{2} & \leq & 1600 & (labor) \\ 50 & 𝑥_{1} & + & 20 & 𝑥_{2} & \leq & 48000 & (machine) \\ 𝑥_{1} & \geq & 0 \\ 𝑥_{2} & \geq & 0 \end{aligned}

Compact Formulation

\begin{aligned} max & \sum_{𝑗 = 1}^{𝑛} 𝑃_{𝑗} 𝑥_{𝑗} \\ 𝑠 . 𝑡 . & \sum_{𝑗 = 1}^{𝑛} 𝐴_{𝑖 𝑗} 𝑥_{𝑗} \leq 𝑅_{𝑖} \forall 𝑖 = 1, \dots, 𝑚 \\ 𝑥_{𝑗} \geq 0 \forall 𝑗 = 1, \dots, 𝑛 \end{aligned}

Where:

$𝑛$ : number of products
$𝑚$ : number of resources
$𝑗$ : indices of products
$𝑖$ : indices of resources
$𝑃_{𝑗}$ : unit sale price of product $𝑗$
$𝑅_{𝑖}$ : supply limit of resource $𝑖$
$𝐴_{𝑖 𝑗}$ : unit of resource $𝑖$ to produce one unit of product $𝑗$
$𝑖 = 1, \dots, 𝑚$
$𝑗 = 1, \dots, 𝑛$
$𝑥_{𝑗}$ : production quantity for product $𝑗$ , $𝑗 = 1, \dots, 𝑛$

model.py

from pyomo.environ import *
from pyomo.dataportal import DataPortal

model = AbstractModel()

# Sets
model.Products = Set()
model.Resources = Set()

# Parameters
model.Profit = Param(model.Products)
model.Supply = Param(model.Resources)
model.Consumption = Param(model.Resources, model.Products)

# Variables
model.x = Var(model.Products, domain=NonNegativeReals)

# Objective: Maximize profit
def objective_rule(model):
    return sum(model.Profit[j] * model.x[j] for j in model.Products)
model.OBJ = Objective(rule=objective_rule, sense=maximize)

# Constraints: Do not exceed resource supply
def constraint_rule(model, i):
    return sum(model.Consumption[i, j] * model.x[j] for j in model.Products) <= model.Supply[i]
model.ResourceConstraint = Constraint(model.Resources, rule=constraint_rule)

# Load data from .dat file
data = DataPortal()
data.load(filename='data.dat', model=model)

# Create an instance of the model
instance = model.create_instance(data)

# Create solver
solver = SolverFactory('glpk')
solver.options['tmlim'] = 60

# Solve with solver timeout (optional)
results = solver.solve(instance, tee=True)

# Display results
instance.display()

data.dat

set Products := Desk Table ;
set Resources := Wood Labor Machine ;

param Profit :=
Desk   700
Table  900 ;

param Supply :=
Wood    3600
Labor   1600
Machine 48000 ;

param Consumption:
          Desk Table :=
Wood        3     5
Labor       1     2
Machine    50    20 ;

Output:

  Variables:
  x : Size=2, Index=Products
      Key   : Lower : Value            : Upper : Fixed : Stale : Domain
      Desk :     0 : 884.210526315789 :  None : False : False : NonNegativeReals
      Table :     0 : 189.473684210526 :  None : False : False : NonNegativeReals

Objectives:
  OBJ : Size=1, Index=None, Active=True
      Key  : Active : Value
      None :   True : 789473.6842105257

Constraints:
  ResourceConstraint : Size=3
      Key     : Lower : Body              : Upper
        Labor :  None : 1263.157894736841 :  1600.0
      Machine :  None : 47999.99999999997 : 48000.0
        Wood :  None : 3599.999999999997 :  3600.0

)

Example

Multi Period (Inventory)

We produce and sell products

For the coming 4 days, the demand will be

Days 1: 100
Days 2: 150
Days 3: 200
Days 4: 170

The unit production costs are different for different days

Days 1: $9
Days 2: $12
Days 3: $10
Days 4: $12

Prices are all fixed. So maximizing profit is equivalent to minimizing costs.

We may store products and sell them later. Inventory cost is $1 per unit per day.

E.g., producing 620 units on day 1 to fulfill all demands:

$ 9 \times 620 + $ 1 \times 150 + $ 2 \times 200 + $ 3 \times 170 = $ 6, 640

𝐼_{𝑡 - 1} + 𝑥_{𝑡} - 𝑑_{𝑡} = 𝐼_{𝑡}

$𝐼_{𝑡 - 1}$ : inevntory at the end of day $𝑡 - 1$ (i.e., beginning inventory for day $𝑡$ )
$𝑥_{𝑡}$ : units produced on day $𝑡$
$𝑑_{𝑡}$ : demand (or sales) on dat $𝑡$
$𝐼_{𝑡}$ : inventory at the end of day $𝑡$

Inventory costs are calculated according to ending inventory

$𝑥_{𝑡}$ : production quantity of day $𝑡$ , $𝑡 = 1, \dots, 4$
$𝑦_{𝑡}$ : ending inventory of day $𝑡$ , $𝑡 = 1, \dots, 4$

Objective function:

min 9 𝑥_{1} + 12 𝑥_{2} + 10 𝑥_{3} + 12 𝑥_{4} + 𝑦_{1} + 𝑦_{2} + 𝑦_{3} + 𝑦_{4}

Inventory balancing constraints:

Day 1: $𝑥_{1} - 100 = 𝑦_{1}$
Day 2: $𝑦_{1} + 𝑥_{2} - 150 = 𝑦_{2}$
Day 3: $𝑦_{2} + 𝑥_{3} - 200 = 𝑦_{3}$
Day 4: $𝑦_{3} + 𝑥_{4} - 170 = 𝑦_{4}$

Demand fulfillment constraints:

$𝑥_{1} \geq 100$
$𝑦_{1} + 𝑥_{2} \geq 150$
$𝑦_{2} + 𝑥_{3} \geq 200$
$𝑦_{3} + 𝑥_{4} \geq 170$

Non-negativity constraints:

$𝑦_{𝑡} \geq 0$
$𝑥_{𝑡} \geq 0$

Complete Formulation

\begin{aligned} min & 9 𝑥_{1} + 12 𝑥_{2} + 10 𝑥_{3} + 12 𝑥_{4} + 𝑦_{1} + 𝑦_{2} + 𝑦_{3} + 𝑦_{4} \\ 𝑠 . 𝑡 . & 𝑥_{1} - 100 = 𝑦_{1} \\ 𝑦_{1} + 𝑥_{2} - 150 = 𝑦_{2} \\ 𝑦_{2} + 𝑥_{3} - 200 = 𝑦_{3} \\ 𝑦_{3} + 𝑥_{4} - 170 = 𝑦_{4} \\ 𝑥_{1} \geq 100 \\ 𝑦_{1} + 𝑥_{2} \geq 150 \\ 𝑦_{2} + 𝑥_{3} \geq 200 \\ 𝑦_{3} + 𝑥_{4} \geq 170 \\ 𝑥_{𝑡}, 𝑦_{𝑡} \geq 0 \forall 𝑡 = 1, \dots, 4 \end{aligned}

Simplification (Redundant Constraints)

Inventory balancing & non-negativity imply fulfillment
- E.g., in day 1, $𝑥_{1} - 100 = 𝑦_{1}$ and $𝑦_{1} \geq 0$ means $𝑥_{1} \geq 100$

\begin{aligned} min & 9 𝑥_{1} + 12 𝑥_{2} + 10 𝑥_{3} + 12 𝑥_{4} + 𝑦_{1} + 𝑦_{2} + 𝑦_{3} + 𝑦_{4} \\ 𝑠 . 𝑡 . & 𝑥_{1} - 100 = 𝑦_{1} \\ 𝑦_{1} + 𝑥_{2} - 150 = 𝑦_{2} \\ 𝑦_{2} + 𝑥_{3} - 200 = 𝑦_{3} \\ 𝑦_{3} + 𝑥_{4} - 170 = 𝑦_{4} \\ 𝑥_{𝑡}, 𝑦_{𝑡} \geq 0 \forall 𝑡 = 1, \dots, 4 \end{aligned}

No need to have ending inventory in period 4 (costly but useless)

\begin{aligned} min & 9 𝑥_{1} + 12 𝑥_{2} + 10 𝑥_{3} + 12 𝑥_{4} + 𝑦_{1} + 𝑦_{2} + 𝑦_{3} + 𝑦_{4} \\ 𝑠 . 𝑡 . & 𝑥_{1} - 100 = 𝑦_{1} \\ 𝑦_{1} + 𝑥_{2} - 150 = 𝑦_{2} \\ 𝑦_{2} + 𝑥_{3} - 200 = 𝑦_{3} \\ 𝑦_{3} + 𝑥_{4} - 170 = 0 \\ 𝑥_{𝑡} \geq 0 \forall 𝑡 = 1, \dots, 4 \\ 𝑦_{𝑡} \geq 0 \forall 𝑡 = 1, \dots, 3 \end{aligned}

Compact Formulation

\begin{aligned} min & \sum_{𝑡 = 1}^{4} (𝐶_{𝑡} 𝑥_{𝑡} + 𝑦_{𝑡}) \\ 𝑠 . 𝑡 . & 𝑦_{𝑡} - 1 + 𝑥_{𝑡} - 𝐷_{𝑡} = 𝑦_{𝑡} \forall 𝑡 = 1, \dots, 4 \\ 𝑦_{0} = 0 \\ 𝑥_{𝑡}, 𝑦_{𝑡} \geq 0 \forall 𝑡 = 1, \dots, 4 \end{aligned}

Where:

$𝐷_{𝑡}$ : demand on day $𝑡$
$𝑦_{𝑡}$ : ending inventory of day $𝑡$
$𝐶_{𝑡}$ : unit production cost on day $𝑡$

Example

Personnel Scheduling

Scheduling employees

Each employee must work for 5 consecutive days and then take 2 consecutive rest days

Number of employees required for each day

Mon	Tue	Wed	Thu	Fri	Sat	Sun
110	80	150	30	70	160	120

Seven shifts

Mon to Fri
Tue to Sat
…

Minimize number of employees hired

Let $𝑥_{𝑖}$ be the number of employees assigned to work from day $𝑖$ for 5 consecutive days

Objective function:

min 𝑥_{1} + 𝑥_{2} + 𝑥_{3} + 𝑥_{4} + 𝑥_{5} + 𝑥_{6} + 𝑥_{7}

Demand Fullfilment Constraints

110 employeeds needed Monday

𝑥_{1} + 𝑥_{4} + 𝑥_{5} + 𝑥_{6} + 𝑥_{7} \geq 110

An employee that works on Tues ( $𝑡_{2}$ ) or Wed ( $𝑡_{3}$ ) does not work on Mon ( $𝑡_{1}$ )

80 employees needed Tuesday

𝑥_{1} + 𝑥_{2} + 𝑥_{5} + 𝑥_{6} + 𝑥_{7} \geq 80

An employee that works on Wed ( $𝑡_{3}$ ) or Thu ( $𝑡_{4}$ ) does not work on Tue ( $𝑡_{2}$ )

120 employees needed Sunday

𝑥_{3} + 𝑥_{4} + 𝑥_{5} + 𝑥_{6} + 𝑥_{7} \geq 120

An employee that works on Mon ( $𝑡_{1}$ ) or Tue ( $𝑡_{2}$ ) does not work on Sun ( $𝑡_{7}$ )

Non-Nagativity Constraints

𝑥_{𝑖} \geq 0 \forall 𝑖 = 𝑖, \dots, 7

Complete Formulation

\begin{aligned} min & 𝑥_{1} + & 𝑥_{2} + & 𝑥_{3} + & 𝑥_{4} + & 𝑥_{5} + & 𝑥_{6} + & 𝑥_{7} \\ 𝑠 . 𝑡 . & 𝑥_{1} + & 𝑥_{4} + & 𝑥_{5} + & 𝑥_{6} + & 𝑥_{7} & \geq & 110 \\ 𝑥_{1} + & 𝑥_{2} + & 𝑥_{5} + & 𝑥_{6} + & 𝑥_{7} & \geq & 80 \\ 𝑥_{1} + & 𝑥_{2} + & 𝑥_{3} + & 𝑥_{6} + & 𝑥_{7} & \geq & 150 \\ 𝑥_{1} + & 𝑥_{2} + & 𝑥_{3} + & 𝑥_{4} + & 𝑥_{7} & \geq & 30 \\ 𝑥_{1} + & 𝑥_{2} + & 𝑥_{3} + & 𝑥_{4} + & 𝑥_{5} & \geq & 70 \\ 𝑥_{2} + & 𝑥_{3} + & 𝑥_{4} + & 𝑥_{5} + & 𝑥_{6} & \geq & 160 \\ 𝑥_{3} + & 𝑥_{4} + & 𝑥_{5} + & 𝑥_{6} + & 𝑥_{7} & \geq & 120 \\ 𝑥_{𝑖} \geq 0 & \forall 𝑖 = 1, \dots, 7 \end{aligned}

model.py

from pyomo.environ import *
from pyomo.dataportal import DataPortal

model = AbstractModel()

# Sets
model.Days = Set(ordered=True)           # Days 1..7 (Mon..Sun)
model.Shifts = Set(ordered=True)         # Shifts starting on days 1..7

# Parameters
model.Demand = Param(model.Days)         # Daily staffing requirements
model.Cover = Param(model.Days, model.Shifts, within=Binary)  # Coverage matrix (1 if shift j covers day i)

# Variables
model.x = Var(model.Shifts, domain=NonNegativeReals)

# Objective: Minimize total employees hired
def obj_rule(model):
    return sum(model.x[s] for s in model.Shifts)
model.OBJ = Objective(rule=obj_rule, sense=minimize)

# Constraints: Cover daily demand
def demand_rule(model, d):
    return sum(model.Cover[d, s] * model.x[s] for s in model.Shifts) >= model.Demand[d]
model.DemandConstraint = Constraint(model.Days, rule=demand_rule)

# Load data from .dat file
data = DataPortal()
data.load(filename='data.dat', model=model)

# Create an instance of the model
instance = model.create_instance(data)

# Create solver
solver = SolverFactory('glpk')
solver.options['tmlim'] = 60

# Solve with solver timeout (optional)
results = solver.solve(instance, tee=True)

# Display results
instance.display()

data.dat

set Days := Mon Tue Wed Thu Fri Sat Sun ;
set Shifts := s1 s2 s3 s4 s5 s6 s7 ;

param Demand :=
Mon 110
Tue 80
Wed 150
Thu 30
Fri 70
Sat 160
Sun 120 ;

# Each shift covers 5 consecutive days starting on the shift's day
# 1 if shift s_j covers day d_i, 0 otherwise

param Cover:
        s1 s2 s3 s4 s5 s6 s7 :=
Mon     1  0  0  1  1  1  1
Tue     1  1  0  0  1  1  1
Wed     1  1  1  0  0  1  1
Thu     1  1  1  1  0  0  1
Fri     1  1  1  1  1  0  0
Sat     0  1  1  1  1  1  0
Sun     0  0  1  1  1  1  1 ;

Output:

  Variables:
  x : Size=7, Index=Shifts
      Key : Lower : Value            : Upper : Fixed : Stale : Domain
        s1 :     0 : 3.33333333333333 :  None : False : False : NonNegativeReals
        s2 :     0 :             40.0 :  None : False : False : NonNegativeReals
        s3 :     0 : 13.3333333333333 :  None : False : False : NonNegativeReals
        s4 :     0 :              0.0 :  None : False : False : NonNegativeReals
        s5 :     0 : 13.3333333333333 :  None : False : False : NonNegativeReals
        s6 :     0 : 93.3333333333333 :  None : False : False : NonNegativeReals
        s7 :     0 :              0.0 :  None : False : False : NonNegativeReals

Objectives:
  OBJ : Size=1, Index=None, Active=True
      Key  : Active : Value
      None :   True : 163.33333333333323

Constraints:
  DemandConstraint : Size=7
      Key : Lower : Body               : Upper
      Fri :  70.0 :  69.99999999999993 :  None
      Mon : 110.0 : 109.99999999999993 :  None
      Sat : 160.0 :  159.9999999999999 :  None
      Sun : 120.0 :  119.9999999999999 :  None
      Thu :  30.0 :  56.66666666666663 :  None
      Tue :  80.0 : 149.99999999999994 :  None
      Wed : 150.0 : 149.99999999999994 :  None