Single machine scheduling - due dates constraints

I am trying to program the single machine scheduling/bottleneck optimization, with python pulp, without making for loops to better understand.

I have 2 tasks called task1 and task2 to schedule that respectively last 1H and 3H, with due dates respectively at 1H and 4H.

Here are two solutions cases: first one is bad, the second one is good.

# Ko, because the task 1 due date is outdated  
# hours   1 2 3 4 
# task1         -
# task2   - - - 
# Ok, because due dates are respected
# hours   1 2 3 4 
# task1   - 
# task2     - - - 

I want the pulp solver to find solution 2 above.

Here is my Pulp code, solution is close, maybe you can help me write the "due dates" type constraints, I can't do it.There are no FOR loops, it's done on purpose to try to understand...

import pulp as p
# Tasks
tasks = ["task1","task2"]
duration = {1,3}
due = {1,4}
starts = {1,2,3,4}
# Create problem
prob = p.LpProblem('single_machine_scheduling', p.LpMinimize)  
# Each possible "time slots" to select for tasks
task1_1 = p.LpVariable("task1_1", 0, None, p.LpBinary) 
task1_2 = p.LpVariable("task1_2", 0, None, p.LpBinary) 
task1_3 = p.LpVariable("task1_3", 0, None, p.LpBinary) 
task1_4 = p.LpVariable("task1_4", 0, None, p.LpBinary) 
task2_1 = p.LpVariable("task2_1", 0, None, p.LpBinary) 
task2_2 = p.LpVariable("task2_2", 0, None, p.LpBinary) 
task2_3 = p.LpVariable("task2_3", 0, None, p.LpBinary)  
task2_4 = p.LpVariable("task2_4", 0, None, p.LpBinary) 
# Theses constraints should be used for due dates constraints eventually...
start_1 = p.LpVariable("start_1", 0, None, p.LpContinuous) 
start_2 = p.LpVariable("start_2", 0, None, p.LpContinuous) 
start_3 = p.LpVariable("start_3", 0, None, p.LpContinuous) 
start_4 = p.LpVariable("start_4", 0, None, p.LpContinuous) 
# Objective : Minimize timespan
prob += 1 * task1_1 + 1 * task1_2 + 1 * task1_3 + 1 * task1_4 + 3 * task2_1 + 3 * task2_2 + 3 * task2_3 + 3 * task2_4
# Constraints
# Only one task1 and one task2 can be selected
prob +=  task1_1 + task1_2 + task1_3 + task1_4 == 1
prob +=  task2_1 + task2_2 + task2_3 + task2_4 == 1
# Due dates constraints ( How to ?)
# Solve
prob.solve()
# Print variables
for v in prob.variables():
print(v.name, "=", v.varValue)
# Print objective
print("Minimized timespan = ", p.value(prob.objective))
start_3 = 0.0
start_4 = 0.0
task1_1 = 0.0
task1_2 = 0.0
task1_3 = 0.0
task1_4 = 1.0
task2_1 = 0.0
task2_2 = 0.0
task2_3 = 0.0
task2_4 = 1.0
Minimized timespan =  4.0

Maybe you can point me in the right direction? I also can't write them in the mathematical model, The ones I found include additional parameters like weight, but I don't want that here..

Edit: Oops sorry, I have learned that it's called "Total Tardiness minimization", and the Moore and Hodgson algorithm .

I just don't think this is a problem for pulp, because there's nothing to optimize. A simple pair of loops provides the most straightforward solution:

import itertools
tasks = [("task1", 1, 1), ("task2", 3, 4)]
# For each permutation
for tasklist in itertools.permutations(tasks):
time = 1
start = []
order = []
for task in tasklist:
start.append( time )
order.append( task )
time += task[1]
if time-1 > task[2]:
# We violated a due date constraint.
break
else:
print("Order",order,"succeeds")

Output:

Order [('task1', 1, 1), ('task2', 3, 4)] succeeds