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Un proyecto, contempla la siguiente lista de activ...

Erstellt am: 19. Oktober 2025

Beantwortet mit GPT-5 Thinking von Chat01

Frage

Un proyecto, contempla la siguiente lista de actividades, junto con sus tiempos requeridos de terminación.

Actividad Tiempo Precedentes inmediatos
A 1 _ _
B 4 A
C 3 A
D 7 A
E 6 B
F 2 C,D
G 7 E,F
H 9 D
I 4 G,H

text
a. Determine el diagrama de Ruta Crítica. 
b. Muestre los tiempos de inicio temprano, terminación temprana, inicio tardío y terminación tardía. 
c. Muestre la ruta crítica. 
d. ¿Qué ocurriría si la actividad F, se modificara para que tarde cuatro días, en lugar de dos?

C.R.: Para el logro del objetivo 9, Ud. debe responder correctamente todas las preguntas con los siguientes aspectos:
El procedimiento desarrollado se debe justificar detalladamente.
El resultado final debe estar correcto y simplificado.
Debe trabajar de manera ordenada y secuencialmente.

Python

python
# Let's compute CPM for the given network.
from collections import defaultdict, deque

activities = {
    'A': {'dur': 1, 'pred': []},
    'B': {'dur': 4, 'pred': ['A']},
    'C': {'dur': 3, 'pred': ['A']},
    'D': {'dur': 7, 'pred': ['A']},
    'E': {'dur': 6, 'pred': ['B']},
    'F': {'dur': 2, 'pred': ['C','D']},
    'G': {'dur': 7, 'pred': ['E','F']},
    'H': {'dur': 9, 'pred': ['D']},
    'I': {'dur': 4, 'pred': ['G','H']},
}

# compute topological order
# build successors for later
succ = defaultdict(list)
in_deg = defaultdict(int)
for a, info in activities.items():
    in_deg[a] = len(info['pred'])
    for p in info['pred']:
        succ[p].append(a)

# Kahn
order = []
dq = deque([a for a in activities if in_deg[a]==0])
while dq:
    n = dq.popleft()
    order.append(n)
    for s in succ[n]:
        in_deg[s]-=1
        if in_deg[s]==0:
            dq.append(s)
order


['A', 'B', 'C', 'D', 'E', 'F', 'H', 'G', 'I']

Python

python
# forward pass
ES = {}
EF = {}
for a in order:
    preds = activities[a]['pred']
    if not preds:
        ES[a] = 0
    else:
        ES[a] = max(EF[p] for p in preds)
    EF[a] = ES[a] + activities[a]['dur']
ES, EF


({'A': 0, 'B': 1, 'C': 1, 'D': 1, 'E': 5, 'F': 8, 'H': 8, 'G': 11, 'I': 18},
 {'A': 1, 'B': 5, 'C': 4, 'D': 8, 'E': 11, 'F': 10, 'H': 17, 'G': 18, 'I': 22})

Python

python
# backward pass
project_duration = EF['I']  # 22
LS = {}
LF = {}
for a in reversed(order):
    # successors; if none, LF = project_duration
    su = succ[a]
    if not su:
        LF[a] = project_duration
    else:
        LF[a] = min(LS[s] for s in su)
    LS[a] = LF[a] - activities[a]['dur']
LS, LF, project_duration


({'I': 18, 'G': 11, 'H': 9, 'F': 9, 'E': 5, 'D': 2, 'C': 6, 'B': 1, 'A': 0},
 {'I': 22, 'G': 18, 'H': 18, 'F': 11, 'E': 11, 'D': 9, 'C': 9, 'B': 5, 'A': 1},
 22)

Python

python
slack = {a: LS[a]-ES[a] for a in activities}
slack


{'A': 0, 'B': 0, 'C': 5, 'D': 1, 'E': 0, 'F': 1, 'G': 0, 'H': 1, 'I': 0}

Python

python
activities2 = {k: v.copy() for k,v in activities.items()}
activities2['F'] = {'dur': 4, 'pred': ['C','D']}
# rebuild succ/in_deg for this network
succ2 = defaultdict(list)
in_deg2 = defaultdict(int)
for a, info in activities2.items():
    in_deg2[a] = len(info['pred'])
    for p in info['pred']:
        succ2[p].append(a)
order2 = []
dq = deque([a for a in activities2 if in_deg2[a]==0])
while dq:
    n = dq.popleft()
    order2.append(n)
# forward
ES2={}; EF2={}
for a in order2:
    preds = activities2[a]['pred']
    ES2[a] = 0 if not preds else max(EF2[p] for p in preds)
    EF2[a] = ES2[a] + activities2[a]['dur']
# backward
project_duration2 = EF2['I']
LS2={}; LF2={}
for a in reversed(order2):
    su = succ2[a]
    if not su:
        LF2[a] = project_duration2
    else:
        LF2[a] = min(LS2[s] for s in su)
    LS2[a] = LF2[a] - activities2[a]['dur']
slack2 = {a: LS2[a]-ES2[a] for a in activities2}
project_duration2, ES2, EF2, LS2, LF2, slack2

Gedanke

activities2 = {k: v.copy() for k,v in activities.items()} activities2['F'] = {'dur': 4, 'pred': ['C','D']}

rebuild succ/in_deg for this network

succ2 = defaultdict(list) in_deg2 = defaultdict(int) for a, info in activities2.items(): in_deg2[a] = len(info['pred']) for p in info['pred']: succ2[p].append(a) order2 = [] dq = deque([a for a in activities2 if in_deg2[a]==0]) while dq: n = dq.popleft() order2.append(n)

forward

ES2={}; EF2={} for a in order2: preds = activities2[a]['pred'] ES2[a] = 0 if not preds else max(EF2[p] for p in preds) EF2[a] = ES2[a] + activities2[a]['dur']

backward

project_duration2 = EF2['I'] LS2={}; LF2={} for a in reversed(order2): su = succ2[a] if not su: LF2[a] = project_duration2 else: LF2[a] = min(LS2[s] for s in su) LS2[a] = LF2[a] - activities2[a]['dur'] slack2 = {a: LS2[a]-ES2[a] for a in activities2} project_duration2, ES2, EF2, LS2, LF2, slack2

Thought for 23s

Python

python
order2


['A']

Python

python
in_deg2


defaultdict(int,
            {'A': 0,
             'B': 1,
             'C': 1,
             'D': 1,
             'E': 1,
             'F': 2,
             'G': 2,
             'H': 1,
             'I': 2})

Python

python
succ2


defaultdict(list,
            {'A': ['B', 'C', 'D'],
             'B': ['E'],
             'C': ['F'],
             'D': ['F', 'H'],
             'E': ['G'],
             'F': ['G'],
             'G': ['I'],
             'H': ['I']})

Python

python
# Let's recompute Kahn step-by-step to see.
from collections import deque
in_deg2c = in_deg2.copy()
dq = deque([a for a in activities2 if in_deg2c[a]==0])
seq=[]
while dq:
    n = dq.popleft()
    seq.append(n)
    for s in succ2[n]:
        in_deg2c[s]-=1
        if in_deg2c[s]==0:
            dq.append(s)
seq, in_deg2c


(['A', 'B', 'C', 'D', 'E', 'F', 'H', 'G', 'I'],
 defaultdict(int,
             {'A': 0,
              'B': 0,
              'C': 0,
              'D': 0,
              'E': 0,
              'F': 0,
              'G': 0,
              'H': 0,
              'I': 0}))

Python

python
order2 = seq
# forward
ES2={}; EF2={}
for a in order2:
    preds = activities2[a]['pred']
    ES2[a] = 0 if not preds else max(EF2[p] for p in preds)
    EF2[a] = ES2[a] + activities2[a]['dur']
project_duration2 = EF2['I']
# backward
LS2={}; LF2={}
for a in reversed(order2):
    su = succ2[a]
    if not su:
        LF2[a] = project_duration2
    else:
        LF2[a] = min(LS2[s] for s in su)
    LS2[a] = LF2[a] - activities2[a]['dur']
slack2 = {a: LS2[a]-ES2[a] for a in activities2}
project_duration2, ES2, EF2, LS2, LF2, slack2


(23,
 {'A': 0, 'B': 1, 'C': 1, 'D': 1, 'E': 5, 'F': 8, 'H': 8, 'G': 12, 'I': 19},
 {'A': 1, 'B': 5, 'C': 4, 'D': 8, 'E': 11, 'F': 12, 'H': 17, 'G': 19, 'I': 23},
 {'I': 19, 'G': 12, 'H': 10, 'F': 8, 'E': 6, 'D': 1, 'C': 5, 'B': 2, 'A': 0},
 {'I': 23, 'G': 19, 'H': 19, 'F': 12, 'E': 12, 'D': 8, 'C': 8, 'B': 6, 'A': 1},
 {'A': 0, 'B': 1, 'C': 4, 'D': 0, 'E': 1, 'F': 0, 'G': 0, 'H': 2, 'I': 0})

Gedanke

Reviewing Critical Path

The project duration is 23 days now, and the critical path shifted from A-B-E-G-I to A-D-F-G-I. This change happened because extending F makes A-D-F-G-I longer compared to the original path. I’ll double-check the lengths of the paths and present a simple diagram, possibly in text form or ASCII.

Explaining Project Scheduling

I need to show the ES, EF, LS, and LF times for all activities before and after the modification, as the question asks for the impact of changes, even though it doesn’t ask for the full recomputation. I should provide a detailed procedure explaining the forward and backward passes, slack, and critical path. I'll structure it into clear tables for both cases.

Thought for 1m 41s

Gedanke

Recomputing and compiling values

The Python environment resets between calls, but that's not an issue — I can recompute the values quickly without showing the process to the user. Instead, I’ll embed the table manually with the recorded values. I'll create a nice markdown table with the necessary details, while verifying LS and LF for accuracy.

Finalizing values and project duration

I’ll re-list all the project details systematically one more time to check the values. For example, I see that for H, LS should be 9 given that LF is 18, so the slack is 1. After reviewing, I can confirm the total project duration is 22 days and the correct values for each task.

Re-listing and verifying durations

I’m reaffirming the project details from earlier, including the new durations where F is 4. The critical path is A-B-E-G-I, with a total project duration of 22 days. I’ve calculated new values: for instance, F now starts at 8, ending at 12. Let’s just ensure correctness across all tasks.

Confirming durations and dependencies

I’m reviewing the project timeline with F’s new durations (from 4). The critical path remains A-B-E-G-I, with the total duration now 23 days. Checking LS values, it appears that LS for A is 0 and LF is 1. There’s a small discrepancy for B's LS value, so I’ll need to reverify earlier entries.

Thought for 9s

Antwort

¡Vamos paso a paso (CPM/PERT, actividades en nodos) y con todo el procedimiento!

a) Diagrama de precedencias (texto)

text
Inicio
  └─ A(1)
       ├─ B(4) ── E(6) ──┐
       ├─ C(3) ── F(2) ──┤
       └─ D(7) ── H(9) ──┤
                 └─ F(2) ─┘
                    └─ G(7) ── I(4) ── Fin

Interpretación: las flechas van “de predecesora → sucesora”; el número entre paréntesis es la duración (días).

b) Cálculos CPM con F = 2 días

Metodología (justificación breve)

  1. Pase hacia adelante (Forward Pass):
    • ESi=max⁡(EF_predecesoras)ES_i = \max(EF\_\text{predecesoras}) (o 0 si no tiene).
    • EFi=ESi+DuriEF_i = ES_i + Dur_i.
  2. Pase hacia atrás (Backward Pass): (tomando como LFLF del último nodo el tiempo del proyecto)
    • LFi=min⁡(LS_sucesoras)LF_i = \min(LS\_\text{sucesoras}) (o tiempo del proyecto si no tiene).
    • LSi=LFi−DuriLS_i = LF_i - Dur_i.
  3. Holgura: Slacki=LSi−ESi=LFi−EFiSlack_i = LS_i - ES_i = LF_i - EF_i.
    • Ruta crítica: actividades con holgura 0.
    • Duración del proyecto: EFEF de la última actividad.

Resultados (tiempos en días; origen en 0)

Act.Dur.PredecesorasESEFLSLFHolgura
A1—01010
B4A15150
C3A14695
D7A18291
E6B5115110
F2C,D8109111
G7E,F111811180
H9D8179181
I4G,H182218220

Duración total del proyecto: 22 días (EF de I).

c) Ruta crítica

Las actividades con holgura = 0 son: A → B → E → G → I.
Ruta crítica: A–B–E–G–I (1 + 4 + 6 + 7 + 4 = 22).

d) ¿Qué pasa si F tarda 4 días (en lugar de 2)?

Repetimos el cálculo cambiando solo DurF=4Dur_F = 4.

Nuevos resultados clave

  • Forward pass (nodos afectados aguas abajo de F):
    • ESF=max⁡(EFC,EFD)=max⁡(4,8)=8ES_F = \max(EF_C, EF_D) = \max(4, 8) = 8 → EFF=8+4=12EF_F = 8 + 4 = 12
    • ESG=max⁡(EFE,EFF)=max⁡(11,12)=12ES_G = \max(EF_E, EF_F) = \max(11, 12) = 12 → EFG=12+7=19EF_G = 12 + 7 = 19
    • ESI=max⁡(EFG,EFH)=max⁡(19,17)=19ES_I = \max(EF_G, EF_H) = \max(19, 17) = 19 → EFI=19+4=23EF_I = 19 + 4 = 23
  • Backward pass (tiempo del proyecto ahora 23).

Tabla (solo filas que cambian respecto al caso base)

Act.Dur.ESEFLSLFHolgura
B415261
D718180
E65116121
F48128120
G7121912190
H981710192
I4192319230

Nueva duración del proyecto: 23 días.
Nueva ruta crítica: A–D–F–G–I (1 + 7 + 4 + 7 + 4 = 23).

Conclusión del cambio: Aumentar F de 2→4 días desplaza la ruta crítica desde A–B–E–G–I a A–D–F–G–I, incrementando el plazo total 1 día (de 22 a 23).

Resumen (todo en orden y simplificado)

  • Diagrama de precedencias presentado.
  • Tiempos ES/EF/LS/LF calculados y tabulados.
  • Ruta crítica (caso base): A–B–E–G–I, 22 días.
  • Si F = 4 días: ruta crítica pasa a A–D–F–G–I, 23 días.
  • Justificación: reglas CPM (pases adelante/atrás y holguras) aplicadas rigurosamente.

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