This document provides steps for crashing a project schedule to reduce its duration. It begins by explaining the two types of activity durations: normal and crash. It then outlines 9 steps to determine the maximum possible time reduction for a project: 1) draw the network diagram, 2) find the critical path using normal durations, 3) find the critical path using crash durations, 4) calculate the difference in critical paths, 5) calculate the slope for each activity, 6) select the activity with the lowest slope for crashing, 7) calculate the possible crash amount, 8) redraw the network and recalculate duration and cost, 9) compare to crashed critical path and repeat steps if different. It provides two notes: the critical path may change between

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Project Management
Project Time-Cost Trade-Off
‫مالحظات‬‫عن‬ ‫مهمة‬‫تعجيل‬‫المشاريع‬
Dr. Mahmoud Abbas Mahmoud
Assistant Professor
2017

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‫مالحظات‬‫عن‬ ‫مهمة‬‫تعجيل‬‫الم‬‫شاريع‬
‫من‬ ‫نوعين‬ ‫هناك‬‫ا‬‫ألوقا‬‫ت‬‫لتنفيذ‬‫أنشطة‬‫أي‬‫مشر‬‫و‬‫ع‬
1-)‫(األعتيادي‬ ‫الطبيعي‬ ‫التنفيذ‬Normal‫الطبيعي‬ ‫الزمن‬ ‫ويتضمن‬nD‫الطبيعية‬ ‫وكلفته‬nC
2-‫ال‬ ‫التنفيذ‬‫معجل‬)‫(المقلص‬Crash‫المعجل‬ ‫الزمن‬ ‫ويتضمن‬cD‫المعجلة‬ ‫وكلفته‬cC
‫لتعجيل‬ ‫الحل‬ ‫خطوات‬ ‫وتكون‬‫تنفيذ‬‫المشروع‬:‫يلي‬ ‫كما‬
1-‫إرسم‬‫للمشروع‬ ‫الشبكي‬ ‫المخطط‬.
2-‫أو‬‫الطبيعية‬ ‫األزمنة‬ ‫باستخدام‬ ‫الحرج‬ ‫المسار‬ ‫جد‬‫الرسم‬ ‫على‬(Normal Network)‫و‬‫أو‬‫الزمن‬ ‫جد‬
‫حالة‬ ‫في‬ ‫المشروع‬ ‫وكلفة‬ ‫للمشروع‬ ‫الحرج‬‫التنفيذ‬‫ب‬‫الطبيعية‬ ‫األزمنة‬.
3-‫أو‬‫جد‬‫المعجلة‬ ‫األزمنة‬ ‫باستخدام‬ ‫الحرج‬ ‫المسار‬‫الرسم‬ ‫على‬(Crash Network)‫و‬‫أو‬‫جد‬‫الزمن‬
‫حالة‬ ‫في‬ ‫المشروع‬ ‫وكلفة‬ ‫للمشروع‬ ‫الحرج‬‫التنفيذ‬‫ب‬‫المعجلة‬ ‫األزمنة‬.
4-‫أو‬‫الحرج‬ ‫الزمن‬ ‫بين‬ ‫الفرق‬ ‫جد‬‫للمشروع‬‫في‬ ‫للحالتين‬‫الخطوتين‬2‫و‬3‫وه‬‫ذا‬‫الفرق‬‫سيكون‬‫هو‬‫أقصى‬
‫من‬ ‫تخفيضة‬ ‫يمكن‬ ‫ما‬‫ال‬‫وقت‬‫لتنفيذ‬ ‫الكلي‬‫المشروع‬(Project crash limit).
5-‫أو‬‫الميل‬ ‫جد‬(Slope)‫العالقة‬ ‫باستخدام‬ ‫المشروع‬ ‫أنشطة‬ ‫من‬ ‫نشاط‬ ‫لكل‬:
𝐒𝐥𝐨𝐩𝐞 =
𝐂𝐫𝐚𝐬𝐡 𝐜𝐨𝐬𝐭 – 𝐍𝐨𝐫𝐦𝐚𝐥 𝐜𝐨𝐬𝐭
𝐍𝐨𝐫𝐦𝐚𝐥 𝐝𝐮𝐫𝐚𝐭𝐢𝐨𝐧 – 𝐂𝐫𝐚𝐬𝐡 𝐝𝐮𝐫𝐚𝐭𝐢𝐨𝐧
=
𝐂𝐜 – 𝐂𝐧
𝐃𝐧 – 𝐃𝐜
=
𝝙 𝐂
𝝙 𝐃
6-‫حدد‬‫لل‬ ‫المرشح‬ ‫النشاط‬‫تقليص‬‫الـ‬ ‫على‬(Normal Network)‫وهو‬‫النشاط‬‫الحرج‬‫الذي‬‫يقابله‬
‫ميل‬ ‫أقل‬)‫الزمن‬ ‫لوحدة‬ ‫تعجيل‬ ‫كلفة‬ ‫(أقل‬.
7-‫إحسب‬‫التقليص‬ ‫مقدار‬‫الممكن‬‫النشاط‬ ‫ذلك‬ ‫لزمن‬(‫وهو‬‫بين‬ ‫الفرق‬‫ال‬‫زمن‬‫ل‬ ‫الحالي‬‫المرشح‬ ‫لنشاط‬
‫و‬‫ال‬‫زمن‬‫النشاط‬ ‫لذلك‬ ‫المعجل‬)‫ال‬ ‫كان‬ ‫فاذا‬‫فرق‬(1)‫قلص‬‫وقت‬‫النشاط‬‫كان‬ ‫إذا‬ ‫أما‬ .ً‫ة‬‫مباشر‬‫الفرق‬
‫من‬ ‫أكثر‬(1‫قيم‬ ‫فأحسب‬ )‫ال‬‫ـ‬Free Float(FF‫األ‬ ‫لكافة‬ )‫نشطة‬‫حرجة‬ ‫الغير‬‫الى‬ ‫تتحول‬ ‫قد‬ ‫والتي‬
‫حرجة‬‫أثناء‬‫التقليص‬‫و‬‫المرشح‬ ‫النشاط‬ ‫زمن‬ ‫قلص‬‫ب‬‫الذي‬ ‫الحد‬‫يساوي‬‫ال‬‫قيمة‬‫األقل‬( ‫بين‬‫مقد‬‫ار‬
‫الممكن‬ ‫التقليص‬‫للنشاط‬‫و‬‫أعلى‬Free Float‫ل‬‫أل‬‫حرجة‬ ‫الغير‬ ‫نشطة‬‫المذكورة‬).‫و‬‫الحد‬ ‫هذا‬ ‫يسمى‬
‫ا‬‫لـ‬(Compression Limit)‫و‬‫التالية‬ ‫بالمعادلة‬ ‫ذلك‬ ‫تمثيل‬ ‫يمكن‬:
Compression Limit = Min.[(‫الممكن‬ ‫التقليص‬ ‫,)مقدار‬ Max. FF of noncritical activities]
8-‫الحرج‬ ‫المسار‬ ‫وحدد‬ ‫الجديدة‬ ‫باألوقات‬ ‫المشروع‬ ‫شبكة‬ ‫أرسم‬‫من‬‫و‬ ‫جديد‬‫أوجد‬‫الجديد‬ ‫التنفيذ‬ ‫وقت‬
‫و‬‫الكلية‬ ‫الكلفة‬‫الجديدة‬‫للمشروع‬‫العالقة‬ ‫باستخدام‬:
‫الكلية‬ ‫الكلفة‬‫الجديدة‬( + ‫السابقة‬ ‫الكلية‬ ‫الكلفة‬ = ‫التقليص‬ ‫بعد‬‫التق‬ ‫بسبب‬ ‫المضافة‬ ‫الكلفة‬‫ليص‬)

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9-‫الحرج‬ ‫الزمن‬ ‫مع‬ ‫للتنفيذ‬ ‫الجديد‬ ‫الوقت‬ ‫قارن‬‫المعجلة‬ ‫األزمنة‬ ‫تطبيق‬ ‫حالة‬ ‫في‬ ‫للمشروع‬‫(المس‬‫تخرج‬
‫الخطوة‬ ‫من‬3)‫أعاله‬‫(من‬ ‫الخطوات‬ ‫كرر‬ ‫وبخالفه‬ ‫توقف‬ ‫له‬ ‫مساويا‬ ‫كان‬ ‫إذا‬ .6‫الى‬9)‫أخرى‬ ‫مرة‬.
:‫مهمتين‬ ‫مالحظتين‬
‫أ‬-‫دائما‬ ‫ليس‬‫للشبكتين‬ ‫نفسة‬ ‫الحرج‬ ‫المسار‬ ‫يكون‬‫باألوقات‬‫والمعجلة‬ ‫الطبيعية‬‫إنه‬ ‫كما‬‫قد‬‫ي‬‫تغير‬‫خطوة‬ ‫من‬
‫ألخرى‬‫مع‬‫األنشطة‬ ‫أوقات‬ ‫تغير‬‫وفي‬ ‫األصلي‬ ‫الحرج‬ ‫المسار‬ ‫مع‬ ‫أخرى‬ ‫حرجة‬ ‫مسارات‬ ‫تدخل‬ ‫(فقد‬
)ً‫ا‬‫حرج‬ ‫المسار‬ ‫ذلك‬ ‫بقاء‬ ‫على‬ ‫نحافظ‬ ‫ان‬ ‫يجب‬ ‫األحوال‬ ‫كل‬‫ما‬ ‫وهذا‬‫اإل‬ ‫يجب‬ً‫ا‬‫دائم‬ ‫اليه‬ ‫نتباه‬.
‫ب‬-‫حرج‬ ‫مسار‬ ‫من‬ ‫أكثر‬ ‫هناك‬ ‫كان‬ ‫أذا‬‫ف‬‫الوقت‬ ‫يخفض‬ ‫أن‬ ‫يجب‬‫المقدار‬ ‫وبنفس‬ ً‫ا‬‫مع‬ ‫المسارات‬ ‫لتلك‬
‫في‬ ‫ميل‬ ‫أقل‬ ‫يقابلها‬ ‫التي‬ ‫ولألنشطة‬‫كل‬‫من‬ ‫مسار‬‫المسارات‬ ‫تلك‬)‫الثاني‬ ‫المثال‬ ‫(الحظ‬ ‫الحرجة‬.
‫المثال‬ ‫نأخذ‬ ‫أعاله‬ ‫الخطوات‬ ‫ولتوضيح‬‫ين‬‫ال‬‫تاليين‬:
EXAMPLE- 1
A project consists of the following activities. The normal and crash durations
(days) and costs ($) are given in Table 1 below. It is required to complete the
project with minimum possible time and cost.
:‫مالحظة‬‫بالسؤال‬ ‫المطلوب‬ ‫كان‬ ‫لو‬ ‫ولكن‬ ‫ممكنين‬ ‫وكلفة‬ ‫وقت‬ ‫ألدنى‬ ‫التقليص‬ ‫هو‬ ‫المثال‬ ‫هذا‬ ‫في‬ ‫المطلوب‬
‫لذلك‬ ‫الوصول‬ ‫عند‬ ‫الحل‬ ‫ويتوقف‬ ‫التقليص‬ ‫عن‬ ‫نتوقف‬ ‫فاننا‬ ‫محدد‬ ‫زمن‬ ‫لغاية‬ ‫المشروع‬ ‫إنجاز‬ ‫زمن‬ ‫تقليص‬
‫الالزمة‬ ‫الكلية‬ ‫الكلفة‬ ‫ونحسب‬ ‫الحد‬‫المشروع‬ ‫لتنفيذ‬‫الحد‬ ‫ذلك‬ ‫عند‬.
TABLE 1
Activity
(i, j)
Normal Crash
Time Cost Time Cost
(1, 2) 8 1000 7 2000
(1, 3) 4 1500 2 2000
(2, 4) 2 500 1 900
(2, 5) 2 1000 1 1500
(3, 4) 2 1000 1 2000
(4, 5) 3 800 2 1000
∑ ---- 5800 ---- 9400
SOLUTION
Calculate the slope for each activity.
Activity
(i, j)
Normal Crash
ΔC =
Cc - Cn
ΔD =
Dn - Dc
Slope =
ΔC/ ΔD
$/day
Time (Dn) Cost (Cn) Time (Dc) Cost (Cc)
(1, 2) 8 1000 7 2000 1000 1 1000
(1, 3) 4 1500 2 2000 500 2 250
(2, 4) 2 500 1 900 400 1 400
(2, 5) 2 1000 1 1500 500 1 500
(3, 4) 2 1000 1 2000 1000 1 1000
(4, 5) 3 800 2 1000 200 1 200
∑ ---- 5800 ---- 9400

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Normal Network Crash Network
Project completion time = 13 days Project completion time = 10 days
Cost = 5800 $ Cost = 9400 $
CP is (1 – 2 – 4 – 5) CP is (1 – 2 – 4 – 5)
Project crash limit = Project completion time (with normal times) - Project
completion time (with crashed times)
Project crash limit = 13 – 10 = 3 days
Now we start compression of the project times (Normal times). From (Normal
Network), there is only one critical path (1 – 2 – 4 – 5). We will select one of the
critical activities.
Activity (4, 5) is selected because it has the smallest slope.
The time of this activity will be reduced from 3 to 2 days
The new network will be as follows
The new:
Project completion time = 12 days
Cost = 5800 + (1 × 200) = 6000 $
CP is (1 – 2 – 4 – 5)

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Now we will start new compression of the project times. We will select a new
critical activity with smallest slope. Therefore, we will select the activity (2, 4).
The time of this activity will be reduced from 2 to 1 day.
The new network will be as shown.
The new:
Project completion time = 11 days
Cost = 6000 + (1 × 400) = 6400 $
CP is (1 – 2 – 4 – 5)
Now we will start new compression of the project times. We will select a new
critical activity with smallest slope.
Therefore, we will select the activity (1, 2)
The time of this activity will be reduced from 8 to 7 days
The new network will be as shown.
Now the new:
Project completion time = 10 days
Cost = 6400 + (1 × 1000) = 7400 $
CP is (1 – 2 – 4 – 5)
We can see that the new project completion time (10 days) is equal to the (crash)
completion time. Therefore, it is no longer possible to reduce the time of the
project, and the total cost will be 7400 $.

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EXAMPLE- 2
A project consists of the following activities. The normal and crash points for each
activity are given in Table 2 below. It is required to complete the project with
minimum possible time and cost.
TABLE 2
Activity
(i, j)
Normal Crash
Time (days) Cost ($) Time (days) Cost ($)
(1, 2) 8 100 6 200
(1, 3) 4 150 2 350
(2, 4) 2 50 1 90
(2, 5) 10 100 5 400
(3, 4) 5 100 1 200
(4, 5) 3 80 1 100
Solution
Calculate the slope for each activity.
Activity
(i, j)
Normal Crash
ΔC =
Cc - Cn
ΔD =
Dn - Dc
Slope =
ΔC/ ΔD
$/day
Time
(days)
Cost
($)
Time
(days)
Cost
($)
(1, 2) 8 100 6 200 100 2 50
(1, 3) 4 150 2 350 200 2 100
(2, 4) 2 50 1 90 40 1 40
(2, 5) 10 100 5 400 300 5 60
(3, 4) 5 100 1 200 100 4 25
(4, 5) 3 80 1 100 20 2 10
∑ = 580 ∑ = 1340
Normal Network Crash Network
Project completion time = 18 days Project completion time = 11 days
Cost = 580 $ Cost = 1340 $
CP is (1 – 2 – 5) CP is (1 – 2 – 5)
Crash limit = 18 – 11 = 7 days

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Selected critical activity (1, 2) [smallest slope]. The time of this activity can be
reduced from 8 to 6 days.
Compression limit = Min. [(8 - 6), Max. FF of noncritical activities]
From the previous (Normal Network) we can calculate the free floats of the
noncritical activities as follows.
The Max. FF is of activity (4, 5) = 5
Compression limit = Min. [2, 5] = 2
Therefore, we will reduce the time of activity (1, 2) from 8 days to 6 days.
The result will be as shown
The new:
Project completion time = 16 days
Cost = 580 + 2(50) = 680 $
CP is (1 – 2 – 5)
Select a new critical activity (2, 5) [smallest slope]. The time of this activity can
be reduced from 10 to 5 days.
Compression limit = Min. [(10 - 5), Max. FF of noncritical activities]
The new Max. FF is of activity (4, 5) = 4
Compression limit = Min. [5, 4] = 4
Therefore, we will reduce the time of activity (2, 5) only from 10 days to 6 days.
The result will be as shown
The new:
Project completion time = 12 days
Cost = 680 + 4(60) = 920 $
There are two critical paths, they are:
CP1 is (1 – 2 – 5) and CP2 is (1 – 3 – 4 – 5)

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Now referring to the new network we can see that in order to make any useful
decrease to the activity (2 – 5), the activity (1 – 3) or activity (3 – 4) or activity (4
– 5) must be decreased also.
The best choice is which it has the lowest cost.
(2 – 5), (1 – 3) the cost = 60 + 100 = 160 $
(2 – 5), (3 – 4) the cost = 60 + 25 = 85 $
(2 – 5), (4 – 5) the cost = 60 + 10 = 70 $ ……. The lowest cost
Therefore, we will decrease the activities (2 – 5) and (4 – 5) together and the
cost of this step is = 70 $
The new:
Project completion time = 11 days
Cost = 920 + 70 = 990 $
Critical paths are: CP1 is (1 – 2– 5) and CP2 is (1 – 3 – 4 – 5)
We can see that the new project completion time (11 days) is equal to the (crash)
completion time. Therefore, it is no longer possible to reduce the time of the
project, and the total cost is 990 $.
EXERCISE
For the information of a project consists of the activities listed in Table 3. It is
required to complete the project with minimum possible time and cost.
TABLE 3
Activity
(i, j)
Normal Crash
Time
(days)
Cost
($)
Time
(days)
Cost
($)
(1, 2) 3 15 2 20
(2, 3) 3 15 1 35
(2, 4) 6 10 3 40
(2, 5) 15 70 13 90
(3, 4) 2 10 1 25
(4, 5) 4 10 2 30