_VIP_Cost Management -20-10-2018.pdf

PM
‫باهلل‬ ‫إال‬ ‫توفيقي‬ ‫وما‬
Cost Management
(Planning , Control & Improvements)
Dr. Attia Gomaa
Industrial Engineering Professor & Consultant
Mechanical Eng. Department – Shoubra Faculty of Eng. - Benha University
& Engineering and Science Services - American University in Cairo
Facebook: Attia Gomaa & Group: Project Management - Dr Attia Gomaa
October 2018

PM 2
Cost Management Best Practice - Dr. Attia Gomaa - 2018
Course Contents:
1. Cost Management
2. Cost Estimation
3. Cost Analysis
4. Cash Flow Analysis
5. Financial Analysis
6. Project Cost Analysis
7. Project Cost Control
8. Case Studies
Cost Management
(Planning , Control & Improvements)

PM 3
Rules of the Course
This is an Open Discussion Course:
Team Approach
Remember … You can’t do it alone!
Let us be a teamwork.
– Share Knowledge
– Share Experiences
– Share Best Practices
– Share Questions
–
‫المعرفة‬ ‫تبادل‬
–
‫الخبرات‬ ‫تبادل‬
–
‫الجيدة‬ ‫التجارب‬ ‫تبادل‬
–
‫األسئلة‬ ‫تبادل‬

PM 4
Project Triple Constraints:

PM 5
1) Material
2) Machine
3) Manpower
4) Method
5) Money
(Cost)
Resources Analysis (5 M’s):
Project Constraints:
1. Scope of Work (Specs)
2. Quality (Service Quality)
3. Time (Schedule)
4. Cost (Budget)

PM 6
Cost Management
(1)
Resource
Planning
(2)
Cost
Estimation
(3)
Cost
Budgeting
(4)
Cost
Control
Cost Planning
(5)
Cost
Forecasting
2017
2018
2019

PM 7
• Resource planning: determining what resources and
quantities of them should be used.
• Cost estimating: developing an estimate of the costs and
resources needed to complete a project.
• Cost budgeting: allocating the overall cost estimate to
individual work items to establish a baseline for measuring
performance.
• Cost control: controlling changes to the project budget.
COST MANAGEMENT - Definitions

PM 8
(1) Resource Planning (2) Cost Estimating
1. Work breakdown structure (WBS)
2. Main activities
3. Resource allocation
4. Resource limits (constraints)
5. Resource requirements
6. Cost rates
7. Cost estimation
8. Cost analysis
(3) Cost Budgeting (4) Cost Control
9. Cost baseline
10.Cash in / cash out
11.Budget plan
12.Budget limits (constraints)
13.Planned performance
14.Actual performance
15.Change requests
16.Performance reports
Source: PMBOK Standard
Cost Management

PM 9
Discuss briefly the resource estimation sheet for the following item:
• Project scope of Work: Office Building (250 m2 * 5 floors)
• Main Item: Plain Concerete
• Activity: Concrete Casting For Foundation (250 kg/cm2)
• BOQ: 300 m3 Duration: 3 day
Resource Estimation:
• Resource Allocation (Standard Information):
Materials Rates:
 Cement = 0.4 ton/m3 700 LE/ton
 Sand = 0.5 m3/m3 30 LE/m3
 Gravel = 0.8 m3/m3 90 LE/m3
 Water = 160 liter/m3 0.1 LE/liter
Labor Rates:
 Concrete Crew = 5 man/day 80 LE/man-day
 Carpenter Crew = 3 man/day 90 LE/man-day
Equipment Rates:
 Mixer =1 Eq./day 250 LE/day
 Loader =1 Eq./day 200 LE/day
Margin Factor = Overhead + Profit = 50% Direct Cost

PM 10
Activity
ID
Activity
Description
BOQ
(units)
Dur.
Day
Resource
Type
Resource
Name
Resource
Units
Resource
Quantity
Resource
Price/unit
Item Cost
LE
C250
Concrete
Casting
For
Foundation
(strength
250
kg/cm2)
by
Concrete
Mixer
300
m3
3
day
Materials
Cement
0.4
ton/m3
120
ton
700
LE/ton 84000
114900
Sand
0.5
m3/m3
150
m3
30
LE/m3
4500
Gravel
0.8
m3/m3
240
m3
90
LE/m3
21600
Water
160
liter/m3
48000
liter
0.1
LE/liter
4800
Labor Concrete
Crew
5
man/day
15
man-day
80
LE/man-d
1200
2010
Carpenter
Crew
3
man/day
9
man-day
90
LE/man-d
810
Equipment
Mixer
1
Eq./day
3
Eq.-day
250
LE/day 750
1350
Loader
1
Eq./day
3
Eq.-day
200
LE/day 600
Total Direct Cost = 118,260
Margin Factor = Overhead + Profit = 50% Direct Cost 59,130
Total Income (LE) 177,390
Unit Price (LE/m3) 591
Project scope of Work: Office Building (250 m2 * 5 floors) Main Item: Foundation Activity
Resource Estimation Sheet:

PM 11
Activity
ID
Activity
Description
BOQ
(units)
Dur.
Day
Resource
Type
Resource
Name
Resource
Units
Resource
Quantity
Resource
Price/unit
Item Cost
LE
C250
Concrete
Casting
For
Foundation
(strength
250
kg/cm2)
by
Concrete
Mixer
300
m3
3
day
Materials
Cement
0.4
ton/m3
120
ton
700
LE/ton 84000
114900
Sand
0.5
m3/m3
150
m3
30
LE/m3
4500
Gravel
0.8
m3/m3
240
m3
90
LE/m3
21600
Water
160
liter/m3
48000
liter
0.1
LE/liter
4800
Labor Concrete
Crew
5
man/day
15
man-day
80
LE/man-d
1200
2010
Carpenter
Crew
3
man/day
9
man-day
90
LE/man-d
810
Equipment
Mixer
1
Eq./day
3
Eq.-day
250
LE/day 750
1350
Loader
1
Eq./day
3
Eq.-day
200
LE/day 600
Total Direct Cost = 118,260
Margin Factor = Overhead + Profit = 50% Direct Cost 59,130
Total Income (LE) 177,390
Unit Price (LE/m3) 591
Project scope of Work: Office Building (250 m2 * 5 floors) Main Item: Foundation Activity
Resource Estimation Sheet:
ID Item Name
Cost Duration
Responsibility
C250 Plain Concrete
118260 LE 3 Days
Eng. Attia Gomaa

PM 12
Cost
Tracking
Schedule
Responsibility
Quality
QA/QC
WBS
Work Breakdown Structure (WBS)
For example; MIS Project for Resource Planning
20% 27% 53%
100%
37%
100%
41% 22%
Project
Main Items

PM 13
Requirements
Design
Resource Module
Cost Module
Resource Module
Cost Module
For example; MIS Project for Resource Planning

PM 14
Project: Computer Lab 30 Users
Client: AUC
WBS - LAB
Electrical
Works
Mechanical
Works
Civil Works
IT Works
Finishing

PM 15
Cost Breakdown Structure
Unit Profit
- (-%)
Labour
Cost
- (-%)
Machine
Cost
- (- %)
Office
Overhead
- (-%)
Overhead
- (-%)
Technical Cost
- (-%)
Unit Price
- $/unit
Unit Cost
- (100%)
Critical
Resources
Technical Direct Cost
- (-%)
Materials
Cost
- (- %)
Technical
Overhead
- (-%)
Base

PM 16
The cost classification matrix for a project (First year 2016) is as follows:
Unit Price = 300 $/unit (Market Price) Planned Capacity = 100,000 units/year
Case Study:
Cost Elements
Total Cost
1,000,000 $
Direct
Costs
Raw Materials 10
Technical labors 5
Equipment & Tools 10
Overheads
Technical Overhead 2
Office Overhead 1
Total 1,000,000 $ 28
Based on this information, discuss briefly the Cost Breakdown Structure

PM 17
Total Profit
2 (7.1 %)
Labour
Cost
10 (35.7%)
Machine
Cost
5 (17.8 %)
Office
Overhead
1 (3.5%)
Overhead
3 (10.7%)
Factory Cost
27 (96.4%)
Total Price
30 M$
Total Cost
28 (100%)
Critical
Resources
25 (89.2%)
Materials
Cost
10 (35.7 %)
Technical
Overhead
2 (7.1 %)
Base
Annual Level
Planned Capacity
= 100,000
units/year

PM 18
Unit Profit
20 (7.1 %)
Labour
Cost
100 (35.7%)
Machine
Cost
50 (17.8 %)
Office
Overhead
10 (3.5%)
Overhead
30 (10.7%)
Factory Cost
270 (96.4%)
Unit Price
300 $/unit
Unit Cost
280 (100%)
Critical
Resources
250 (89.2%)
Materials
Cost
100 (35.7 %)
Technical
Overhead
20 (7.1 %)
Base
Unit Level
Planned Capacity
= 100,000
units/year
Price Policy:
Target Price = 300 $/unit
Worst Price = 270 $/unit

PM 19
• Planned Cost
• Actual Cost
• Variance Analysis
Each Indicator:
• Value ($)
• Ratio (%)
• Factor
Profit
Ratio
Value
Added
Margin
Factor
Markup
Factor
Breakeven
Point
Cost Analysis
Main Indicators:
- Profit = Price - Cost  Total Productivity
- Value Added Factor = Price / Mat. Cost  Material Productivity
- Margin Factor = Price / Direct Cost  Direct Resource Productivity
- Markup Factor = Price /Factory Cost  Factory Productivity
- Breakeven Point = F / (p-v)  Margin of Safety

PM 20
Price
TC
Profit
Total
Cost
Value
Added
1.5 to 2.5
 2.0
External
Resource
Cost
Margin
Factor
1.3 to 1.7
 1.5
Direct
Cost
Markup
Factor
1.2 to 1.3
 1.25
Factory
Cost
Or
Site Cost
Contribution
Margin
1.4 to 1.6
 1.5
Variable
Cost
Cost Analysis

PM 21
Price Structure Analysis: (Cost KPIs - Objectives)
Contract Price
Margin = Price - Direct
Direct Cost
Total Cost = Direct Cost + Overhead Profit
Value Added = Price - Outsource
Outsource Cost
or rough cut estimation by using the value added
or rough cut estimation by using the margin factor
or rough cut estimation by using the markup factor
Markup (Cost Plus)
= Office O/h + Profit
Labor
Mat. Jobsite OH
Subs
Equip
Factory Cost = Direct + Tech. O/h
For same scope of work
For different scope of work
For Cost Plus Contract

PM 22
The project cost information is as follows:
• Project : Office Building 10 Floors
• Bill of Quantity (BOQ): 6000 m2
----------------
• Direct Material Cost: 18,000,000 LE
• Direct Machine Cost: 6,000,000 LE
• Direct Labor Cost: 6,000,000 LE
----------------
• Site Overhead 3,000,000 LE
• Office Overhead: 1,500,000 LE
• Unit Price: 7,000 LE/m2
Case Study:
Actual Cost
(2018)
Inflation Rate
15% Annually
Cost
Estimation
(2019 , 2020)

PM 23
Total Profit
7.5 (21.7%)
Labour
Cost
6 (17.4%)
Machine
Cost
6 (17.4 %)
Office
Overhead
1.5 (4.3%)
Overhead
4.5 (13.0%)
Site Cost
33 (95.6%)
Total Price
42 M.LE
Total Cost
34.5 (100%)
Critical
Resources
30 (86.9%)
Materials
Cost
18 (52.2 %)
Site
Overhead
3 (8.6%)
Base
Project Level
Actual Cost
(2018)

PM 24
Unit Profit
1250 (21.7%)
Labour Cost
1000 (17.4%)
Machine Cost
1000 (17.4 %)
Office
Overhead
250 (4.3%)
Overhead
750 (13.0%)
Site Cost
5500 (95.6%)
Unit Price
7000 LE/unit
Unit Cost
5750 (100%)
Critical
Resources
5000 (86.9%)
Materials Cost
3000(52.2 %)
Site
Overhead
500 (8.6%)
Base
Unit Level
Actual Cost
(2018)
Inflation Rate
15%

PM 25
Unit Profit
1437 (21.7%)
Labour Cost
1150 (17.4%)
Machine Cost
1150 (17.4 %)
Office
Overhead
287 (4.3%)
Overhead
862 (13.0%)
Site Cost
6325 (95.6%)
Unit Price
8050 LE/unit
Unit Cost
6613 (100%)
Critical
Resources
5750 (86.9%)
Materials Cost
3450 (52.2 %)
Site
Overhead
575 (8.6%)
Base
Unit Level
Cost
Estimation
(2019)
Inflation Rate
15%

PM 26
Contract Price = 7000 LE/m2
Margin =7000-5000= 2000 LE
Direct Cost =5000
Total Cost = Direct Cost + Overhead = 5750
Profit
1250 LE
Value Added = 7000-3000= 4000 LE
Outsource Cost
3000
Markup (Cost Plus)
=7000-5500= 1500 LE
Labor
Mat. Jobsite OH
Subs
Equip
Site Cost = Direct + Tech. O/h = 5500
Actual Cost
(2018)

PM 27
Contract Price = 7000 LE/m2
Margin Factor =7000/5000= 1.4
(1 Direct : 1.4 Income)
Direct Cost =5000
Total Cost = Direct Cost + Overhead = 5750
Profit
21.7%
Value Added Factor = 7000/3000= 2.33
Material Productivity = (1 Material : 2.33 Income)
Outsource Cost
3000
Markup Factor
=7000/5500= 1.27
(Cost Plus)
Labor
Mat. Jobsite OH
Subs
Equip
Site Cost = Direct + Tech. O/h = 5500
Actual Cost
(2018)

PM 28
Unit Profit
-- (21.7%)
Labour Cost
--- (17.4%)
Machine Cost
-- (17.4 %)
Office
Overhead
-- (4.3%)
Overhead
-- (13.0%)
Site Cost
-- (95.6%)
Unit Price
--- LE/unit
Unit Cost
--- (100%)
Critical
Resources
-- (86.9%)
Materials Cost
-- (52.2 %)
Site
Overhead
-- (8.6%)
Base
Given:
Raw Material Cost
= 4000 LE/unit
Cost Estimation #1:

PM 29
• Direct Material Cost: 18 M.LE
• Direct Machine Cost: 6 M.LE
• Direct Labor Cost: 6 M.LE
----------------
• Technical Overhead 10% Direct Cost
• Office Overhead: 5% Direct Cost
• Profit Ratio 22% Cost
Total Cost = Direct + Overhead = 30*1.15 = 34.5 M.LE
Price = Total Cost + Profit = 34.5 * 1.22 = 42 M.LE
Cost Estimation #2:

PM 30
• Bill of Quantity (BOQ): 600*10 = 6000 m2
----------------
• Direct Machine Cost: 6 M.LE
• Direct Labor Cost: 6 M.LE
----------------
• Margin Factor 40 % Direct
(Margin Factor = Overhead + Profit = Price - Direct)
Price = Direct Cost * Margin Factor = 30 * 1.4 = 42 M.LE
Cost Estimation #3:

PM 31
----------------
----------------
• Value Added Factor 2.333
(Value Added Factor = Price / Material)
Price = Material Cost * Value Added Factor = 18*2.333 = 42 M.LE
Cost Estimation #4:

PM 32
• Project : Office Building One Floor
----------------
• Direct Material Cost: 600,000 LE
• Direct Machine Cost: 100,000 LE
• Direct Labor Cost: 300,000 LE
----------------
• Site Overhead 100,000 LE
• Office Overhead: 50,000 LE
• Unit Price: 5,000 LE/m2
Example: Actual Cost

PM 33
• Item (Main Activity): Plain Concrete
• Bill of Quantity (BOQ) 300 m3
• Direct Material Cost 120,000 LE
• Direct Machine Cost 54,000 LE
• Direct Labor Cost 24,000 LE
• Factory Overhead 33,000 LE
• Office Overhead 16,500 LE
• Unit Price 1,100 LE/m3
Example:

PM 34
• Item (Main Activity): Power Cable
• Bill of Quantity (BOQ) 1000 m
• Direct Material Cost 1,500,000
• Direct Machines 500,000
• Direct Labors 400,000
• Tech. Overhead 15% Direct Cost
• Office Overhead 5% Direct Cost
• Profit 20% Total Cost
Brainstorming:

PM 35
The project cost information is as follows: (Main Contractor)
• Project : Gas Pipe Line
• Bill of Quantity (BOQ) 100 Km
• Direct Material Cost 1,000,000 $
• Direct Sub-Contactor $ 400,000
• Direct Engineers $ 100,000
-----------
• Tech. Overhead 7% Total Cost
• Office Overhead 3% Total Cost
• Profit 20% Total Cost
Brainstorming:

PM 36
Total Profit
0.333 (20%)
Engineers
Cost
0.1 (6 %)
Subcontractor
Cost
0.4 (24 %)
Office
Overhead
0.050 (3%)
Overhead
0.167 (10%)
Site Cost
1.617(97%)
Total Price
2.0 M.$
Total Cost
1.667 (100%)
Critical
Resources
1.5 (90%)
Materials
Cost
1.0 (60 %)
Site
Overhead
0.117(7%)
Base
Project Level

PM 37
Total Profit
333 (20%)
Engineers
Cost
100 (6%)
Subcontractor
Cost
400 (24 %)
Office
Overhead
50 (3%)
Overhead
167 (10%)
Site Cost
1,617(97%)
Total Price
20,000 $/km
Total Cost
1,667 (100%)
Critical
Resources
1,500 (90%)
Materials
Cost
1,000 (60 %)
Site
Overhead
117 (7%)
Base
Unit Level

PM 38
Break-Even Analysis

PM 39
Total costs = fixed costs + variable costs
– Total fixed cost (TFC) is the cost of fixed inputs, inputs that do not
vary with output (e.g., rent)
– Total variable cost (TVC) is the cost of all inputs that vary with output
(e.g., wages, raw materials)
Breakeven Quantity of Sales:
BE
Total Fixed Costs
Q =
Price Var.Cost per unit


PM 40
• Determines at what level cost and revenue are in equilibrium
• Break-even point
– Obtained directly by mathematical calculations
– Usually presented in graphic form known as break-even chart
Break-Even Analysis:
Breakeven quantity =
(Total Fixed Cost) / (unit price – unit variable cost)
Fixed cost
Variable cost
Total cost line
Total revenue line
Profit
Breakeven point
Total cost = Total revenue
Profit = 0
Production Volume (units/period)
Cost
Loss
Cost-Volume-Profit Chart

PM 41
Breakeven - Margin of Safety

PM 42
Planned Capacity 60 units
Annual Fixed Cost 300
Annual Variable Cost 360
Revenue $ 900
Example:
Planned Capacity = 60 units
Unit variable cost = 360/60 = 6 $/unit
Unit Price = 900/60 = 15
BEP = F / (p-v) = 300 / (15-6) = 33.3 = 34
Margin of safety = 60 – 34 = 26 units

PM 43
Example:
Demand (units) 500
Unit price ($) 50
Unit Variable cost ($) 15
Fixed Cost ($) 10,000
Based on this information, discuss the profit & breakeven analysis
Profit = Revenue – Total Cost
= 500*50 – (10000 + 500*15) =
BEP:
Q = F / (p-v) = 10000 / (50-15) = 285.7
Q % = 285.7 / 500 = 57.14 %
Safety Margin = 100 – 57.14 = 42.86% > 30% (Excellent)

PM 44
Scenario Analysis
Worst-Case
Scenario
Most-Likely-
Case Scenario
Best-Case
Scenario
Demand (units) 400 500 600
Unit price ($) 48 50 53
Unit Variable cost ($) 17 15 12
Fixed Cost ($) 11,000 10,000 8,000
Total Cost
Revenue
Profit ($)
Profit (%)
BEQ (units)
BEQ (%)
Safety Margin (%)
Comment

PM 45
A company is building a workshop for producing metal windows. There is a
considerable degree of uncertainty in cost terms.
• Annual fixed costs = $ 300,000 to 500,000
• Variable cost per unit = $ 500 to 600
• Selling price per unit = $ 1,000 to 1,200
• Annual Production = 900 to 1,100 units
Worst-Case
Scenario
Most-Likely-
Case Scenario
Best-Case
Scenario
Demand (units) 900 1000 1100
Unit price ($) 1000 1100 1200
Unit Variable cost ($) 600 550 500
Fixed Cost ($) 500,000 400,000 300,000
Example:

PM 46
Year 1 2 3 4 5
Income:
Sales Quantity, units 4000 5000 6000 7000 8000
Unit Price; $ 3000 3250 3500 3750 4000
Variable Cost ($1000):
Production Cost 1000 1250 1500 1750 2000
Technical Overhead 500 625 750 875 1000
Fixed Cost ($1000):
Depreciation 5000 5000 5000 5000 5000
Maintenance (3% of investment) 900 900 900 900 900
Insurance (1% of investment) 300 300 300 300 300
Production Cost 1000 1250 1500 1750 2000
Office Overhead 500 625 750 875 1000
1) Economical Analysis (Unit Cost , Profit Ratio for each period , …)
2) Risk Analysis (Breakeven Analysis for each period )
3) Risk Analysis (Sensitivity Analysis for Market Price Change ± 10%)
4) Risk Analysis (Sensitivity Analysis for variable Cost Change ± 10%)
Based on this information, discuss the following:

PM 47
A manager is trying to decide between two
machines (A or B) to produce a certain product:
• Alt. 1: Machine A:
– Fixed Costs:
• Annual Depreciation: $120,000
• Annual Maintenance: $20,000
– Variable costs:
• Material: $2.25 / unit
• Labor: $6.25 / unit
• Alt. 2: Machine B:
– Fixed Costs:
• Annual Depreciation : $165,000
• Annual Maintenance: $35,000
– Variable costs:
• Material: $2.25 / unit
• Labor: $2.25 / unit
Based on this information, select the best machine.
Example: • Alt. 1: Machine A:
– Fixed Costs: 140,000
– Variable costs: 8.5 /unit
– TC(A) = 140,000 + 8.5 Q
• Alt. 2: Machine B:
– Fixed Costs: 200,000
– Variable costs: 4.5 /unit
– TC(B) = 200,000 + 4.5 Q
At BEQ:
TC(A) = TC(B)
140,000 + 8.5 Q = 200,000 + 4.5 Q
Q = 15,000
<15000 15000 >15000
A A,B B
B
A
15000
BEP

PM 48
EGR 312 - 22 48
Two alternatives exist for a machining process.
Alternative 1 has an initial cost of $10,000 and a salvage value of
$1000 after 5 years. Alternative 1 also has a variable cost of $1/unit
of product produced and an annual maintenance of $1000.
Alternative 2 has an initial cost of $15,000 and a salvage value of
$2,000 after 7 years. Alternative 2 also has a variable cost of
$0.80/unit of product produced and an annual maintenance cost of
$1200.
Example:
Based on this information, select the best machine.

PM 49
Total Cost
‫الكلية‬ ‫التكاليف‬
Capital Cost
‫اإلستثمار‬ ‫التكليف‬
‫ية‬
Running Cost
‫التشغيل‬ ‫التكاليف‬
‫ية‬
CAPEX and OPEX Analysis

PM 50
Assets:
The economic resources owned by a business that
can be used for future operations
‫األصول‬
:
Fixed Assets: ‫ثابته‬ ‫أصول‬
• Land
• Building
• Equipment
Current Assets: ‫متداولة‬ ‫أصول‬
• Cash
• Accounts Receivable
• Inventories
• Raw Materials
• Supplies
Intellectual Assets:
‫ملموسة‬ ‫غير‬ ‫أصول‬
)
‫فكرية‬
(
• Brand , Know-how

PM 51
CAPEX and OPEX Analysis
CAPEX Ratio + OPEX Ratio = 100%
Capital Expenses = CAPEX Ratio (Best CAPEX ≥ 25%)
= Fixed Asset Depreciation / Annual Total cost
Operating Expenses = OPEX Ratio (Best OPEX ≤ 75%)
= 100 – CAPEX

PM 52
Product / Project: xxxxxxx
Unit Price= -- LE/unit Planned Capacity = --- units/year
Product Cost Analysis: (Information)
Cost Classification Matrix
Cost Elements
Annual Fixed
Cost
Annual Variable
Cost
Direct Costs
1) Materials
2) Technical labors
3) Equipment & Tools
4) Sub-Contractor
Overheads
5) Technical Overhead Costs
6) Office Overhead Costs
Cost Analysis?
Cost
Account

PM 53
Unit Price= 300 $/unit (Market Price) Planned Capacity = 100,000 units/year
Cost Elements
Annual Fixed Cost
1,000,000 $
Annual Variable Cost
1,000,000 $
Direct
Costs
Raw Materials - 10
Technical labors 3 2
Equipment & Tools 9 1
Overheads
Technical Overhead 1 1
Office Overhead 1 -
a) Cost breakdown structure
b) Break even ratio & Margin of safety
c) Sensitivity analysis for material cost and unit price change (±20%)
Case Study:

PM 54
Cost Elements
Annual Fixed
Cost
1,000,000 $
Annual Variable
Cost
1,000,000 $
Total Cost
1,000,000 $
Direct
Costs
Raw Materials - 10 10
Technical labors 3 2 5
Equipment &
Tools
9 1 10
Overhe
ads
Technical
Overhead
1 1 2
Office Overhead 1 - 1
Total 1,000,000 $ 14 14 28
Unit Cost = 28,000,000 / 100,000 = 280 $/unit
Unit Profit = 300 – 280 = 20 $/unit (7.14%) << 20% “Low Profit Ratio”

PM 55
Total Profit
2 (7.1 %)
Labour
Cost
10 (35.7%)
Machine
Cost
5 (17.8 %)
Office
Overhead
1 (3.5%)
Overhead
3 (10.7%)
Factory Cost
27 (96.4%)
Total Price
30 M$
Total Cost
28 (100%)
Critical
Resources
25 (89.2%)
Materials
Cost
10 (35.7 %)
Technical
Overhead
2 (7.1 %)
Base
Annual Level
Planned Capacity
= 100,000
units/year

PM 56
Unit Profit
20 (7.1 %)
Labour
Cost
100 (35.7%)
Machine
Cost
50 (17.8 %)
Office
Overhead
10 (3.5%)
Overhead
30 (10.7%)
Factory Cost
270 (96.4%)
Unit Price
300 $/unit
Unit Cost
280 (100%)
Critical
Resources
250 (89.2%)
Materials
Cost
100 (35.7 %)
Technical
Overhead
20 (7.1 %)
Base
Unit Level
Planned Capacity
= 100,000
units/year
Price Policy:

PM 57
Total Fixed Cost = 14,000,000 $ Unit Variable Cost = 140 $/unit Unit Price= 300 $/unit
Breakeven %
BEP = Total Fixed Cost / (Unit Price - Unit Variable Cost)
= 14,000,000 / (300 – 140) = 87,500 unit
Planned Capacity = 100,000 units/year
BE % = 87,500 / 100,000 = 87.5%
Margin of Safety = 12.5% <<< 30 % (Very Low  High Risk)
Corrective Action: Try to reduce fixed cost by 20%
Fixed cost
Variable cost
Total cost line
Total revenue line
Profit
Breakeven point
87,500 units
Cost
Loss

PM 58
Sensitivity analysis for material cost and unit price change (±20%)
Deviation -20% -10%
0%
Base
10% 20%
Unit price -40 -10 20 50 80
Unit Material 40 30 20 10 0
Profit Matrix ($/unit)
Scenario Profit $/unit
Pessimistic 80
Most Likely 20
Optimistic -40

PM 59
Cost Elements
Annual Fixed Cost
1,000,000 $
Annual Variable Cost
1,000,000 $
Direct
Costs
Raw Materials - 10
Technical labors 2 3
Equipment & Tools 6 4
Overheads
Technical Overhead 1 1
Office Overhead 1 -
a) Cost breakdown structure
b) Break even ratio & Margin of safety
c) Sensitivity analysis for material cost and unit price change (±20%)
Case Study:

PM 60
Cost Elements
Annual Fixed
Cost
1,000,000 $
Annual Variable
Cost
1,000,000 $
Total Cost
1,000,000 $
Direct
Costs
Raw Materials - 10 10
Technical labors 2 3 5
Equipment &
Tools
6 4 10
Overhe
ads
Technical
Overhead
1 1 2
Office Overhead 1 - 1
Total 1,000,000 $ 10 18 28
Unit Cost = 28,000,000 / 100,000 = 280 $/unit
Unit Profit = 300 – 280 = 20 $/unit (7.14%) << 20% “Low Profit Ratio”

PM 61
Total Profit
2 (7.1 %)
Labour
Cost
10 (35.7%)
Machine
Cost
5 (17.8 %)
Office
Overhead
1 (3.5%)
Overhead
3 (10.7%)
Factory Cost
27 (96.4%)
Total Price
30 M$
Total Cost
28 (100%)
Critical
Resources
25 (89.2%)
Materials
Cost
10 (35.7 %)
Technical
Overhead
2 (7.1 %)
Base
Annual Level
Planned Capacity
= 100,000
units/year

PM 62
Unit Profit
20 (7.1 %)
Labour
Cost
100 (35.7%)
Machine
Cost
50 (17.8 %)
Office
Overhead
10 (3.5%)
Overhead
30 (10.7%)
Factory Cost
270 (96.4%)
Unit Price
300 $/unit
Unit Cost
280 (100%)
Critical
Resources
250 (89.2%)
Materials
Cost
100 (35.7 %)
Technical
Overhead
20 (7.1 %)
Base
Unit Level
Planned Capacity
= 100,000
units/year
Price Policy:

PM 63
Total Fixed Cost = 10,000,000 $ Unit Variable Cost = 180 $/unit Unit Price= 300 $/unit
Breakeven %
BEP = Total Fixed Cost / (Unit Price - Unit Variable Cost)
= 10,000,000 / (300 – 180) = 83,333 unit
Planned Capacity = 100,000 units/year
BE % = 83,333 / 100,000 = 83.3%
Margin of Safety = 16.7% <<< 30 % (Very Low  High Risk)
Corrective Action: Try to reduce fixed cost by 20%
Fixed cost
Variable cost
Total cost line
Total revenue line
Profit
Breakeven point
83,333 units
Cost
Loss

PM 64
Sensitivity analysis for material cost and unit price change (±20%)
Deviation -20% -10%
0%
Base
10% 20%
Unit price -40 -10 20 50 80
Unit Material 40 30 20 10 0
Profit Matrix ($/unit)
Scenario Profit $/unit
Pessimistic 80
Most Likely 20
Optimistic -40

PM 65

PM 66

PM 67

PM 68

PM 69

PM 70

PM 71
Product Name : Microwave Product Code: FMW-25KC-S
Unit Price= 2000 LE/unit Planned Capacity = 20,000 units/year
Product Cost Analysis:
Cost Classification Matrix
Cost Elements Fixed Cost Variable Cost
Direct
Cost
1) Materials 20,000,000
2) Technical labors 3,000,000 2,000,000
3) Equipment & Tools 2,000,000 1,000,000
4) Sub-Contractor - -
Overhead
Cost
5) Technical Overhead Costs 1,500,000 1,500,000
6) Office Overhead Costs 1,500,000 500,000
Cost Analysis?

PM 72
Cash Flow Analysis
Time is money;
there is a difference between $1000 today and $1000 after 3 years.
If the interest rate (i) 10% annual; then:
$ 1000 today = $ 1331 after 3 years
How
Financial Analysis
(Long Term  Project Life)
F = P (1 + i)n
P = F / (1 + i)n

PM 73
Cash Flow
Cash Out
Initial
Investment
Running
Cost
• Equipment
• Building
• Know-How
• … etc.
• Production
• Salaries
• Marketing
• Maintenance
• … etc.
Cash In
Sales
Revenue
Salvage
Value
• Quantity
• Price
• … etc.
• Equipment
• Building
• … etc.
Cash Flow Analysis:
CAPEX OPEX CAPEX

PM 74
Year 0 1 2 3 4 … …
Equipment (Assets)
Building (Assets)
Production Cost
……
……
……
Cost Elements (1000 LE):
Sales Quantities, units
Unit Market Price
Salvage Value 4000
Cash Out Analysis
Cash In Analysis
Income (1000 LE):
Analysis Tools?
Comments?
Salvage Value
‫المصروفات‬ ‫قائمة‬
:
‫اإلرادات‬ ‫قائمة‬
:
Source:
Technical Study
Source:
Market Study
Project:
Feasibility Study: ‫الجدوي‬ ‫دراسة‬
:

PM 75
Year 0 1 2 3 4 5 6
Equipment (Assets) 7000
Building (Assets) 3000
Production Cost 2000 2600 2600 2600 2600 2600
Salaries Cost 300 300 300 300 300 300
Marketing Cost 200 250 250 250 250 250
Overhead 100 100 100 100 100 100
Sales Quantities, units 400 550 600 600 600 600
Unit Market Price 10 10 10 10 10 10
Salvage Value 4000
Salvage Value 4000
Cost Elements (1000 LE): Cash Out Analysis ‫المصروفات‬ ‫قائمة‬
:
Cash In Analysis
Income (1000 LE): ‫اإلرادات‬ ‫قائمة‬
:
Project: ABC Manufacturing Company
:
1) Cash Flow Analysis
2) Financial Analysis (Payback Period , IRR , … )

PM 76
Cash Flow Analysis
i = Interest rate (%) n = Life time (year)
P = Present value  Value at time = 0
F= Future value  Value at time > 0
A= Annual uniform value  Equal value over time
• Interest rate  Certain bank  i% given (10%)
• Minimum Attractive Rate of Return (MARR)
• Internal Rate return  Cash In/out for Certain project
P
A F
Cost
(Cash Out)
Revenue
(Cash In)
(-)
(+)
F = P (1 + i)n
0
(F , P , i , n )

PM 77
Analysis Tools?
Financial
&
Economical
Analysis
1) Financial Analysis (Project Life):
• Net Cash Flow
• Payback Period
• Internal Rate of Return IRR
---------------------------
• Net Present Value NPV at i %
• CAPEX% & OPEX%
• Return On Investment
• ……
1
)
‫المالي‬ ‫التحليل‬
)
‫عمر‬ ‫إلجمالي‬
‫المشروع‬
:)
•
‫المالي‬ ‫التدفق‬ ‫صافي‬
•
‫فترة‬
‫المال‬ ‫رأس‬ ‫استرداد‬
•
‫الداخلي‬ ‫المعدل‬
‫المالي‬ ‫للعائد‬
---------------------------
•
‫للمشروع‬ ‫الحالية‬ ‫القيمة‬ ‫صافي‬
•
‫المصروفات‬ ‫نسبة‬
‫مالية‬ ‫الرأس‬
/
‫التشغيلية‬
•
‫اإلستثمار‬ ‫علي‬ ‫العائد‬ ‫متوسط‬
•
…..
‫للمشروع‬
‫واإلقتصادي‬
‫المالي‬
‫التحليل‬
2) Economical Analysis (Annually):
• Unit Cost
• Annual Profit%
---------------------------
• Cost Breakdown Structure
• CAPEX% & OPEX%
2
)
‫االقتصادي‬ ‫التحليل‬
)
‫سنويا‬
:)
•
‫تكلفة‬
‫الوحدة‬
•
‫السنوي‬ ‫الربح‬ ‫نسبة‬
---------------------------
•
‫التكاليف‬ ‫عناصر‬ ‫هيكل‬
•
‫المصروفات‬ ‫نسبة‬
‫مالية‬ ‫الرأس‬
/
‫التشغيلية‬
3) Risk Analysis (Financial / Economical) :
• Break Even Point
• Sensitivity Analysis for Material
Cost & Unit Price Change
•
3
)
‫المخاطر‬ ‫تحليل‬
)
‫مالي‬ ‫المحتملة‬
/
‫إقتصادي‬
:)
•
‫إنتاج‬ ‫حجم‬
‫نقطة‬
‫التعادل‬
•
‫الخامات‬ ‫تكلفة‬ ‫تغير‬ ‫إحتمال‬ ‫حساسية‬ ‫تحليل‬
‫وخالفه‬ ‫الوحدة‬ ‫وسعر‬
•
:

PM 78
Project Financial Analysis:
• Payback Period (Breakeven Period); (years)
•
‫المال‬ ‫رأس‬ ‫إسترداد‬ ‫فترة‬
• Net Present Value (NPV)  (i% given)
•
‫للمشروع‬ ‫الحالية‬ ‫القيمة‬
• Profitability Index  (i% given)
•
‫الربجية‬ ‫معامل‬
• Internal Rate of Return (IRR)  (i% Unknown)
•
‫الداخلي‬ ‫العائد‬ ‫معدل‬
)
‫االستثمار‬ ‫علي‬ ‫العائد‬ ‫معدل‬
(
• Minimum Attractive Rate of Return (MARR)  Given
•
‫للمستثمر‬ ‫جاذب‬ ‫عائد‬ ‫معدل‬ ‫أقل‬
Most Common

PM 79
Example:
Payback period = 2 + 25 / 75 = 2.333 years
Firm recovers its initial investment in 2.333 years
Year
Net Cash
Flow
Cumulative
Cash Flow
0 –100 –100
1 25 –75
2 50 –25
3 75 50
Payback Period
Definition:
The length of time until the original investment has been recouped by the project
• Ignore time value of money
• Does not consider profitability
• A shorter payback period is better

PM 80
Example:
Payback Period = (2 to 3) years
= 2 + 300/350 = 2.857 years = 2 years & 11 months
$1000
$350
years
1 2 3 4
$350
$350 $350
Year 0 1 2 3 4
Cost -1000
Revenue +350 +350 +350 +350
Cumulative
Cash Flow
-1000 - 650 - 300 + 50 +400
1400
1000
Payback Period

PM 81
Example:
Payback Period = (2 to 3) years
= 2 + 300/500 = 2.6 years = 2 years & 7.2 months
$1000
$300
years
1 2 3 4
$500
$400 $400
Year 0 1 2 3 4
Cost -1000
Revenue +300 +400 +500 +400
Cumulative
Cash Flow
-1000 - 700 - 300 + 200 +600
1600
1000
Payback Period

PM 82
Net Present Value (NPV)
Converting Cash Flows to Present Value
End of project
Time zero:
Initial Investment = $105,000
TIME
Year 1 Year 2 Year 3
$38,463 $38,463 $38,463
= ??
= ??
= ??
Annual Savings
• A higher NPV is better
• Higher the discount rate lower the NPV
 
 



n
t t
t
i
F
P 1
0
1
NPV

PM 83
Net Present Value
 
63
64
.
0
*
350
71
.
0
*
350
80
.
0
*
350
89
.
0
*
350
000
,
1
12
.
1
350
12
.
1
350
12
.
1
350
12
.
1
350
000
,
1 4
3
2














Investment
NPV
Interest rate
of i =12%.
-$1,000
$350
years
1 2 3 4
$350
$350 $350
Example:
P = F / (1 + i)n
‫؟‬
(NPV +)  IRR > 12%
NPV (cash inflow)
Loan
or
MARR
-
0
+
IRR < i
IRR = i
IRR > i
NPV

PM 84
Profitability Index:
• Allows a comparison of the costs and benefits of different projects to be
assessed and thus allow decision making to be carried out
NPV (cash inflow)
Profitability Index = ---------------------------
Initial Capital Cost
PI < 1.0 Losses  Rejected (or Service projects)
PI = 1.0 Break even point  short term projects
PI > 1.0 Profit  Accepted )Best ≥ 1.20)
-$1,000
$350
years
1 2 3 4
$350
$350 $350
Profitability Index = 1063 / 1000 = 1.063 > 1.0 Accepted
Cost Benefits
(B/C)
B
C
Interest rate
of 12%
NPV (cash inflow)

PM 85
Internal Rate of Return (IRR)
Definition:
The rate of return that yields a Net Present Value of zero.
(rate at which NPV=0)
Example:
-$1,000
$350
years
1 2 3 4
$350
$350 $350
15
.
0
0
000
,
1
)
1
(
350
)
1
(
350
)
1
(
350
1
350
4
3
2











IRR
IRR
IRR
IRR
IRR
Example:
P = F / (1 + i)n
14.96%
IRR
(i% Unknown)

PM 86
Return On Investment (ROI)
Year 0 1 2 3 Total to end Year 3
Cost (Outflows) 100 30 30 35 195
Income (Inflows) 0 100 95 75 265
ROI is a ratio comparing net gains to net costs.
(measure of investment profitability)
For example:

PM 87
A
Year 0 1 2 3
ROI = 20%
IRR = 10%
Cost (Outflows) 10 2 2 2
Income (Inflows) 6 6 6
Net Flows -10 4 4 4
ROI & IRR
B
Year 0 1 2 3
ROI = 20%
IRR = 11%
Net Flows -10 5 5 2
C
Year 0 1 2 3
ROI = 20%
IRR = 9%
Net Flows -10 2 5 5

PM 88
You can purchase a building for $350,000. The investment will generate
$16,000 in cash flows (i.e. rent) during the first three years. At the end
of three years you will sell the building for $450,000.
What is the IRR on this investment?
0 350 000
16 000
1
16 000
1
466 000
1
1 2 3
  





,
,
( )
,
( )
,
( )
IRR IRR IRR
IRR = 12.96% ≥ MARR
Example:
350
16
years
1 2 3
16
450
16 P = F / (1 + i)n
Feasibility Study For Small Projects

PM 89
Period Cash 2 Years 3 Years
0 200,000 75,000 60,000
1 - 75,000 60,000
2 - 75,000 60,000
3 60,000
You need to buy a new car, three alternatives are available :
1. Cash Flow Diagram
2. Internal Rate of Return (IRR)
Example:

PM 90
2
)
1
(
75
1
75
75
200
i
i 




3
2
)
1
(
60
)
1
(
60
1
60
60
200
i
i
i 






75
years
1 2
0
75 75
P= 200
=
B
P= 200
=
60
years
1 2 3
0
60 60 60
C
i = IRR = 13.066%
i = IRR = 13.7009%
PV (A) < PV (B) < PV (C) & IRR (B) < IRR (C)
The best choice is A then B then C

PM 91
Initial Cost: $4000
Annual M&O Costs: $500
Salvage Value $500
1- Pump (A)
Initial Cost: $4500
Salvage Value $500
Based on this information, discuss & select the best pump.
- Centrifugal Pump 1.0 HP - Interest rate is 10%
- Average Annual Costs are constant - Life 10 years
Assumptions / Calculations:
2- Pump (B)
Example:

PM 92
Initial Cost: $4000
Capital Maintenance
after 6 years $3000
Salvage Value $500
1- Pump (A)
Initial Cost: $5000
Capital Maintenance
after 6 years $2000
Salvage Value $500
Based on this information, discuss & select the best pump.
- Centrifugal Pump 1.0 HP - Interest rate is 10%
- Average Annual Costs are constant - Life 10 years
Assumptions / Calculations:
2- Pump (B)
PV: A= 8,572.9 B= 8,394.0 (Ranking: B < A)
Example:

PM 93
A company needs a mobile crane for the next 10 years. Two alternatives are
available to buy or to lease the crane.
1) The buy decision includes a first cost of 800,000 LE with a market value
after 10 years of 100,000 LE and annual O&M cost of 30,000 LE.
2) To lease a similar crane, the annual charge will be 120,000 LE
including everything.
If the company earns 20% annually which alternative is to be selected.
Example:
By using IRR method.

PM 94
800
30
100
120
680
100
90
IRR = 5.1 < 20% (Rent)
or NPV(20%) = - 305.9
Net
800
30
100
120
MARR 20%
Cash Out
Cash In
0
0
IRR

PM 95
800
30
100
120
MARR 20%
Buy
Rent
NPV (i=20%) =
909.6
NPV (i=20%) =
603.7 (min.)
0
0
NPV

PM 96
‫المشروع‬ ‫وصف‬
‫المكان‬
:
‫الجديدة‬ ‫القاهرة‬
/
‫األندلس‬ ‫منطقة‬
‫األرض‬ ‫مساحة‬
:
500
‫م‬
2
‫مباني‬ ‫مساحة‬
:
6000
‫م‬
2
‫المواصفات‬
:
‫بدروم‬
+
‫أرضي‬
+
3
‫أدوار‬
+
‫رووف‬
‫األرضي‬
:
‫شقة‬
180
‫م‬
2
+
‫دوبلكس‬
330
‫م‬
2
‫الدور‬
:
‫عدد‬
2
‫شقة‬
*
180
‫م‬
2
=
360
‫م‬
2
‫التكاليف‬ ‫عناصر‬ ‫متوسط‬
)
2017
)
‫األرض‬ ‫قيمة‬
:
6000
‫جنيه‬
/
‫م‬
2
‫التنفيذ‬ ‫تكاليف‬
:
6000
‫جنيه‬
/
‫م‬
2
‫اإلشراف‬ ‫نسبة‬
:
10
%
‫التنفيذ‬ ‫من‬
‫التسويق‬ ‫نسبة‬
:
5
%
‫المبيعات‬ ‫من‬
‫الحديد‬ ‫سعر‬
:
12300
‫جنيه‬
/
‫طن‬
‫سعر‬
‫األسمنت‬
:
830
‫جنيه‬
/
‫طن‬
‫المبيعات‬ ‫عناصر‬ ‫متوسط‬
)
2017
)
‫األرضي‬
:
‫شقة‬
:
7300
‫جنيه‬
/
‫م‬
2
‫دوبلكس‬
:
5500
‫جنيه‬
/
‫م‬
2
‫األدوار‬
:
‫شقة‬
:
7500
‫جنيه‬
/
‫م‬
2
‫الجراج‬
:
50000
‫جنيه‬
‫التضخم‬ ‫معامل‬
1
1
%
‫سنويا‬
‫عقاري‬ ‫إلستثمار‬ ‫جدوي‬ ‫دراسة‬
‫التنفيذ‬ ‫مدة‬
=
12
‫شهر‬

PM 97
Project: Car Rent
Initial Investment = 500,000 LE
Project Life = 10 years Salvage Value 100,000
Annual Running Cost = 8% Initial investment
Annual Maintenance Cost = 3% Initial investment
Annual Risk and Insurance = 1% Initial investment
Annual Working Days = 200 day/year
Daily Rent Price = 700 LE/day

PM 98
Year 0 1 2 3 4 5 6
Production Cost 2000 2600 2600 2600 2600 2600
Salaries Cost 300 300 300 300 300 300
Marketing Cost 200 250 250 250 250 250
Overhead 100 100 100 100 100 100
Sales Quantities, units 400 550 600 600 600 600
Unit Market Price 10 10 10 10 10 10
Salvage Value 4000
Salvage Value 4000
Cost Elements (1000 LE): Cash Out Analysis ‫المصروفات‬ ‫قائمة‬
:
Cash In Analysis
Income (1000 LE): ‫اإلرادات‬ ‫قائمة‬
:
:
1) Cash Flow Analysis
2) Financial Analysis (Payback Period , IRR , … )

PM 99
(1000 LE):
Year 0 1 2 3 4 5 6
Total Cost 10000 2600 3250 3250 3250 3250 3250
Salvage Value 4000
Sales Revenue 4000 5500 6000 6000 6000 6000
Net Cash Flow -10000 1400 2250 2750 2750 2750 6750
Cash Flow Analysis:
Payback Period = 4 + 850 / 2750 = 4.31 years = 4 year + 4 months
Accumulative Net Cash -10000 -8600 -6350 -3600 -850 +1900 +8650
1) Financial Analysis
-15000
-10000
-5000
0
5000
10000
0 2 4 6 8
Manual

PM 100
Internal Rate of Return (IRR) = 16.7% ≥ MARR
(1000 LE):
Year 0 1 2 3 4 5 6
Total Cost 10000 2600 3250 3250 3250 3250 3250
Salvage Value 4000
Sales Revenue 4000 5500 6000 6000 6000 6000
Net Cash Flow -10000 1400 2250 2750 2750 2750 6250
Cash Flow Analysis:
P = F / (1 + i)n
NPV = 0 = -10000 + {1400/(1+i)1 +2250/ (1+ i)2 +2750/ (1+ i)3
+2750/ (1+ i)4 +2750/ (1+ i)5 +6250/ (1+ i)6 }
1) Financial Analysis
Or Trail & error method:
i= 10%  NPV = + 2,312.13
i= 20%  NPV = - 1,154.93
i ?  NPV = 0
i = IRR
= 16.7%
NPV
i %
0
Manual
F = P (1 + i)n

PM 101
Year 1 2 3 4 5 6
Annual Depreciation 1000 1000 1000 1000 1000 1000
Production 2000 2500 2500 2500 2500 2500
Salaries 300 300 300 300 300 300
Marketing 200 250 250 250 250 250
Overhead 100 100 100 100 100 100
Total Cost 3600 4150 4150 4150 4150 4150
Sales Quantities 400 550 600 600 600 600
Unit Price 10 10 10 10 10 10
Sales Revenue 4000 5500 6000 6000 6000 6000
Unit Cost 9.0 7.5 6.9 6.9 6.9 6.9
Unit Price 10 10 10 10 10 10
Profit % 11.1% 32.5% 44.6% 44.6% 44.6% 44.6%
Annual Cost (1000 LE):
Annual Revenue (1000 LE):
2) Economical Analysis:
Cost Analysis (1000 LE):
Annual Depreciation = (10000 – 4000)/6 = 1000
(Straight Line Method)

PM 102
Sensitivity Analysis for Market Price Change ± 10%
3) Risk Analysis:
Indicator
Change in Market Price
- 10% 0% + 10%
Payback Period ; years 5.13 4.31 3.61
IRR 10.5% 16.7% 22.6%
Financial
Risk:
Economical
Risk
(Year #1):
Indicator
- 10% 0% + 10%
Unit Cost 9.0 9.0 9.0
Unit Price 9.0 10 11
Profit % 0 % 11.1% 22.2%
Year 0 1 2 3 4 5 6
Unit Market Price (Base) 10 10 10 10 10 10
+10% 11 11 11 11 11 11
-10% 9 9 9 9 9 9

PM 103
(1000 LE):
Year 0 1 2 3 4 5 6
Total Cost 10000 2600 3250 3250 3250 3250 3250
Salvage Value 4000
Sales Revenue 4400 6050 6600 6600 6600 6600
Net Cash Flow -10000 1800 2800 3350 3350 3350 7350
Cash Flow Analysis:
Payback Period = 3 + 2050 / 3350 = 3.61 years = 3 year + 7 months
Accumulative Net Cash -10000 -8200 -5400 -2050 1300 4650 12000
Financial Risk Change in Market Price +10% = 11 LE/unit
-15000
-10000
-5000
0
5000
10000
15000
0 1 2 3 4 5 6 7

PM 104
(1000 LE):
Year 0 1 2 3 4 5 6
Total Cost 10000 2600 3250 3250 3250 3250 3250
Salvage Value 4000
Sales Revenue 3600 4950 5400 5400 5400 5400
Net Cash Flow -10000 1000 1700 2150 2150 2150 6150
Cash Flow Analysis:
Payback Period = 5 + 850 /6150 = 3.13 years = 3 year + 1 month
Accumulative Net Cash -10000 -9000 -7300 -5150 -3000 -850 5300
Financial Risk Change in Market Price -10% = 9 LE/unit
-15000
-10000
-5000
0
5000
10000
15000
0 1 2 3 4 5 6 7 8

PM 105
Sensitivity Analysis for Operating Cost Change ± 10%
3) Risk Analysis:
Indicator
- 10% 0% + 10%
Payback Period ; years 4.31
IRR 16.7%
Financial
Risk:
Economical
Risk
(Year #1):
Indicator
- 10% 0% + 10%
Unit Cost 9.0
Unit Price 10
Profit % 11.1%
Year 0 1 2 3 4 5 6
Operating Cost (Base) 2600 3250 3250 3250 3250 3250
+10% 2860 3575 3575 3575 3575 3575
-10% 2340 2925 2925 2925 2925 2925

PM 106
Year 0 1 2 3 4 ….
Income:
Sales Quantity, units
Unit Price, $
Initial Investment ($1000):
Equipment (Assets)
Building (Assets)
Cost Elements ($1000):
Direct Production Cost
Study
Assumptions:
Project: Manufacturing Company
Feasibility Study Guidelines (Dr. Attia Gomaa)
• Annual Maintenance Cost = 3% of initial investment
• Annual Insurance = 1% of initial investment
• Salvage Value after 5 years = 50% of initial investment
• Technical Overhead = 50% of Direct Production Cost
• Office Overhead = 25% of Direct Production Cost

PM 107
Year 0 1 2 3 4 5
Income:
Unit Price, $ 3000 3250 3500 3750 4000
Direct Production Cost 2000 2500 3000 3500 4000
Study
Assumptions:
1) Financial Analysis (Cash Flow, Payback Period , IRR , … )
3) Risk Analysis (Sensitivity Analysis for Market Price & Production Cost Change ± 10%)

PM 108
Year 0 1 2 3 4 5
Income:
Unit Price, $ 3000 3250 3500 3750 4000
Salvage Value 15000
Direct Production Cost 2000 2500 3000 3500 4000
Office Overhead 500 625 750 875 1000
Maintenance (3% of investment) 900 900 900 900 900
Insurance (1% of investment) 300 300 300 300 300

PM 109
Year 0 1 2 3 4 5 6
Income:
Sales Quantity, units 500 550 600 650 700 750
Unit Price, $ 10.0 10.5 11.0 11.5 12.0 12.5
Direct Production Cost 2000 2500 3000 3500 4000 4500
Study
Assumptions:

PM 110
Project (B)
Project (A)
Item
7
5
Initial Investment; M$
2
2
Annual Running Cost; M$/year
5
4
Annual Revenue; M$/year
2
1
Salvage Value; M$
4
4
Project Life; years
A manager is trying to decide between two projects (A or B):
2. Payback Period
4. Which offer should you accept and why?
5. Sensitivity analysis for annual revenue , salvage value & initial investment (±10%)
Minimum Attractive Rate of Return (MARR) 20%
Example:

PM 111
Project (A)
Cash Flow Diagram: Project (B)
Payback Period = 5/2 = 2.5 years
= 2 years & 6 months
IRR = 26.40 % > 20% MARR (Accepted)
Payback Period = 7/3 = 2.33 years
IRR = 31.56 % > 20% MARR (Accepted)
NPV = 0 = - 5 +2/(1+i) + 2/(1+i)^2
+ 2/(1+i)^3 + 3/(1+i)^4
NPV = 0 = - 7 +3/(1+i) + 3/(1+i)^2
+ 3/(1+i)^3 + 5/(1+i)^4
IRR (B) > IRR (A) > MARR 20%
Project B is better than project A
7
1 2
0 3 4
2
5
2
7
0 1 2 3 4
3 2
5
1 2
0 3 4
2
4
1
5
0 1 2 3 4
2 1
Net cash flow =
Cash inflows - Cash outflows

PM 112
Project (B)
Payback Period = 7/2.5 = 2.8 years
IRR = 22.96 %
NPV = 0 = - 7 +2.5/(1+i) + 2.5/(1+i)^2
+ 2.5/(1+i)^3 + 4.5/(1+i)^4
7
1 2
0 3 4
2
5-0.5 2
7
0 1 2 3 4
2.5 2
Project (B)
Payback Period = 7/3.5 = 2 years
IRR = 39.94 %
NPV = 0 = - 7 +3.5/(1+i) + 3.5/(1+i)^2
+ 3.5/(1+i)^3 + 5.5/(1+i)^4
7
1 2
0 3 4
2
5+0.5 2
7
0 1 2 3 4
3.5 2
Sensitivity Analysis for Project (B) Annual revenue (±10%)

PM 113
Financial Risk Analysis:
Indicator
Change in Annual Revenue
- 10% 0% + 10%
Payback Period ; years 3 3 2
IRR 22.96 % 31.56 % 39.94 %
Sensitivity Analysis for Project (B)
Indicator
Change in Salvage Value
- 10% 0% + 10%
IRR 31.02 31.56 % 32.09
Indicator
Change in Initial Investment
- 10% 0% + 10%
IRR 37.70 31.56 % 26.37
Scenarios
analysis (IRR%):
Optimistic 39.94
Most likely 31.56
Pessimistic 22.96

PM 114
Sensitivity Analysis Graph for Project (B)
IRR%
-5
5
10
20
30
40
60
80
-30 -20 -10 0 10 20 30
Base
Annual
revenue
Initial
Invest
% change from base
Deviation -20% -15% -10% -5%
0%
Base
5% 10% 15% 20%
Annual revenue 22.96 31.56 % 39.94
Salvage value 31.56 %
Initial Investment 31.56 %
IRR Matrix
Sketch

PM 115
Project (B)
Project (A)
Item
8
6
3
3
6
5
3
2
Salvage Value; M$
5
5
Project Life; years
2. Payback Period
5. Sensitivity analysis for annual revenue , salvage value & initial investment (±10%)
Example:

PM 116
Project (A) Project (B)
Payback Period = 3 years
IRR = 25.25% > MARR 20%
Payback Period = 2.67 years
IRR = 30.7% > MARR 20%
Sensitivity analysis:
Deviation -10%
0%
Base
10%
Annual revenue 4.5 5 5.5
IRR% 15.12 25.2 % 35
Payback Period 4.0 3 2.4
Scenarios analysis (IRR%):
Optimistic 35%
Most likely 25.5
Pessimistic 15.2%
Deviation -10%
0%
Base
10%
IRR% 21.9 30.72 39.2
Payback Period 3.3 2.67 2.22
Optimistic 39.2%
Most likely 30.72%
Pessimistic 21.9%

PM 117
Project (B)
Project (A)
Item
5
4
2
1
5
3
1
1
Salvage Value; M$
4
4
Project Life; years
2. Payback Period
5. Sensitivity analysis for market price change (±10%)
Loan 15%
MARR = 20%
Example:

PM 118
Scenario Analysis
Item
Best
Base
Worst
25%
50%
25%
Probability; %
4
5
5.1
3
4
4.1
6
5
4.9
3
2
1.9
Salvage Value; M$
4
4
4
Project Life; years
Example:
2. Payback Period
Risk Analysis

PM 119
Scenario Analysis
Item
Best
Base
Worst
25%
50%
25%
Probability; %
4
5
5.1
3
4
4.1
6
5
4.9
3
2
1.9
Salvage Value; M$
4
4
4
Project Life; years
2
4
4
Payback Period (years)
72.7
6.4
0
Risk Analysis
E (Payback Period) = 4 * 0.25 + 4 * 0.50 + 2 * 0.25= 3.5 years
E (IRR) = 0 * 0.25 + 6.4 * 0.50 + 72.7 * 0.25 = 21.4 %
Estimation:

PM 120
Scenario Analysis
Item
Best
Base
Worst
25%
50%
25%
Probability; % (by default)
4
5
6
3
4
5
6
5
4
3
2
1
Salvage Value; M$
4
4
4
Project Life; years
Example:
2. Payback Period
Risk Analysis

PM 121
Scenario Analysis
Item
Best
Base
Worst
25%
50%
25%
Probability; %
4
5
6
3
4
5
6
5
4
3
2
1
Salvage Value; M$
4
4
4
Project Life; years
2
4
N/A
Payback Period (years)
72.7
6.4
N/A
Infeasible
Risk Analysis
E (Payback Period) =
E (IRR) =
Estimation:

PM 122
Project (B)
Project (A)
Item
One year
One year
Construction Period
5
4
2
1
5
3
1
1
Salvage Value; M$
4
4
Life after construction; years
2. Payback Period
Loan 15%
MARR = 20%
(Construction Period)
Example:

PM 123
Project (B)
Project (A)
Item
One year
Six months
Construction Period
6
5
3
2
6
4
2
2
Salvage Value; M$
6
5
Life after construction; years
2. Payback Period
MARR = 20%
(Construction Period)
Example:

PM 124
IRR = 27.7% > MARR 20%
IRR = 32.3% > MARR 20%
Deviation -10%
0%
Base
10%
IRR% 20.4 27.7 34.7
Optimistic 34.7
Most likely 27.7
Pessimistic 20.4
Deviation -10%
0%
Base
10%
IRR% 25.4 32.3 38.5
Optimistic 38.5
Most likely 32.3
Pessimistic 25.4

PM 125
Project (B)
Project (A)
Item
Two years
One year
Construction Period
$ 100,000 Down payment (t=0)
200,000 after one year
100,000 after two years
$ 300,000 Final payment
Construction
investment
50,000 first year
100,000 second year
150,000 third year
200,000 forth year
$ 100,000 / year
Running cost after
construction
150,000 first year
300,000 second year
500,000 third year
400,000 forth year
$ 250,000 / year
Revenue after
construction
Four years
Four years
Life after construction
$ 100,000
$ 100,000
Salvage value
Based on this information, discuss & select the best alternative.
Bank Interest Rate 10% Annual
Example: (Construction Period)

PM 126
Project (A)
Item
One year
Construction
Period
$ 300,000 Final payment
Construction
investment
$ 100,000 / year
Running cost
after
construction
$ 250,000 / year
Revenue after
construction
Four years
Life after
construction
$ 100,000
Salvage value
100 100
100
250
300
0 1
100
100
150
300
0 1
IRR = 21.7 < 25%
(Rejected)

PM 127
Project (B)
Item
Two years
Construction
Period
200,000 after one year
100,000 after two years
Construction
investment
50,000 first year
100,000 second year
150,000 third year
200,000 forth year
Running cost
after
construction
150,000 first year
300,000 second year
500,000 third year
400,000 forth year
Revenue after
construction
Four years
Life after
construction
$ 100,000
Salvage value
350
100
200
100
100
200
300
IRR = 25.25% > 25%
(Accepted)
0

PM 128

PM 129
Lake Lake
x
x
5 km
5 km
5 km
5 km
There are two ways (A and B) to construct a power line:
Standard Information
Example:
Project (B)
Project (A)
Item
Around lake
Under lake
Method
15 km
5 km
Length

PM 130
Project (B)
Project (A)
Item
Around lake
Under lake
Method
15 km
5 km
Length
$ 5,000/km
$ 25,000/km
First Cost
$ 200/km
$ 400/km
Annual Maintenance Cost
15 year
15 year
Expected Life
$ 3000/km
$ 5000/km
Salvage Value
$ 400/km
$ 500/km
Annual Power Losses
Sensitivity analysis for first cost change (±20%) and line length (±10%).
Interest rate 12% (Loan)
Example:

PM 131
Project (A)
Time
First
Cost
Annual
Maintenance
Power
Losses
Salvage
Value
Net
0 125 -125
1 2 2.5 -4.5
2 2 2.5 -4.5
3 2 2.5 -4.5
4 2 2.5 -4.5
5 2 2.5 -4.5
6 2 2.5 -4.5
7 2 2.5 -4.5
8 2 2.5 -4.5
…… …… …… …… ……
15 2 2.5 25 +20.5
NPV (12%) = 151.08 “Cost”

PM 132
Project (B)
Time
First
Cost
Annual
Maintenance
Power
Losses
Salvage
Value
Net
0 75 -75
1 3 6 -9
2 3 6 -9
3 3 6 -9
4 3 6 -9
5 3 6 -9
6 3 6 -9
7 3 6 -9
8 3 6 -9
…… …… …… …… ……
15 3 6 45 +36
NPV (12%) = 128.08 “Cost”

PM 133
Project (B)
Project (A)
Item
Around lake
Under lake
Method
16 km
8 km
Length
$ 5,000/km
$ 25,000/km
First Cost
$ 200/km
$ 400/km
10 year
10 year
Expected Life
$ 3000/km
$ 5000/km
Salvage Value
$ 400/km
$ 500/km
Annual Power Losses
NPV(A) = 227,803 NPV(B) = 126,515
For this non-profitable project, we should select project (B),
because it has lower NPV and also lower first cost.
Example:

PM 134
Project (B)
Project (A)
Item
Around lake
Under lake
Method
15 km
5 km
Length
$ 5,000/km
$ 25,000/km
First Cost
$ 200/km
$ 400/km
15 year
15 year
Expected Life
$ 3000/km
$ 5000/km
Salvage Value
$ 400/km
$ 500/km
Annual Power Losses
$ 25,000
$ 25,000
Annual Revenue
Sensitivity analysis for first cost change (±20%) and line length (±10%).
Example:

PM 135

PM 136
Capitalization Rate (Cap Rate)
is an indirect measurement of how an investment pays
for itself Average national cap rates for limited-service
hotels over the past decade ranged between 9-13%.

PM 137
Project (B)
Project (A)
Item
5.0
4.5
1.2
1.1
3.0
2.5
-
-
Salvage Value; M$
4
4
Project Life; years
2. Profitability Index
3. Capitalization Rate
Bank interest rate 12%
Example:

PM 138
5.0
1 2
0 3 4
1.2
3.0
5.0
0 1 2 3 4
1.8
4.5
1 2
0 3 4
1.1
2.5
4.5
0 1 2 3 4
1.4
PI(B) > PI(A) & CR(B)>CR(A)
Capitalization Rate =
Annual Net Income / Initial Capital Cost
= 1.4 / 4.5 = 0.311 = 31.1%
= 1.8/5.0 = 0.36 = 36%
NPV (cash inflow) =
= 1.4/(1.12) + 1.4/(1.12)^2
+ 1.4/(1.12)^3 + 1.4/(1.2)^4 = 4.25
Profitability Index =
NPV (cash inflow) / Initial Capital Cost
= 4.25 /4.5 = 0.94
= 5.47 /5.0 = 1.094
NPV (cash inflow) =
1.8/(1.12) + 1.8/(1.12)^2
+ 1.8/(1.12)^3 + 1.8/(1.12)^4 = 5.47

PM 139
Hotel Cap Rate
Net Operating Income $1,953,930
Total Development Cost $12,192,731
Cap Rate 16%

PM 140
Project (B)
Project (A)
Item
5.0
4.5
1.2
1.1
3.0
2.5
-
-
Salvage Value; M$
4
4
Project Life; years
Example:

PM 141
5.0
1 2
0 3 4
1.2
3.0
5.0
0 1 2 3 4
1.8
4.5
1 2
0 3 4
1.1
2.5
4.5
0 1 2 3 4
1.4
PI(B) > PI(A) & CR(B)>CR(A)
= 1.4 / 4.5 = 0.311 = 31.1%
= 1.8/5.0 = 0.36 = 36%
NPV (cash inflow) =
= 1.4/(1.12) + 1.4/(1.12)^2
+ 1.4/(1.12)^3 + 1.4/(1.12)^4 = 4.25
= 4.25 /4.5 = 0.94
= 5.47 /5.0 = 1.094
NPV (cash inflow) =
1.8/(1.12) + 1.8/(1.12)^2
+ 1.8/(1.12)^3 + 1.8/(1.12)^4 = 5.47

PM 142
Project (B)
Project (A)
Item
5.0
4.5
1.2
1.1
3.0
2.5
1.2
1.1
Salvage Value; M$
4
4
Project Life; years
Example:

PM 143
Project (A)
Cash Flow Diagram:
Project (B)
NPV (cash inflow) =
= 1.4/(1.12) + 1.4/(1.12)^2
+ 1.4/(1.12)^3 + 2.5/(1.2)^4 = 4.95
= 4.95/4.5 = 1.10
= 6.23/5.0 = 1.25
5.0
1 2
0 3 4
1.2
3.0
1.2
5.0
0 1 2 3 4
1.8 1.2
4.5
1 2
0 3 4
1.1
2.5
1.1
4.5
0 1 2 3 4
1.4 1.1
PI(B) > PI(A)
NPV (cash inflow) =
1.8/(1.12) + 1.8/(1.12)^2
+ 1.8/(1.12)^3 + 3.0/(1.12)^4 = 6.23

PM 144
Project (A)
Cash Flow Diagram:
Project (B)
NPV (cash inflow) = 4.95
Annual Net Income = (A/P, i, n)
= {i (1+i)n } / {(1+i)n -1}
= 4.95 *0.12*(1.12)^4 / (1.12^4-1) = 1.63
5.0
1 2
0 3 4
1.2
3.0
1.2
5.0
0 1 2 3 4
1.8 1.2
4.5
1 2
0 3 4
1.1
2.5
1.1
4.5
0 1 2 3 4
1.4 1.1
CR(B) > CR(A)
NPV (cash inflow) = 6.23
= 6.23 *0.12*(1.12)^4 / (1.12^4-1)
= 2.05
= 1.63 / 4.5 = 0.36 = 36%
= 2.05 / 5.0 = 0.41 = 41%

PM 145
Economic Service Life for a Lift Truck
Element
Life; Year
0 1 2 3 4 5 6 7
Initial Cost; $ 20000
O&M; $ 1000 + 15% Annual
Capital Maintenance; $ 2000
Book Value; $ 15000 11000 8000 6000 5000 4000 2000
Interest rate 15%
Replacement Analysis:
Based on this information; What is the economic life of a lift truck?
Example:
Annual Equivalent Cost (AEC)

PM 146
AEC1 = x (A/P,15%,1)
20,000
- 14,000 (P/F,15%,1)
i = 15%
0
20
1
1
15

PM 147
AEC2 = x (A/P,15%,2)
20,000
+ 1,000 (P/F,15%,1)
- 9,850 (P/F,15%,2)
i = 15%
0
20
1
1
2
1.15
11

PM 148
AEC3 = x (A/P,15%,3)
20,000
+ 1,000 (P/F,15%,1)
+ 1,150 (P/F,15%,2)
- 6,700 (P/F,15%,3)
i = 15%
0
20
1
1
2
1.15
3
8
1.3

PM 149
AEC4 = x (A/P,15%,4)
20,000
+ 1,000 (P/F,15%,1)
+ 1,150 (P/F,15%,2)
+ 1,300 (P/F,15%,3)
- 4,550 (P/F,15%,4)
0
20
1
1
2
1.15
3 4
1.3
6
1.45
i = 15%

PM 150
AEC5 = x (A/P,15%,5)
20,000
+ 1,000 (P/F,15%,1)
+ 1,150 (P/F,15%,2)
+ 1,300 (P/F,15%,3)
+ 1,450 (P/F,15%,4)
- 1,400 (P/F,15%,5)
0
20
1
1
2
1.15
3 4 5
1.3
1.45
1.6
2
5
i = 15%

PM 151
AEC6 = x (A/P,15%,6)
20,000
+ 1,000 (P/F,15%,1)
+ 1,150 (P/F,15%,2)
+ 1,300 (P/F,15%,3)
+ 1,450 (P/F,15%,4)
+ 3,600 (P/F,15%,5)
- 2,250 (P/F,15%,6)
0
20
1
1
2
1.15
3 4 5
1.3
1.45
1.6
1.75
2
4
i = 15%

PM 152
AEC7 = x (A/P,15%,7)
20,000
+ 1,000 (P/F,15%,1)
+ 1,150 (P/F,15%,2)
+ 1,300 (P/F,15%,3)
+ 1,450 (P/F,15%,4)
+ 3,600 (P/F,15%,5)
+ 1,750 (P/F,15%,6)
- 100 (P/F,15%,7)
0
20
1
1
2
1.15
3 4 5 6
1.3
1.45
1.6
1.75
7
2 1.9
2
i = 15%

PM 153
Economic Life for truck 6 Years
AEC
Year
9000
1
8225
2
7592
3
7003
4
6780
5
6405
6
6852
7
Year
AEC
Summary:

PM 154
Economic Service Life for a compressor
Life; Year
0 1 2 3 4 5 6 7
Initial Cost; $ 18000
O&M; $ 1000 + 15% Annual
Capital Maint.; $ 3000 4500
Book Value; $ 10000 7500 5625 1780
Interest rate 15%
Based on this information; What is the economic life of this
compressor?
Example:

PM 155
Economic Service Life
• N = 1
AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000
= $11,700

PM 156
• N = 2
AEC2 = [$18,000 + $1,000(P/A,15%, 15%, 2)](A/P, 15%, 2)
- $7,500 (A/F, 15%, 2)
= $8,653

PM 157
N = 3, AEC3 = $7,406
N = 4, AEC4 = $6,678
N = 5, AEC5 = $6,642
N = 6, AEC6 = $6,258
N = 7, AEC7 = $6,394
Minimum cost
Economic Service Life

PM 158
Economic Service Life for a car
Element
Life; Year
0 1 2 3 4 5 6
Initial Cost 150000
O&M 2000 first year (+500 Annual)
Capital Maintenance 10000
Book Value 120000 100000 80000 60000 80000 60000
Interest rate 15%
Based on this information; What is the economic life of this car?
Replacement Analysis:
Example :

PM 159
ID Activity Length (m)
A Line A 400
B Line B 300
C Line C 1000
D Line D 1200
E Line E 1600
F Line F 400
G Line G 800
Resources: Crew productivity = 200 m/day Average material cost = 10 $/m
Item : Water Pipeline Network
A B
C
D
E
F
G
Source
Source
Source
Process
Process
Project Sketch
Constraints: Maximum duration = 17 day Maximum number of crew = 4
Cost : Average crew cost rate= 100 $/day Margin factor= 30% from direct cost
(Predecessor, Duration, Resources)?
(Target Schedule)?
(Target Resource Plan)?
Project Cost Analysis

PM 160
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
A
B
C
D
E
F
G
Number of
Crews
(Target Schedule)
Maximum duration = 17 day Maximum number of crew = 4

PM 161
ID Name Pred.
Duration
(Day)
Number
of Crews
Material
Cost, $
A Line A - 2 2 4000
B Line B - 3 1 3000
C Line C - 5 2 10000
D Line D A,B 6 2 12000
E Line E C 8 2 16000
F Line F C 2 2 4000
G Line G D,E 4 2 8000
Based on this information, discuss & analysis
the following:
1) Network (Logic & Risk)
2) Gantt Chart (Target Schedule)
3) Workers Profile (Target Resource Plan)
4) Cost Analysis (Target Cost Plan)
5) Cash In/ Out Analysis (Payment / Cost)
A B
C
D
E
F
G
Source
Source
Source
Process
Process
Payment Plan:
1) 25% Down Payment
2) 25% After One Week
3) 25% After Two Weeks
4) 25% Final Payment
Project Sketch
Cost : Average crew cost rate= 100 $/day Margin factor= 30% from direct cost

PM 162
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
2
2
2
2
2
2
2
2
2
8
C
E
2
2
2
2
C
F
2
2
2
2
2
4
D,E
G
2
2
2
2
4
4
4
4
4
4
4
4
3
3
3
4
4
Number of
Crews
(Target Schedule)

PM 163
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
2
2
2
2
2
2
2
2
2
8
C
E
2
2
2
2
C
F
2
2
2
2
2
4
D,E
G
2
2
2
2
4
4
4
4
4
4
4
4
4
4
3
3
3
Number of
Crews
(Target Schedule)

PM 164
Network Analysis: A
B
C
D
G
E
F
Start
Finish
Day
Pred.
(FS=0)
ID
2
-
A
3
-
B
5
-
C
6
A,B
D
8
C
E
2
C
F
4
D,E
G
EF
d
ES
Activity ID
LF
TF
LS
3
B
6
D
2
A
5
C
8
E
2
F
4
G
0
Start
0
Finish

PM 165
Network Analysis:
0 3 3
B
4 4 7
3 6 9
D
7 4 13
0 2 2
A
5 5 7
0 5 5
C
0 0 5
5 8 13
E
5 0 13
5 2 7
F
15 10 17
13 4 17
G
13 0 17
Planned Duration = 17 Critical Path: C-E-G
It is recommended that “Delay risk analysis” be considered for critical and near-
critical path work in the schedule because such work carries the greatest
likelihood of affecting timely completion
EF
d
ES
Activity ID
LF
TF
LS
0
Start
0
17
Finish
17

PM 166

PM 167

PM 168

PM 169
Cash In/ Out Analysis

PM 170
ID Activity Pred.
Normal Crashing
Material
Cost, $
Duration
(Day)
Number of
Crews
Duration
(Day)
Number of
Crews
A Line A - 2 2 1 4 4000
B Line B - 3 1 2 2 3000
C Line C - 5 2 - - 10000
D Line D A,B 6 2 4 4 12000
E Line E C 8 2 6 3 16000
F Line F C 2 2 - - 4000
G Line G D,E 4 2 3 3 8000
Cost : Average Crew Cost Rate = 100 $/day Margin Factor = 30% direct cost

PM 171
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
2
2
2
2
2
2
2
2
2
8
C
E
2
2
2
2
C
F
2
2
2
2
2
4
D,E
G
2
2
2
2
4
4
4
4
4
4
4
4
4
4
3
3
3
Number of
Crews
(Target Schedule)
Normal : 17 day & 57 crew-day

PM 172
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
2
2
2
2
2
2
2
2
2
8
C
E
2
2
2
2
C
F
3
3
3
2
4
D,E
G
3
3
3
4
4
4
4
4
4
4
4
4
4
3
3
3
Number of
Crews
(Target Schedule)
#2 : 16 day & 58 crew-day

PM 173
ID Activity Pred.
Normal Crashing
Material
Cost, $
Duration
(Day)
Number of
Crews
Duration
(Day)
Number of
Crews
A Line A - 2 2 1 4 4000
B Line B - 3 1 2 2 3000
C Line C - 5 2 - - 10000
D Line D A,B 6 2 4 4 12000
E Line E C 8 2 6 3 16000
F Line F C 2 2 - - 4000
G Line G D,E 4 2 3 3 8000
Constraints: Maximum duration = 17 day Number of crew = Open
Cost : Average Crew Cost Rate = 100 $/day Margin Factor = 30% direct cost

PM 174
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
3
3
3
3
3
3
2
8
C
E
2
2
2
2
C
F
3
3
3
2
4
D,E
G
3
5
5
5
5
5
5
5
5
4
4
3
3
3
Number of
Crews
(Target Schedule)
Maximum duration = 17 day Maximum number of crew = Open

PM 175
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
3
3
3
3
3
3
2
8
C
E
2
2
2
2
C
F
3
3
3
2
4
D,E
G
3
5
5
5
5
5
5
5
5
2
2
5
5
3
Number of
Crews
(Target Schedule)

PM 176
Day
Crews
Day
Pre.
ID
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
2
2
2
2
-
A
1
1
1
1
3
-
B
2
2
2
2
2
2
5
-
C
2
2
2
2
2
2
2
6
A,B
D
3
3
3
3
3
3
2
8
C
E
2
2
2
2
C
F
3
3
3
2
4
D,E
G
3
5
5
5
5
5
5
5
5
5
5
3
2
2
Number of
Crews
(Target Schedule)

PM 177
Example; Building
Direct Cost $
Predecessor
Duration (weeks)
ID
15,000
-
5
A
30,000
-
3
B
33,000
A
8
C
42,000
A
7
D
57,000
-
7
E
61,000
C, E
4
F
72,000
F
5
G
Based on this information, discuss &
analysis the following:
1) Network
2) Gantt Chart
3) Cost Analysis
4) Cost Smoothing
5) Cash Flow (In/ Out) Analysis
Payment Conditions:
1) 25% Down Payment
2) 25% After 8 Weeks
3) 25% After 16 Weeks
4) 25% Final Payment
Project Cost Analysis
Planned Duration = 22 weeks Margin Factor = 30% from direct cost

PM 178
Logic Diagram:
E
Start
Finish
A
B
D
C
F G

PM 179
Network Analysis:
5
A
0
Start
Finish
8
C
7
E
4
F
7
D
5
G
3
B

PM 180
Weeks
Pred
Day
ID
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
-
5
A
-
3
B
A
8
C
A
7
D
-
7
E
C, E
4
F
F
5
G

PM 181
Project Co$t Control
Cost
Control
Actual
Performance
Planned
Performance
KPIs report
(Earned Value Analysis)

PM 182
Project: Office Building
Item (main activity): Excavation works
Brainstorming: Project Control
Actual Performance (Status Report after 6 days):
Actual Quantity = 320 m3
Actual Cost = 3300 $
Planned Performance (Baseline):
Bill of Quantity (BOQ) = 500 m3
Unit Price = 10 $/m3
Duration = 10 day
Control

PM 183
Unit Cost = 10 $/m3
Duration = 10 day
Daily Performance = 500 / 10 = 50 m3/day
Control
Planned Performance after 6 days:
Panned Quantity = 50*6 = 300 m3
Planned Cost = 10*300 = 3000 $
Analysis:
Quantity Variance = 320 – 300 = + 20 Ahead
Cost Variance = 320*10 – 3300 = - 100 Over budget

PM 184
Project Cost Control
Cost control is a Part of Cost Management.
70% of projects are
over budget and/or
behind schedule
52% of all projects
finish at 189% of their
initial budget
30% of completed
projects have a cost
variance of >10%
50% of completed
projects have a
schedule variance
of >10%
(Earned Value Analysis)
(1)
(2)
(3)
(4)

PM 185
Earned Value Management (EVM)
EVM is a cost management technique where the value of completed
work is compared to actual costs and the scheduled costs
EVM is a project management tool that integrates
the project performance with schedule and cost elements.
Earned Value Management Systems measure progress

PM 186
Schedule
Variance
Cost
Variance
Schedule
Performance
Index
Cost
Performance
Index
Planned Value = PV = BCWS
(Budget Cost for the Work Schedule)  Base line
Actual Cost = AV = ACWP
(Actual Cost for the Work Performed)  Actual
Earned Value = EV = BCWP
(Budget Cost for the Work Performed)  Invoice
Earned Value Analysis

PM 187
• Schedule Variance SV = BCWP – BCWS
(>0 is ahead, 0 is on target, <0 is behind)
• Schedule Performance Index SPI = BCWP/BCWS
(>1 is ahead, 1 is on target, <1 is behind)
• Cost Variance CV = BCWP – ACWP
(>0 is under-run, 0 is on target, <0 is overrun)
• Cost Performance Index CPI = BCWP/ACWP
(>1 is under-run, 1 is on target, <1 is over-run)
Earned Value Analysis

PM 188
Duration = 10 day
Control

PM 189
Actual Quantity = 320 m3 WP
Actual Cost = 3300 $ ACWP
Actual Cost Rate (AC) = 10.3 $/m3 AC
Bill of Quantity (BOQ) = 500 m3 WS
Unit Price = 10 $/m3 BC
Duration = 10 day
Control
Actual Value (ACWP) = 3300 $
Planned Value (BCWS) = 3000 $
Earned Value (BCWP) = 3200 $
After 6 days:

PM 190
Project: Office Building Item: Excavation works
Actual Value (ACWP) = 3300 $
Planned Value (BCWS) = 3000 $ Control
Earned Value (BCWP) = 3200 $
• Cost variance:
• BCWP – ACWP =3200 – 3300 = -100 $ (Overrun)
• Cost Performance Index:
CPI = BCWP/ACWP = 3200/3300 =0.97 = 97%
• Schedule variance:
• BCWP – BCWS = 3200 – 3000 = +200 $ (ahead)
• Schedule Performance Index:
SPI = BCWP/BCWS = 3200/3000 =1.07 = 107%

PM 191
Project: Gas Pipeline Activity: Excavation
Bill of Quantity (BOQ) = 500 m3 Unit Price = 10 LE/m3 Duration= 10 day
Status Report after 6 day:
Actual Quantity = 320 m3 Actual Cost = 3300 LE
Example:
6 day
$
BAC = 5000
CV =
- 100
SV =
+ 200
10 day
Ahead of
schedule
•Over –run cost
•Ahead of schedule

PM 192
Duration = 10 day
Brainstorming: Earned Value Analysis
Based on this information, calculate and analyze the following:
a) What is the cost variance?
b) What is the Schedule variance?
c) What is the Cost Performance Index?
d) What is the Schedule Performance Index?
e) How would you describe this project?

PM 193
Project: Office Building Item: Excavation works
Actual Value (ACWP) = 3300
Planned Value (BCWS) = 3600
Earned Value (BCWP) = 3000
• Cost variance:
• BCWP – ACWP = 3000 – 3300 = -300
• Cost Performance Index:
CPI = BCWP/ACWP = 3000 / 3300 = 91% (Overrun)
• Schedule variance:
• BCWP – BCWS = 3000 – 3600 = - 600
• Schedule Performance Index:
SPI = BCWP/BCWS = 3000 / 3600 = 83% (Behind)
After 6 days:

PM 194
Duration = 10 day
Brainstorming: Earned Value Analysis
Based on this information, calculate and analyze the following:
a) What is the cost variance?
b) What is the Schedule variance?
c) What is the Cost Performance Index?
d) What is the Schedule Performance Index?
e) How would you describe this project?

PM 195
A construction project has a total budget (BAC) of $300,000 and the
total duration of 12 months. After 4 months, the cost control revealed
the following figures:
• Actual Cost of Work Performed (ACWP) = $ 110,000
• Budgeted Cost of Work Scheduled (BCWS) = $ 100,000
• Budgeted Cost of Work Performed (BCWP) = $ 90,000
a) Calculate the CPI and SPI
b) Calculate the Estimate At Completion (EAC) if the project will
continue to perform in the same way
c) Calculate the EAC if the project cost performance improved in the
remaining period by 10%
Example:

PM 196
A construction project has a total budget of $300,000 and the total duration of
12 months. After 4 months, the cost control revealed the following figures:
b) Calculate the EAC if the project will continue to perform in the same way
CPI = BCWP / ACWP = 90,000 / 110,000 = 0.818 < 1 Over budget
SPI = BCWP / BCWS = 90,000 / 100,000 = 0.90 < 1 Behind schedule
Estimate At Completion (EAC) = BAC/CPI = 300000/0.818= 366,748
Improve  Corrective Actions  Improve CPI by 10%
Improve CPI by 10%  New CPI = 0.818 *1.1 = 0.8998 (Expected)
New (EAC) = AC + {(BAC – EV) / CPI}
=110,000 + (300,000 – 90,000) / 0.8998 = $ 343,385

PM 197

PM 198
b) Calculate the Estimate At Completion (EAC) if the project will
continue to perform in the same way
Next:

PM 199

PM 200
b) Calculate the EAC if the project will continue to perform in the same way
CPI = BCWP / ACWP = 190,000 / 215,000 = 0.884 < 1 (Overrun)
SPI = BCWP / BCWS = 190,000 / 200,000 = 0.95 < 1 (Behind)
Estimate At Completion (EAC) = BAC/CPI = 300,000/0.884 = 340,909
 Corrective Actions  Improve CPI by 10%
New CPI = 0.884 *1.1 = 0.968
New (EAC) = AC + {(BAC – EV) / CPI}
= 215,000 + (300,000 – 190,000)/0.968 = 335,743

PM 201
total duration of 12 months. After 13 months (closeout), the cost control
revealed the following figures:
CPI = BCWP / ACWP = 310,000 / 340,000 = 0.912 < 1 (overrun by 8.8%)
SPI = BCWP / BCWS = 310,000 / 300,000 = 1.03 > 1 (Ahead by 3%)
Closing:

PM 202
Summary Report:
total duration of 12 months.
Period (months) 4 8 13
Planned Value (PV = BCWS) 100,000 200,000 300,000
Actual Value (AV = ACWP) 110,000 215,000 340,000
Earned Value (EV =BCWP) 90,000 190,000 310,000
CPI = BCWP / ACWP
SPI = BCWP / BCWS
Estimate At Completion (EAC)
= BAC/CPI

PM 203
Period #1 #2 #3 #4 #5 #6 #7 #8 #9 #10
BAC $1,230 $1,230 $1,230 $1,230 $1,400 $1,400 $1,400 $1,400 $1,400 $1,400
EV $100 $200 $300 $450 $750 $800 $1,125 $1,200 $1,400 $1,400
AC $100 $205 $315 $600 $800 $1,000 $1,200 $1,350 $1,475 $1,525
PV $100 $220 $325 $550 $725 $925 $1,175 $1,275 $1,450 $1,500
Period #1 #2 #3 #4 #5 #6 #7 #8 #9 #10
CV $0 ($5) ($15) ($150) ($50) ($200) ($75) ($150) ($75) ($125)
SV $0 ($20) ($25) ($100) $25 ($125) ($50) ($75) ($50) ($100)
CPI 1.00 0.98 0.95 0.75 0.94 0.80 0.94 0.89 0.95 0.92
SPI 1.00 0.91 0.92 0.82 1.03 0.86 0.96 0.94 0.97 0.93
ETC $1,130 $1,056 $977 $1,040 $693 $750 $293 $225 $0 $0
EAC $1,230 $1,261 $1,292 $1,640 $1,493 $1,750 $1,493 $1,575 $1,475 $1,525
VAC $0 ($31) ($62) ($410) ($93) ($350) ($93) ($175) ($75) ($125)
Excel File:
($1000)
Analysis:

PM 204

PM 205

PM 206

PM 207

PM 208
BCWS
ACWP
BCWP
Late
Item
Planned
Value
(BCWS)
Actual
Cost
(ACWP)
Earned
Value
(BCWP)
#1 3000000 3080000 2800000
#2 250000 210000 175000
#3 500000 468000 520000
#4 100000 88000 80000
Item BCWS ACWP BCWP SV SPI CV CPI
Comments
Schedule Cost
#1 3000000 3080000 2800000 -200000 0.93 -280000 0.91 Behind 7% Over run 9%
#2 250000 210000 175000 -75000 0.70 -35000 0.83 Behind 30% Over run 17%
#3 500000 468000 520000 20000 1.04 52000 1.11 Ahead 4% Under run 11%
#4 100000 88000 80000 -20000 0.80 -8000 0.91 Behind 20% Over run 9%
Total 3850000 3846000 3575000 -275000 0.93 -271000 0.93 Behind 7% Over run 7%
EVM Report:
Project: A Construction Project

PM 209

PM 210
√
X

PM 211

PM 212
Activity
ID
Planned
Cost
Planned Date Actual
Performance at
period #10
Start Finish
A 4000 1 4 Completed
B 10000 3 6 Completed
C 12000 6 9 In Progress
D 4000 9 12 In Progress
G 6000 9 12 Not Started
H 6000 11 14 Not Started
I 10000 14 17 Not Started
Project Status Report:
Case Study:
Based on this information; discuss the earned value analysis using the
50-50 rule?

_VIP_Cost Management -20-10-2018.pdf

_VIP_Cost Management -20-10-2018.pdf

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