W9_Zakiya_Replacing Transportation Equipment


Problem Statement

Transportation equipment that was purchased in 2004 for $200,000 must be replaced. What is the estimated cost of the replacement, based on the following equipment cost index? If I only have budget of $250,000 then on which year should I change the equipment?

Year

Index

Year

Index

2004

223

2007

257

2005

238

2008

279

2006

247

2009

293

Alternative Solution

How much is the estimated cost of the replacement car for year:

  • 2005
  • 2006
  • 2007
  • 2008
  • 2009

Selection Criteria

With the budget of $250,000, when should the transportation equipment change?


Analysis & Comparison of Alternatives

We will be using equation: Cn = Ck (Īn / Īk)

Where:

k              = reference year for which cost or price of item is known = 2004
n              = year for which cost or price is to be estimated (n>k) = 2005 – 2009
Cn           = estimated cost or price of item in year n
Ck           = cost or price of item in reference year k = $200,000
Īn            = ratio of index value in year n (see index table)
Īk            = ratio of index value in year k = 223

Year

Cost

2005

$200,000 x (238/223) = $213,452.91

2006

$200,000 x (247/223) = $221,524.66

2007

$200,000 x (257/223) = $230,493.27

2008

$200,000 x (279/223) = $250,224.22

2009

$200,000 x (293/223) = $262,780.27

Selection of preferred Alternatives

Since the budget is only $250,000 then the alternatives chosen is year 2007, which will cost $230,493.27 for transportation equipment replacement

Reference 

  • Sullivan G. William, Wick M. Elin, Koelling Patrik C., Engineering Economy Chapter 3,Question 3-08, Page 129, Pearson International.
  • Sullivan G. William, Wick M. Elin, Koelling Patrik C., Engineering Economy Chapter 3,equation (3-1), Page 103, Pearson International.


W8_Zakiya_Material Selection


Problem Statement

In the design of a tractor radiator, an engineer has a choice of using either a brass-copper alloy casting or a plastic molding. Either material provides the same service. However, the brass-copper alloy casting weighs 30 pounds, compared to 25 pounds for the plastic molding. Every pound of extra weight has been assigned a penalty of $6.50 to account for the increased fuel consumption during the life cycle of the car. The brass-copper alloy casting costs $3.85 per pound, whereas the plastic molding costs $7.90 per pound. Machining costs per casting are $6.50 for the brass-copper alloy. Which material should the engineer select, and what is the difference in the unit costs?


Alternative Solution

Engineer must choose 2 different types of material:

  • Brass-copper alloy
  • Plastic

Selection Criteria

Which material that gives the lowest cost


Analysis & Comparison of Alternatives

1. Brass-copper alloy

Casting Cost = 30 pounds x $3.85           = $ 115.5

Penalty          =   5 pounds x $6.50           = $   32.5

Machining Cost                                           = $     6.5  +

Total Cost                                                     = $ 154.5

2. Plastic

Casting Cost = 25 pounds x $7.90 = $197.5

Difference Unit Cost : $197.5 – $154.5 = $43

Selection of preferred Alternatives

Based on above calculation, it concluded that brass-copper alloy has lower cost compare to plastic. Difference cost between 2 materials is $43. So, brass-copper alloy is the chosen material.

Reference

  • Sullivan G. William, Wick M. Elin, Koelling Patrik C., Engineering Economy Chapter 2,Question 2-33, Page 77, Pearson International.

W7_Zakiya_Which Sites is More Profitable?


 

Problem Statement

A municipal solid-waste site for a city must be located at Site 1 or Site 2. After sorting, some of the solid refuse will be transported to an electric power plant where it will be used as fuel. Data for the hauling of refuse from each site to the power plant are shown in below table. If the power plant will pay $9.50 per cubic yard of sorted solid waste delivered to the plant, where should the solid-waste site be located? Use the city’s viewpoint and assume that 250,000 cubic yards of refuse will be hauled to the plant for one year only. One site must be selected.

 

Site 1

Site 2

Average hauling distance

5 miles

4 miles

Annual rental fee for solid-waste site

$10,000

$125,000

Hauling cost

$2.00 / yd3-mile

$2.00 / yd3-mile

Alternative Solution

Which one is the best alternative?

  • Site 1
  • Site 2

Selection Criteria

Which site with the lowest cost for municipal solid-waste location?

 

Analysis & Comparison of Alternatives

Cost Type

Fixed

Variable

Site 1

Site 2

Rent

X

$10,000

$125,000

Hauling


X

$2 x 5 x 250,000 = $ 2,500,000

$2 x 4 x 250,000 = $ 2,000,000

Cost / Year


$ 2,510,000

$ 2,125,000

Revenue


90 x 250,000 = $ 2,375,000

90 x 250,000 = $ 2,375,000

Profit

2,375,000 – 2,510,000 = -$135,000

2,375,000 – 2,125,000 = $250,000

Selection of preferred Alternatives

Site 2 has larger fixed cost, but has smaller total cost. Therefore site 2 is more profitable than site 1. Then alternative 2 is the chosen one.

 

Reference

 

  • Sullivan G. William, Wick M. Elin, Koelling Patrik C., Engineering Economy Chapter 2,Question 2-4, Page 74, Pearson International.


 

 

 

W6_Zakiya_The Most Economical Decision


Problem Statement

Tyler just wracked his new Nissan and the accident was his fault. The owner of the other vehicles got 2 estimates for the repairs: one was for $803 and the other was for $852.Tyleris thinking of keeping the insurance companies out of the incident to keep his driving record “clean”. Tyler’s deductible on his comprehensive coverage insurance is $500 and he doesn’t want his premium to increase because of the accident. In this regards, Tyler estimates that his semi annual premium will rise by $60 if he files a claim against his insurance company. In view of the above information, Tyler’s initial decision is to write a personal check for $803 payable to other owner of the vehicle. Did Taylor make the most economical decision? In your answer, be sure to state your assumptions and quantify your thinking.


Alternative Solution

This solution it to pay for the lowest estimate cost for repair the car

  • Claim to insurance company
  • Pay with personal check

Selection Criteria

Which solution will gives Tyler the most economical decision?

Analysis & Comparison of Alternatives

1. Claimed to insurance company

Repair cost estimates                  = $ 803

Deductible coverage                   = $ 500  –

Additional Cost need to pay     = $ 303

Premium rise                                 = $   60  +

Total payment                             = $ 363

2. Pay with personal check

Personal Check            = $ 803

Selection of preferred Alternatives

Tyler should choose claimed to insurance company, even though he will have a record, but he will have the most economical payment. He only needs to pay additional cost of $363, since $500 has been covered by the insurance company. And it is $440 cheaper than paid with personal check.

Reference

  • Sullivan G. William, Wick M. Elin, Koelling Patrik C., Engineering Economy Chapter 1,Question 1-4, Page 36, Pearson International.

W7_Mohammed_when should i start building my own house


Problem definition

 

I bought my own land two years ago through bank. Now it’s the time to think about building this land, this analysis will guide me through to the best time / year to start this capital investment in the contraction of my own house considering the material prices & available fund.

 

Development of the feasible alternatives

 

As estimation its need not less than 100,000USD as to complete one story building given that the land is already available, Here is the evaluation of my possible choices to start the construction of my house, my strategy is to save much I the beginning and then pay the balance in installments, the plan could be started in one of the below options:

1. End of 2012

2. End of 2013

3. End of 2014

 

Analysis and comparison of the alternatives

 

Considering that the construction will take around 6 months to one year as an average, the above three options have been evaluated according to the following factors:

-My savings:

1. By end of 2012 ——- I may start with 100,000SDG = 40,000USD

2. By end of 2013 ——- I may start with 140,000SDG = 55,000USD

3. By end of 2014 ——- I may start with 180,000SDG = 70,000USD

-Inflation trends in Sudan based on the central bank of Sudan records were given as 16%.

 

Selection of the preferred alternatives

Considering the time value of money and the inflation rate in Sudan from the central bank of Sudan, by using the relating present and future equivalent values of single cash flows, as follows:

F = P (1+i)ᴺ

 

Given that:

F:         Future single sum

P:         Is the present Value

i :          inflation or interest rate

N:         Period

 

Solution

Back to the above options:

 

P:         The nowadays estimation of 100,000USD is needed to complete one story building given

i:           an average of 16% as per the Central Bank of Sudan records

 

1. End of 2012:

F = P (1+i) ᴺ

   = 100,000 (1+0.16)¹

   = $116,000

Considering that I’m planning to save $40,000 by end of 2012 then the difference is going to be $76,000

 

2. End of 2013:

F = P (1+i) ᴺ =

                           = 100,000 (1+0.16)²

                           = $134,560

Considering that I’m planning to save $55,000 by end of 2013 then the difference is going to be $79,560

 

3. End of 2014:

F = P (1+i) ᴺ =

                           = 100,000 (1+0.16)³

                           = $156,089

Considering that I’m planning to save $70,000 by end of 2012 then the difference is going to be $86,089

 

 

Conclusion

 

It’s clear that I should plan for 2012 since the difference is the least among the other options of 2013 and 2014.

 

References

*  Engineering Economics-Fifteen Edition, chapter 4, Relating Present and Future Equivalent Values of Single Cash Flows, page113

 

W3_Luis_Selection of Property for Investment


1)      Problem recognition, definition, evaluation

I am now in the midst of choosing a property to invest in Singapore. My budget is S$ 550,000 to S$610,000 . Investing in property is almost a sure thing especially in Singapore where the land is scarce and with the uncertainties, real estate investment is a good investment against inflation, currency and stock market.

2)      Development of the feasible alternatives

I have shortlisted the following properties for my analysis:

Option 1:

Sycamore Tree

Size: 420 square feet

Price: S$ 608,000.00

PSF: S$1,447

Freehold

1 Block of 17 shop units & 96 residential units, swimming pool, BBQ Pit, No Security.

Located on the 3rd floor.

Option 2:

Centra Suites

Size: 463 square feet

Price: S$ 610,000.00

PSF: S$ 1,317

Freehold

1 Block of 8 Storeys, 62 Apartment Units, swimming pool, BBQ Pit, Gymnasium, 24 hour
security.

Located on the 4th floor.

3)      Development of the outcomes and cash flows for each alternative

Centra Suites

Sycamore Tree

Price

$610,000.00

$608,000.00

Size (square feet)

463

420

Per Square Feet (PSF)

$1,317.49

$1,447.62

Downpayment

5.00%

$30,500.00

$30,400.00

Option to Purchase within 8 weeks Cash

5.00%

$30,500.00

$30,400.00

Option to Purchase within 8 weeks CPF

10.00%

$61,000.00

$60,800.00

Sales & Purchase Agreement

3.00%

$12,900.00

$12,840.00

Legal Fees

$1,000.00

$1,000.00

Foundation Work

10.00%

$61,000.00

$60,800.00

Reinforce Concrete

10.00%

$61,000.00

$60,800.00

Total Cash Outflow

$257,900.00

$257,040.00

Bank Loan

60.00%

$366,000.00

$364,800.00

Monthly Mortgage Annual Interest Rate of 2.5% for 25 years

$1,641.94

$1,636.55

Monthly Mortgage Annual Interest Rate of 2.5% for 30 years

$1,446.14

$1,441.40

Estimated Monthly Rental:
Best

$2,400.00

$2,600.00

Most Likely

$2,000.00

$2,400.00

Worst

$1,800.00

$2,200.00

P90

$2,161.33

$2,485.33

Yield (yearly rental/property price)

4.25%

4.91%

Estimated Monthly Net Cashflow against Mortgage 30 years

$715.19

$1,043.93

4)      Selection of a criterion (or criteria)

The criteria for the selection of my alternatives on the order of priorities are:

  1. Access to MRT Public Transport
  2. Access to Eateries
  3. Monthly Cashflow
  4. Access to Amenities (shopping complexes)
  5. Cost of property
  6. Rental Yield

5)      Analysis and comparison of the alternatives

Table 1 shows the attributes with the data. Based on this, I applied the Compensatory Model where I need to convert the data into a single dimension to a common measurement scale.

Table 1:

Attribute

Centra Suites

 

Sycamore Tree

No Investment

Access to MRT Public Transport (m)

200

600

0

Access to Eateries

Good

Good

Fair

Monthly Cashflow

$715.00

$1,043.00

0

Access to Amenities (shopping complexes)

Poor

Excellent

Fair

Cost of Property

$610,000.00

$608,000.00

0

Rental Yield

4.25%

4.91%

0

I applied the nondimensional scaling with the following results as shown in Table 2.

Table 2:

Nondimensional Scaling

Value

 

Rating Procedure

Dimensionless Value

Access to MRT Public Transport (m)

0

Worst-“Outcome”/Worst-Best

1.0

200

0.7

600

0.0

Access to Eateries

Fair

1

“Outcome”-Worst/Best-Worst

0.0

Good

2

1.0

Monthly Cashflow

0

“Outcome”-Worst/Best-Worst

0.0

715

0.7

1,043

1.0

Access to Amenities (shopping complexes)

Poor

1

“Outcome”-Worst/Best-Worst

0.0

Fair

2

0.5

Excellent

3

1.0

Cost of Property

0

Worst-“Outcome”/Worst-Best

1.0

608,000

0.0

610,000

0.0

Rental Yield

0

“Outcome”-Worst/Best-Worst

0.0

4.25

0.9

4.91

1.0

Next I again converted the data from Table 1 to Table 3.

Table 3:

Attribute

Centra Suites

 

Sycamore Tree

No Investment

Access to MRT Public Transport (m)

0.7

0.0

1.0

Access to Eateries

1.0

1.0

0.0

Monthly Cashflow

0.7

1.0

0.0

Access to Amenities (shopping complexes)

0.0

1.0

0.5

Cost of Property

0.0

0.0

1.0

Rental Yield

0.9

1.0

0.0

Sum

3.3

 

4.0

2.5

Based on Table 3, the results pointed me that Sycamore Tree property is the best investment option for me.

I further applied the Additive Weighting Technique based on the order of priorities.

Table 4:

Attribute

Ordinal Ranking

Weightage

Access to MRT Public Transport (m)

6

0.29

Access to Eateries

5

0.24

Monthly Cashflow

4

0.19

Access to Amenities (shopping complexes)

3

0.14

Cost of Property

2

0.10

Rental Yield

1

0.05

21

1.00

I then factored in the weightage and translated the value from Table 3 to Table 5 as shown.

Table 5:

Attribute

Centra Suites

 

Sycamore Tree

No Investment

Access to MRT Public Transport (m)

0.20

0.00

0.29

Access to Eateries

0.24

0.24

0.00

Monthly Cashflow

0.13

0.19

0.00

Access to Amenities (shopping complexes)

0.00

0.14

0.07

Cost of Property

0.00

0.00

0.10

Rental Yield

0.04

0.05

0.00

Sum

0.61

 

0.62

0.45

Again the results pointed to me to invest in Sycamore Tree. Interestingly, after using the Additive Weighting Technique, there is not much of a difference to choose between Centra Suites and Sycamore Tree.

6)      Selection of the preferred alternative

Based on the results, Sycamore Tree is the best investment option.

7)      Performance monitoring and post-evaluation results

In my opinion, the results of Centra Suites and Sycamore Tree are too close to call. Since the results are too near to differentiate, I most likely need to add further attributes for my decision making analysis such as:

a)    Extra facilities offered by the property

b)   Potential for capital gain for the property in the future

c)    Monthly maintenance cash outflow

d)  Potential for rental gain in the future

References

William G. Sullivan, Elin M. Wicks, C. Patrick Koelling. Engineering Economy 15th Edition. Prentice Hall. n.d. Print.

W6_Mohammed_Run WayExtension


Problem definition

Should our company extend the under construction Run Way?

Development of the feasible alternatives

Our company received a request from the government to extend the under construction runway for another 1km. Our company has to right to accept or reject this request. But in other hand our company is looking to the benefit and cost of this project ie.operate a large size jet aircrafts instead of the small size aircraft operated now; this will save mainly save the time and the cost of the flight, but the cost of this extension is rather high. Here I’m using the conventional Benefit–Cost Ratio method to make go/no-go decision on the RW extension.

 

 

Analysis and comparison of the alternatives

-The cost of extending the runway additional 1000 m includes supply and process of earthworks, asphalt works, fence extension, and extension of AFL system and Compensation of lands. The cost estimation as per current market price for runway extension is about $6,000,000 including but not limited to (land Compensation Cost, EPCC cost: is excavation, BF, asphalt works, marking system, Fence System, AFL system).

-An annual estimated saving to company is estimated 800,000USD plus per year, moreover saving a lot of time per flight, considering using Jet Aircrafts (50 seats) or even Boeing 737 (126 seats) which will need only 45min to reach the field airport from Khartoum airport, instead of using only Dash8, Fokker 50 which need about 2 hours to reach our field airport.

-There is about 300,000USD for maintenance per year (maintenance of asphalt, marking, and lighting).

-Lifetime is 30 years

-International finance interest rate 6%

(A/P,6%,30) = 0.07265

B-C      = AW (benefit of the proposed project) / AW (total costs of the proposed project)

            = AW (B) / CR+AW (O&M)

Where:

AW (B) = Annual worth of benefit of the proposed project

CR         = Capital recovery a mount

O&M   = Operating and maintaince cost of the proposed project

At the interest rate of 6% per year the conventional B-C ratio of the proposed R/W extension is

B-C= 800,000 / [6,000,000 (A/P, 6%, 30) + 300,000] = 1.04

Selection of the preferred alternatives

Based on the above results, which shows the B-C > 1 its recommended that the company should take the decision of go and do the R/W extension.

 

 

Conclusion

The cost Benefit Ratio method is an interesting procuders for making go/no-go decision on independent projects. Applying this method to the proposed project company can easily decide to go on for the extension of the under construction R/W.

 

 

References

*  The benefit-Cost Ratio Method P429, Engineering E economics 15th Edition