Cost Estimation and Analysis - Desklib

This text covers various topics related to cost estimation and analysis, including formulas and techniques for break-even analysis, opportunity cost, life cycle costs, and cost capacity factor. It also discusses the relevance of payback period, present value, and future worth in project feasibility analysis. The subject is Economics and the course code is not mentioned. The text is relevant for students and professionals interested in cost estimation and analysis.

ECONOMICS
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Question 4
The general form of total cost equation is shown below.
Total Cost = Fixed Cost + Variable Cost per unit *(Number of Units) ........................... Eq.1
The given total cost equation is shown below.
Total Cost = $100 + $10*(Number of Units) ................................ Eq.2
Comparing equation 1 and equation 2, we ger
a) Fixed Cost = $ 100
b) Variable Cost = $ 10 per unit
c) Incremental cost is the extra cost incurred which would be equal to the variable cost and
hence is $ 10.
Question 5
The sum of all coats during the life-cycle would be referred to as life cycle costs and would
include the various intangible costs besides the tangible costs. The tangible costs would be
often represented in the financial statements of the business but the intangible costs with
regards to social and environmental costs are often unrepresented in traditional financial
statements.
Question 6
The cost of the best rejected opportunity is known as opportunity cost. Typically with regards
to capital, there may be a lot of alternative uses and it is expected that the capital is deployed
in the manner that maximises the returns. In order to pursue this best opportunity, the
business has to forego the returns from the next best opportunity which is known as
opportunity cost.
Question 11
The necessary condition for break-even is that at this activity level, the cost and revenue must
be equal.
Revenue ($) = a0 +a1Dk
Cost ($) = b0

For breakeven, a0 +a1Dk = b0
a1Dk = (b0-a0)
D (million units)=(b0-a0)/a1k
Question 12
The formula for breakeven derived above has been stated below.
D=(b0-a0)/a1k
As per the question, a0=5, a1=8, b0=32 and k =3
Substituting the above, D= (32-5)/(8*3) = 1.125 million units
Question 13
For the given cost function, k is a constant which does not change with activity level x and
hence k indicates the fixed cost. a and b highlight the variable costs considering that they
need to be multiplied/divided by x to reach the underlying cost. The constant a highlights the
variable cost per unit is directly proportional to the underlying output level x and hence
increases linearly with the same. However, the constant b highlights an inversely proportional
relationship with underlying output which may be attributed to economies of scale.
Question 15
The relevant formula to be used is shown below.
Cn=Ck (In/Ik)
where C is the cost , n is the year for which cost has to be estimated and k is the base year and
I is the index for the year.
a) For 2019, Ck = $ 1,000, IK=2, Tn=2.1
C2019 = 1000*(2.1/2) = $1,050

For 2020, Ck = $ 1,000, IK=2, Tn=2.2
C2020 = 1000*(2.2/2) = $1,100
b) Assuming that the rate of increase remains the same, then estimation of costs for 2021 and
2022 can be done which would be $ 1,150 and $ 1,200.
c) The given data should not be used for estimation of costs in 2035 as it is a recommended
practice to shift the base periodically so as to reflect the costs more accurately.
Question 16
With regards to cost estimation, the largest contributor of error would be omission related
errors or bias related errors especially when these have been committed from the past and
have potentially accumulated.
Question 20
It is apparent that the power of the capacitor is doubled and hence the cost estimate would
also become twice i.e. $100,000*2 = $ 200,000.
Question 21
The relevant formula is indicated below.
C2/C1 = (Q2/Q1)X
Putting the respective input values in the formula indicated above, we get the following
(20/10)=(4*100/100)X
Solving the above X = 0.5
Hence, the cost capacity factor is 0.5
Question 22

With regards to a thermal power plant, the cost capacity factor is lower than equal to 0.79.
Question 23
A cost capacity factor of less than 1 would indicate that economies of scale is realised and
hence the cost of setting an incremental unit capacity would be progressively lower. A cost
capacity factor of more than 1 indicates diseconomies of scale and thereby higher cost per
unit incremental capacity.
Question 24
For the given scenario, it would be prudent to use the load capacity technique. The
measurement of the cost capacity factor would be contingent on the maximum load that the
bridge would have to carry and construction would be accordingly carried out.
Question 26
If the fourth item’s time reduced by half, the slope parameter for the group’s learning curve
should have been 25 percent for the second unit (+1) and 50 percent reduction by the fourth
unit
Question 28
The cash flow diagram for the given scenario is summarised as follows.

Question 33
The payback period does make sense since the original investment is recovered within 4.67
years. The PV also makes sense considering that the net present value of the project can be
negative. Similarly, AW and FW are also relevant as they may be positive or negative
depending on the underlying pattern and quantum of cash flows involved.
Question 34
The given project is not feasible considering the fact that the net present value of the project
is negative as has been indicated in the computation for Q29.
Question 35
The present worth would be set equal to the upfront investment of $ 50,000 in order to solve
the equation and obtain the IRR.
Question 36
The positive future worth would be for diamonds while the negative present worth would be
for atmospheric sciences. Considering the above combination, it is apparent that no
investment would be made as the respective signs for the two cases should have been reverse.