DEPARTMENT OF MECHANICAL ENGINEERING
MG 6863 ENGINEERING ECONOMICS
FORMULA SHEET
UNIT I
1. 𝑇𝐸𝐶𝐻𝑁𝐼𝐶𝐴𝐿 𝐸𝐹𝐹𝐼𝐶𝐼𝐸𝑁𝐶𝑌 =
2. 𝐸𝐶𝑂𝑁𝑂𝑀𝐼𝐶 𝐸𝐹𝐹𝐼𝐶𝐼𝐸𝑁𝐶𝑌 =
𝑂𝑈𝑇𝑃𝑈𝑇
𝐼𝑁𝑃𝑈𝑇
𝑊𝑂𝑅𝑇𝐻
𝐶𝑂𝑆𝑇
× 100
× 100
3. Prime Cost = Direct Material Cost +Direct Labour Cost+ Direct Expenses
4. Factory Cost = Prime Cost + Factory overhead
5. Cost of Production = Factory cost + Office & Administrative overhead.
6. Cost of goods sold = Cost of production+ Opening finished stock-Closing
finished stock.
7. Cost of Sales = Cost of goods sold + Selling and Distribution overhead.
8. Sales = Cost of sales + Profit
9. Selling Price/Unit = Sales/Quantity sold
BREAK EVEN ANALYSIS
s = Selling price/unit
v = variable cost/unit
Q = Volume of production
FC = Fixed cost /period
TC = Total Cost of the firm
S = The total Sales Revenue
1. The total sales revenue(S) of the firm S = s × Q
2. The total cost of the firm
TC = Total variable cost + Fixed Cost
TC = v × Q + FC
3. Profit = Total sales – Total cost
4. Break even quantity =
= ( s × Q) – (v × Q + FC )
𝐹𝐼𝑋𝐸𝐷 𝐶𝑂𝑆𝑇
(𝑆𝐸𝐿𝐿𝐼𝑁𝐺 𝑃𝑅𝐼𝐶𝐸/𝑈𝑁𝐼𝑇 −𝑉𝐴𝑅𝐼𝐴𝐵𝐿𝐸 𝐶𝑂𝑆𝑇/𝑈𝑁𝐼𝑇 )
𝐵𝑅𝐸𝐴𝐾 𝐸𝑉𝐸𝑁 𝑄𝑈𝐴𝑁𝑇𝐼𝑇𝑌 =
𝐹𝐶
𝑠−𝑣
5. 𝐵𝑅𝐸𝐴𝐾 𝐸𝑉𝐸𝑁 𝑆𝐴𝐿𝐸𝑆 =
𝐹𝐼𝑋𝐸𝐷 𝐶𝑂𝑆𝑇
× 𝑆𝐸𝐿𝐿𝐼𝑁𝐺 𝑃𝑅𝐼𝐶𝐸/𝑈𝑁𝐼𝑇
𝑆𝐸𝐿𝐿𝐼𝑁𝐺 𝑃𝑅𝐼𝐶𝐸/𝑈𝑁𝐼𝑇 − 𝑉𝐴𝑅𝐼𝐴𝐵𝐿𝐸 𝐶𝑂𝑆𝑇 /𝑈𝑁𝐼𝑇
𝐵𝑅𝐸𝐴𝐾 𝐸𝑉𝐸𝑁 𝑆𝐴𝐿𝐸𝑆 =
𝐹𝐶
𝑠−𝑣
×𝑠
6. Contribution = Sales – Variable costs
7. Contribution/Unit = Selling price/unit – Variable cost/unit
8. Margin of Safety = Actual sales – Break Even Sales
9. 𝑀𝐴𝑅𝐺𝐼𝑁 𝑂𝐹 𝑆𝐴𝐹𝐸𝑇𝑌 =
𝑃𝑅𝑂𝐹𝐼𝑇
𝐶𝑂𝑁𝑇𝑅𝐼𝐵𝑈𝑇𝐼𝑂𝑁
× 𝑆𝐴𝐿𝐸𝑆
10. P/V RATIO = CONTRIBUTION/SALES
11. BREAK EVEN POINT (BEP) =
𝐹𝐼𝑋𝐸𝐷 𝐶𝑂𝑆𝑇
𝑃
𝑅𝐴𝑇𝐼𝑂
𝑉
DEPARTMENT OF MECHANICAL ENGINEERING
MG 6863 ENGINEERING ECONOMICS
FORMULA SHEET
UNIT II
Notations used:
P = Principle amount
F = Future amount at the end of the year ‘n’
n = Number of interest periods
i = Interest rate
A = Equal amount deposited at the end of every interest period
G = Uniform amount which will be added/subtracted period after period to/from
the amount of deposit A1 at the end of the period 1.
Formula :
1. To find the future worth of money F = P × (1+i)n = P(F/P, i, n)
𝐹
2. To find the present worth of money P =
𝑛 = F(P/F, i, n)
(1+𝑖)
3. Equal payment series compound amount
F=A
4. Equal payment series sinking fund A = F
(1+𝑖)𝑛 −1
𝑖
𝑖
(1+𝑖)𝑛 −1
= A (F/A i, n)
= A (F/A,i,n)
5. Equal payment Series Present worth amount
P=A
(1+𝑖)𝑛 −1
𝑖(1+𝑖)𝑛
= A (P/A,i,n)
6. Equal payment series capital recovery amount
A=P
𝑖(1+𝑖)𝑛
(1+𝑖)𝑛 −1
= P (A/P,i,n)
7. Uniform Gradient series amount equivalent amount
A = A1 + G
𝑖
(1+𝑖)𝑛 −𝑖𝑛−1
𝑖(1+𝑖)𝑛 −1
= A1 + G (A/G,i,n)
8. Effective Interest Rate R = (1 + )𝑐 – 1 where i = nominal interest rate,
𝐶
C = Number of interest periods in a year.
DEPARTMENT OF MECHANICAL ENGINEERING
MG 6863 ENGINEERING ECONOMICS
FORMULA SHEET
UNIT III
Present Worth Method of Comparison:
Revenue Dominated Cash Flow Diagram:
0
S
R1
R2
R3
.
1
2
3
.
Rj
Rn
j
n
P
Pw(i) = - P + R1[ 1/(𝟏 + 𝒊)𝟏 ] + R2[ 1/(𝟏 + 𝒊)𝟐 ] + …. +Rj [ 1/(𝟏 + 𝒊)𝒋 ] +
Rn [ 1/(𝟏 + 𝒊)𝒏 ] + S [ 1/(𝟏 + 𝒊)𝒏 ]
Where P = Initial investment
R1, R2, …Rj = Net Revenue at the end of the 1,2,…jth period
S = Salvage Value at the end of the nth year.
In this method the expenditure is assigned a (-) sign with arrow pointing
downwards and the revenue assigned a (+) sign with arrow pointing
upwards
Cost Dominated Cash Flow Diagram:
0
1
C1
2
C2
.
.
S
j
n
Cj
Cn
P
Pw(i) = P + C1[ 1/(1 + 𝑖)1 ] + C2[ 1/(1 + 𝑖)2 ] + …. +Cj [1/(1 + 𝑖)𝑗 ] +
Cn [1/(1 + 𝑖)𝑛 ] - S [1/(1 + 𝑖)𝑛 ]
Where P = Initial investment
C1, C2, …Cj = Net Cost at the end of the 1,2,…jth period
S = Salvage Value at the end of the nth year.
In the above formula the expenditures is assigned with (+) sign with the
arrow pointing downwards
In the above formula the revenue is assigned with (-) sign with the arrow
pointing upwards.
Future Worth Method :
Revenue Dominated Cash Flow Diagram:
In this method the expenditure is assigned a (-) sign with arrow pointing
downwards and the revenue assigned a (+) sign with arrow pointing
upwards
S
R1
0
1
R2
2
R3
.
3
.
Rj
Rn
j
P
FW(i) = - 𝑷(𝟏 + 𝒊)𝒏 + 𝑹𝟏(𝟏 + 𝒊)𝒏−𝟏 + R2(𝟏 + 𝒊)𝒏−𝟐 +
… 𝑹𝒋(𝟏 + 𝒊)𝒏−𝟏 + 𝑹𝒏 + 𝑺
Where P = Initial investment
R1, R2, …Rj = Net Revenue at the end of the 1,2,…jth period
S = Salvage Value at the end of the nth year.
n
Cost dominated cash flow diagram
S
0
P
1
2
3
j
C1
C2
C3
Cj
Cn
FW(i) = 𝑷(𝟏 + 𝒊)𝒏 + 𝑪𝟏(𝟏 + 𝒊)𝒏−𝟏 + C2(𝟏 + 𝒊)𝒏−𝟐 +
… 𝑪𝒋(𝟏 + 𝒊)𝒏−𝟏 + 𝑪𝒏 − 𝑺
Where P = Initial investment
C1, C2, …Cj = Net Cost at the end of the 1,2,…jth period
S = Salvage Value at the end of the nth year.
In the above formula the expenditures is assigned with (+) sign with the arrow
pointing downwards
In the above formula the revenue is assigned with (-) sign with the arrow
pointing upwards.
Annual equivalent method
In the annual equivalent method first the revenue of each alternative will be
computed. The alternative with the maximum annual equivalent revenue in the
case of revenue comparison or with the minimum annual equivalent cost in the
case of cost dominated comparison will be selected as the best alternative
Revenue Dominated Cash flow diagram
0
R1
R2
R3
1
2
3
S
.
Rj
Rn
j
n
Steps :
1. In this method the first step is to find the net present worth using the
formula
PW(i) = - P + R1[ 1/(𝟏 + 𝒊)𝟏 ] + R2[ 1/(𝟏 + 𝒊)𝟐 ] + …. +Rj [ 1/(𝟏 + 𝒊)𝒋 ] +
Rn [ 1/(𝟏 + 𝒊)𝒏 ] + S [ 1/(𝟏 + 𝒊)𝒏 ]
Where P = Initial investment
R1, R2, …Rj = Net Revenue at the end of the 1,2,…jth period
S = Salvage Value at the end of the nth year.
In this method the expenditure is assigned a (-) sign with arrow pointing
downwards and the revenue assigned a (+) sign with arrow pointing
upwards
2. The annual equivalent revenue is computed using the following formula
A = PW(i)
𝑖(1+𝑖)𝑛
(1+𝑖)𝑛 −1
A = PW(i) (A/P,i,n)
A = - P (A/P,i,n) + A + S (A/F,i,n)
Where (A/P,i,n) is called equal payment series capital
recovery factor.
3. The above steps 1 and 2 are repeated for all the alternatives
4. Finally the alternative with the maximum annual equivalent revenue
should be selected as the best alternative.
Cost Dominated Cash flow diagram
S
0
P
1
C1
2
C2
3
C3
j
Cj
Cn
Steps :
1. In this method the first step is to find the net present worth using the
formula
FW(i) = 𝑷(𝟏 + 𝒊)𝒏 + 𝑪𝟏(𝟏 + 𝒊)𝒏−𝟏 + C2(𝟏 + 𝒊)𝒏−𝟐 +
… 𝑪𝒋(𝟏 + 𝒊)𝒏−𝟏 + 𝑪𝒏 − 𝑺
Where P = Initial investment
C1, C2, …Cj = Net Cost at the end of the 1,2,…jth period
S = Salvage Value at the end of the nth year.
In the above formula the expenditures is assigned with (+) sign with the
arrow pointing downwards
In the above formula the revenue is assigned with (-) sign with the arrow
pointing upwards.
2. The annual equivalent revenue is computed using the following formula
A = PW(i)
𝑖(1+𝑖)𝑛
(1+𝑖)𝑛 −1
A = PW(i) (A/P,i,n)
A = P (A/P,i,n) + A - S (A/F,i,n)
Where (A/P,i,n) is called equal payment series capital
recovery factor.
3. The above steps 1 and 2 are repeated for all the alternatives
4. Finally the alternative with the minimum annual equivalent revenue
should be selected as the best alternative.
(OR)
Alternate Approach
1. Step 1 : Find the future worth of the cash flow diagram for the
Given alternatives.
2. Step 2 : The annual equivalent cost is calculated using the formula
A=F
𝑖
(1+𝑖)𝑛 −1
(or) A = F (A/F,i,n) where (A/F,i,n) is called
equal payment series sinking fund factor.
UNIT V
1. Straight line method of depreciation:
Depreciation Dt = (P-F)/n
Book value = Bt-1 – Dt = P-t [(P-F)/n]
Where P = First cost of the asset
F = Salvage value, n = number of years,
Dt = depreciation amount for the period “t”
Bt = Book value at the end of the period “t”
2. Declining Balance method of Depreciation :
Depreciation Dt = K x Bt-1
Book value Bt = (1-K) Bt-1
Where K = a fixed percentage
For double declining balance method K = 2/n
3. Sum of Years Digits Method of Depreciation :
Sum of years = n(n+1)/2
Rate = year/ sum of years
Dt = Rate (P-F)
Bt = Bt-1 - Dt
Dt =
𝑛−𝑡+1
𝑛(𝑛+1)
2
(𝑃 − 𝐹)
Bt = (𝑃 − 𝐹)
(𝑛−𝑡) (𝑛−𝑡+1)
𝑛
(𝑛+1)
+𝐹
4. Sinking fund Method of Depreciation :
A = (P-F) [A/F,i,n]
Dt = (P-F) x [A/F,i,n] (F/P,i,n)
Bt = P – (P-F) (A/F,i,n) (F/A,i,n)
5. Service Output method of Depreciation :
Depreciation = (P-F)/ X
(𝑷−𝑭)
Depreciation =
(𝒙)
𝑿
X = Maximum capacity
x = quantity of service rendered for a period.