# Engineering economics homework

## 1. The fixed assets of a project are 6.5 million yuan. If the depreciation life is 6 years and the residual value rate is 4%, try to calculate the depreciation rate of each year by using the average life method, the double declining balance method, the number of years and the depreciation method respectively.

Solution:
(1) Average life method: annual depreciation rate = (1 - estimated net residual value rate) / depreciation life = (1-4%) / 6 = 16%
(2) Double declining balance method: annual depreciation rate = 2 / 6 * 100% = 1 / 3 ≈ 33.3%

(3) Years and depreciation method: annual depreciation rate = remaining years / years and

## Figure 3-21 shows the cash flow of a project

Calculation: present value of capital, final value and equivalent capital at the end of the fourth year

```Solution:
F=-70(F/A,8%,3)(F/P,8%,5)+100(F/P,8%,4)+85(F/P,8%,3)+80(F/A,8%,3)=162.86
P=F(P/F,8%,7)=162.86*0.5835=95.03
Q=P(F/P,8%,4)=95.03*1.36=129.24
```

## 4. The annual interest rate is 8%, which is calculated once a quarter to calculate the actual interest rate.

``` i=(1+r/n)n  -1=(1+8%/4)4-1=(F/P,r/n,n)-1=1.082-1=0.082
```

## 5. When a company purchases a set of equipment with a loan, the total expenditure is 5 million yuan. It will be repaid every six months within six years after the loan. If the annual nominal interest rate is 12%, Q:

(1) If the interest is compounded every six months, what is the installment amount of the company
(2) If the interest is compounded once a month, how much is the installment

```Solution:
(1)
A=500*(A/P,6%,12)= 500*0.1193=59.65
(2)
set up X Is the installment value, then X(A/F,10%/12,6)Is the monthly payment value,
X(A/F,12%/12,6)(P/A,12%/12,6*12)=500
X*0.1626*50.5685=500
X ≈ 60.81
```

## 6. An airline borrowed 40 million yuan from the bank to buy Boeing aircraft. If the service life of the aircraft is 12 years and the annual interest rate is 6%, how much is the annual income of the aircraft without considering the overhaul.

```Solution:
A=4000(A/P,6%,12)
=4000*0.1193=477.2 Ten thousand yuan
```

## 7. An enterprise borrows 200000 yuan from the bank for a technical transformation project, with a loan period of 10 years and an annual interest rate of 6%. Try to design two repayment schemes for the enterprise.

```Solution:

Two options:
(1)lump-sum
F=P(F/P,6%,10)=20*1.791=35.817 Ten thousand yuan
(2)Installment repayment
A=P(A/P,6%,10)=20*0.1359=2.718 Ten thousand yuan
```

## 8. The fixed asset investment of a project at the beginning of the first year is 2 million yuan, and the working capital is 300000 yuan when it is put into operation. If the project is put into operation at the beginning of the second year and reaches the design capacity, the annual operating cost is 500000 yuan, and the service life of the project is 10 years. At that time, the residual value is 100000 yuan, if the annual interest rate is 10%. Try to calculate the present value of the total cost of the project.

```Solution:
F=200+30(P/F,10%,1)+50(P/A,10%,10)(P/F，10%，1)-10(P/F,10%,11)
=200+30*0.9091+50*6.1446*0.9091-10*0.3505=503.07 (Ten thousand yuan)
```

## 9. The initial investment of a project is 2000 yuan, and the net cash flow in subsequent years is as follows: 3000 yuan in the first year and 5000 yuan in the second to tenth years. The calculation period of the project is 10 years. Calculate the static investment payback period and the dynamic payback period when the discount rate is 8%.

```Solution:
Static payback period 4.4 year
Dynamic payback period 5.6 year
```

## 10. The net cash flow of mutually exclusive schemes A, B and C is shown in table 5-16. Try to choose the best scheme

```solution
NPVA= －450＋180(P/A,10%,5)= －450＋180×3.791=232.38
NPVB= －200－220(P/F,10%,1)＋220(P/A,10%,4) (P/F,10%,1)= －200－220×0.9091＋220×0.9091×3.170= －200－200.002＋634.006=234.004
NPVC= －150＋70(P/A,10%,5)= －150＋70×3.791=115.37
programme B The net present value of is the largest, so B The scheme is excellent.
```

## 11. See table 5-19 for the cash flow of two mutually exclusive schemes with different service lives. If ic=10%, try to choose the best scheme

```Solution:
The two evaluation methods with different service lives can be compared and selected by the research cycle method. The selected study period is 5 years. Calculate the net present value of the two schemes
NPVA=-250+80(P/A,10%,5)=-250+80×3.791=53.28
NPVB=-200+60(P/A,10%,5)=-200+60×3.791=27.46
therefore A The scheme is excellent.
```

## 12. See table 5-20 for the cash flow of schemes a and B. the benchmark rate of return is known to be 10%. Try to choose the best scheme

```Solution:
The net annual value method is used for comparison
Ask first A,B NPV of the two schemes:
NPVA=-1000+300(P/A,10%,6)=
-1000+300×4.3553=306.59
NPVB=-1200+500 (P/A,10%,3)=
-1200+500×2.487=43.5
Net annual value:
NAVA=306.59×(A/P,10%,6)=306.59×0.2296=70.393
NAVB=43.5 ×(A/P,10%,3)=43.5 ×0.4021=17.491
NAVA >NAVB ，And NPVA and NPVB  Both are greater than zero, so the scheme A Is the best scheme.
```

## 14. There are three schemes A, B and C. see table 5-21 for relevant information. Assuming that the investment limit is 7.5 million yuan, try to select the optimal scheme combination

```Solution:
Combination number     Scheme combination      Investment amount     net present value
1               0                 0              0
2               A                200          50.25
3               B                300          60.85
4               C                400          55.45
5               AB              500          111.1
6               AC              600          105.7
7               BC              700          116.3
8               ABC            900          166.550

choose BC Combination scheme.
```

## 15. It is known that company A has 80 million assets, 30% debt ratio, 8% annual interest rate and 4 million ordinary shares. The company plans to raise 30 million yuan to expand its production scale by issuing additional ordinary shares or long-term loans.

1. If additional ordinary shares are issued, it is planned to issue 1.5 million shares at 20 yuan / share.
2. For long-term loans, the interest rate is 9%.
Assuming that the expected EBIT after capital increase is 15 million, the income tax rate is 25%. Please calculate the no difference in earnings per share and make a financing decision.

```(1)Original debt of the enterprise: 8000 * 30% = 2400 Million, with an annual debt of 24 million * 8% = 192 ten thousand
(2)Earnings per share = (EBIT – I – T)/ N= (EBIT – I)*(1 – T%)/ N
= (EBIT – Debt interest – Income tax)/Number of ordinary shares
(3)Earnings per share after borrowing 30 million long-term loans = (EBIT – 192 – 3000*9%)*(1-25%)/ 400
(4)Earnings per share after issuing 30 million additional ordinary shares = (EBIT – 192)*(1-25%)/ (400+150)
(5)If the two equations are equal, there are:
(EBIT – 192 – 3000*9%)*(1-25%)/ 400
= (EBIT – 192)*(1-25%)/ (400+150)
(EBIT – 462)*75% /400 = (EBIT – 192)*75% /550
EBIT = 1182 Ten thousand yuan
E Earnings per share without difference=
EPS1 = EPS2 = (1182 – 462)*75% /400 = (1182 – 192)*75% /550
= 1.35 element/thigh
```

## 16. The design capacity of a project is 2.4 million units, the annual fixed cost is 15 million yuan, the unit product price is 190 yuan, the unit convertible cost is 170 yuan, and the business tax and surtax rate are 6%. Try to make a break even analysis of the project

```Solution:
From the meaning of the question: S=(190-190*6%)Q
C=1500+170Q
S=C Break even point (190)-190*6%)Q= 1500+170Q
(1)Breakeven point expressed by output         QBEP=174.42(10000 pieces)
(2)Breakeven point expressed by production capacity utilization:
RBEP =F/Q(P-V-T)=174.42/240=72.7%
(3)Breakeven point expressed in sales price
PBEP =F/Q+V+T=1500/240+170+190*6%=187.65((10000 yuan)
(4)Breakeven point expressed in sales
SBEP =P*F/(P-V-T)=    190*1500/(190-170-190*6%)=33139.5(Ten thousand yuan)
```

## 17. The total annual cost of a project is C=1/2x 2-4x+8, and the price of finished products is P=6-1/8x, where x is the output, which is the output at break even

```Solution:
According to the meaning of the title, S=C=(6-1/8x)x=6x-1/8x2  =1/2x2  -4x+8
x1=0.845，x2=15.15
Breakeven Output is 0.845 To 15.15 between.
When the production level reaches the maximum profit point,
P'(x)=10x-80=0
x=8 When the output level reaches 8, the product gets the maximum profit.
```

## explain

Title Source: after class exercises in engineering economics textbook