1999-2-19-eco

Each of the methods given in the left-hand column is closely linked to one of the items listed in the right-hand column. Match each method with the corresponding item.

I. Discounted cash flow

II. Declining balance

1999-1-29-eco

In a manufacturing industry, break-even point occurs when

- the annual rate of production equals the assigned value
- the total annual product cost equals the total annual sales
- the annual profit equals the expected value
- the annual sales equals the fixed costs

2000-1-29-eco

- end of the project life
- break-even point
- start-up
- end of the design stage

2001-1-22-eco

- 5 years
- 7 years
- 12 years
- 10 years

2002-1-13-eco

- 15%
- 10%
- 1.5%
- 150%

2005-73-eco

- 8
- 4
- 10
- 20

2006-64-eco

- 750,000
- 683,750
- 621,500
- 332,750

2008-18-eco

- 10
- 20
- 1
- 5

2008-66-eco

- 113,600
- 42,000
- 52,500
- 10,500

2016-20-eco

Option \(P\) | Option \(Q\) | |
---|---|---|

Installed cost of the system ($ in million) |
150 | 120 |

Cost of cooling water for condenser ($ in million/year) |
6 | 8 |

Cost of steam for reboiler ($ in million/year) |
16 | 20 |

- 40
- 42.4
- 92
- 128

2016-22-eco

- The profitability measure 'return on investment' does not consider the time value of money
- A cost index is an index value for a given time showing the cost at that time relative to a certain base time
- The 'six-tenths factor rule' is used to estimate the cost of an equipment from the cost of a similar equipment with a different capacity
- Payback period is calculated based on the payback time for the sum of the fixed and the working capital investment

0900-9-eco

1989-19-i-eco

An equipment costs Rs. 1,70,000 and will have a scrap value of Rs. 25,000 at the end of its useful life of 10 years. If the interest is compounded at 10% per year, what are: (i) the present worth of renewable perpetuity, and (ii) the capitalised cost (all in Rs.)?

(i) ____________

{#1}

(ii) ____________

{#2}

1997-28-eco

For a chemical plant, the fixed capital investment is Rs. \(4\times 10^9\) and the working capital is 20% of the total capital investment. The annual total product costs are Rs. \(2\times 10^9\) whereas the annual depreciation costs are Rs. \(2\times 10^8\). If the annual sales are Rs. \(3\times 10^9\) and the income-tax rate is 40%, then determine (by neglecting start-up costs):

(a) the percent of total capital investment returned annually as gross profit.

{#1}

(b) the pay-out time (year).

{#2}

2000-12-eco

A plant is designed to produce \(1.2 \times 10^8\) kg/yr of an agrochemical. The estimated fixed capital investment is $ \(1.5\times 10^9\). The working capital is $ \(2\times 10^8\) and start-up cost (only in the first year of commissioning and to be
accounted for in the first year) is $ \(1.5\times 10^8\).

The following cost data are available:

Raw materials: $ 0.8/kg product

Labour and utilities, etc: $ 0.27/kg product

Selling price of product: $ 10/kg

Other costs (on per year basis) including maintenance, insurance, etc. @10% of fixed capital.

Indirect cost of administration,
R& D, marketing, etc. @ 20% of sale proceeds.

The plant will be fully depreciated over a period of 5 years using the straight-line method.

The rate of income tax is 40%.

Calculate:

(a) the net profit at the end of first year (in $) {#1}

(b) the payout period (in years, rounded to 1 decimal) {#2}

2003-82-83-eco

A process has fixed capital investment of $ 150 millions, working capital $ 30 millions and salvage value zero. Annual revenues from sales are $ 250 millions, manufacturing costs $ 145 millions and other expenses 10% of the revenue. Assume project life span of 11 years, tax life of 5 years and interest rate to be 10%. Tax rate is 40% and straight line depreciation, i.e. 20% per year, is applicable.

(i) Discounted value (to present time) of the profit before tax (for the total plant life period) in $ is {#1}

(ii) Discounted value of the depreciation benefit over the tax life in $ is {#2}

2010-50-51-eco

A plant produces phenol. The variable cost in rupees per tonne of phenol is related to the plant capacity \(P\) (in tonnes/day) as \(45,000 + 5P\). The fixed charges are $ 100,000 per day. The selling price of phenol is $ 50,000 per tonne.

(i) The optimal plant capacity (in tonnes per day) for minimum cost per tonne of phenol, is

{#1}

(ii) The break-even capacity in tonnes per day, is

{#2}

PI-2010-50-51-eco

A company is engaged in producing and selling a single product. The fixed cost of the product is \(F\) per period. The selling price for the product is \(S\) per unit. The variable cost is \(V\) per unit, which is half of the selling price, i.e., \(S/2\) per unit. The company has computed its Break Even Sales in monetary units. Not being satisfied with this Break Even Sales, the company has decided to increase the selling price from \(S\) to \(1.5S\). The company has again computed the new Break Even Sales in monetary units keeping the fixed cost (\(F\)) and variable cost (\(S/2\) per unit) of the product same.

(i) The ratio of new to old Break Even Sales is

{#1}

(ii) The firm desires to make a profit equal to fixed cost of the product. In this scenario, the ratio of new to old Required Sales Volume is

{#2}

1987-18-i-eco

A plant is producing 1000 ton/year of a product. The overall yield is 70% on mass basis. The raw material costs Rs. 100/ton, and the product sells at Rs.350/ton. A process modification has been devised that will increase the yield to 75%. The additional
investment required is Rs. 400,000. Is the modification worth making if the minimum acceptable rate of return is 20%? (Yes/No)

- Yes
- No

1988-19-ii-eco

A single effect evaporator costs Rs. 100,000 and a two effect evaporator costs Rs. 200,000 for identical duty. The annual steam cost for single effect is Rs. 25,000, and for double effect it is Rs. 5,000. Depreciation is calculated on a straight line
basis, assuming a ten-year service life and zero salvage value. Should one effect or two effects be used if the value of money is 15 percent?

- one
- two

1990-19-ii-eco

For a project having a life of ten years the following cash flow pattern is expected:

End of year | Net cash flow (Rs.) |
---|---|

0 | \(-\)50,00,000 |

1-10 | 20,00,000 |

10 | \(-\)1,50,00,000 |

- recommended
- not recommended

2002-2-15-eco

- 0
- 250
- 375
- 500

2003-81-eco

Two pumps under consideration for installation at a plant have the following capital investments and salvage values. Pump \(A\): \(C_I\) = $ 40,000; \(C_S\) = $ 3,900. Pump B: \(C_I\) = $ 50,000; \(C_S\) = $ 20,000. Using capitalized cost, determine what should be the common life of the pumps for both to be competitive (economically equivalent). Interest rate is 10% per annum. Maintenance and operational costs are negligible.

- 3 years
- 5 years
- 6 years
- 8 years

2004-84-eco

- zero and -28.9
- -40 and -28.9
- -40 and 12.75
- zero and 12.75

2010-45-eco

- $ 6.1 million
- $ 6.5 million
- $ 6.9 million
- $ 7.6 million

2012-44-eco

- 33500
- 43800
- 54200
- 64500

PI-2011-45-eco

- $ 130,950
- $ 118,340
- $ 102,460
- $ 86,500

1989-19-ii-eco

The annual production costs for a plant are Rs. 36.4 lakhs, while the sum of the annual fixed charges, overhead costs and general expenses are Rs. 26.0 lakhs. What is the break-even point, in units of production per year if the total annual sales are
Rs. 72.8 lakhs and the product sells at Rs. 520 per unit?

1991-20-ii-eco

A heat exchanger with an initial investment of Rs. 300,000 has a 6 years life. How much (Rs.) can be spent on an improved design which has a life of 12 years and is expected to save Rs. 10,000 per year?

Annual compound interest rate = 8%.

1999-21-eco

A new equipment made of material \(A\) costs, post installation, Rs. 3,00,000 and is expected to have a scrap value of 10% of this cost at the end of a useful life of 10 years. Similar equipment made of material \(B\) costs Rs. 1,50,000, but is likely
to have no scrap value. Assume that both types of equipments could be replaced at a cost that is 20% more than the original value. On the basis of equal capitalized costs for both types of equipments, estimate what should be the useful life (in year)
for equipment made of material \(B\). The company has to pay an annual interest on the investment at a rate of 15%.

2014-44-eco

2002-20-eco

An engineer has purchased a pump for which the installed cost is Rs. 200,000. If no annual maintenance is carried out on the pump, it will have a service life of 5 years, with no salvage value. On the other hand, if annual maintenance is carried out at a cost of Rs. 10,000 per year, then the pump will have a service life of 7 years, with a salvage value of Rs. 10,000.

Derive the formula for capitalized cost, in terms of \(C_I\) (initial purchase cost of the equipment), \(C_S\) (salvage value), \(n\) (lifetime), \(C_M\) (annual end-of-year maintenance cost) and \(i\) (interest rate). Assume that the cost of replacing the equipment at the end of its service life is the same as its initial cost.

If the interest rate is 10%, determine based on capitalized cost analysis, whether annual maintenance on the pump should be carried out or not (neglect depreciation in your analysis).

Last Modified on: 02-May-2024

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