ECONOMICS: SUMMARY OF EQUIPMENT PURCHASE COST

Equipments MUR (Rs)

Cane preparation and milling

Cane carriers 883,380

Bagasse carrier 559,469

Cane Knives 1,995,676

Tongaat Shredder 672000

Tramp Iron Separator 30,950

3 roller mills 21, 634,399

Juice clarification unit

Heater 1 2,753,582

Heater 2 2, 414,493

Cooler 906,550

Flash Tank 496,360

Clarifier 5,098,500

Rotary drum filter 6,165,965

Milk of Lime Tank 204,673

Liming Tank 116,635

Juice Mixer/Blender 247,200

MEE 4 effects 37,684,325

Barometric Condenser 92 700

Fermentation unit

Absorption Tower 3,090,000

Scrubber column packing 1,409,000

Centrifuges (x2) 983,276

Fermentation Tanks (x5) 32, 059,223

Heat exchangers (x5) 8,114,243

Holding Tank 6,358,340

Pre-Fermentor 1,552,931

Pumps (x7) 4,099,275

Distillation unit

Rectifying column 543,060

Stripping column 543,060

Reboiler 3,508,660

Pre-heater 1,086,639

Condenser 1,258,300

Dehydration unit

U-tube preheater 2,771,029

Bayonet vaporizer 993,491

Bayonet condenser 665,732

Molecular sieves 558,817

Vacuum pump 815,944

Adsorption towers (x2) 3,176,848

Storage tanks (x5) 56,954,365

EQUIPMENT PURCHASE COST

(EPC)

193,249,332

ECONOMICS: CALCULATING TOTAL CAPITAL INVESTMENT

Parameter Range Factor

chosen

Price (Rs) Justification

DIRECT COSTS

Purchase equipment

(f1)

- - - 193,249,332 -

Installation (f2) EPC (x) 0.25-0.55 0.25 48312332.98 Includes structural supports and installation

Instrumentation &

control (f3)

EPC (x) 0.06-0.30 0.30 57974799.58 Intensive control throughout the plant

Piping (f4) EPC(x) 0.10-0.80 0.45

86962199.37 Average value taken

Electrical (f5) EPC(x) 0.10-0.40 0.10 19324933.19 Sinnot (1999) reports the contribution of this element

as 10% of the EPC

Building, process &

auxiliary (f6)

EPC(x) 0.10-0.70 0.22 42514853.02 Peters and Timmerhaus (1991) reports the

contribution of this element as 22% of the EPC for a new plant at an existing

site.

Service facilities

and yard

improvement (f7)

EPC (x) 0.10-1.0 0.10 19324933.19 A new plant requires these services but a minimum is

taken so as priority is given to other more important services.

Land (f8) EPC (x) 0.04-0.08 0.08 15459946.55 Land will be fully required for setting of the plant.

TOTAL DIRECT COST = (f1 + f2 + f3 +f4 + f5 + f6 + f7)483,123,329.82

INDIRECT COSTS

Engineering and supervision (f9)

DC (x) 0.05-0.30 0.30 144936998.9 Maximum expertise is necessary since it is a new

plant.

Construction and contractor’s fee

(f10)

DC (x) 0.06-0.30 0.18 86962199.37 An average is taken as a better estimate.

Contingency (f11) FCI (x) 0.05-0.15 0.1 0.1 FCI A proper average safety margin is deemed

necessary.

TOTAL INDIRECT COST = (f9 + f10 + f11 ) 231899198.3+ 0.1 FCI

Fixed capital investment (FCI) = Direct cost + Indirect cost

FCI = (483,123,329.82 + 231,899,198.3 + 0.1 FCI)

Hence, FCI = Rs 794,469,475.7

Working capital (WC) = 0.15 TCI

Total capital investment = FCI + WC

TCI = Rs 794,469,475.7 + 0.15 TCI

Hence, TCI = Rs 934,669,971.4

ECONOMICS: CALCULATING TOTAL PRODUCT COST

Total product cost = manufacturing cost + General Expenses

MANUFACTURING COST

Manufacturing cost = Direct Production cost + fixed charges + plant overheads

1. DIRECTION PRODUCTION COST

Raw materials

Price of raw materials: The price of sugarcane was fixed at 1700 Indian rupees per tonne of

sugarcane (N.Deshmukh, TNN, 2013) (

http://articles.timesofindia.indiatimes.com/2012-07-23/pune/32803260_1_sugarcane-farmers-sugar-

factories-raghunath-patil)

Converting Indian currency to Mauritian currency and obtaining: 1 tonne of sugarcane = Rs 1190

Amount of sugarcane used = 103.102 tonnes/h

Total amount = 103.102 * 24 * 150 = 371,167.2 tonnes

371,167.2 tonnes of sugarcane = Rs 445,536,000

Operating labour and direct supervisory and clerical labour

Product capacity = 167 tons/day

Operating labour = 85 (from the graph; the graph chosen in this section as that in case

multiple small units for increasing capacity).

Number of operating days = 150

Annual operating labour requirement = 85*5*150 = 63750 hours

Considering 9 working hours per employee,

Total number of employees required per day = 63750150∗9

= 47.22 = 48

Figure 1: Labour requirement for chemical process industries

(Source: Plant design and economics for chemical engineers, Peters M.S and Timmerhaus K.D)

The technical team in charge of the operation of the BioEPP will comprise of:

The General Manager

Operations Manager

Process Manager

Maintenance Manager

Shift supervisors and Operators

The salary were taken from PRB 2012-2013

Occupation Quantity Salary per person/Rs Justification

OPERATING LABOUR

General Manager 1 102,000 Responsible for all business operations of the ethanol plant including product operations,

purchasing, marketing and personnel.

Operation Manager 1 48 600Responsible for optimizing plant runtime as well as downtime while ensuring a quality end product.

Process Manager 1 48 600 Work with department managers and supervisors in operations,

maintenance, logistics, environmental, health, safety and

quality

Maintenance Manager 1 44 100 In charge of coordinating and supporting plant maintenance performance personnel and

contractors to achieve efficient and safe completion of the workload.

Senior electrical engineer

1 26 250

New plant, therefore one of each category is enough.

Electrical engineer 1 18 900

Chief technician 1 16 500 Decision making during shut down or major technical problems

Supervisor 2 14 300 Supervising daily operations including ethanol storage, loading,

inventory control measures and safety compliance and sanitation.

Laboratory attendant 1 10 250Run multiple tests in accordance

to the standard operation procedures and in accordance with the frequency established

by the Process Manager.

Driver 2 10 000For emergencies and running odd

jobs on working premises.

Watchman 2 9 000Assuring plant security

Store attendant 1 8 500

Keep track of equipment and chemicals.

General Worker 3 7 400 Ensure proper up-keeping of the working premises.

DIRECT SUPERVISORY AND CLERICAL LABOUR

Senior Technician 5 16500

One major technician per unit

Technician 10 15800 1 technician per unit per shift

Foreman 5 15380 Extensive supervision required for floor work

Chef de quarts 5 20000 Verify all the controls and instrumentations regarding the plant. Take necessary steps in

case of cropping up of technical problems.

Plumbers and pipe fitters

4 102503 fitters per shift

Account clerk 1 11250To record all the transactions

TOTAL 48 882 150

Total salary = (412 500 + 469650) = Rs 882 150 [for 48 staff working 9 hrs/day during 1 month]

Cost of labour per hour = 882 150 / (48 x 30 x 9) = Rs 68.08

Annual labour cost = 68.08 x 63750 = Rs 4,339,280

Utilities

Since the plant is self-sufficient in terms of steam and electricity produced by the

cogeneration plant for partly internal consumption, utilities cost (10-15% of TPC) is

taken as 15 % of the total product cost.

Utilities cost = 0.15 * TPC

Maintenance and repairs

Maintenance is a planned programme in any process plant in order to keep all the

equipment in a good working condition. The maintenance cost includes cost for material

tools and supervision and can be in the range of 2-10% of the fixed capital investment

(FCI).

Available data

FCI = Rs 794,469,475.70

% chosen = 2

Hence cost associated with maintenance and repair = 0.002 FCI = Rs 15,889,390

Operating supplies

Name Use Quantity

(Unit/day)

Price/Unit Cost/Yr

Lime For Juice Treatment 1903 kg $110/ton 31,400

Activated dry yeast Source for

fermentation

0.5 kg $2000/ton $ 150

Sulphuric Acid

(98%)

pH control 100 L $250/metric ton $6900

Di Ammonium

phosphate

Nutrient for yeast

propagation

80 Kg $800/metric ton $9600

Phosphoric acid Nutrient for yeast

propagation

30 L $800/ metric ton $ 6800

Magnesium sulphate Nutrient for yeast

propagation

0.25 kg $100/metric ton $3.75

Caustic soda For cleaning and pH

control

100 Kg $400/ton $6 000

TOTAL $ 60,854

In addition, corrosion inhibitors will be needed to prevent corrosion in storage tanks as well as

the use for gasoline for denaturation.

Gasoline

The gasoline price in Mauritius is found to be 51.30 MUR/L

Finding the required amount of gasoline:

1 day = 2.5173 m3

150 days = 378 m3 = 378, 000 L

Cost of gasoline = 51.30 MUR * 378000 L

= 19,391,400 MUR

Corrosion inhibitor

According to ChemWorld.com based in USA, the cost of corrosion inhibitor is as follows:

Quantity/ gallons Cost/ USD

55 709.99

30 549.99

15 399.99

5 149.99

Finding the amount of corrosion inhibitor required:

1 day = 2.27 * 10-3 m3

150 days = 34.05 m3 = 340.5 L

1 US gallon = 3.7854 L

Number of gallons needed = 340.5

3.7854 = 90 gallons

Since corrosion inhibitor is sold in 55/30/15/5 gallons by the company, 3 * 30 gallons are selected to be bought.

30 gallons = 549.99 USD

Cost of corrosion inhibitor = 3 * 549.99

= 1650 USD

= 51,402 MUR

Total cost for operating supplies = Rs (1,880,389 + 19,391,400 + 51,402)

= Rs 21,323,191

Laboratory charges

Laboratory charges are taken normally as 10-20% of the operating labour cost, thus;

Laboratory charges = 0.20 x operating labour

= 0.20 x Rs 4,339,280

= Rs 867,856

Patents and royalties

This item is taken as 6% of the Total Product Cost (TPC)

2. FIXED CHARGES

Depreciation

The straight line method is used to calculate the overall depreciation of the equipment. The fixed %

factor is then evaluated by the double declining method. The depreciation of the equipment over the

whole service life is calculated using the declining balance method. However, the yearly depreciation

of building is taken as a % of the building value.

a. Equipment depreciation

Straight line method (TimmerHaus, 1991),

d = V−Vs

n

Where

d = annual depreciation / USD

V = original value of total equipment / USD

Vs = salvage value of total equipment / USD

n = number of service life / years

Available data:

V = 193,249,332

Vs = 25 million rupees (Local sugar factory)

n = 30 (BP Biofuels 2012, Brazil)

d = Rs 5,608,312

Double declining balance method (TimmerHaus, 1991),

f = 2 x dv

Where

f = fixed % factor

f = 0.0580 (3 s.f)

Declining balance method (TimmerHaus, 1991),

Vs = V (1-f)n

= Rs 32,184,375

For year 1; vs = Rs 1,072,813

b. Building depreciation

Available data:

Building Cost = 42514853.02

% chosen = 2.5

Depreciation on building = 0.025x Building cost = Rs 1,062,871

Total depreciation = 1,062,871 + 1,072,813

= Rs 2,135,684

Local taxes

Since the location of the property is meant to be in the northern part of Mauritius, taxes

are lower as compared to more developed and highly populated areas, thus a factor of 2%

of the FCI can be used.

Local taxes = 0.020 * FCI

= 0.020 * Rs 794,469,475.70

= Rs 15,889,389

Insurance

Insurance rates depend on the type of process being carried out in the manufacturing

operation and on the extent of available protection facilities. On an annual basis, these

rates amount to about 1 percent of the fixed-capital investment (FCI).

FCI = Rs 794,469,475.70

% chosen = 1

Hence cost associated with maintenance and repair = 0.001 FCI = Rs 7,944,695

Rent

This item is taken as 10% of the cost of building cost

Given building cost = Rs 42,514,853

Cost for rent = 0.1 x 42514853 = Rs 4,251,485

3. PLANT OVERHEAD

Plant overhead is taken as 1 % of the total product cost (TPC)

GENERAL EXPENSES

1. ADMINISTRATIVE COSTS

Administrative costs = 0.15 [supervision costs + maintenance costs + operating labour]

(P.Jeetah, 2012)

= 0.15 [132,713,887.2 + 14,549,374.3 + 4,339,280]

= Rs 22, 740, 381

2. DISTRIBUTION AND SELLING COSTS

Usually amounts to 1-20% of the total product cost. (TPC). Percentage chosen = 1 %

Hence, distribution and selling costs = (0.01 TPC )

3. RESEARCH AND DEVELOPMENT COSTS

Cost for this item is taken as 5 % of the Total Product Cost (TPC)

Calculating Total Product Cost (TPC)

Manufacturing cost = Direct Production cost + fixed charges + plant overheads

= Rs 518,176,970 + 0.22 TPC

Total product cost = manufacturing cost + General Expenses

TPC = (Rs 518,176,970 + 0.22 TPC) + (Rs 24,774,850 + 0.06 TPC)

Solving for TPC, we have:

Total Product Cost (TPC) = Rs 754,099,750

Calculating the total revenue

UNIT TOTAL (RS)

Ethanol (Rs/L) 38.00 997,500,000

Carbon Dioxide (Rs/ton) 2291.25 19,590,187.50

Total income 1,017,090,188

Total income = Rs 1,017,090,188

Gross earnings = Total income – Total product cost

= (1,017,090,188- 754,099,750)

= Rs 262,990,437

Net earnings = Gross earning – tax paid

Whereby excise tax = 15% (MRA)

Net earnings = Rs [262,990,437 – (0.15 x 262,990,437) ] = Rs 223,541,871

Calculating Simple Payback Period

With constant revenue of Rs 223,541,871:

Simple payback period = TotalCapital Investment

Net earning=934,669,971

223,541,871=4.18

Simple payback is found to be 4.18 years.

PROFITABILITY FACTORS

Turnover ratio

The asset turnover ratio is a measure of how efficiently a company's assets generate revenue.

For production to be profitable, turnover ratio ≥ 1.

Turnover ratio = Annual Sales

¿Capital Investment

= 1017090188794469475.7

= 1.28

Thus production is said to be still profitable, increasing the chances of investing in it.

Finding internal rate of return, IRR

Total profit alone cannot be used as a determining factor for profitability (Dr D.Surroop, 2013),

another important profitability factor is the internal rate of return (IRR) which helps to compare

the time span taken to recover all investments. IRR is given by the following formula:

IRR = Net earning costs

TotalCapital Investment

To find the net earnings, the tax paid on earnings should be deducted from the gross earnings:

Net earnings = Gross earnings – excise tax

Whereby excise tax = 15% (MRA)

Net earning cost = 262990437 – (0.15 * 262990437)

= 223541871.5 MUR

Total Capital Investment (TCI) = 934669972 MUR

IRR = 2235418712934669972

= 0.239

A higher rate of return indicates less time will be taken to recover initial investments in the

project.

Net Present Value, NPV analysis

NPV is a calculation that compares the amount invested today to the present value of the future

cash receipt from the investment. If NPV<0; project should not go ahead, NPV=0; all

investments are recovered over service life and if NPV>0; project is profitable.

Finding NPV

NPV = Present value - Initial Investment

= ∑i=1

i=n

P V n - Co

Whereby P V n = Cn* 1

(1+r )n

Co = initial investment (TCI)

Cn = net cash flow at year n

r = rate of return

n = service life

Whereby n is assumed to be 10 years for a bio-ethanol plant (Technical and Economic Feasibility of Production of Ethanol from Sugarcane by L.A.M Van Der Wielen et al, 2005)

The discounted flows at an IRR of 0.239 are as shown below:

Year Annual revenue (MUR) PV (MUR)1 262990437 212231696.4

2 262990437 171269698.9

3 262990437 138213614

4 262990437 111537552.9

5 262990437 90010132.43

6 262990437 72637633.98

7 262990437 58618132.5

8 262990437 47304479.37

9 262990437 38174429.54

10 262990437 30806534.39

∑i=1

i=n

P V n-

970803904.5

Co= 934669972 MUR

NPV = ∑i=1

i=n

P V n - Co

= 970803904.5 – 934669972

= + 36,133,934 MUR

NPV being positive indicates the profitability of the investments in the bio-ethanol plant.

Finding IRR when NPV = 0 to know break-even point:

Year

Annual revenue

(MUR)

PV

IRR= 0.100

PV

IRR= 0.150

PV

IRR= 0.200

PV

IRR= 0.239

PV

IRR = 0.300

PV

IRR= 0.400

1262990437 239082215.5

228687336.

5 219158697.5

212231696.

4 202300336.2 187850312.1

2262990437 217347468.6

198858553.

5 182632247.9

171269698.

9 155615643.2 134178794.4

3262990437 197588607.8

172920481.

3 152193539.9 138213614 119704340.9 95841995.99

4262990437 179626007.1

150365635.

9 126827949.9

111537552.

9 92080262.25 68458568.57

5262990437 163296370.1

130752726.

9 105689958.3

90010132.4

3 70830970.96 48898977.55

6262990437 148451245.5

113698023.

4 88074965.24

72637633.9

8 54485362.28 34927841.11

7262990437 134955677.8

98867846.4

2 73395804.37 58618132.5 41911817.14 24948457.93

8262990437 122686979.8

85972040.3

6 61163170.31

47304479.3

7 32239859.34 17820327.09

9262990437 111533618

74758295.9

7 50969308.59

38174429.5

4 24799891.8 12728805.07

10262990437 101394198.2

65007213.8

8 42474423.82

30806534.3

9 19076839.84 9092003.62

∑i=1

i=n

P V n-

1615962388 1319888154 1102580066

970803904.

5 813045323.9 634746083.5

TCI -934669972 934669972 934669972 934669972 934669972 934669972

681292416.9

385218182.

7 167910094.5

36133933.0

7 -121624647.5 -299923888

Since IRR lies between the +NPV and –NPV, finding IRR when NPV = 0;

When NPV = 36133933.07, IRR = 0.239

When NPV = -121624647.5, IRR = 0.300

Finding IRR when NPV = 0,

IRR = (0.3−0.239 )(0−36133933.07)(−121624647.5−36133933.07)

+ 0.239

= 0.253

Discounted payback period

One of the main disadvantages of the simple payback period is that it ignores the time value of

money. To counter this limitation, an alternative method known as the discounted payback

period can be followed. In discounted payback period we have to calculate the present value of

each cash inflow taking the start of the first period as zero point.

Initial Investment = Total Capital Investment

= Rs 934669,971

Taking Rate, K = 0.1

Discounting Factor = (1+k )−n

Where n = number of years

Year Cash Inflow/ Rs

Discounting Factor

(Rate = 0.1 %)

Discounted Cash Flow/ Rs

Cumulative Discounted

Cash flow/ Rs

0 -934669971.4 1 -934669971.4 -934669971.4

1 223541871.5 0.909090909 203219883.1 -731450088.3

2 223541871.5 0.826446281 184745348.3 -546704740

3 223541871.5 0.751314801 167950316.6 -378754423.3

4 223541871.5 0.683013455 152682106 -226072317.3

5 223541871.5 0.620921323 138801914.6 -87270402.69

6 223541871.5 0.56447393 126183558.7 38913156.02

Discounted payback period = 5 + ( 87270402.69126183558.7 ) = 5.69 years