Project Management - Problem Solving

Description
breakeven analysis, fixed costs, variable costs, cost of production, production costs, different types of depreciation, methods of depreciation,

Project Management Problem Solving
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Break Even Analysis
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Break even analysis is a planning technique based on volume-cost relationships. Break Even analysis is a useful tool in strategic planning. It was developed by Rautenstrauch who showed that the assumptions of linearity in the relationships was not restrictive in practice. Basically costs of production may be divided into two types as follows:
? Fixed

Costs ? Variable costs

Fixed Costs
Fixed Costs (FC) are those Production Costs which are independent of volume. ? Examples of Fixed Costs include rent, rates, insurance costs and staff salary costs (independent of production/sales level bonuses).
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Variable Costs
Variable Costs (VC) are those production costs which are a function of volume. ? In simple Break Even analysis, Variable Costs are assumed to be a linear function of volume. ? Examples of Variable Costs include material costs, direct labour costs (where output is proportional to direct labour input).
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Cost of Production
Total Cost (TC) =

the sum of
Fixed Cost (FC) and Variable Cost (VC)
T.C. = F.C. + V.C.

Production Costs
= F.C. = F.C. where V.C. = a = constant; V
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T.C.

+ V.C. + aV aV = volume of output

Total Revenue (T.R.) = Turnover = p * V p = selling price; V = Volume of output

Break Even Analysis
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In simple Break Even analysis an assumption is made that there is no storage of final product. Everything produced is sold.
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T.C. = F.C. + a V ? T.R. = p * V ? T.C. = T.R. at Break Even point

Profit/ Loss
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T.R. - T.C. = Profit (Loss) Profit = pV - F.C. - a V = (p - a) V - F.C. Profit = 0 when V = F.C./(p-a) Note (p - a) > 0, normally

Profit and Loss
(p - a) is the marginal contribution to overhead per unit, being the difference between price of goods sold per unit and the variable cost per unit. ? It is a very old dictum of business that you should always cover your marginal costs, i.e. p > a.
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Break Even Analysis
At Break Even, V = F.C./(p - a). ? Below Break Even volume a loss is made, whereas above Break Even volume a profit is made.
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BEA – Problem #1
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An engineer has designed two different production processes to manufacture the same product. Details are as follows: Design I Design II Fixed Cost $10,000 $20,000 Variable Cost $ 100 $ 50 per Unit Selling price $ 200 $ 200 per Unit

BEA – Problem #1
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The decision as to which production process to use depends on the volume to be produced. Break Even Volumes for Design I Design II = 133 =100 and

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I.e. Both processes lose money at volumes below their respective breakeven volumes. In particular, both processes run at a loss at volumes of less than 100 units with process II losing more than process I at the same volume.

BEA – Problem #1: Analysis
You are indifferent to Design I or II when ? 10,000 + 100Q = 20,000 + 50Q or 50Q = 10,000 or Q = 200 ? So although Design II is profitable in the range 133 < Q < 200 Design I is more profitable in that range. ? For output levels in the range 100 to 133 using Process I gives rise to a profit whereas the use of Process II results in a loss. ? In the region of outputs between 133 and 200 units both processes are profitable but with Process I giving rise to more profit than Process II. ? In an output range about 200 units, Process II is more profitable than Process I.

BEA – Problem #1: Analysis
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In summary up to an output level of 200 units the manufacturer should choose to use Process I to minimise loss or maximise profit, whereas with output levels in excess of 200 units use should be made of Process II to maximise profit. It is possible to incorporate non linear variable costs and non constant selling prices in the Break even model. A further modification of the simple breakeven model is to consider the demand for the product to be a random variable with a particular distribution and associated parameters.

BEA – Problem #2
A manufacturing company is planning the process for making a new product. A consultant has suggested that companies making similar products have costs that depend on the type of process as follows. Annual fixed cost Variable cost Job shop 100,000 50 Batch 250,000 40 Assembly Line 1,000,000 15
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The company has forecast demand for the product at 25,000 units a year. Which process will probably suit it best? Within what ranges is each type of process best?

BEA – Problem #2
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A job shop process is best for demand, d, from zero until: 100,000 + 50 * D = 250,000 + 40 * D i.e. D = 15,000 A batch process is best for demand from 15,000 until: 250,000 + 40 * D = 1,000,000 + 15 * D i.e. D = 30,000 After this a mass production process is best. With a forecast demand of 25,000 a year the company should look at batch processes.

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Break-even analysis for Worked Example
Total annual cost (£1000s) 1600 Assembly line 1400 1200 Batch Job shop

1000
800 600 400 200

0 10 20 30 Annual demand (1000s )

40

50

Cost of Different Technologies
Total Cost Manual Mechanised Automated

Manual

Mechanised

Automated

Throughput Volume

Depreciation
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Is the reduction in the value or the effective economic life of a product arising from the passage of time, use or abuse, wear and tear, influence of the elements or the cessation of demand for use.

Types of Depreciation
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Types of depreciation ? Physical Causes ? Custom or Usage ? Abnormal Occurrence ? Technological Development and Changes

Depreciation Types
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Physical Causes
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Normal physical wear and tear, due to friction, pull, impact, fatigue, twisting etc. Lack of maintenance and timely repairs Passage of time

Depreciation Types
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Custom or Usage
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There is a policy in Govt., every year the fixed assets like car is depreciated in a fixed percentage (It is usually 20 %). This basically covers depreciation due to usage.

Depreciation Types
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Abnormal Occurrence
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Accidents Defect in material Excessive wear and tear Contingent occurrences, Ex: appearance of hairline crack in pressure vessels adequately tested.

Depreciation Types
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Technological Development and Changes
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Technologically absolute machines

Methods of Depreciation
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Methods of depreciation
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Straight line method Reducing balance method Production based method
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Per unit Per Hour

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Repair provisioning method ? Annuity method ? Sinking fund method ? Endowment policy method ? Revaluation method ? Sum of digits method

Straight Line Method
This method provides for depreciation by means of equal periodic charges over the life of the product. ? Linear method of depreciation ? This method is also called fixed installation method or proportional method. ? This method assumes that the fixed asset wear out at precisely the same amount over the life.
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Straight Line Depreciation
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Annual Depreciation Cost (ADC)
=(I–S)/N ? I – Original cost of the asset ? S – Salvage value (Scrap value) ? N – Life of the asset in years.

Depreciation
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Advantages: ? Easy and simple to operate ? Uniform annual charge afford better comparative costs Disadvantages: ? Unrealistic ? Usually maintenance and repair costs increase as the asset is aged. ? This is considered same from for the life of the asset

Depreciation: Exercise #1
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A melting unit in a steel foundry was purchased at 30,000 Rs. And spent 5000 Rs for erection and commissioning. The estimated residual value of the asset after ten years is 7000 Rs.
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Calculate the annual rate of depreciation? Determine the depreciation amount collected at the end of 7 years after the purchase of melting unit? Value of Asset after 7 years?

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Initial Investment (I)= 30,000+5,000 = 35,000 Rs. Salvage value (S) = 7000 Rs. Life of the asset = 10 Years ADC = (I-S)/N = (35,000 – 7000)/10 = 2,800 Rs/Year Depreciation amt collected at the end of 7 Years = 7 X 2,800 = 19,600 Rs. The value of the asset after 7 years = 35,000 – 19,600 = 15,400 Rs.

Reducing balance method
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This is accelerated method of depreciation. This is also known as diminishing balance or percentage on book value method. Non-linear method of depreciation Value of the asset is depreciated at a much higher rate at the beginning and during the end of the life it is very less. This is the reason it is called accelerated method.

Reducing balance method
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Percentage depreciation (P) = 1 – (S/I)1/N
I – Original cost of the asset ? S – Salvage value (Scrap value) ? N – Life of the asset in years.
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Reducing balance method
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Advantages
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It is simple to understand and use Uses mathematical relation to arrive at the appropriate depreciation percentage. More logical, more depreciation during the initial period, during which minimum maintenance and repair costs.

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Disadvantages
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It is not simple to fix the percentage (P) Uses a standard percentage for all conditions.

Reducing balance method
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An old car was purchased for 32,000 Rs. The life is estimated as 10 years and scrap value is 18,000 Rs. Using the reducing balance method calculate the depreciation? ? Estimate the depreciation fund at the end of 2 years? I = 32,000 ; S = 18,000; N=10 P = 1 – (18,000/32,000)1/10 = 0.1294 The value of the car at the end of 1st year = 32,000(1-0.1294) = 27,859.20 Depreciation amt for the first year = 32,000 – 27,859.20 = 4140.80 The value of the car at the end of 2nd year = 27,859.20(1-0.1294) = 24,237.50 Depreciation amt for the first year = 27,859.20 – 24,237.50 = 3,621.70 Depreciation fund at the end of 2 years = 4140.80 + 3621.70 = 7,762.50 Using the St line depreciation method ADC = (32,000 – 18000)/10 = 1,400 Rs/Year

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Comparison
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In the following graph, the same problem has been worked out for the straight line method and reducing balance and the results are compared for the study.
Comparison of St line with Reducing balance
35000

Cost of the asset

30000 25000 20000 15000 10000 5000 0 1 2 3 4 5 6 7 8 9 10 No of Years Series1 Series2 Series3

Production Based Depreciation
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This is based on the number of units produced or no. of hours the asset has been used (Run). If the number of units produced is less the depreciation is less. This more accurate when compared to the previous methods of depreciation. Rate of depreciation = (Value of the asset / Numbers of units of production) Rate of depreciation = (Value of the asset / Numbers of production hours)

Production Based Depreciation
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Advantages
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Depreciation is based on the usage. ? No assumptions, values are on actual
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Disadvantage
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Very difficult to know the exact production forecast for the future period ? Does not consider the repair and maintenance costs.

Production Based Depreciation:
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Example: A machine costing 200,000 and having a salvage value 100,000 after 10 years of service. The estimated rate of production is 8 units per hour. Assume 50 Weeks for the year and 46 hours a week.
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Rate of depreciation = (Value of the asset / Numbers of units of production)
= 100,000/(10X50X46X8) = 0.5435 Rs/Unit

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Rate of depreciation = (Value of the asset / Numbers of production hours)
= 100,000/(50X46X10) = 4.347 Rs/Hour

Thank You
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Thank You…

Decision Matrix Activity
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Use COWS method to arrive at decision
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Criteria. Develop a hierarchy of decision criteria, also known as decision model. ? O Options. Identify options, also called solutions or alternatives. ? W Weights. Assign a weight to each criterion based on its importance in the final decision. ? S Scores. Rate each option on a ratio scale by assigning it a score or rating against each criterion

Decision Metrics
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A decision matrix allows decision makers to structure, then solve their problem by:
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and prioritizing their needs with a list a criteria; then ? evaluating, rating, and comparing the different solutions; and ? selecting the best matching solution



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