Present Worth Method of Comparison-Engineering Economics-Lecture Slides, Slides of Microeconomics

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Module No. 08

Present Worth Method of

Comparison

Engineering

Economics

Introduction

 In this method of comparison, the cash flows of each alterative will be reduced to time zero by assuming an interest rate i.  The best alternative will be selected depending on the type of decision by comparing the present worth amounts of the alternatives.  Alternatives may be mutually exclusive because:  The alternatives being considered may require different amounts of capital investment  The alternatives may have different useful lives  The subject of this section will help:  Analyze and compare feasible alternatives  Select the preferred alternative

Rule For Choosing Among Alternatives

 In case the decision to select the alternative with the minimum cost, the alternative with the least present worth amount will be selected.

 In the decision is to select the alternative with the maximum profit, the alternative with the maximum present worth will be selected.

Revenue-Dominated cash flow analysis

 Revenue-dominated cash flow analysis is given as:

PW = - P + R 1 [1/(1 +i)^1 ] + R 2 [1/(1 +i)^2 ] + ………….…+ Rn[1/(1 +i)n] + S [1/(1 +i)n] P = Initial investment Rn = Net revenue at the end of nth year. S = Salvage value at the end of nth year.

 In this formula expenditures are assigned negative sign and revenues are assigned positive signs.

 If there are more alternatives which are to be compared, the alternative with the maximum present worth amount should be selected as the best alternative.

Practice Questions

 An industry is planning to expand its production operation, having different technologies for meeting the goal. The initial outlay and annual revenues with respect to each of the technologies are summarized in the given table.

 Suggest the best technology which is to be implemented based on the present worth method of comparison assuming 20% interest rate, compounded annually.

Initial outlay Annual revenue Life (Years)

Technology 1 (^) 12,00,000 4,00,000 10

Technology 2 (^) 20,00,000 6,00,000 10

Technology 3 (^) 18,00,000 5,00,000 10

 An engineer has two bids for an elevator to be installed in a new building. The details of the bids for the elevators are as follows:

 Determine which bid should be accepted based on PW comparison assuming 15% interest rate compounded annually

Practice Questions

Bid Engineer’s estimates Initial cost Service life Annual Opt & Main. cost

Alpha Company 4,50,000 15 27,

Beta Company 5,40,000 15 28,

 A granite company is planning to buy a fully automated granite cutting machine.

 If it is purchased under down payment, the cost of the machine is Rs. 16,00,000.

 If it is purchased under installment basis, the company has to pay 25% of the cost at the time of purchase and remaining amount in 10 annual equal installments of Rs. 2,00,000 each.

 Suggest the best alternative for the company using the present worth basis at i = 18% , compound annually.

Practice Questions

 A finance company advertises two investment plans.

 In plan 1, the company pays Rs. 12,000 after 15 years for every Rs. 1,000 investment now.

 In plan 2, for every Rs. 1,000 invested, the company pays Rs. 4,000 at the end of the 10th year and Rs. 4,000 at the end of 15th year.

 Select the best investment plan from the investor’s point of view at i = 12%, compound annually.

Practice Questions

 A small business with an initial outlay of Rs. 12, yields Rs.10,000 during the first year of its operation and the yield increases by Rs.10,000 from its second year of operation up to its 10th year of operation.

 At the end of the life of the business, the salvage value is zero.

 Find the present worth of the business by assuming an interest rate of 18%, compounded annually.

Practice Questions