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COURSE INSTRUCTOR : PROF. DR. SHAHAB KHUSHNOOD. Chapter#10 Techniques for making better engineering management decisions. Lecture No. 08 Course: Engineering Management. MED DEPARTMENT, U.E.T TAXILA. INTRODUCTION.

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## Chapter#10 Techniques for making better engineering management decisions

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**COURSE INSTRUCTOR :**PROF. DR. SHAHAB KHUSHNOOD Chapter#10Techniques for making better engineering management decisions Lecture No. 08 Course: Engineering Management MED DEPARTMENT, U.E.T TAXILA**INTRODUCTION**• During middle of 1940s, a significant growth of quantitative management techniques occurred. • The increase in the use of computers and the modern problem complexities have enhanced the importance of many of these techniques. For Example: • Linear Programming. • Non-Linear Programming.**Today, the mostly management techniques which are in**practice: • Linear Programming • Decision Trees • Exponential Smoothing • Discounted Cash Flow Analysis**Optimization Techniques**There are two widely known optimization techniques known as: • Lagrangian Multiplier Technique (Technique of undetermined Multipliers) • Linear Programming Technique (Advanced form of simplex method technique)**1-LAGRANGIAN MULTIPLIER TECHNIQUE**• This technique is demonstrated for a two variable function. • However on similar lines it can be extended fornvariables. • Assume that we have defined function f(y1,y2) Subject to the constraint function k(y1,y2)=0**1-LAGRANGIAN MULTIPLIER TECHNIQUE**• With the aid of these two functions the Lagrangian function , L(y1,y2,λ) formulated as follows:**1-LAGRANGIAN MULTIPLIER TECHNIQUE**• Subject to the following necessary conditions for estimating a relative maximum or minimum value: • Expressions for y1,y2 and λ can be obtained by solving the simultaneous equations.**Solution:**The Lagrange function is formulated as follows Taking the partial derivative with respect to y1,y2 and λresults in**2y1+2λ = 0**2y2- λ=0 2y1-y2=0 So Y1=8/5 Y2=-8/10 λ =-8/5 Thus the critical point of f, subjected to specified condition, is (8/5,-8/10) .**2-LINEAR PROGRAMMING**• In real life situations the objective and constraint functions will be rather complex. • However the simplest form of linear programming problem formulation may be expressed as follows**Example 2**For above equations following values for symbols are defined Write down the resulting equations by assuming that the objective function is to be maximized.**(1)**(2) (3) (4) (5) (6)**From Plot of equations 1-6:**• Company profit is optimum at point F. (F point satisfies all the constraints) • F point shows maximum profit. The optimum values of Z1 and Z2 at point F are 6.5 and 2. substitute these values in equation.1.**Discounted Cash Flow Analysis**• In various engineering investment decisions, the time value of money plays an important role. • Therefore it is necessary for the engineers to have some knowledge of engineering economics.**Discounted cash flow analysis**The basics of engineering economics are • Simple Interest • This is the interest which is calculated on the original sum of money, called the original principal, for the period in which the lent or borrowed sum is being utilized. • The simple interest,St,is given by: (1) Where M is the principal amount lent or borrowed i is the interest rate per period ( this is normally a year) k is the interest periods (these are usually years)**Simple Interest (Cont.…)**The total amount of money, Mt, after the specified lent or borrowed period is given by (2)**Solution:**(1) (2)**2. Compound Interest**• In this case at the end of each equal specified period, the earned interest is added to the original principal or amount lent or borrowed at the beginning of that period • Thus, this new principal, or amount, acts as a principal for the next period and the process continues.**To calculate compound amount, Mck, the resulting formula is**developed as follows:**3.Present Worth**• The present worth of a single payment is given by: • In simple terms, this formula is used to obtained the present worth, M, of money,Mck, after k periods, discounted at the periodic interest rate of i. • Sometime above equation can be written as**Formula For Uniform Periodic Payments**• For using formula it is assumed that at the end of each of the K period or years, the depositor adds D amount of money. The money is invested at the interest rate i, compounded annually or periodically. • Thus the total amount of money (1) where**It should be noted in this equation that the D amount of**money is first time deposited at the end of first year or period.**By substituting k=3, D=2000 and i= 0.15 in equation (1):**From above equation the total amount of money after three year period is:**Example # 8**Given Data: (From Example 7)**Solution:**Substitute the data in given equation:**PRESENT VALUE OF UNIFORM PERIODIC PAYMENTS**• In present value of uniform periodic payment we wish to find the present value of uniform periodic payments after K periods or years instead of total amount. • Thus the present worth, PW, of uniform periodic payments is given by • Multiply both side with (1+i)-1,The resulting present worth formula**Solution:**In this example following values are given: Substitute these values in equation:**Example 9 Cont.**• Thus decision to purchase the machine will be profitable investment.**Depreciation Techniques**• The term “Depreciation” means a decline in value. • In order to take into consideration the change in the value of the product, the depreciation charges are made during the useful life of the engineering products. • The three depreciation techniques are: 1-Declining-Balance Depreciation Method 2-Straight-line Depreciation Method 3-Sum Of Digits Depreciation Method**Depreciation Techniques (Cont..)**1. Declining-Balance Depreciation Method • This method dictates the accelerated write-off of the product worth in its early productive years and corresponding lower write-off near the end of useful life years. • The depreciation rate αd is given by where**The declining balance techniques requires a positive value**of s. The product book value, Vbv(M), at the end of year M is given by:**The annual depreciation charge, DC(M), at the end of year M**is given by**Solution:**From example statement Substitute in equation

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