Example (1) A concrete aggregate mix is required to contain at least 31% sand by volume for proper batching. One source of material, which has 25% sand and 75% coarse aggregate, sells for $3 per cubic meter (m3). Another source, which has 40% sand and 60% coarse aggregate, sells for $4.40/m3. Determine the least cost per cubic meter of blended aggregates. Solution Solution : The least cost of blended aggregates will result from maximum use of the lower-cost material. The higher-cost material will be used to increase the proportion of sand up to the minimum level (3 1%) specified. Let x = Portion of blended aggregates from $3.00/m3 source 1 - x = Portion of blended aggregates from $4.40hi3 source Sand Balance X(0.25) + (1 -x)(0.40) = 0.31 0.25x + 0.4-0.4x = 0.31 0.31 – 0.4 = -0.09 0.25 -.04 -0.15 Thus the blended aggregates will contain : 60% of $3.0m3 material, 40% of $4.40/m3 material The least cost per cubic meter of blended aggregates is: 0.6($3.00)+ 0.4($4.4) = 1.8+1.76 = $3.56 / m3
Example (2) A machine part is manufactured at a unit cost of 40¢ for material and 15 ¢ for direct labor. An investment of $500,000 in tooling is required. The order calls for 3 million pieces. Halfway through the order, a new method of manufacture can be put into effect that will reduce the unit costs to 34 ¢ for material and 10 ¢ for direct labor--but it will require $100,000 for additional tooling. This tooling will not be useful for future orders. Other costs are allocated at 2.5 times the direct labor cost. What, if anything, should be done? SolutionSolution : Since there is only one way to handle first 1.5 million pieces, our problem concerns the second half of the order. Alternative A: Continue with Present Method Material cost 1,500,000 pieces x 0.40 = $600,000 Direct labor cost l,500,900 pieces x 0.15 = 225,000 Other costs 2. 5 0 x direct labor cost = 562,500 Cost of remaining 1,500,000 pieces = $ 1,378,500
Cont. Alternative B: Change the manufacturing method Additional tooling cost = $100,000 Material cost 1,500,000 pieces x 0.34 = $510,000 Direct labor cost l,500,900 pieces x 0.10 = 150,000 Other costs 2. 5 x direct labor cost = 375,000 Cost of remaining 1,500,000 pieces = $ 1,135,000 Before making a final decision, one should examine closely the other cost to see that the other costs to see that they do, in fact, vary as the direct labor cost varies. Assuming they do, the decision would be to change the manufacturing method.
Example (4) In the design of a cold storage warehouse, the specifications call for a maximum heat transfer through the warehouse walls of 30,000 joules per hour per square meter of wall when there is a 30°C temperature difference between the inside surface and the outside surface of the insulation. The two insulation materials being considered are as follows: Insulation MaterialCost per Cubic Meter Conductivity(J-m/m2-°C-hr) Rock wool $12.50 140 Foamed insulation 14.0 110 The basic equation for heat conduction through a wall is: Q = K( T)/ (L) Where : Q = heat transfer, in J/hr/m2 of the wall K = conductivity in J-m/m2-°C-hr T = difference in temp. between two surfaces, in °C L = thickness of insulting material in meters. Which insulation material should be selected.
Cont. Solution : There are 2 steps to solve the problem : first, the required thickness of each of the alternative material must be calculated, then, since the problem is one providing fixed output, (heat transfer through the wall is limited to a fixed max. amount), the criteria is to minimize the input. Required insulation thickness : Rock wall 30,000 = 140(3)/L = 0.14 m Foamed insulation 30,000 = 110(30)/L = 0.11m Cost of insulations per square meter of wall thickness in meters. unit cost/m3 X insulation Rock wool unit cost = $ 12.5 X 0.14m = $ 1.75 /m2 Foamed insulation unit cost = $ 14.0X 0.11m = $ 1.54 / m2. The foamed insulation is lesser cost alternative. However, there is an intangible constrain that must be considered. How thick is the available wall space? Engineering economy and time value of money are needed to decide. what the max. heat transfer should be. What the max. heat transfere should be. What is the cost of more insulation versus the cost of cooling the warehouse over its life?
Summary Classifying to solve. Others problems : Many problems are simple and thus easy to solve, others are of intermediate difficulty and need considerable thought / calculation to properly evaluate. There intermediate problems tends to have a substantial economic component. Hence, are good candidates for economic analysis. Complex problems, on other hand, o
Summary Classifying to solve. Others problems : Many problems are simple and thus easy to solve, others are of intermediate difficulty and need considerable thought / calculation to properly evaluate. There intermediate problems tends to have a substantial economic component, hence, are good candidates for economic analysis. Complex problems, on other hand, often contain people elements along with political and economic components. Economic analysis is still very important but the best alternative must be selected considering all criteria not just the economic.
The Rational decision making process : Rational Decision – Making Process uses a logical method to select the best alternative from among the feasible alternatives. The following nine steps can be followed sequentially, but decision makers often repeat some steps undertake some simultaneously, and skip others altogether. 1- recognize the problem 2- define goal or objective. What is the task? 3- assemble relevant data: What are the facts? Are more data needed, and is it worth more than the cost to obtain it ? 4- Define feasible alternative. 5- Select the criterion for choosing the best alternative, possible criteria include political, economic, environmental, and humanitarian. The single criterion may be a composite or several criteria. Cont.
6 - Mathematically model the various interrelations. 7- Predict the outcomes for each alternative. 8- choose best alternative. 9- audit the alternative. Engineering DM refers to solving substantial engineering problems in which economic aspects dominate and economic efficiency is the criteria for choosing from among possible alternatives. It is particular case of general decision making process, some of unusual aspects of engineering decision making are as follows : 1- cost – accounting systems, while an important source of cost data, contain allocations of indirect costs that may be incorporate for use in economic analysis. 2-the various consequences – costs & benefits- of an alternative may be of three types : (a)Market consequences – there are established market prices. (b)Extra – market consequences – there are no direct market process, but prices can be assigned by indirect costs. (c)Intangiable consequences – valued by judgment, not by monetary prices. Cont.
3- the economic criteria for judging alternatives can be reduced to three cases : (a)For fixed input : maximize the benefits or other outputs. (b)For fixed output : minimize the costs or other outputs. (c)When neither input or output is fixed, maximize the differences between benefits, and costs or more simply stated, maximize profit. The third case states the general rule from which both first and second cases may derived. 4- to choose among the alternatives, the market consequences and extra market consequences are organized into a cash flow diagram, we will see in the next chapter, that engineering economic calculation can be used to compare differing cash flows.. These outcomes are compared against the selection criteria. From this comparison plus the consequences not included in Monterey analysis, the best alternative is selected. 5- an essential part of engineering decision making is post audit of results. This step helps to ensure that projected benefits are obtained and to encourage realistic estimates in analysis. Cont.