Engineering Economics: Present Economy Principles and Analysis

Fundamentals of Present Economy Analysis

  • Definition of Present Economy:

    • Present economy involves the quantitative analysis of engineering problems associated with manufacturing a product or rendering a service, evaluated strictly on the basis of present or immediate costs.
    • It applies specifically when the effects of time—such as interest on capital and depreciation of equipment—are negligible.
    • Present economy studies are employed when the operational alternatives being compared yield identical or equivalent final results and the time frame involved in the analysis is relatively short.

  • Conditions Governing Present Economy Studies:

    • Interest effects are assumed to be negligible due to short duration.
    • Depreciation impacts on equipment are non-significant over the study period.
    • Operational outcomes of all evaluated alternatives are completely equivalent.
    • Primary evaluation criterion is immediate operational cost minimization.

  • Primary Operational Applications:

    • Present economy evaluations are categorized into seven distinct situational domains:
    1. Selection of material.
    2. Selection of method to be used.
    3. Selection of design.
    4. Selection of project location or site.
    5. Comparison of proficiency among workers.
    6. Economy of tool and equipment maintenance.
    7. Economy in the utilization of personnel (number of laborers).

Seven Situational Categories in Present Economy

  • Selection of Material:

    • In product manufacturing, multiple satisfactory materials are frequently available to satisfy required technical specifications.
    • The economic analysis focuses on selecting the material that minimizes total cost—combining raw material unit cost and all associated processing costs—while maintaining acceptable product quality and functionality.

  • Selection of Method:

    • Alternative engineering methods or equipment setups can accomplish identical operational outcomes.
    • Examples include manual labor versus mechanized ditch digging, or selecting among delivery trucks of differing payload capacities.
    • Present economy determines the specific method that accomplishes the required output at the lowest overall monetary expenditure.

  • Selection of Design:

    • Engineers design machines or components by generating multiple candidate configurations and systematically eliminating unsuitable options to find the most economical solution.
    • Product packaging design balances structural fitness with aesthetic appeal to the consumer market.
    • Early design selection requires extreme care because modifications post-production are cost-prohibitive, particularly after mass-producing large package inventories.

  • Site Selection:

    • Evaluating candidate locations for industrial plants, factories, or civil infrastructure requires assessing variable location-dependent costs.
    • Key factors include land acquisition cost, site-specific construction costs, variable transportation expenses, and proximity to borrow pits in highway or dam projects (which directly determines earth transportation costs).

  • Comparison of Proficiency of Workers:

    • Worker efficiency varies significantly across industrial and service operations, directly impacting operational cost structures.
    • In specialized fields, above-average proficiency is essential:
    • In academic instruction, candidate selection favors individuals with superior academic performance records, all other factors being equal.
    • In the entertainment industry, highly popular actors command high compensation and star roles based on market value, even when less-known actors are available.
    • In standardized industrial processes, worker proficiency differences are directly quantified into precise monetary cost differentials.

  • Economy of Tool and Equipment Maintenance:

    • Regular operational maintenance preserves peak equipment performance and minimizes operational expenses over time.
    • Empirical guidelines govern scheduled maintenance protocols, such as automotive preventive maintenance standards:
    • Vehicle washing and greasing scheduled every 2000 km2000\,\text{km}.
    • Engine oil replacement scheduled every 1000 km1000\,\text{km}.
    • Spark plug replacement scheduled every 10,000 km10,000\,\text{km}.
    • Daily maintenance routines ensure optimal functional efficiency (e.g., carpenters maintaining sharp chisels and saws at the start of each work day; barbers sharpening razors and scissors to maximize customer comfort).

  • Economy in the Utilization of Personnel:

    • Team size optimization determines the exact worker complement required for maximum productivity on a given operational stage.
    • Increasing crew size beyond optimal capacity triggers the law of diminishing returns.
    • Over-staffing leads to worker idle time while waiting for dependent steps to complete; economy dictates maintaining only the number of laborers that can remain productively engaged at all times.

Detailed Case Examples and Quantitative Solutions

  • Example 1: Selection of Material (Warehouse Floor Joists):

    • Problem Description: A warehouse floor structural design can support design loads using either guijo or apitong timber floor joists over one floor bay. Joist span length is 12 ft12\,\text{ft} (3.7 m3.7\,\text{m}).
    • Configuration Options:
    • Option A (Guijo): 2′′×8′′2'' \times 8'' guijo joists spaced 0.30 m0.30\,\text{m} on centers. Total required = 2121 joists per bay. Board foot unit cost = P29\text{P}29.
    • Option B (Apitong): 2′′×10′′2'' \times 10'' apitong joists spaced 0.40 m0.40\,\text{m} on centers. Total required = 1616 joists per bay. Board foot unit cost = P21\text{P}21.
    • Quantitative Analysis:
    • Guijo Board Foot Calculation:       Total bd ft (Guijo)=21×(2′′×8′′12′′)×12 ft=336 bd ft\text{Total bd ft (Guijo)} = 21 \times \left(\frac{2'' \times 8''}{12''}\right) \times 12\,\text{ft} = 336\,\text{bd ft}
    • Guijo Total Cost Calculation:       Total Cost (Guijo)=336 bd ft×P29/bd ft=P9,744.00\text{Total Cost (Guijo)} = 336\,\text{bd ft} \times \text{P}29/\text{bd ft} = \text{P}9,744.00
    • Apitong Board Foot Calculation:       Total bd ft (Apitong)=16×(2′′×10′′12′′)×12 ft=320 bd ft\text{Total bd ft (Apitong)} = 16 \times \left(\frac{2'' \times 10''}{12''}\right) \times 12\,\text{ft} = 320\,\text{bd ft}
    • Apitong Total Cost Calculation:       Total Cost (Apitong)=320 bd ft×P21/bd ft=P6,720.00\text{Total Cost (Apitong)} = 320\,\text{bd ft} \times \text{P}21/\text{bd ft} = \text{P}6,720.00
    • Cost Differential Calculation:       Cost Difference=P9,744.00−P6,720.00=P3,024.00\text{Cost Difference} = \text{P}9,744.00 - \text{P}6,720.00 = \text{P}3,024.00
    • Decision Outcome: Apitong joists are more economical, resulting in a direct cost savings of P3,024.00\text{P}3,024.00 per bay.

  • Example 2: Selection of Method (Machine Part Production Jig):

    • Problem Description: Manufacturing contract for 100,000100,000 units of a machine part requires selecting between two alternative production jigs.
    • Parameter Specifications:
    • Jig A: Service life = 100,000100,000\text{ pieces}; First cost = P20,000\text{P}20,000; Operator wage = P25/hr\text{P}25/\text{hr}; Output rate = 100100\text{ pieces/hr}.
    • Jig B: Service life = 100,000100,000\text{ pieces}; First cost = P19,000\text{P}19,000; Operator wage = P22/hr\text{P}22/\text{hr}; Output rate = 8080\text{ pieces/hr}.
    • Quantitative Analysis:
    • Evaluation for Jig A:       Operating Time (Jig A)=100,000 pieces100 pieces/hr=1,000 hrs\text{Operating Time (Jig A)} = \frac{100,000\,\text{pieces}}{100\,\text{pieces/hr}} = 1,000\,\text{hrs}Operator Wages (Jig A)=1,000 hrs×P25/hr=P25,000.00\text{Operator Wages (Jig A)} = 1,000\,\text{hrs} \times \text{P}25/\text{hr} = \text{P}25,000.00First Cost (Jig A)=P20,000.00\text{First Cost (Jig A)} = \text{P}20,000.00Total Cost (Jig A)=P25,000.00+P20,000.00=P45,000.00\text{Total Cost (Jig A)} = \text{P}25,000.00 + \text{P}20,000.00 = \text{P}45,000.00
    • Evaluation for Jig B:       Operating Time (Jig B)=100,000 pieces80 pieces/hr=1,250 hrs\text{Operating Time (Jig B)} = \frac{100,000\,\text{pieces}}{80\,\text{pieces/hr}} = 1,250\,\text{hrs}Operator Wages (Jig B)=1,250 hrs×P22/hr=P27,500.00\text{Operator Wages (Jig B)} = 1,250\,\text{hrs} \times \text{P}22/\text{hr} = \text{P}27,500.00First Cost (Jig B)=P19,000.00\text{First Cost (Jig B)} = \text{P}19,000.00Total Cost (Jig B)=P27,500.00+P19,000.00=P46,500.00\text{Total Cost (Jig B)} = \text{P}27,500.00 + \text{P}19,000.00 = \text{P}46,500.00
    • Cost Differential Calculation:       Cost Difference=P46,500.00−P45,000.00=P1,500.00\text{Cost Difference} = \text{P}46,500.00 - \text{P}45,000.00 = \text{P}1,500.00
    • Decision Outcome: Select Jig A, which yields total savings of P1,500.00\text{P}1,500.00 over Jig B.

  • Example 3: Selection of Method (Irrigation Canal Digging Bidding):

    • Problem Description: A construction company based in Manila is preparing a project bid for digging an irrigation canal in a nearby province. Canal dimensions are 2 m2\,\text{m} wide, 1 m1\,\text{m} deep, and 2 km2\,\text{km} (2000 m2000\,\text{m}) long. Total profit and contingencies allowance is fixed at P60,000.00\text{P}60,000.00.
    • Parameter Specifications:
    • Manual Method: Crew consists of 2020 laborers at P300/day\text{P}300/\text{day} each, digging average 0.5 m3/hr0.5\,\text{m}^3/\text{hr} per person; 22 foremen at P500/day\text{P}500/\text{day} each; standard working shift = 8 hrs/day8\,\text{hrs/day}.
    • Machine Method: Ditch digging machine rented at P3000/day\text{P}3000/\text{day} (charged during transit and operation); machine dig rate = 1 m31\,\text{m}^3 per 4 minutes4\,\text{minutes}; fuel/oil cost = P350/day\text{P}350/\text{day} (operating days only); transit time = 1 day1\,\text{day} each way (2 days2\,\text{days} total); freight transport cost = P2500\text{P}2500 each way (P5000\text{P}5000 total); machine operator wage = P500/day\text{P}500/\text{day} (all days); auxiliary crew = 22 laborers at P300/day\text{P}300/\text{day} each (operating days only).
    • Volume of Excavation:     Total Volume=2 m×1 m×(2 km×1000 m/km)=4000 m3\text{Total Volume} = 2\,\text{m} \times 1\,\text{m} \times \left(2\,\text{km} \times 1000\,\text{m/km}\right) = 4000\,\text{m}^3
    • Quantitative Analysis (Manual Labor Option):     Total Work Hours Required=4000 m30.5 m3/hr=8000 hrs\text{Total Work Hours Required} = \frac{4000\,\text{m}^3}{0.5\,\text{m}^3/\text{hr}} = 8000\,\text{hrs}Total Man-Days Required=8000 hrs8 hrs/day=1000 days\text{Total Man-Days Required} = \frac{8000\,\text{hrs}}{8\,\text{hrs/day}} = 1000\,\text{days}Project Duration (20 Laborers)=1000 man-days20 laborers=50 days\text{Project Duration (20 Laborers)} = \frac{1000\,\text{man-days}}{20\,\text{laborers}} = 50\,\text{days}Foremen Wages=2×P500/day×50 days=P50,000.00\text{Foremen Wages} = 2 \times \text{P}500/\text{day} \times 50\,\text{days} = \text{P}50,000.00Laborers Wages=20×P300/day×50 days=P300,000.00\text{Laborers Wages} = 20 \times \text{P}300/\text{day} \times 50\,\text{days} = \text{P}300,000.00Profit and Contingencies=P60,000.00\text{Profit and Contingencies} = \text{P}60,000.00Total Bid Price (Manual)=P50,000.00+P300,000.00+P60,000.00=P410,000.00\text{Total Bid Price (Manual)} = \text{P}50,000.00 + \text{P}300,000.00 + \text{P}60,000.00 = \text{P}410,000.00
    • Quantitative Analysis (Machine Option):     Total Digging Minutes=4000 m3×4 min/m3=16,000 min\text{Total Digging Minutes} = 4000\,\text{m}^3 \times 4\,\text{min/m}^3 = 16,000\,\text{min}Total Digging Hours=16,000 min60 min/hr=266.67 hrs\text{Total Digging Hours} = \frac{16,000\,\text{min}}{60\,\text{min/hr}} = 266.67\,\text{hrs}Operating Days Required=266.67 hrs8 hrs/day=33.333→34 days\text{Operating Days Required} = \frac{266.67\,\text{hrs}}{8\,\text{hrs/day}} = 33.333 \rightarrow 34\,\text{days}Total Paid Machine Days=34 operating days+2 transit days=36 days\text{Total Paid Machine Days} = 34\,\text{operating days} + 2\,\text{transit days} = 36\,\text{days}Machine Rental Cost=P3000/day×36 days=P108,000.00\text{Machine Rental Cost} = \text{P}3000/\text{day} \times 36\,\text{days} = \text{P}108,000.00Fuel Cost=P350/day×34 days=P11,900.00\text{Fuel Cost} = \text{P}350/\text{day} \times 34\,\text{days} = \text{P}11,900.00Freight Charge=P2500/way×2 ways=P5,000.00\text{Freight Charge} = \text{P}2500/\text{way} \times 2\,\text{ways} = \text{P}5,000.00Operator Wages=P500/day×36 days=P18,000.00\text{Operator Wages} = \text{P}500/\text{day} \times 36\,\text{days} = \text{P}18,000.00Laborers Wages=2×P300/day×34 days=P20,400.00\text{Laborers Wages} = 2 \times \text{P}300/\text{day} \times 34\,\text{days} = \text{P}20,400.00Profit and Contingencies=P60,000.00\text{Profit and Contingencies} = \text{P}60,000.00Total Bid Price (Machine)=P108,000.00+P11,900.00+P5,000.00+P18,000.00+P20,400.00+P60,000.00=P223,300.00\text{Total Bid Price (Machine)} = \text{P}108,000.00 + \text{P}11,900.00 + \text{P}5,000.00 + \text{P}18,000.00 + \text{P}20,400.00 + \text{P}60,000.00 = \text{P}223,300.00
    • Cost Differential Calculation:     Cost Difference=P410,000.00−P223,300.00=P186,700.00\text{Cost Difference} = \text{P}410,000.00 - \text{P}223,300.00 = \text{P}186,700.00
    • Decision Outcome: The lowest bid price for the company is P223,300.00\text{P}223,300.00, achieved by utilizing the ditch digging machine, which provides an economic savings of P186,700.00\text{P}186,700.00 compared to manual labor.