SAB4143 Construction Equipment Dr Shaiful Amri Mansur FKA UTM Skudai

Transcription

1 Job Constraints & Obligations Specific operation Job spec requirements Site conditions & obstacles Site location & weather Time allowed Interdependent balancing Equipment mobility Equipment versatility Specific operation Equipment must be capable to do the work There are many alternative equipment Must consider other alternatives Selected equipment must satisfy constraints imposed by contract documents & job conditions 1 2 Job Spec Requirements Contract documents only specify desired end results Contractor choose the equipment Sometimes mentioned in specs: Methods, sequences, equipment, capacity Prescriptive vs Performance Site conditions & obstacles Ground condition Grades of haul roads Work space, including headroom 3 4 Site location & weather Weather Temperature Precipitation Wind Altitude Power sources Site access Time allowed Time allowed by contract Economical timing of activities sequencing Relative effect of overhead costs on total costs Variation of rental rates with time 6 1

2 Inter-dependence / balance Two or more equipment working together Production rates must be compatible Minimize or eliminate idle time Equipment mobility Movement during one work Movement from one work to another Movement from one project to another Assembling/dismantling time Frequency of moving 7 8 Equipment versatility When many works require one same equipment Need equipment stand-by from nearby project Need equipment schedule Versatile = can be used for many works REFLECTIONS 1. Describe the use of construction equipment 1. by functions 2. by operations 2. Discuss job constraints and obligations: 1. Specifications 2. site condition 3. project duration 4. equipment versatility LEARNING OUTCOMES Know the cost elements EQUIPMENT COSTS 1. Students will be able to discuss the components of equipment costs by grouping them according to the cost hierarchy. 2. Students will be able to explain the time value of money by calculating current, future, and compounded values of money. 3. Students will be able to estimate total unit cost of equipment by analyzing the ownership, operating and investment costs

3 Activity Cost vs Value Cost = Value = Direct Costs Sum of all cost components that can be Labour 1. Production equipment 2. Installed Equipment 3. Material 4. Subcontractors Indirect Costs Project Admin Services Cost Hierarchy Labor Utilities Overhead Costs Facilities Insurance, Permits and Fees +Production Equipment +Material and Installed Equipment +Subcontractors Activity Direct Costs +Site Overhead Cost Equipment and Supplies Mobilisation and Demobilisation +HQ Overhead Cost Activity Total Costs 1 Activity Value 16 Estimating Equipment Cost Why needed? To prepare and win bid/tender To monitor & control costs EQUIPMENT ECONOMICS To find the best equipment Rent, lease or buy? When to replace?

4 Estimating Equipment Cost (cont.) A contractor must know Cash Flow Conventions Notation used in the compound interest factors: 1. quantities of work 2. rates of production 3. hourly costs 4. time-value of money P present sum A uniform installment g uniform gradient cash flow increases by the amount g n no. of interest period i interest rate per period F future sum S salvage value Factors Relating P and F If P is borrowed at interest rate i compounded annually, F to repay the loan at end of n time periods is F = P x (1 + i) n F = single compound amount Factors Relating P and F (cont.) A tractor was paid in cash RM10,000 this year. What s the cost of replacing it with the same amount at the end of 4 th year if inflation is expected 6% per year? F = RM10,000 x ( ) 4 = RM189, Factors Relating A and P Loans intallments or uniform series Uniform series capital recovery factor: A = P i (1 + i) n (1 + i) n - 1 Factors Relating A and P (cont.) A RM10,000 pickup truck was purchased by a 20% downpayment and 60 equal monthly. What s the monthly if annual interest rate is 12% compounded monthly? Downpayment = 0.20 x 10,000 = RM30,000 Remaining = RM10,000 RM30,000 = RM120,000 A = 120,000 (0.01)( ) 60 (1.01) 60-1 = RM2,

5 Factors Relating P and A (cont) Uniform series present worth factor P = A (1 + i) n - 1 i(1 + i) n 0 60 At the time of the 0 th payment, the payer decides to pay off the loan. What s the amount of the final payment? P = 2, , (1.01) 10-1 (0.01)(1.01) 10 = RM27, Factors Relating A and F To find the equivalent uniform annual value of a lump sum amount at the end of time n Uniform series sinking fund factor A = F i (1 + i) n Factors Relating A and F (cont.) Learning Outcomes How much to be deposited each year with 10% earning compounded annually to get RM 200,000 for equipment replacement at the end of years? Students will be able to analyze ownership & operating costs of equipment. A = 200,000 (0.10) (1.10) - 1 = RM32, Ownership Costs Whether or not Recover the costs by 1. Capital 2. Interest 3. Insurance 4. Property taxes. License fees Ownership Costs (cont.) 6. Depreciation loss of value during ownership Depreciation accounting methods: 1. Straight-line 2. Double declining-balance 3. Sum-of-the-year digits Depreciable value = Delivered cost Salvage value Useful life usually 3 to 12 yrs 29 30

6 Straight-line method Depreciable value: D k = P S n Example: Equipment costs RM100,000; salvage estimated RM2,000 over useful yrs. D k = 100,000 2,000 = RM1,000 per yr Book value at end of 3 rd yr B 3 = 100,000 (3 x 1,000) = RM, Double Declining-balance (DDB) method Multiplying the book value of preceding year by a constant factor. Depreciation charge for any year k : D k = 2 x B k n Book value at the end of any year k : B k = Example: (1 2) k x P n Learn the manual technique first Equipment costs RM 100,000, salvage estimated RM 2,000 over useful n= yrs. D 1 = 2 x 100,000 = RM 40,000; D 2 = RM 24,000 Book value at end of 3 rd yr B 3 = (1 2) 3 x 100,000 = RM 21, Sum-of-the-year s-digits (SOYD) method Example 1 If useful life n=, SOYD = = 1 (1 st year depreciation /1; 2 nd year 4/1; etc.) Or SOYD = n (n + 1) = ( + 1) = D k = n k + 1 (P - S) = 2 (1 k ) (P S) SOYD n n + 1 Learn the manual technique first B k = n k x n k + 1 x (P S) + S n n + 1 Example, equipment costs RM100,000 and salvage RM2,000 D 1 = 2(1-1 )(100,000 2,000) = RM 2, D 2 = 2(1-2 ) x 7,000 = RM 20, B 3 = 3 x x 7, ,000 = RM 40, Initial costs = RM12,000; Salvage = RM2,000 Useful life = yrs; Usage = 2,000hr/yr Straight-line depreciation: Total depreciation: 12,000-2,000 = RM10,000 Annual depreciation: 10,000 = RM2,000 Hourly depreciation: 2,000 2,000 = RM Example 2 Equipment costs = RM10,000; Salvage value = RM1,000; Est. life yrs, average depreciation DDB depreciation cost for 1 st and 2 nd year. Double rate, 2 x 20% = 40% Depreciation 1 st yr, 0.40 x RM10,000 = RM4,000 Book value at start of 2 nd year, RM10,000 4,000 = RM6,000 Depreciation 2 nd yr, 0.40 x RM6,000 = RM2,400 Year Value DD Depreciation Book value DDB Rate = 20% DD = 40% (777.6)

7 Example 3 Using the same previous example Total cost = RM 10,000 Estimated salvage value = RM 1,000 Total depreciation = RM 9,000 Estimated useful life, yrs SOYD, = 1 SOYD Year P - S SOYD Depreciation Cum Book value 1 for yrs P - S =