Question

4. Yearly labor costs of a highway maintenance group are currently $420,000/year. If labor rates escalate at a 12% rate and general inflation increases at 9%, determine for each of the next 6 years the labor cost in then-current and constant worth dollars. (3.3) of the next be ars 129 rate and

Answer 1

We are allowed to solve only one question at once. So, I am solving only one question that is provided in the proper orientation. For the first year Cost in dollars Labor lost-3 million =3*106 Material cost = 2 million = 2*10^6 Energy cost=15*10^6 Given compounding rate of labor, material and the energy are- Labor cost-5% decrease Material cost-6% increase Energy cost-3 % increase Labor cost of the nth year = (3*10^6)(1-0.5)^n-1 Thus for the fifth year labor cost={3*10^6)*(1-0.5)44 dollars =2.44*10^6 dollars Material cost = 2*10^6*(1+0.6)An-1 Thus material cost for 5th year=2*10^6*(1+0.644 dollars 2.52*10^6 dollars

Energy cost = (1.5*10^9131-0.5)^n-1 Thus energy cost for 5th car=1.5*10^6 (110.3)~ 4 dollars = 1.68*10^6 dollars Part b When we use the interest rate of 10%. Total cost using 10% interest rate. First year total cost = 6.5*10^6 dollars Second year total cost = 6.5*1.1.10^6 dollars Nth year total cost =6.5*1.1^n-1*10^6 dollars Total cost is sum of labor, material and the energy cost. Sum forms a GP with common ratio of 1.1 Surn =alran-1)/(-1) Here a is the first term of the GP (first year expense) and r is the common ratio, and n is the number of terms. When we have uniform interest rate let us assume that the rate is risuch that cumulative labor material and the energy cost is same. Then a+(1+r1)a+1+211)a+(1+3r1)a+(1+4r1)a =a* (ran-1)/(k-1) a- first year expensel We can see that it does not depend on gor in other words it does not depend on the first year investment. We know -0.1 and n=5 5+10r1=6.1 Thus r1=11/100 Thus a uniform interest rate of 11% will have same cumulative value as that of the compound interest of 10%.
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