Question

The years of service for the bridge exceed the original design life. As a result of last year's inspection, engineers at the Minnesota Department of Transportation (MnDOT) decided to replace the bridge. Based on past experience, the engineers identified two significantly different design options: a steel design option, and a concrete design option. The engineers computed preliminary cost estimates for the two competing options.

Steel Design
The steel bridge is estimated to cost $20,000,000 now with a salvage value of $3,000,000 at the end of its 20 years. Maintenance costs on the steel bridge are estimated at $100,000 per year for the first 3 years, and then increase by $20,000 per year for the next 17 years (i.e. $120,000 in the fourth year, $140,000 in the fifth year, etc.).

Concrete Design
The concrete bridge is estimated to cost $17,000,000 with a negative salvage value of $2,000,000 (demolition cost). Maintenance costs on the concrete bridge are estimated at $200,000 per year for the full 20 year life of the bridge.

Questions
(a) Using an effective annual interest rate of 7%, compute the present worth and the equivalent uniform annual cost for both of the options.

(b) Which of these two options is economically most attractive?

(c) Holding everything else fixed (as given in the original problem statement), what effective annual interest rate would make these two options economically equivalent?

PLEASE ANSWER WHAT YOU CAN IN EXCEL PLEASE THANK YOU

Answer 1

Annual interest rate = 7%

PV = C + C1/(1+r) + C2/(1+r)² + C3/(1+r)³ + C4/(1+r)⁴ ….. Ct/(1+r)²⁰ - S/(1+r)²⁰

Where, Ct is cost at time=t

S is salvage value

r is interest rate

PV of cost of Steel Bridge

Year (t)	Cost incurred (C) =Acquisition cost + Maintenance cost - Salvage value	Discounting factor (D) =(1/1+7%)t	PV (C)*(D)
0	20000000	1	20000000.00
1	100000	0.934579439	93457.94
2	100000	0.873438728	87343.87
3	100000	0.816297877	81629.79
4	120000	0.762895212	91547.43
5	140000	0.712986179	99818.07
6	160000	0.666342224	106614.76
7	180000	0.622749742	112094.95
8	200000	0.582009105	116401.82
9	220000	0.543933743	119665.42
10	240000	0.508349292	122003.83
11	260000	0.475092796	123524.13
12	280000	0.444011959	124323.35
13	300000	0.414964448	124489.33
14	320000	0.387817241	124101.52
15	340000	0.36244602	123231.65
16	360000	0.338734598	121944.46
17	380000	0.31657439	120298.27
18	400000	0.295863916	118345.57
19	420000	0.276508333	116133.50
20	-2560000	0.258419003	-661552.65

PV of cost of Steel Bridge = Sum of PV = $21,465,417.00

Equivalent uniform annual cost:

PV = C/(1+r) + C/(1+r)² + C/(1+r)³ + C/(1+r)⁴ ….. C/(1+r)²⁰

where c is Equivalent uniform annual cost

For steel bridge,

21465417 = C/(1+7%) + C/(1+7%)² + C/(1+7%)³ + C/(1+7%)⁴ ….. C/(1+7%)²⁰

Solve for C, C= 2026183.51

Equivalent uniform annual cost for steel bridge = $2,026,183.51

PV of cost of Concrete Bridge

Year (t)	Cost incurred (C) = Acquisition cost + Maintenance cost + demolition cost	Discounting factor (D) =1/(1+7%)t	PV (C)*(D)
0	17000000	1	17000000.00
1	200000	0.934579439	186915.89
2	200000	0.873438728	174687.75
3	200000	0.816297877	163259.58
4	200000	0.762895212	152579.04
5	200000	0.712986179	142597.24
6	200000	0.666342224	133268.44
7	200000	0.622749742	124549.95
8	200000	0.582009105	116401.82
9	200000	0.543933743	108786.75
10	200000	0.508349292	101669.86
11	200000	0.475092796	95018.56
12	200000	0.444011959	88802.39
13	200000	0.414964448	82992.89
14	200000	0.387817241	77563.45
15	200000	0.36244602	72489.20
16	200000	0.338734598	67746.92
17	200000	0.31657439	63314.88
18	200000	0.295863916	59172.78
19	200000	0.276508333	55301.67
20	2200000	0.258419003	568521.81

PV of cost of Concrete Bridge = Sum of PV = $19,635,640.85

Equivalent uniform annual cost:

PV = C/(1+r) + C/(1+r)² + C/(1+r)³ + C/(1+r)⁴ ….. C/(1+r)²⁰

where c is Equivalent uniform annual cost

For steel bridge,

19635640.85 = C/(1+7%) + C/(1+7%)² + C/(1+7%)³ + C/(1+7%)⁴ ….. C/(1+7%)²⁰

Solve for C, C= 1,853,465.59

Equivalent uniform annual cost for concrete bridge = $1,853,465.59

b) Concrete Bridge is economically most attractive as the PV of cost of concrete bridge is less than that of steel bridge.

c)

Let us assume interest rate is r% in order to make these two options economically equivalent.

C_s + C1_s/(1+r) + C2_s /(1+r)² + C3_s /(1+r)³ + C4_s /(1+r)⁴ ….. Ct_s /(1+r)²⁰ – S_s/(1+r)²⁰ = C_c + C1_c /(1+r) + C2_c/(1+r)² + C3_c/(1+r)³ + C4_c/(1+r)⁴ ….. Ct_c/(1+r)²⁰ – S_c/(1+r)²⁰

Where, Ct_s is cost of steel bridge at time t

Ct_c is cost of concrete bridge at time t

S_s is salvage vale of steel bridge

S_c is salvage vale of concrete bridge

20000000+ 100000/(1+r) + 100000/(1+r)² + 100000/(1+r)³ + 120000/(1+r)⁴ ….. 440000/(1+r)²⁰ – 3000000/(1+r)²⁰ = 17000000 + 200000 /(1+r) + 200000 /(1+r)² + 200000 /(1+r)³ + 200000 /(1+r)⁴ ….. 200000/(1+r)²⁰ + 2000000/(1+r)²⁰

Solving above eqn for r,

r=1.4366%

Interest rate is 1.44% in order to make these two options economically equivalent