Question

Question 2 (6 points) Two industrial robot options are currently proposed to be purchased for the assembly company. The followings are the information about two options. Robot 1: The initial cost is $220K. The savings are expected to be $85K in the first year and increase by 3% from year 2 until the end of its life. The salvage value is expected to be $96K and life cycle is expected to be 6 years. Robot 2: The initial cost is $200K. The savings are expected to be $100K in the first year and increase $5K from year 2 until the end of its life. There will be service tune-up fee, $50K, at year 2. The salvage value is expected to be $90K and life cycle is expected to be 6 years. Which robot should the company purchase if MARR is 10%?

Answer 1

The present worth of geometric series = A * [1 - (1+g)^n/(1+i)^n] / (i - g)

Present worth of savings for robot 1 = 85000 * [1 - (1+0.03)^6/(1+0.1)^6] / (0.1- 0.03)

= 85000 * [1 - (1.03)^6/(1.1)^6] / (0.07)

= 85000 * 4.656980

= 395843.31

Net present worth of Robot 1 = -220000 + 395843.31 + 96000 * (P/F,10%,6)

= -220000 + 395843.31 + 96000 * 0.564474

= 230032.81

Net present worth of Robot 2 = -200000 + 100000 * (P/A,10%,6) + 5000 * (P/G,10%,6) - 50000 * (P/F,10%,2) + 90000 * (P/F,10%,6)

= -200000 + 100000 *4.355261 + 5000 *9.684171 - 50000 *0.826446 + 90000 *0.564474

= 293427.31

As NPV of robot 2 is more, it should be selected