QUESTION 9 points She A production engineer is considering two robots for purchase by a fiber optic manufacturing company, Robot will have a first cost of $275,000, an annual maintenance and operation (MBO) cost of $225,000, and a $260,000 salvage value. Roboty will have a first cost of $290,000, an annual M&O cost of $230,000, and a $240.000 salvage value. Which should be selected on the basis of an annual worth comparison at an interest rate of 10 per year if both of them have 5 years of lives. Please select correct answers from the gven options for the following 1) AW Equation for Robot X: 21 AW Equation for Roboty 3) The best alternative is 10% End-of-Period Compound Interest Factors Single Payment Uniform Payment Series Compound Present Capital Present Sinking Compound Amount Worth Recovery Worth Fund Amount Factor Factor Factor Factor Factor Factor F/P P/F A/P P/A A/F F/A 1 1.100 9091 1.1000 900 1.0000 1.OXXO 1.210 8264 5762 1.736 4762 2.100 1.331 7513 ,4021 2.487 3021 3310 4 1464 .680 3155 3.170 2155 4.641 5 1.611 6209 2638 3.791 .1638 6.105 6 1.772 5645 4.355 .1296 7.716 7 1.949 5132 2054 4.868 1054 9.487 1. 4665 1874 5335 0974 11436 2296 Fund Factor A/F Amount Factor FIA Amount Worth Recovery Factor Factor Factor N F/P P/F A/P 1 1.100 9091 1.1000 2 1.210 8264 5762 3 1.331 7513 4021 4 .6830 3155 5 1611 .6209 2638 6 1.772 5645 2296 2 1.949 5132 2054 8 2.144 4665 .1874 9 2.358 .4241 1736 10 2.504 3855 1627 Robot DAWY-275,000 (PIA 10%,5) - 225.000 260,000 5,104,5) Wy-290,000 (NP, 10%.5)-230,000 240,000 FIA 10%) Both are OK because As of Xand Yare the same DAW - 290.000 (NP,10%.5) - 230,000 (AP.10%.5) . 240.000 (NF.10%.5) DAW 275,000 (AP,10%.5) - 225,000 + 260,000 WWF, 104.5 Wy--290,000 (NP.10%.5) - 230,000 240,000 (AF,10%.5) Robot Y DNy-275.000 (A/P 10% 5) - 225.000(AJP, 10%,5) 280,000 INF, 104.5) Worth Factor P/A 99 1.736 2.487 3.170 3.791 4.355 4.868 5.335 5.759 6.145 1.0X200 -4762 .3021 2155 1638 .1206 1054 0874 ,0736 0627 1.OXXO 2.100 3.310 4.641 6.105 7.716 9.487 11.436 13.579 15.037 QUESTION 9 points She A production engineer is considering two robots for purchase by a fiber optic manufacturing company, Robot will have a first cost of $275,000, an annual maintenance and operation (MBO) cost of $225,000, and a $260,000 salvage value. Roboty will have a first cost of $290,000, an annual M&O cost of $230,000, and a $240.000 salvage value. Which should be selected on the basis of an annual worth comparison at an interest rate of 10 per year if both of them have 5 years of lives. Please select correct answers from the gven options for the following 1) AW Equation for Robot X: 21 AW Equation for Roboty 3) The best alternative is 10% End-of-Period Compound Interest Factors Single Payment Uniform Payment Series Compound Present Capital Present Sinking Compound Amount Worth Recovery Worth Fund Amount Factor Factor Factor Factor Factor Factor F/P P/F A/P P/A A/F F/A 1 1.100 9091 1.1000 900 1.0000 1.OXXO 1.210 8264 5762 1.736 4762 2.100 1.331 7513 ,4021 2.487 3021 3310 4 1464 .680 3155 3.170 2155 4.641 5 1.611 6209 2638 3.791 .1638 6.105 6 1.772 5645 4.355 .1296 7.716 7 1.949 5132 2054 4.868 1054 9.487 1. 4665 1874 5335 0974 11436 2296 Fund Factor A/F Amount Factor FIA Amount Worth Recovery Factor Factor Factor N F/P P/F A/P 1 1.100 9091 1.1000 2 1.210 8264 5762 3 1.331 7513 4021 4 .6830 3155 5 1611 .6209 2638 6 1.772 5645 2296 2 1.949 5132 2054 8 2.144 4665 .1874 9 2.358 .4241 1736 10 2.504 3855 1627 Robot DAWY-275,000 (PIA 10%,5) - 225.000 260,000 5,104,5) Wy-290,000 (NP, 10%.5)-230,000 240,000 FIA 10%) Both are OK because As of Xand Yare the same DAW - 290.000 (NP,10%.5) - 230,000 (AP.10%.5) . 240.000 (NF.10%.5) DAW 275,000 (AP,10%.5) - 225,000 + 260,000 WWF, 104.5 Wy--290,000 (NP.10%.5) - 230,000 240,000 (AF,10%.5) Robot Y DNy-275.000 (A/P 10% 5) - 225.000(AJP, 10%,5) 280,000 INF, 104.5) Worth Factor P/A 99 1.736 2.487 3.170 3.791 4.355 4.868 5.335 5.759 6.145 1.0X200 -4762 .3021 2155 1638 .1206 1054 0874 ,0736 0627 1.OXXO 2.100 3.310 4.641 6.105 7.716 9.487 11.436 13.579 15.037
