Question

Machine F costs $1,480,000, has a life of 8 years and will return a benefit of $875,500 in the first year, and increasing by 6% for the life of the machine. The first year costs for Machine F is $545,000, increasing by an inflation rate of 3% per year. The salvage value at the end of life of the machine is 12% of the original investment cost. If MARR is 10%, what is the NPW for the machine? Please show all work ....with excel functions..explaining which excel functions were used and why

                               

$980,457

                               

$1,012,544

                               

$1,031,391

                               

$1,234,764

Answer 1

Answer: NPW = $1031391
Calculations:
	0	1	2	3	4	5	6	7	8
Annual benefit (with 6% increase from Year 2		875500	928030	983712	1042735	1105299	1171616	1241913	1316428
Annual costs (with 3% increase from Year 2)		545000	561350	578191	595536	613402	631804	650759	670281
Net annual benefit		330500	366680	405521	447198	491896	539812	591155	646147
Initial cost of the machine	1480000
Salvage value (1480000*12%)									-177600
Annual project cash flows	-1480000	330500	366680	405521	447198	491896	539812	591155	823747
PVIF at 10%	1	0.90909	0.82645	0.75131	0.68301	0.62092	0.56447	0.51316	0.46651
PV at 10%	-1480000	300455	303041	304674	305442	305429	304710	303356	384284
NPW	1031391									Answer: NPW = $1031391 Calculations: 0 1 2 3 4 5 6 7 8 Annual benefit (with 6% increase from Year 2 875500 928030 983712 1042735 1105299 1171616 1241913 1316428 Annual costs (with 3% increase from Year 2) 545000 561350 578191 595536 613402 631804 650759 670281 Net annual benefit 330500 366680 405521 447198 491896 539812 591155 646147 Initial cost of the machine 1480000 Salvage value (1480000*12%) -177600 Annual project cash flows -1480000 330500 366680 405521 447198 491896 539812 591155 823747 PVIF at 10% 1 0.90909 0.82645 0.75131 0.68301 0.62092 0.56447 0.51316 0.46651 PV at 10% -1480000 300455 303041 304674 305442 305429 304710 303356 384284 NPW 1031391