Question

Score: 0 of 1 pt 4 of 4 (3compete) HW Score: 75%, 3 of 4 pts Problem 7-40 (algorithmic) Queston ep nts. Indvdal ndes wll useanegy as officiently an tsoomcal to de so and there are several incerties to mprove the efficency of eney comnton To a coder te se of a ewwa pump The pump is to oper 800 hours per year Punp A costs $1,900 has an overall officency of 81 2% and delivers 106 h The other available atemat pp ,cos $00, has an oveal effeey of 445% and delivers 12 4 hp Both puns hvea e of e pars and we he0746 kW PumpAwill ne t depreciation over fe years with an estmated SV of ze Pump B wl use the MACRS depreciation methad with a class e of tee years Aer e years, pumpA has an achlaket sae of $400, and pump B h an acal maket vale of 5220 Using the R mthod on he aer tax cash fows and a before t MAR of 15% is the incremental investtinp Aconomcally tfble? The offecive ineome tax e is 23% The cost of electicty 04 e years Click the ieon to view the GOS Recovery Raes )fr the 3-year property class asold at that se emenber 1 d the pmps ane sbyect to a study period of Tools The R of the nemental iesent is 19%Round to one decmal place)

individual industries will use energy as efficiently as it is economical to do so, and there are several incentives to improve the efficiency of energy consumption. To illustrate, consider the selection of a new water pump. The pump is to operate 800 hours per year. Pump a cost $1900, has an overall efficiency of 81.2% and it delivers 10.6 hp. The other available alternative, pump B, cost $900, has an overall efficiency of 44.45% and delivers 12.4 hp. Both pumps have a useful life of five years and will be sold at that time. (Remember one horse power equals .746 kW) pump a will use SL depreciation over five years with an estimated SV of zero. pump be will use the MACRS depreciation method with a class life of three years. After five years, Pompei has an actual market value of $400, and pump be has an actual market value of $220. Using the IRR method on the after-tax cash flows and a before tax MARR of 15%, is the incremental investment in pump a economically justifiable? The effective income tax rate is 23%. The cost of electricity is four cents per kilowatt hour, and the pumps are subject to a study period of 5 years.

I did not realize the original picture was blurry till after it was uploaded. Use the typed information please, it has all the information from the problem.

the problem is asking to compute the IRR between pump A and B. all the information provided is typed above.

Answer 1

Since, Non of the option is able to match the MARR of 15% (before Tax)/11.55% of After Tax, it is not economically justifiable to invest in pump

Clipboard Font Alignment Number tyles M4 A G М P Pump A Pump B 1,900.00 Cost 900.00 Average Use Per Year (Hours) 800.00 800.00 Power (HP) 10.60 12.40 Total HP 8,480.00 9,920.00 0.44 0.81 Efficiency 4,409.44 6,885.76 Output in HP 0.75 0.75 KWH in HP Total KWh 3,289.44 5,136.78 Generation 1 Cost of Eelectricity per 0.04 KWH 0.04 22 23 MARR (Before Sheetd4 Sheeti 2 Sheeti Sheet15 Sheet3 Sheet? Sheets Sheetl1 Sheet! Sheeti0 1008 0925 P4

M42 G H B C MARR (Before 15% Tax) 15% MARR (AFTER TAX @23%) 11.55% 11.55% Calculation of PUMP A Calculation of Depreciation Year 2 1900 Investment 20% 20% 20% 20 % 20% Depreciation Rate 380 380 380 380 380 Depreciation 1,900 1,520 1,140 760 380 Accumulated Depreciation

M69 G K L M C A CASH FLOW CALCULATION OF THE PROJECT 45 46 Year 47 5 31 4 1 48 49 5,136.78 5,136.78 5,136.78 5,136.78 5,136.78 Total Saving in KWh 50 51 0.04 0.04 0.04 Cost of Electricity (per KWh) 0.04 0.04 52 53 0.04 0.04 0.04 Contribution D-(B-C) 0.04 0.04 54 55 205.47 Total value (AD) 205.47 205.47 205.47 205.47 56 57 380.00 380.00 380.00 380.00 380.00 Depreciation 58 59 380.00 380.00 380.00 380.00 Total Fixed Cost 380.00 60 61 (174.53) (174.53) (174.53) (174.53) (174.53) Total 62 63 (40.14) (40.14) (40.14) (40.14) (40.14) Tax @23 % 64 65 (134.39) (134.39) (134.39) (134.39) (134.39) Post Tax 66 67 380.00 380.00 380.00 380.00 Add: Depreciation 380.00 68 69 Sheet14 Sheet15 Sheet13 Sheet2 Sheet3 Sheets Sheet1 2 Sheeti1 Sheeti Sheeti 0

M93 A C D F. G H. J K L M Cash Flow from Income Side 245.61 70 245.61 245.61 245.61 245.61 71 Salvage Value (Net of Tax) 72 308.00 73 Total Cash Flow 74 245.61 245.61 245.61 245.61 553.61 75 76 77 IRR Cash Outflow -1900 78 Inflow 79 245.61 1 80 245.61 2 81 245.61 Year 3 82 245.61 4 83 553.61 51 84 85 6% IRR 86 87 88 89 Calculation of PUMP B 90 91 Calculation of Depreciation 92 93

A C D E G K L M 0 Calculation of PUMP B 91 92 Calculation of Depreciation 93 | 94 Year 95 3 1 4. 96 97 Investment 1900 98 99 Depreciation Rate 33% 45% 15% 7% 100 133 101 102 Depreciation 627 855 285 103 Accumulated Depreciation 627 1,482 1,767 1,900 104 105 106 CASH FLOW CALCULATION OF THE PROJECT 107 Year 108 4 1 2 109 110 3,289.44 3,289.44 3,289.44 3,289.44 3,289.44 Total Saving in KWh 111 112 113 114 Cost of Electricity (per KWh 0.04 0.04 0.04 0.04 0.04 Sheet13 Sheet2 Sheet3 Sheetl4 Sheet15 Sheet12 Sheet10 Sheet11 Sheet! Sheets 100%

Number Editing Styles Cells M137 A B D E G J K L M N 114 115 Contribution D-(B-C) 0.04 0.04 0.04 0.04 0.04 116 117 Total value (A*D) 131.58 131.58 131.58 131.58 131.58 118 Depreciation 119 627.00 855.00 285.00 133.00 120 121 Total Fixed Cost 855.00 627.00 133.00 285,00 122 (723.42) (153.42) 123 Total (495.42) (1.42) 131.58 124 125 Таx @23% (113.95) (166.39) (35.29) (0,33) 30.26 126 127 (557.04)(118.14) Post Tax (381.48) (1.10) 101.31 128 129 Add: Depreciation 855.00 285.00 133.00 627.00 130 Cash Flow from Income Side 166.86 131 245.52 297.96 131.90 101.31 132 133 Salvage Value (Net of Tax) 169.40 134 297.96 270.71 Total Cash Flow 131.90 245.52 135 166.86 136 137 Sheeti4 Sheets Sheet15 Sheetl2 Sheetl3 Sheeti Sheet2 Sheets Sheet1 0 Sheeti 1

Filter Select- Editing Formatting as Table Styles Styles Clear Format Painter Alignment Number Cells Font Clipboard G138 NCENWAN G P J K L о E D A в C IRR -900 Cash Outflow Inflow 1 245.52 -1 12 297.96 2 166.86 3 3 131.90 41 14 270.71 5 46 8% IRR 47 48 49 51 152 53 54 155 156 157 158