1. EQUIPMENT COSTS Peters Timmerhaus & West

2. PURCHASED EQUIPMENT p.243 PT&W Cost a/Cost b=(size a/size b) 0.6

3. Lang Factor (in this example) = 4.3 We will say I F = 5.0*E

4. SHOW ME THE MONEY! CHE462 “Figures don’t lie, but liars sure can figure!” BECKMAN 2012

5. CASH FLOW & ROI (standard) Operating Cost Gross Sales $/yr S Operating cost $/yr bye-bye C Net sales, $/yr S-C depreciation Profit before tax, $/yr S-C-d depreciation $/yr d=I F /N F Income tax  Income tax $/yr bye-bye  (S-C-d) Profit after tax, $/yr (  S-C-d) Total Profit after tax (  S-C-d)+d I F+ I W Total Investment = ROI = ($/yr)/$ I F =  I W =(C/12)*N W Total Investment. $

6. EXAMPLE of ROI (standard) S (sales) = $100 million/year C (operating cost) = $60 million/year E (equipment cost) = $10 million N (project life) = 20 years N W (working capital months) = 3    (income tax rate ) = 0.4 Then I F = 5*$10 = $50 d = $50/20 yr = $2.5/yr I W = $60/yr*3mo/12mo/yr = $15 ROI = ((100-60-2.5)*(1-.4) +2.5)/(50+15) = 0.385/yr = 38.5%/yr

7. EXAMPLE MAX ROI (standard) S (sales) = $100 million/year C (operating cost) = $40 million/year E (equipment cost) = $15 million N (project life) = 20 years N W (working capital months) = 3    (income tax rate ) = 0.4 Then I F = 5*$15 = $75 d = $75/20 yr = $3.75/yr I W = $40/yr*3mo/12mo/yr = $10 ROI = ((100-40-3.75)*(1-.4) +3.75)/(75+10) =0.435/yr =43.5%/yr By observation, if E is increased from $10 million to $15 million, maybe C will decrease from $60 million/yr to $40 million/yr so:

8. Lets EXPENSE the Investment we will BORROW I F i(1+i) N (1+i) N -1 R/I F = =.05*(1.05) 20 /[1.05 20 -1] = 0.0802 /yr Total loan payback = N*R/IF = 20*.0802 = 1.60 or 60% more than you borrowed! If interest rate is 5% and project life is 20 years, then

9. CASH FLOW & ROI (I F Expensed) Operating Cost Gross Sales $/yr S Operating cost $/yr bye-bye C+R Net sales, $/yr S-C-R depreciation Profit before tax, $/yr S-C-R-d depreciation $/yr d=I F /N F Income tax  Income tax $/yr bye-bye  (S-C-R-d) Profit after tax, $/yr (  S-C-R-d) Total Profit after tax (  S-C-R-d)+d I W Total Investment = ROI = ($/yr)/$ I F =  is  R  I F * i(1+i) N /[(1+i) N -1] I W =[(C+R)/12]*N W Total Investment. $

10. EXAMPLE of ROI with I F expensed S (sales) = $100 million/year C (operating cost) = $60 million/year E (equipment cost) = $10 million N (project life) = 20 years N W (working capital months) = 3    (income tax rate ) = 0.4 i=5% Then I F = 5*$10 = $50 R= 0.0802*$50 = $4.0/yr d = $50/20 yr = $2.5/yr I W = $60/yr*3mo/12mo/yr = $15 ROI = ((100-60-4.0-2.5)*(1-.4) +2.5)/(0+15) = 1.51/yr = 151%/yr If E=$15 million and C= $40 million/yr, then ROI=295% wow

11. RULE # 1 IN BUSINESS NEVER BUY something when you can either rent or borrow (or steal?) !!!!!

12. CASH FLOW & ROI (d Payback into I F ) Operating Cost Gross Sales $/yr S Operating cost $/yr bye-bye C Net sales, $/yr S-C depreciation Profit before tax, $/yr S-C-d depreciation $/yr d=I F /N F Income tax  Income tax $/yr bye-bye  (S-C-d) Profit after tax, $/yr (  S-C-d) Total Profit after tax (  S-C-d) I F+ I W Total Investment = ROI = ($/yr)/$ I F =  n*d I W =(C/12)*N W Total Investment. $ OK here (where n is the nth year of the project I F = 0 after n = N F )

13. EXAMPLE of ROI ( d payback) S (sales) = $100 million/year C (operating cost) = $60 million/year E (equipment cost) = $10 million N (project life) = 20 years N W (working capital months) = 3    (income tax rate ) = 0.4 Then I F = 5*$10 = $50 initially d = $50/20 yr = $2.5/yr I W = $60/yr*3mo/12mo/yr = $15 ROI = ((100-60-2.5)*(1-.4) +2.5)/(50-2.5+15) = 0.40/yr = 40%/yr For year 1

14. I F with d payback as a function on time (“book value”)

15. ROI as a function of time (with d payback into I F )