EGR_102_Lab_03B_Solution

EGR_102_Lab_03B_Solution - interest_rate_less_9_APR = 0 0 1...

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Lab 03B Homework Name: Solution Section: 1 Spring 2012 Command Window Results: >> i = 0.01*[9.5 9.2 6.9 7.5 8.8] i = 0.0950 0.0920 0.0690 0.0750 0.0880 >> n = [9 8 8 9 6] n = 9 8 8 9 6 >> P = 1000*[390 440 520 380 290] P = 390000 440000 520000 380000 290000 >> A = P.*(i.*(1+i).^n)./((i+1).^n-1) A = 1.0e+004 * 6.6380 8.0089 8.6746 5.9572 6.4262 >> max_annual_payment = max(A) max_annual_payment = 8.6746e+004 >> min_annual_payment = min(A) min_annual_payment = 5.9572e+004 >> annual_payment_less_70K = A < 70000
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annual_payment_less_70K = 1 0 0 1 1 >> interest_rate_less_9_APR = i < 0.09
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Unformatted text preview: interest_rate_less_9_APR = 0 0 1 1 1 &gt;&gt; less_70K_and_less_9 = A &lt; 70000 &amp; i &lt; 0.09 less_70K_and_less_9 = 0 0 0 1 1 Problem Solution: Based on these results, the loans from banks 4 and 5 are viable. Bank 4 offered a $380,000 loan, and bank 5 offered a $290,000 loan. We have $200,000 on hand and each machine would cost $90,000. &gt;&gt; (380000+200000)/90000 % loan from bank 4 ans = 6.4444 &gt;&gt; (290000+200000)/90000 % loan from bank 5 ans = 5.4444 Based on these results I would recommend using bank 4 which means the company can purchase 6 new machines....
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This note was uploaded on 04/02/2012 for the course EGR 100 taught by Professor Hinds during the Fall '08 term at Michigan State University.

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EGR_102_Lab_03B_Solution - interest_rate_less_9_APR = 0 0 1...

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