Homework_Solutions

# Homework_Solutions - Total processing time by machine...

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Solutions Chapter 5 3. FC = \$9200/month VC = \$.70/unit Price = \$.90/unit a. Q BEP = FC = \$9,200 = 46,000 units Price – VC \$.90 - \$.70 b. Profit = Price * Q - (FC + VC x Q) 1. P 61,000 = \$.90(61,000) - [\$9,200 + \$.70(61,000)] = \$3,000 2. P 87,000 = \$.90(87,000) - [\$9,200 + \$.70(87,000)] = \$8,200 c. Q = Specified profit + FC = \$16,000 + 9,200/month = 126,000 units . Price – VC \$.90/unit - \$.70/unit d. Total Revenue = Price * Q, so Q = Total Revenue = \$23,000 = 25,556 units R \$.90/unit e. 1. 4. FC Rev VC A: \$40,000 \$15/unit \$10/unit B: \$30,000 \$15/unit \$11/unit a. units 000 , 8 unit / 10 \$ unit / 15 \$ 000 , 40 \$ Q A , BEP = - = units 500 , 7 unit / 11 \$ unit / 15 \$ 000 , 30 \$ Q B , BEP = - = b. Profit = Q(Price – VC) – FC [A’s Profit] [B’s Profit] Q(\$15 – \$10) – \$40,000 = Q(\$15 – \$11) – \$30,000 Solving, Q = 10,000 units 0 Volume (units) TR = \$90,000 @ Q = 100,000 units TC = \$79,200 @ Q = 100,000 units TR TC 100,000 \$100,000 \$50,000 \$9,200 Cost

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c. P A = 12,000(\$15 – \$10) – \$40,000 = \$20,000 [A is higher] P B = 12,000(\$15 – \$11) – \$30,000 = \$18,000 10. a. Given: 10 hrs. or 600 min. of operating time per day. 250 days x 600 min. = 150,000 min. operating time per machine per year.
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Unformatted text preview: Total processing time by machine Product A B C 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 4 60,000 60,000 30,000 Total 186,000 208,000 122,000 machine 1 81 . 000 , 150 000 , 122 machine 2 38 . 1 000 , 150 000 , 208 machine 2 24 . 1 000 , 150 000 , 186 ≈ = = ≈ = = ≈ = = C B A N N N You would have to buy two “A” machines at a total cost of \$80,000, or two “B” machines at a total cost of \$60,000, or one “C” machine at \$80,000. b. Total cost for machine A: 186,000 min ÷ 60 = 3,100 hrs. 3,100 hrs x \$10 + \$80,000 = \$111,000 machine B: 208,000 ÷ 60 = 3,466.67 hrs. 3,466.67 hrs. x \$11 + \$60,000 = \$98,133 machine C: 122,000 ÷ 60 = 2,033.33 hrs. 2,033.33 hrs. x \$12 + \$80,000 = \$104,400 13. a. 11/hr. b. Increase capacity of operation 3 by about 10%. Beyond that, Operation 1 would become the limiting (bottleneck) operation....
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## This note was uploaded on 04/09/2008 for the course ECO 101 taught by Professor Ghosh during the Spring '08 term at Scranton.

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Homework_Solutions - Total processing time by machine...

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