3. A small business company is considering updating the current production line.

Question

3. A small business company is considering updating the current production line. There are two plans. For plan A, the fixed cost will be $40,000 and the variable cost will be $25 per unit after the update. For plan B, the fixed costs will be $49,000 and the variable cost will be $23 per unit after the update. Please answer the following questions (a) Suppose the selling price is $30, what is the break-even volume for each plan? Which plan has a lower breaK-even volume (b) Suppose the selling price is $30. Also, the company aims to achieve a profit of $7,000 after the update. What selling will be required to achieve the profit for each plan? Which plan has a lower volume? (Extra Credit) 4. Here is an Activity on Arrow (AOA) network diagram for a project. The lettered A through F represents a sub-project, as shown in the accompanying diagram. The probabiistic time for each activity is as follows Time 3-4-6 2-5-6 3-5-7 Activit Time 4-5-6 2-5-6 3-4-6 Activity (1) What is the expected time and standard deviation of path A-C-D? (2) What is the probability that path A-C-D can finish in 13.5 weeks?

Explanation / Answer

3.

(a) Breakeven volume for plan A = Fixed cost / (Selling price - Variable cost) = 40000/(30 - 25) = 8000

Breakeven volume for plan B = Fixed cost / (Selling price - Variable cost) = 49000/(30 - 23) = 7000

Plan B has a lower breakeven volume.

(b) Selling required (Sales volume) for plan A = (Fixed cost + Required profit)/(Selling price - Variable cost) = (40000+7000)/(30-25) = 9400 units

Selling required (Sales volume) for plan B = (Fixed cost + Required profit)/(Selling price - Variable cost) = (49000+7000)/(30-23) = 8000 units

Plan B has a lower volume.

4.

(1) For each activity, Calculate the expected mean time () = (a + 4m + b)/6, and Variance 2 = (b-a)2/36.

Expected time of path A-C-D = 4.17+5+5 = 14.17

Variance of path A-C-D = 0.25+0.444+0.111 = 0.806

Standard deviation of path A-C-D = SQRT(0.806) = 0.898

(2)

For target time of 13.5 weeks, z = (13.5-14.17)/0.898 = -0.7428

Corresponding probability = NORMSDIST(-0.7428) = 0.2288

Probability that path A-C-D can finish in 13.5 weeks = 22.9 %

Activity Immediate predecessor a m b Mean time () Variance  (2) A 3 4 6 4.17 0.25 B 2 5 6 4.67 0.44 C A 3 5 7 5.00 0.44 D C 4 5 6 5.00 0.11 E B 2 5 6 4.67 0.44 F E 3 4 6 4.17 0.25

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