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Capacity Planning

IE 214: Operations Management. Lecture. 5. Capacity Planning. KAMAL. EXERCISES S7.1 – S7.5. Solution:. S7.5. EXERCISE S7.6. Solution:. Expected production = effective capacity * efficiency Design EP = 93,600  0.95 = 88,920 Fabrication EP = 156,000  1.03 = 160,680

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Capacity Planning

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  1. IE 214: Operations Management Lecture 5 Capacity Planning KAMAL

  2. EXERCISES S7.1 – S7.5 Solution: S7.5

  3. EXERCISE S7.6 Solution: Expected production = effective capacity * efficiency Design EP = 93,600  0.95 = 88,920 Fabrication EP = 156,000  1.03 = 160,680 Finishing EP = 62,400  1.05 = 65,520

  4. EXERCISE S7.7 Solution: Design capacity = 2,000 students Effective capacity = 1,500 students Actual output = 1,450 students

  5. EXERCISE S7.8 Solution: Actual or expected output = Effective capacity*Efficiency 5.5 cars  0.880 = 4.84 cars. In one 8-hour day, one bay accommodates (8 hr * 4.84 cars per hr) = 38.72 cars To do 200 cars per day it requires, (200 cars) / (38.72 cars/bay) = 5.17 or 6 bays

  6. EXERCISE S7.11 Solution: Stat.1_A 0.05 hr/unit Stat.2 0.2 hr/unit Stat.3 0.08333 hr/unit Stat.1_B 0.05 hr/unit (a) System process time = bottleneck time = 0.2 hr/unit (b) Bottleneck time = 0.2 hr/unit (c) Process cycle time = 0.05 hr+0.2 hr+0.08333 hr = 0.3333 hr = 20 min (d) Weekly capacity= total time in a week/bottleneck time = (10hr*5days)/0.2 = 250 units/week

  7. EXERCISE S7.14

  8. EXERCISE S7.11 Solution: Station A 15 min/unit Station B 10 min/unit Station D 15 min/unit Station C 20 min/unit (a) System process time = bottleneck time = 20 min/unit (b) Hourly capacity = time in an hour/bottleneck time = 60min/20 = 3 units/hr * Process cycle time = 25 min + 15 min = 40 min

  9. EXERCISE S7.22 Given: • F = 15,000$ • V = 0.01$/ copy • P = 0.05$/ copy Solution: (a) BEP$ = [F / 1-(V/P)] = 15000 / 1- (0.01/0.05) = 18,750$ (b) BEPx = [F / P-V] = 15000 / (0.05 – 0.01) = 375,000 copies

  10. EXERCISE S7.23 Solution: Profit = (Selling Price – Variable Cost) * Number of Sales – Fixed Cost Profit A = 30000 * (1- 0.5) – 14000 = 1,000 $ Profit B = 50000 / (1 – 0.6) – 20000 = 0 $ Thus, the company should stay as is.

  11. EXERCISE S7.26 Given:

  12. EXERCISE S7.26 Solution: • Find the contribution [1- v/p] for each category • 1-(0.75/1.5) = 0.5 • Find the revenue for each category • P*no. sales estimated = 1.5 * 30,000 • = 45,000 $ • 3. Find the percent of sales (W) • W = revenue of category / total revenue • = 45,000 / 295,000 = 0.15254237 • 4. Find the weighted contribution • = [1-(V/P)] * W = 0.5*0.1525437 • = 0.076271186

  13. EXERCISE S7.26 Solution:

  14. EXERCISE S7.26 Solution: • a) Find the beak-even point in dollar per month using • BEP$ = F / ∑ [ (1-(Vi/Pi)) * Wi ] • = 7600 $ / month • b)BEP$ = 7600 / 30 = 253.3333 $ / day • Daily number of meals = (0.338983051*253.3333)/10 • = 8.58757 ≈ 9 meals / day

  15. EXERCISE S7.28 Solution: Option A: EMVA = (90,000 × .5) + (25,000 × .5) = $57,500 Option B: EMVB = (80,000 × .4) + (70,000 × .6) = $74,000 Option C: EMVC = 18,000 + 38,500 = $56,500 Option B has the highest EMV.

  16. HW S7.13 S7.15

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