Capacity PlanningIE 214: Operations Management

KAMAL

Lecture5

EXERCISES S7.1 – S7.5

Solution:Actual output 6,000

Utilization = = = 0.857 85.7%Design capacity 7,000

S7.1

Actual output 4,500 Efficiency = = = 0.692 or 69.2%

Effective capacity 6,500S7.2

Expected output = (Effective capacity) × (Efficiency)

= (6,500)(.88) = 5,720

S7.3

Actual (expected) output 800 Efficiency = = = 88.9%

Effective capacity 900S7.4

Actual output 400Efficiency = or 0.80 =

Effective capacity Effective capacity

400Thus, effective capacity = 500

0.80=

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

Finishing EP = 62,400 1.05 = 65,520

EXERCISE S7.7

Solution: Design capacity = 2,000 studentsEffective capacity = 1,500 studentsActual output = 1,450 students

Actual output 1,450Utilization = = = 72.5%

Design capacity 2,000

Actual output 1,450Efficiency = = = 96.7%

Effective capacity 1,500

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

EXERCISE S7.11

Solution:

(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

Stat.1_A0.05 hr/unit

Stat.1_B0.05 hr/unit

Stat.20.2 hr/unit

Stat.30.08333 hr/unit

EXERCISE S7.14

EXERCISE S7.11

Solution:

(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

Station A15 min/unit

Station C20 min/unit

Station B10 min/unit Station D

15 min/unit

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

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.

EXERCISE S7.26

Given:

Item sales forecast P VBeer 30000 1.5 0.75Meal 10000 10 5

Dessert & Wine 10000 2.5 1exp.Sandwich 20000 6.25 3.25

EXERCISE S7.26

Solution: 1. Find the contribution [1- v/p] for each category

1-(0.75/1.5) = 0.5

2. 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

EXERCISE S7.26

Solution:

Item sales forecast P V 1-(V/P) revenue W Weighted Contribution

Beer 30000 1.5 0.8 0.5 45000 0.15254237 0.076271186

Meal 10000 10 5 0.5 100000 0.33898305 0.169491525

Dessert & Wine 10000 2.5 1 0.6 25000 0.08474576 0.050847458exp.Sandwich 20000 6.25 3.3 0.48 125000 0.42372881 0.203389831

Sum 295000 1 0.5

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

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.

HW

S7.13

S7.15