# Operation Management Exam

Topics: Project management, Following, Productivity Pages: 5 (494 words) Published: May 6, 2012
EXAMINATION I NAME: __________________________________________________________

1. David Upton is president of Upton Manufacturing, a producer of Go-Karts tires. Upton makes 2500 tires per day with the following resources:

Labor: 50 hours per day @15.00 per hour Raw Material: 20,000 lbs. per day at @ \$3 per lb
Energy: \$5,000 per day
Capital: \$10,000 per day

a) What is the labor productivity for these tires?

The labor productivity for these tires is … 50 tires/labor hr

Productivity = 2,500/50

= 50 tires per labor hr.

b) What is the multifactor productivity for these tires?

Productivity = 2,500 units/50(\$15) + 20,000 lbs (\$3) + 5,000 + 10,000

= 2,500/75,750

= .033 tires per dollar

c) What is the percent change in the multifactor productivity if the energy bill can be reduced by \$2000 without cutting production or changing any other inputs?

The percent change is 3%

= 2,500/\$750 = \$60,000 + \$3,000 + \$10,000

= 2,500/\$73,750 = .034

.034 / .033 = 1.03 = 3%

2. Draw an AON network for the project described below.

Activity Immediate Predecessor (s)

A-
B-
CA
DA
EB,C
FB,C
GD,E
HD,E
IH
JF,G
K G,I

[pic]

3. A PERT project has an ending completion time of 62 weeks with a variance of 16 weeks. Determine the following probabilities:

a) The project is completed in 68 weeks.

Z = (68 wks – 62wks) / 4

= 6 / 4

= 1.5 = .93319 = 93.32 %

b) The project takes longer than 70 weeks.

Z = (70 Wks – 62 wks) / 4

= 8 / 4

= 2. = .97725 = 97.73%

c) The project is completed between 60 and 68 weeks.

4. The following table provides the information necessary to construct a project network and project crash data:
|  |  |Activity |Times(WKS) |Activity |(\$) | |Activity |Predecessor |Normal |Crash |Normal |Crash | |  |  |  |  |  |  | |a |- |16 |8 |2,000 |4,400 | |b |- |14 |9 |1,000 |1,800 | |c |a |8 |6 |500 |700 | |d |a |5 |4 |600 |1,300 | |e |b |4 |2 |1500 |3,000 | |f |b |6 |4 |800 |1,600 | |g |c |10 |7 |3,000 |4,500 | |h |d,e |15 |10 |5,000 |8,000 |

| | | | | | | |

Determine the maximum time that the project can be crashed and the corresponding crash costs.

Activity Max Reduction Time

A8 4,400 – 2,000 / 16-8 = 2,400/8 = \$300

B51,800 – 1,000 / 14-9 = 800 / 5 = \$160

C2700 – 500 / 8-6 = 200 / 2 = \$100

D11,300 – 600 / 5-4 = 700 / 1 = \$700

E23,000 – 1,500 / 4-2 = 1,500 = \$750

F21,600 – 800 / 4-2 = 800 / 2 = \$400

G34,500 – 3,000 / 10-7 = 1,500 / 3 = \$500

H58,000 – 5,000 / 15-10 = 3,000 / 5 = \$600...