Insert TableChartTextShape Media Comment MGMT 3450-Operations Management (Sectio
ID: 386100 • Letter: I
Question
Insert TableChartTextShape Media Comment MGMT 3450-Operations Management (Section 602) Utah Valley University Fall 2018 Problem Set #1-Handwritten Section Reminder: The handwritten section may be completed in groups of up to 4 people Due: 10:00 am (In class) Monday, September 17th, 2018 Reminder: This portion of the assignment may be done in groups of up to 4 individuals. If you choose to work in a group make sure all group names are on your submission. Show your work! Don't forget the 2 points for professionalism 1. (2 points) Calculate the percent change in productivity for the following situations: (Round answers to 4 decimals (i.e. 0.1234 or 12.34%). a. If amount of labor increased by 14% and your manager needs to get 25% more product through the system. b. Currently, there are two machines producing the same parts. If a third machine is purchased to produce the same parts, and the number of parts produced increases by 33%. 2. (2 points) If your manager comes to you and says, "I want you to get 20% more product out the door and productivity needs to increase by 43%." what overall percent change needs to happen to your labor, machinery, etc.? 3. BP oil, due to high costs, is considering moving some of their UK operations to the Philippines (a more cost-friendly labor force). The current exchange rate can be seen in the image below: 69.70 Philippine Piso (2 points) If the current wage rate for Filipinos is 125Php (Philippine pesos)/ hr. and the current wage rate in the UK is 13E/hr. Assuming that the UK workers are twice as productive as the Filipino workers what is the "relative" wage rate (in E/hr.) in the Philippines? a. Based on "relative" wage rates alone, where should the operations be located?Explanation / Answer
Question 4:
Part a.
Reliabilities of components are as follow:
RA = 0.9, RB = 0.92, RC = 0.87, RD = 0.99
As all components must function in order the humdinger to operate effectively, the components are arranged in series. The reliability of the series system is given as follows:
Reliability of series system = RA x RB x RC x RD = 0.9 x 0.92 x 0.87 x 0.99 = 0.7132
Reliability of System = 71.32%
Cost of failure = $1300
Expected cost of failure = probability of failure x cost of failure = (1 – reliability of system) x $1300
Expected cost of failure = (1 – 0.7132) x 1300 = $372.90
Expected cost of failure of the series system = $372.90
Part b.
Option 1: Add backup component to component A with reliability of 90%
Let, reliability of backup component to A = RAB = 0.90
Since backup will be arranged in parallel with the component. The reliability of parallel sub-system is determined as follows:
RSAB = Reliability of sub-system with backup component = (success of component) + {[1 – (success of component)] x (success of backup component)}
RSAB = RA + (1 – RA)RAB = (0.90) + (1 – 0.90)(0.90) = 0.99
Since the sub-system and remaining components are in series, the improved reliability is given as follows:
Reliability of Alternative 1 system = RSAB x RB x RC x RD = 0.99 x 0.92 x 0.87 x 0.99 = 0.7845
Reliability of Alternative 1 system = 78.45%
Total cost of Alternative 1 system = expected cost of failure + component cost
= [( 1 – reliability) x 1300] + $75
= [(1 – 0.7845) x 1300] + 75 = 280.19 + 75 = $355.19
Total cost of Alternative 1 system = $355.19
Option 2: Add backup component to component C with reliability of 75%
Let, reliability of backup component to C = RCB = 0.75
Since backup will be arranged in parallel with the component. The reliability of parallel sub-system is determined as follows:
RSCB = RC + (1 – RC)RCB = (0.87) + (1 – 0.87)(0.75) = 0.9675
Since the sub-system and remaining components are in series, the improved reliability is given as follows:
Reliability of Alternative 2 system = RA x RB x RSCB x RD = 0.90 x 0.92 x 0.9675 x 0.99 = 0.0.7931
Reliability of Alternative 2 system = 79.31%
Total cost of Alternative 1 system = expected cost of failure + component cost
= [( 1 – reliability) x 1300] + $65
= [(1 – 0.7931) x 1300] + 65 = 268.99 + 65 = $334
Total cost of Alternative 2 system = $334
Option 3: Add backup component to component A and C with reliability of 90% and 75% respectively
Since backup will be arranged in parallel with the component. The reliability of parallel sub-system is determined as follows:
Reliability Subsystem with backup for component A = RSCB = 0.99
Reliability Subsystem with backup for component C = RSCB = 0.9675
Since the sub-systems and remaining components are in series, the improved reliability is given as follows:
Reliability of Alternative 3 system = RSAB x RB x RSCB x RD = 0.99 x 0.92 x 0.9675 x 0.99 = 0.8724
Reliability of Alternative 3 system = 87.24%
Total cost of Alternative 3 system = expected cost of failure + component cost
= [(1 – reliability) x 1300] + ($75 + $65)
= [(1 – 0.8724) x 1300] + (75 + 65) = 165.89 + 140 = $305.89
Total cost of Alternative 3 system = $305.89
Summary of options:
Option
1 (backup for A)
2 (backup for C)
3(backup for A and C)
Reliability of system
78.45%
79.31%
87.24
Total Cost of system
$355.18
$334
$305.89
Thus, lowest cost option is alternative 3 – backup for component A and C
Recommended - option 3
Option
1 (backup for A)
2 (backup for C)
3(backup for A and C)
Reliability of system
78.45%
79.31%
87.24
Total Cost of system
$355.18
$334
$305.89