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8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Home / My courses / ENGR-301-2201-CC / Topic 4 / Assignment 3 (Due Wednesday July 22 @23:55)

Started on State Completed on Time taken Grade

Tuesday, 21 July 2020, 8:05 PM Finished Tuesday, 21 July 2020, 8:34 PM 28 mins 19 secs 12.00 out of 20.00 (60%)

Question 1 Complete Not graded

Make sure you review the Faculty of Engineering and Computer Science Expectations of Originality and standards of academic integrity, available at https://www.concordia.ca/ginacody/students/academic-services/expectation-of-originality.html? utm_source=redirect&utm_campaign=expectation-originality.html I certify that I have conformed to the Faculty’s Expectations of Originality and standards of academic integrity.

Note: By answering this question, you effectively sign the form and submit it electronically. There is thus no need to submit a printed copy. Select one: Yes No

The correct answer is: Yes

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1/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 2 Incorrect Mark 0.00 out of 1.00

Your company intends to purchase a 400 tonne capacity injection moulding machine. A similar machine was purchased eight years ago for $89,500. The price index for heavy duty equipment in the plastics industry was 163.6 eight years ago and is 207.6 now. The current cost of the machine is close to Select one: $108,000 $113,000 $98,000 

$125,000

Price now = $89,500 (207.6/163.6) = $113,570.90

Answer $113,000 is closest

The correct answer is: $113,000

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2/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 3 Incorrect Mark 0.00 out of 1.00

Nadia, John, Saba and Kumar are working together on a project. These are the data for each: Nadia: 60 hours, overhead rate 75%, personal time 10%, $20/hr John: 80 hours, overhead rate 55%, personal time 12%, $40/hr Saba: 80 hours, overhead rate 75%, personal time 10%, $36/hr Kumar: 40 hours, overhead rate 55%, personal time 12%, $20/hr Who has the highest direct labour costs? Select one: John Kumar Nadia 

Saba

Name

Hours

Nadia

Overhead Personal time

Hourly rate

Total direct labour cost

rate

rate

60

1.75

1.1

20

$2,310

John

80

1.55

1.12

40

$5,555.20 (highest)

Saba

80

1.75

1.1

36

$5,544

Kumar

40

1.55

1.12

20

$1,388.80

Projecttotal =

$14,798

The correct answer is: John

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3/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 4 Incorrect Mark 0.00 out of 1.00

Nadia, John, Saba and Kumar are working together on a project. These are the data for each: Nadia: 60 hours, overhead rate 75%, personal time 10%, $20/hr John: 80 hours, overhead rate 55%, personal time 12%, $40/hr Saba: 80 hours, overhead rate 75%, personal time 10%, $36/hr Kumar: 40 hours, overhead rate 55%, personal time 12%, $20/hr The total direct labour costs is close to Select one: $14,300 $14,000 $14,800 

$15,100

Name

Hours

Nadia

Overhead Personal time

Hourly rate

Total direct labour cost

rate

rate

60

1.75

1.1

20

$2,310

John

80

1.55

1.12

40

$5,555.20

Saba

80

1.75

1.1

36

$5,544

Kumar

40

1.55

1.12

20

$1,388.80

Projecttotal =

$14,798 (Answer $14,800 closest)

The correct answer is: $14,800

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4/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 5 Correct Mark 1.00 out of 1.00

Weldex Industries begins to manufacture a newly designed, small, spot welding machine. The labour hours required for making the second and third units are 28.6 and 28.2 hours respectively. The labour hours required to manufacture the first unit are close to Select one: 29.3 hours



29.5 hours 28.8 hours 29.1 hours

Need to find both T1 and b, given that T2 = 28.6 = Tinit 2b and T3 = 28.2 = Tinit 3b

Tinit = 28.6/2b = 28.6/2–0.03474 = 29.3 hours The correct answer is: 29.3 hours

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5/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 6 Correct Mark 1.00 out of 1.00

Weldex Industries begins to manufacture a newly designed, small, spot welding machine. The labour hours required for making the second and third units are 28.6 and 28.2 hours respectively. The labour hours required to manufacture the 50th unit are close to Select one: 29.3 hours 

25.5 hours 23.8 hours 27.9 hours

Need to find both T1 and b, given that T2 = 28.6 = Tinit 2b and T3 = 28.2 = Tinit 3b

Tinit = 28.6/2b = 28.6/2–0.03474 = 29.3 hours

T50 = 29.3 * 50–0.03474 = 25.6 hours

Answer 25.5 is closest

T4 = 29.3 * 4–0.03474 = 27.9 hours

Total labour hours to produce first four units = 29.3 + 28.6 + 28.2 + 27.9 = 114 hours.

The correct answer is: 25.5 hours

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6/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 7 Correct Mark 1.00 out of 1.00

Weldex Industries begins to manufacture a newly designed, small, spot welding machine. The labour hours required for making the second and third units are 28.6 and 28.2 hours respectively. If labour costs are $18/hour, the total labour costs to produce the first four machines are close to Select one: $1900 $2000 $1950 

$2050

T2 = 28.6 = Tinit 2b

Need to find both T1 and b, given that

and

T3 = 28.2 = Tinit 3b

Tinit = 28.6/2b = 28.6/2–0.03474 = 29.3 hours T50 = 29.3 * 50–0.03474 = 25.6 hours T4 = 29.3 * 4–0.03474 = 27.9 hours Total labour hours to produce first four units = 29.3 + 28.6 + 28.2 + 27.9 = 114 hours. Total labour costs = 114 * $18 = $2052

Answer $2050 is closest

The correct answer is: $2050

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7/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 8 Incorrect Mark 0.00 out of 1.00

A 50-mgpd (million gallons per day) water treatment plant was proposed in 2013 for Ureka. A 60-mgpd plant using the same treatment process was built in 2007 in Vancouver for $50,000,000. Given the following additional information and assumptions: Poor site condition in Vancouver required an additional $3,000,000 for foundations, which is not anticipated for Ureka. Assume the power sizing size exponent = 0.67. Assume inflation will persist at an average 4% per annum. Assume the location index for Vancouver is 1.17 and 1.24 for Ureka. A new high-pressure filter process will be added at Ureka at an additional cost of $4,000,000. The construction period for the new plant will include winter construction. The Vancouver plant was not constructed through a winter season. Assume an increase in construction costs for Ureka of 12%. In order to make an order of magnitude estimate of the water treatment plant in Ureka, various multiplication factors need to be calculated. The multiplication factor based on the different capacities is close to Select one: 1.10 1.00 

0.80 0.90

Adjustment for capacity based on the exponential law: Factor is:

(50/60)0.67 = (0.833)0.67 = 0.885

Answer 0.90 is closest

The correct answer is: 0.90

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8/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 9 Incorrect Mark 0.00 out of 1.00

A 50-mgpd (million gallons per day) water treatment plant was proposed in 2013 for Ureka. A 60-mgpd plant using the same treatment process was built in 2007 in Vancouver for $50,000,000. Given the following additional information and assumptions: Poor site condition in Vancouver required an additional $3,000,000 for foundations, which is not anticipated for Ureka. Assume the power sizing size exponent = 0.67. Assume inflation will persist at an average 4% per annum. Assume the location index for Vancouver is 1.17 and 1.24 for Ureka. A new high-pressure filter process will be added at Ureka at an additional cost of $4,000,000. The construction period for the new plant will include winter construction. The Vancouver plant was not constructed through a winter season. Assume an increase in construction costs for Ureka of 12%. In order to make an order of magnitude estimate of the water treatment plant in Ureka, various multiplication factors need to be calculated. The multiplication factor based on the effect of inflation is close to Select one: 

1.22 1.27 1.37 1.42

Adjustment for inflation leads to the cost in 2007 dollars: Factor is:

(1.04)6 = 1.265

Answer 1.27 is closest

The correct answer is: 1.27

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9/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 10 Incorrect Mark 0.00 out of 1.00

A 50-mgpd (million gallons per day) water treatment plant was proposed in 2013 for Ureka. A 60-mgpd plant using the same treatment process was built in 2007 in Vancouver for $50,000,000. Given the following additional information and assumptions: Poor site condition in Vancouver required an additional $3,000,000 for foundations, which is not anticipated for Ureka. Assume the power sizing size exponent = 0.67. Assume inflation will persist at an average 4% per annum. Assume the location index for Vancouver is 1.17 and 1.24 for Ureka. A new high-pressure filter process will be added at Ureka at an additional cost of $4,000,000. The construction period for the new plant will include winter construction. The Vancouver plant was not constructed through a winter season. Assume an increase in construction costs for Ureka of 12%. In order to make an order of magnitude estimate of the water treatment plant in Ureka, various multiplication factors need to be calculated. The multiplication factor based on the different location indexes is close to Select one: 1.03 

0.93 1.07 0.99

Adjustment for location index: Factor is:

1.24/1.17 = 1.06

Answer 1.07 is closest

The correct answer is: 1.07

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10/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 11 Incorrect Mark 0.00 out of 1.00

A 50-mgpd (million gallons per day) water treatment plant was proposed in 2013 for Ureka. A 60-mgpd plant using the same treatment process was built in 2007 in Vancouver for $50,000,000. Given the following additional information and assumptions: Poor site condition in Vancouver required an additional $3,000,000 for foundations, which is not anticipated for Ureka. Assume the power sizing size exponent = 0.67. Assume inflation will persist at an average 4% per annum. Assume the location index for Vancouver is 1.17 and 1.24 for Ureka. A new high-pressure filter process will be added at Ureka at an additional cost of $4,000,000. The construction period for the new plant will include winter construction. The Vancouver plant was not constructed through a winter season. Assume an increase in construction costs for Ureka of 12%. In order to make an order of magnitude estimate of the water treatment plant in Ureka, various multiplication factors need to be calculated. The order of magnitude estimate of the water treatment plant in Ureka is close to Select one: $50M $35M



$85M $65M

Typical cost excluding the additional cost due to poor site conditions in Vancouver $50 million – $3 million = $47 million ($47 million) (50/60)0.67 (1.04)6 (1.24/1.17) = $55.781 million Adjustment for the additional filter at Ureka plant gives $55.781 million + $4 million = $59.781 million Increase of construction cost for Eureka of 12% gives $59,781 * 1.12 = $66.955 million. $65 million is closest answer

The correct answer is: $65M

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11/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 12 Correct Mark 1.00 out of 1.00

Suppose PV for a project was $100,000 and EV was $60,000. The SPI for this project is close to Select one: 1.66 0.60



1.52 Cannot calculate SPI with info given

SPI = EV/PV = 60,000/100,000 = 0.60 The correct answer is: 0.60

Question 13 Correct Mark 1.00 out of 1.00

Activity A is worth $500, is complete, and actually cost $500. Activity B is worth $1000, is 50% complete, and has actually cost $700 so far. Activity C is worth $100, is 75% complete, and has actually cost $90 so far. Total earned value for the project is close to Select one: –$1075 $1290 $1075



$1600

EV = 100% * $500 + 50% * $1000 + 75% * $100 = $1075 The correct answer is: $1075

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12/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 14 Correct Mark 1.00 out of 1.00

Activity A is worth $500, is complete, and actually cost $500. Activity B is worth $1000, is 50% complete, and has actually cost $700 so far. Activity C is worth $100, is 75% complete, and has actually cost $90 so far. CPI for the project is close to Select one: –1.20 1.20 –0.83 0.83



CPI = EV/AC = 1075/(500+700+90) = 0.83 The correct answer is: 0.83

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13/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 15 Correct Mark 1.00 out of 1.00

Activity A is worth $500, is complete, and actually cost $500. Activity B is worth $1000, is 50% complete, and has actually cost $700 so far. Activity C is worth $100, is 75% complete, and has actually cost $90 so far. Which of the following gives the remaining amount to be spent on the project based on current spending efficiency? Select one: Estimated cost to completion



Budget remaining All other answers are wrong Cost Performance Index Cost variance

Estimated cost to completion The correct answer is: Estimated cost to completion

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14/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 16 Correct Mark 1.00 out of 1.00

Activity A is worth $100, is complete, and actually cost $150. Activity B is worth $500, is 75% complete, and has actually cost $400 so far. Activity C is worth $500, is 25% complete, and has actually cost $200 so far. Estimated cost at completion for this project, assuming current variances are typical of future variances, is close to Select one: $750 $1375



$880 $1100

EAC = BAC/CPI BAC = $100 + $500 + $500 = $1100 EV = 100% * $100 + 75% * $500 + 25% * $500 = $600 AC = $150 + $400 + $200 = $750 CPI = EV/AC = 600/750 = 0.80 EAC = 1100/0.8 = $1375 The correct answer is: $1375

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15/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 17 Correct Mark 1.00 out of 1.00

If CPI = 1.3 and SPI = 0.8, what is the status of the project? Select one: Under budget and ahead of schedule 

Under budget and behind schedule Over budget and behind schedule Over budget and ahead of schedule

If CPI > 1 the project is under budget. If SPI < 1 the project is behind schedule. The correct answer is: Under budget and behind schedule

Question 18 Correct Mark 1.00 out of 1.00

During a project, earned value analysis is performed resulting in the following numbers: EV = 523,000, PV = 623,000, AC = 643,000. Which of the following results are correct? Select one: CV = 100,000

and

SV = 120,000

CV = –120,000

and

SV = –100,000

CV = 120,000

and

SV = 100,000



All other answers are wrong CV = –100,000

and

SV = –120,000

CV = EV – AC = 523,000 – 643,000 = –120,000 SV = EV – PV = 523,000 – 623,000 = –100,000 The correct answer is: CV = –120,000

and

SV = –100,000 

https://moodle.concordia.ca/moodle/mod/quiz/review.php?attempt=1393284&cmid=2299438

16/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 19 Correct Mark 1.00 out of 1.00

If earned value is 100 and actual cost is 120, the cost performance index is close to Select one: 1.2 –20 0.83



20

EV = 100. AC = 120. CPI = EV/AC = 0.833 The correct answer is: 0.83

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17/19

8/12/2020

Assignment 3 (Due Wednesday July 22 @23:55): Attempt review

Question 20 Correct Mark 1.00 out of 1.00

You found the following earned value analysis information for a project that was recently closed-out: SPI = 0.7, CPI = 1.0. What does this mean? Select one: The project has been cancelled while it was being executed and at that time it was behind schedule but on budget



The project’s deliverables have all been completed and the project came in on schedule but over budget All other answers are wrong The project’s deliverables have all been completed and the project came in behind schedule but on budget The project’s deliverables have all been completed and the pr...