Please don\'t use excel to these questions. This is Engineering Economic Analysi
ID: 2784400 • Letter: P
Question
Please don't use excel to these questions. This is Engineering Economic Analysis.
* Please show all the calculations and the procedure for solving the questions 1. Installing a new production line has an initial cost of $85,000, and it has no salvage value. The firm's interest is 12%. Following table shows the saving per year which is based on number of shifts per day, and the useful life of the product: i. What is the joint probability distribution for saving per year and useful life? (15 pts) ii. Define optimistic, most likely, and pessimistic scenarios by using both optimistic, both most likely, and both pessimistic estimates. Wha (15 pts) t is the present worth for each scenario Useful life Probability Shifts per dav Saving per year $14,000 $29,000 $43,000 Probability 0.33 0.45 0.22 (vears) 0.5 0.3 0.2 2. Installing new machinery will cost $170,000. The installation will result in following annual saving: (a) Optimistic scenario: $22,000 annual saving with 22% probability, (b) Most likely scenario: $12,000 annual saving with 48% probability, and (c) Pessimistic scenario: $4,000 annual saving with 30% probability. The interest rate can be 5% or 8% (equally likely) and the machinery should have a useful life of 40 years What is the present worth for each estimated value? What is the expected present worth (20 pts) Compute the standard deviation of the present worth. (10 pts)Explanation / Answer
CALCULATION OF JOINT PROBABILITY DISTRIBUTION
If there are two independent events A and B having probability P(A) and P(B) respectively,
The joint probability of occurring of both events P(A,B)=P(A)*P(B)
In this case Probability of shifts per day1 and annual savings of $14,000 P(1)=0.33
Annual savings of$29,000 :P(2)=0.45, Annual savings of $43,000: P(3)=0.22
Let us represent probability of useful life of 4 years as P(A)=0.5
Probability of useful life of 5 years as P(B)=0.3
Probability of useful life of 8 years as P(C)=0.2
Joint probability of 1 shift per day and 4 years useful lifeP(1,A)=0.33*0.5=0.165
Similarly, all possible joint events and their probabilities are calculated and joint probability distribution is given below:
Please note that sum of probabilities =1
Joint Probability Distribution
(P1,A)
0.165
P(2,B)
0.135
P(3,C)
0.044
P(1,B)
0.099
P(1,C)
0.066
P(2,A)
0.225
P(2,C)
0.09
P(3,A)
0.11
P(3,B)
0.066
Total
1
.(ii)Optimistic scenario will be annual saving of $43,000 and 8 years useful life
Probability of this optimistic scenario =P(3,C)=0.044
Pessimistic scenario will be annual saving of $14,000 and useful life of 4 years.
Probability of pessimistic scenario=P(1,A)=0.165
Most likely scenario is the one having highest probability.
Most likely scenario: Annual saving of $29,000 and useful life of 4 years
Probability of this scenario: P(2,A)=0.225
PRESENT WORTH OF EACH SCENARIO
Present Worth of Optimistic Scenario:
Savings =$43,000 for 8 years
Discount rate=12%=0.12
(P/A,i,N) =Uniform Series Present worth factor=UPWF=(((1+i)^N)-1)/(i((1+i)^N))
i=Interest rate=12%=0.12
N=8
UPWF=(((1+i)^N)-1)/(i((1+i)^N))=((1.12^8)-1)/(0.12*(1.12^8))= 4.96764
Present Worth of Optimistic Scenario=$43,000*UPWF=(43000*4.96764)= $ 213,608.51
Present Worth of Pessimistic Scenario:
Savings =$14,000 for 4 years
Discount rate=12%=0.12
(P/A,i,N) =Uniform Series Present worth factor=UPWF=(((1+i)^N)-1)/(i((1+i)^N))
i=Interest rate=12%=0.12
N=4
UPWF=(((1+i)^N)-1)/(i((1+i)^N))=((1.12^4)-1)/(0.12*(1.12^4))= 3.037349
Present Worth of Pessimistic Scenario=$14,000*UPWF=(14000*3.037349)= $ 42,522.89
Present Worth of Most Likely Scenario:
Savings =$29,000 for 4 years
Discount rate=12%=0.12
(P/A,i,N) =Uniform Series Present worth factor=UPWF=(((1+i)^N)-1)/(i((1+i)^N))
i=Interest rate=12%=0.12
N=4
UPWF=(((1+i)^N)-1)/(i((1+i)^N))=((1.12^4)-1)/(0.12*(1.12^4))= 3.037349
Present Worth of Most Likely Scenario=$29,000*UPWF=(29000*3.037349)= $ 88,083.13
Scenario
Probability
Present Worth
Optimistic
0.044
$213,609
Pessimistic
0.165
$42,523
Most Likely
0.225
$88,083
Joint Probability Distribution
(P1,A)
0.165
P(2,B)
0.135
P(3,C)
0.044
P(1,B)
0.099
P(1,C)
0.066
P(2,A)
0.225
P(2,C)
0.09
P(3,A)
0.11
P(3,B)
0.066
Total
1