W18_LUCKY_The Decision to repair or to buy a truck

W18_LUCKY_The Decision to repair or to buy a truck

1.      Problem Recognition

I just received a bad news. My truck has been wrecked by the driver!  I need another truck immediately as I have decided not to walk, ride a bike or take a bus to work.  

2.      Problem Definition

The basic problem is that I need transportation. What do I do? I would like to use cash flow to resolve this problem.

a.      Assumptions

Key assumptions include:

·         Walking, riding a bike or taking a bus are not acceptable options

·         An automobile wholesaler offers to pay me $1500 for the wrecked truck ‘as is.’

·         My insurance company estimates that there is $1500 in damages to my truck

·         My insurance has a collision insurance with $500 deductibility provision

·         Thus I am entitled to a payment of $1000.

·         The odometer reading on my wrecked truck is 24000km

 

3.      Feasible Alternatives

Feasible alternatives include:

1.      Walking to the office – Not an option to be considered (Condition specified at onset)

2.      Riding a bicycle to the office - Not an option to be considered (Condition specified at onset)

3.      Taking a bus to the office - Not an option to be considered (Condition specified at onset)

4.      Repairing the truck

                                                              i.      Spend the $1000 insurance check and $1000 of savings to fix the truck. The total amount paid out of your savings is $1000, and the truck will have 24000 Km of prior use.

                                                            ii.      Spend the $1000 insurance check and $1000 of your savings to fix the truck and then sell the truck for $4500. Spend the $4500 plus $5500 of additional savings to buy the newer truck. The total amount paid out of savings is $6500 and the truck will have 18000 Km.

                                                          iii.      Give the truck to a part-time mechanic, who will repair it for $1100 ($1000   insur­ance and $100 of my savings), but will take an additional month of repair time. I will also have to rent a truck for that time at $400/month (paid out of savings). The total amount paid out of savings is $500, and the truck will have 24000 Km on the odometer.

                                                          iv.      Same as Alternative iii, but then sell the truck for $4500 and use this money plus $5500 of additional savings to buy the newer truck. The total amount paid out of savings is $6000 and the newer truck will have 18000 Km of prior use.

5.      Replacing the truck

                                                              i.      Sell the wrecked truck for $1500 to the wholesaler and spend this money, the $1000 insurance check, and all of your $7500 savings account on a newer truck. The total amount paid out of your savings account is $7500 and the truck will have 18000 Km of prior use.

 

·         Further Assumptions:

 

                                                            ii.      The less reliable repair shop in Alternatives (III) and (IV) will not take longer than an additional month to repair the truck.

                                                          iii.      Each truck will perform at a satisfactory operating condition (as it was originally, intended) and will provide the same total mileage before being sold or salvaged

                                                          iv.      Interest earned on money remaining in savings is negligible.

 

6.      Development of outcomes  for each alternative

The development of outcomes for each alternative using estimated cash flow is as follows:

1.      Alternative 5(i) varies from all others because the truck is not to be repaired at all but merely sold. This eliminates the benefit of the $2500 increase in the value of the truck when it is repaired and then sold. Also this alternative leaves no money in your savings account. There is a cash flow of -$7500 to gain a newer truck valued at $10000.

2.      Alternative 4(i) varies from Alternative 4(ii) because it allows the old truck to be re­paired. Alternative 4 (i) differs from Alternatives 4(iii) and 4 (iv) because it utilizes a more expensive ($900 more) and less risky repair facility. It also varies from Alter­natives 4 (ii) and 5(i) because the truck will be kept. The cash flow is -$2000 and the repaired truck can be sold for $4500.

3.      Alternative 4 (ii) gains an additional $2500 by repairing the truck and selling it to buy the same truck as in Alternative 1. The cash flow is -$7500 to gain the newer truck valued at $10000.

4.      Alternative 4(iii) uses the same idea as Alternative 4(i), but involves a less expensive repair shop. The repair shop is more risky in the quality of its end product, but will only cost $1100 in repairs and $400 in an additional month's rental of a truck. The cash flow is -$1500 to keep the older truck valued at $4500.

5.      Alternative 4(iv) is the same as Alternative 4(iii), but gains an additional $500 by selling the repaired truck and purchasing a newer truck as in Alternatives 5 (i) and 4 (ii). The cash flow is -$7000 to obtain the newer truck valued at $10000.

 

7.      Selection Criteria

The value of the truck is its market value (i.e., $10000 for the newer truck and $4500 for, the repaired truck). Thus, this will be used as the consistent value against which everything is measured.

8.      Analysis and Comparison of the alternatives

Considering all relevant criteria

1. Alternative 5(i) is eliminated, because Alternative 4 (ii) gains the same end result and would also provide the truck owner with $2500 more cash. This is experienced with no change in the risk to the owner.

(Truck value = $10000 savings = $0, total worth = $10000)          .

2. Alternative 4(i) is a good alternative to consider, because it spends the least amount of cash, leaving $5500 in the bank. Alternative 4 (i) provides the same end result as Alternative 4 (iv), but costs $500 more to repair. Therefore, Alternative 4 (i) is elimi­nated.

(Truck value = $4500, savings = $5500, total worth = $10000)

3. Alternative 4 (ii) is eliminated, because Alternative 4 (iv) also repairs the truck but at a lower out-of-savings cost ($500 difference), and both Alternatives 4 (ii) and 4 (iv) have the same end result of buying the newer truck.

(Truck value = $10000 savings = $500 total worth = $10500)

4. Alternative 4(iii) is a good alternative, because it saves $500 by using a cheaper repair facility, provided that the risk of a poor repair job is judged to be small.

(Truck value = $4500, savings = $6000, total worth = $10500)

5. Alternative 4 (iv) repairs the truck at a lower cost ($500 cheaper) and eliminates the risk of breakdown by selling the truck to someone else at an additional $500 gain.

(Truck value = $10000, savings = $500 total worth = $10500)

Thus the options available to me to choose from are alternatives 4 (iii) and 4 (iv).

9.      Selection of preferred alternative

Among the uncertainties that can be found in this problem, the following are the most relevant to the decision.

·         If the original truck is repaired and kept, there is a possibility that it would have a higher frequency of breakdowns (based on personal experience).

·         If a cheaper repair facility is used, the chance of a later breakdown is even greater (based on personal experience).

·         Buying a newer truck will use up most of my savings.

·         Also, the newer truck purchased may be too ex­pensive, based on the additional price paid (which is at least $6000/20,000 Km =30 cents per Km).

·         Finally, the newer truck may also have been in an accident and could have a worse repair history than the presently owned truck.

However, based on the information in all previous steps, alternative 4 (iv) is chosen.

10.  Performance monitoring and post evaluation of results

The newer truck turned out after being "test driven" for 20,000 Km to be a real beauty. Mileage was great, and no repairs were needed. The systematic process of identifying and analyzing alternative solutions to this problem has paid off!

A further review with qualitative factors that may carry their own value (e.g., how much is low mileage or a reliable repair shop worth?) can be done in the future.

 

 

 

Reference

1.      Sullivan, W., Wicks, E., Koelling, P., Kumar, p., & Kumar, N. (2012).Chapter 1 Introduction to Engineering Economy (pp. 29). Engineering economy (15^th edition). England: Pearson Education Limited.

2.      Sullivan, W., Wicks, E., Koelling, P., Kumar, p., & Kumar, N. (2012).Chapter 14 The Time Value of Money (pp. 131 -134). Engineering economy (15^th edition). England: Pearson Education Limited.

3.      Lindeburg, M. (1992).Chapter 13 Engineering Economic Analysis (pp.13-3).Engineer-In-Training Reference Manual (8^th edition).U.S.A: Professional Publications Inc.

4.       Giammalvo, P. (2012, October 22). Integrated portfolio (asset), program (operations) and project management methodology course (cost engineering) slides (An AACE methodology course). Lagos, Nigeria: Lonadek