Question

The XYZ Company’s current production processes have a scrap rate of 15% and a return rate of 3%. Scrap costs (wasted materials) are $12 per unit; warranty/repair costs average $60 per unit returned. The company is considering the following alternatives to improve its production processes:

• Option A: Invest $400,000 in new equipment. The new process will also require an additional $1.50 of raw materials per unit produced. This option is predicted to reduce both scrap and return rates by 40% from current levels.
• Option B: Invest $50,000 in new equipment but spend an additional $3.20 on higher-quality raw materials per unit produced. This option is predicted to reduce both scrap and return rates by 90% from current levels.
• Option C: Invest $2 million in new equipment. The new process will require no change in raw materials. This option is predicted to reduce both scrap and return rates by 50% from current levels.

Required

1. Assume that current production levels of 1 million units will continue. Which option do you recommend? Why?
2. Assume that all of the proposed changes will increase product quality such that production will jump to 1.5 million units. Which option do you recommend? Why?

Answer 1

a. Recommended option

Calculation for investment as follows: Following formula is used to calculate investment:

Investment = Amount to be invested + Scrap rate + Current production units (1) Substitute the "invested amount', "scrap rate" and "units" for current production in the Equation (1) is given below:

Investment = $400,000 + $1.5 x 1,000,000 units

= $1,900,000

Therefore, the investment for option A is $1,900,000

Calculation for savings as follows: Following formulas are used to calculate savings:

Savings = Reduced scrap costs + Reduced warranty/ repair costs (2)

Reducedscrap costs = Reduced scrap rate x Scrap rate x Scrap cost x Production unit

Reduced scrap costs = 40% x 15% x $12 x 1,000,000 units

= $720,000

Repair costs = Reduced scrap rate x return rate x repair cost x Production unit

Reduced warranty/ repair costs = 40% x 3% x $60 x 1,000,000 units

= $720,000

Substitute the "Reduced scrap costs" and "Repair costs" in the Equation (2) is given below:

Savings = $720,000+$720,000 = $41,440,000
Therefore, the savings for option A is $41,440,000

(b).

Following formula is used to calculate investment:

Calculation for investment as follows:
Substitute the invested amount, additional amount and units for current production in the Equation (1) is given below:

Investment = $50,000 + $3.2 x 1,000,000 units

= $3,250,000

Therefore, the investment for option B is $3,250,000
Calculation for savings as follows: Following formulas are used to calculate savings:

Reducedscrap costs = Reduced scrap rate x Scrap rate x Scrap cost x Production unit

Reduced scrap costs = 90% x 15% x $12 x 1,000,000 units

= $1,620,000

Repair costs = Reduced scrap rate x return rate x repair cost x Production unit

Reduced warranty/repair costs = 90% x 3% x $60 x 1,000,000 units

= $1,620,000

Substitute the "Reduced scrap costs" and "Repair costs" in the Equation (2) is given below:

Savings =$1,620,000+$1,620,000 =$3,240,000
Therefore, the savings for option B is $3,240,000

(c)

Following formula is used to calculate investment:
Calculation for investment as follows:

Investment = Amount to be invested

Substitute the invested amount using the above formula.

Investment = $2,000,000

Therefore, the investment for option C is $2,000,000

Calculation for savings as follows:

Following formulas are used to calculate savings:

Reducedscrap costs = Reduced scrap rate x Scrap rate x Scrap cost x Production unit

Reduced scrap costs = 50% x 15% x $12 x 1,000,000 units

= $900,000

Repair costs = Reduced scrap rate x return rate x repair cost x Production unit

Reduced warranty/repair costs = 50% x 3% x $60 x 1,000,000 units

= $900,000

Substitute the "Reduced scrap costs" and "Repair costs" in the Equation (2) is given below: Savings = $9000,000+$9000,000 = $1,800,000
Therefore, the savings for option B is $1,800,000
Conclusion:
Thus, the continuation of 1,000,000 units in the production level, the total cost is not reduced by any other options. But, option "B's" total cost is increased in some level. So, the option B is recommended.

b. Recommended option:

Option A:

Following formula is used to calculate investment:

Calculation for investment as follows:

Substitute the invested amount, scrap rate and units for current production in the Equation (1) is given below:

Investment = $400,000 + $1.5 x 1,500,000 units

= $2,650,000

Calculation for savings as follows:

Following formulas are used to calculate savings:

Reducedscrap costs = Reduced scrap rate x Scrap rate x Scrap cost x Production unit

Reduced scrap costs = 40% x 15% x $12 x 1,500,000 units

= $1,080,000

Repair costs = Reduced scrap rate x return rate x repair cost x Production unit

Reduced warranty/repair costs = 40% x 3% x $60 x 1,500,000 units

= $1,0800,000

Substitute the "Reduced scrap costs" and "Repair costs" in the Equation (2) is given below:

Savings = $1,0800,000+$1,0800,000 = $2,160,000
Therefore, the savings for option A is $2,160,000

Option B:

Following formula is used to calculate investment:

Calculation for investment as follows:
Substitute the invested amount, additional amount and units for current production in the Equation (1) is given below:

Investment = S50,000 + S3.2 x 1,500,000 units

= S4,850,000

Therefore, the investment for option B is $4,850,000.

Calculation for savings as follows:

Following formulas are used to calculate savings:

Reducedscrap costs = Reduced scrap rate x Scrap rate x Scrap cost x Production unit

Reduced scrap costs = 90% x 15% x $12 x 1,500,000 units

= $2,430,000

Repair costs = Reduced scrap rate x return rate x repair cost x Production unit

Reduced warranty/repair costs = 90% x 3% x $60 x 1,500,000 units

= $2,430,000

Substitute the "Reduced scrap costs" and "Repair costs" in the Equation (2) is given below: Savings = $2,430,000+$2,430,000 = $4,860,000
Therefore, the savings for option B is $4,860,000

Option C:

Following formula is used to calculate investment:

Calculation for investment as follows:
Investment = Amount to be invested

Substitute the invested amount using the above formula.
Investment = $2,000,000

Therefore, the investment for option B is $2,000,000

Calculation for savings as follows:

Following formulas are used to calculate savings:

Reducedscrap costs = Reduced scrap rate x Scrap rate x Scrap cost x Production unit

Reduced scrap costs = 50% x 15% x $12 x 1,500,000 units

= $1,350,000

Repair costs = Reduced scrap rate x return rate x repair cost x Production unit

Reduced warranty/repair costs = 50% x 3% x $60 x 1,500,000 units

= $1,350,000

Substitute the "Reduced scrap costs" and "Repair costs" in the Equation (2) is given below: Savings = $1,350,000+$1,350,000 = $2,700,000
Therefore, the savings for option B is $2,700,000

Conclusion:
Thus, the changes will increase the quality of the product. So, the production will jump into of 1,500,000 units in the production level.

Here option "B" and "C" both reduces the total cost. But option "C" reduce the total cost little bit
higher. Therefore, the 'option C is recommended.