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Law Incorporation [45]
4 years ago
5

A Disposal of plant assets LO C1, P1, P2 [The following information applies to the questions displayed below.] Onslow Co. purcha

ses a used machine for $178,000 cash on January 2 and readies it for use the next day at a $2,840 cost. On January 3, it is installed on a required operating platform costing $1,160, and it is further readied for operations. The company predicts the machine will be used for six years and have a $14,000 salvage value. Depreciation is to be charged on a straight-line basis. On December 31, at the end of its fifth year in operations, it is disposed of______________
Business
1 answer:
DerKrebs [107]4 years ago
8 0

Answer:

= $ 41,940

Explanation:

Purchased machine for $178,000 cash on January 2

And readies it for use the next day at a $2,840 cost

On January 3, it is installed costing $1,160

Total Acquisition Cost =        $ 181,640        

Salvage value   $14,000

Useful Life = 6 years

Depreciation Straight Line Method= Cost - Salvage Value/ Useful Life

Depreciation Straight Line Method= $ 181,640 -$14,000/6

                                                         = $ 167,640/6= $ 27,940

After 5 years its Value would be = $ 181,640 -$ 27,940*5

                                                          =      $ 181,640   - 139,700

                                                        = $ 41,940

It must be disposed off to get a value at least equal to $ 41,940 which is its value .

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Videoworld is a discount store that sells color televisions. The monthly demand for color television sets is 100. The cost per o
vladimir1956 [14]

Complete question:

Videoworld is a discount store that sells color televisions. The monthly demand for color television sets is 100. The cost per order from the manufacturer is $600. The carrying cost is $64 per set each year. Assume a year has 360 working days. Determine the following values rounding to the nearest integer (answer them using only numbers without any sign such as the dollar sign, comma, ...):

Q1. The optimal quantity per order: Q2. The minimum total annual inventory costs:

Q3. The optimal number of orders per year:

Q4. The optimal time between orders (in working days):

If the store had an inventory policy that allows shortages with the shortage cost per set estimated at $80, determine the following values:

5) The optimal quantity per order when the store allows shortages

6) The optimal storage level when the store allows shortages

7) The optimal number of orders when the store allows shortages

8)The optimal time between orders (in working days) when the store allows shortages.

Answer:

1) 150

2) $4,800

3) 8

4) 45 days

5) 201

6) 89

7) 6

8) 60 days

Explanation:

We are given:

Monthly demand, = 100

Cost per order, S= $600

Carrying cost, H = $64 per set/ year

Shortage cost, Cs = $80

Yearly demand will be, D= 100*12 =1200

1) The optimal quantity per order:(Q*) = \sqrt{\frac{2*D*S}{H}}

= \sqrt{\frac{2*1200*600}{64}}

= \sqrt{22500} = 150

2) The minimum total annual inventory cost:

Average inventory * H

Where average inventory = Q*/2

= \frac{150}{2} = 75

Therefore,

Average inventory * H

= 75 * 64

= $4,800

3)The optimal number of orders per year:

= \frac{D}{Q*} = \frac{1200}{150} = 8

4) The optimal time between orders:

= \frac{360}{8} = 45 days

5)The optimal quantity per order when the store allows shortages:

Q= \sqrt{\frac{2*D*S*(H+Cs)}{H * Cs}

= \sqrt{\frac{2*1200*600*(64+80)}{64 * 80}

= 201.25 ≈ 201

6) The optimal shortage level when the store allows shortages:

= \frac{Q* H}{H* Cs}

= \frac{201 * 64}{64* 80}

= 89.33 ≈ 89

The optimal shortage level when the store allows shortages = 89

7) The optimal number of orders per year when the store allows shortages:

No. of orders =

\frac{D}{Q} = \frac{1200}{201}

= 5.97 ≈ 6

Optimal number of orders per year = 6

8) The optimal time between orders (in working days) when the store allows shortages:

Time between orders = Number of working days/ Number of orders

= \frac{360}{6} = 60

The optimal time between orders (in working days) = 60 Days

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