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Ratling [72]
3 years ago
12

Based on the following information, calculate the cost of goods sold and ending inventory using FIFO, LIFO, and weighted average

assuming a perpetual inventory system is in place.
Beginning Balance - 90 units at $11
March 3 - Purchase 300 units for $15
April 4 - Sell 240 units for $28
June 30 - Purchase 250 units for $18
August 16 - Sell 180 units for $30
Business
1 answer:
Flauer [41]3 years ago
8 0

Answer:

Cost of Sales :

FIFO = $ 6,030

LIFO = $6,840

Weighted Average = $6,354.60

Ending Inventory :

FIFO =  $4,176

LIFO = $3,150

Weighted Average = $3,636.60

Explanation:

FIFO

This method assumes that the first inventory purchased will be the first to be sold

<em>Cost of Goods Sold :</em>

90 units × $11   =  $990

150 units × $15 = $2,250

150 units × $15 = $2,250

30 units × $18  = $540

Total                 = $ 6,030

<em>Ending Inventory :</em>

232 units × $18 = $4,176

LIFO

This method assumes that the last inventory purchased, will be the last to be sold

<em>Cost of Sales :</em>

240 units × $15 =  $3,600

180 units × $18 =  $3,240

Total = $6,840

<em>Ending Inventory :</em>

90 units × $11  = $ 990

60 units × $15 = $ 900

70 units × $18 = $ 1,260

Total = $3,150

Weighted Average

A new average cost per unit is calculated with every purchase made.

New Average Cost = (90 units × $11 + 300 units × $15) ÷ 390 units

                                = $14.08

Cost of Sale , April 4 =  240 units × $14.08

                                  =   $3,379.20

New Average Cost = (150 units × $14.08 + 250 units × $18.00) ÷ 400 units

                               = $16.53

Cost of Sale, Aug 16 = 180 units × $16.53

                                  = $2,975.40

Total Cost of Sales =  $3,379.20 + $2,975.40

                                = $6,354.60

Ending Inventory = 220 units × $16.53

                             = $3,636.60

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Answer:

max z = 0.3x_{1} + 0.45x_{2} + 0.3y_{1} + 0.45y_{2}

Constraints

x_{1} +  x_{2} \leq  100,000 (Grade 9 oranges)

y_{1} +  y_{2} \leq  120,000 (Grade 6 oranges)

2x_{1} - y_{1} \geq   0 (Avg oranges in bags)

x_{2}  - 2y_{2} \geq 0 (Avg oranges in juice)

x_{1}\geq  0, x_{2} \geq  0, y_{1}\geq 0, y_{2} \geq 0

Explanation:

Let the variables used be x and y

x representing oranges of grade 9

y representing oranges of grade 6

Now, let x_{1} be the oranges used in each bag in lbs, and x_{2} be oranges used in juice of grade 9 each.

Similarly let y_{1} will represent oranges of grade 6 in bags, and y_{2} will represent oranges in juice of grade 6

Now total oranges sold in bags

= x_{1} + y_{1}

And their revenue in $ = 0.5 revenue - 0.2 expense = 0.3

Total profit from bag shall be

0.3 (x_{1} + y_{1}) = 0.3 x_{1} + 0.3y_{1}

Similarly total oranges in juice shall be

= x_{2} + y_{2}

Profit shall be

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Total profit from juice shall be

$0.45 (x_{1} + y_{1}) = 0.45 x_{2} + 0.45 y_{2}

Profit shall maximise as

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x_{1} +  x_{2} \leq  100,000

Constraint 2

Total amount of  grade 6  oranges used shall be max 120,000 lb

y_{1} +  y_{2} \leq  120,000

Constraint 3

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Accordingly,

9x_{1} + 6y_{1} \geq   7x_{1} + 7y_{1}

Simplifying:

2x_{1} - y_{1} \geq   0

Constraint 4

Average quality of oranges sold as juice shall be 8

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Accordingly,

9x_{2} + 6y_{2} \geq   8x_{2} + 8y_{2}

Simplifying:

x_{2}  - 2y_{2} \geq 0

Constraints shall be

x_{1}\geq  0\\x_{2} \geq  0\\y_{1}\geq  0\\y_{2} \geq 0

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