Answer:
Lahdekorpi OY, a Finnish corporation and Three-O Company, a subsidiary incorporated in the United States
Transfer Pricing:
a) The best transfer pricing method in this case is the cost plus method. This gives the transfer price as Cost + 50%.
b) The appropriate transfer price should be $3 ($2 x 1.5).
Explanation:
Transfer pricing arises when controlled entities set prices for exchange of goods and services. When Lahdekorpi OY, a Finnish corporation, sells wooden puzzles to Three-O Company, given their relationship, transfer pricing has arisen. It is the assignment of cost for goods and services exchanged between related parties, like a parent and a subsidiary.
There are many Transfer Pricing methods which entities and the taxing authorities can use to determine the best transfer price. According to the Organisation for Economic Co-operation and Development (OECD) Multinational Entities and tax authorities can use any of these five main transfer pricing methods:
a) Comparable uncontrolled price (CUP) method. The CUP method is grouped by the OECD as a traditional transaction method (as opposed to a transactional profit method)
b) Resale price method
c) Cost plus method
d) Transactional net margin method (TNMM)
e) Transactional profit split method.
Answer:
They should operate Mine 1 for 1 hour and Mine 2 for 3 hours to meet the contractual obligations and minimize cost.
Explanation:
The formulation of the linear programming is:
Objective function:
![C=200M_1+160M_2](https://tex.z-dn.net/?f=C%3D200M_1%2B160M_2)
Restrictions:
- High-grade ore: ![6M_1+2M_2\geq12](https://tex.z-dn.net/?f=6M_1%2B2M_2%5Cgeq12)
- Medium-grade ore: ![2M_1+2M_2\geq8](https://tex.z-dn.net/?f=2M_1%2B2M_2%5Cgeq8)
- Low-grade ore: ![4M_1+12M_2\geq24](https://tex.z-dn.net/?f=4M_1%2B12M_2%5Cgeq24)
- No negative hours: ![M_1,M_2\geq0](https://tex.z-dn.net/?f=M_1%2CM_2%5Cgeq0)
We start graphing the restrictions in a M1-M2 plane.
In the figure attached, we have the feasible region, where all the restrictions are validated, and the four points of intersection of 2 restrictions.
In one of this four points lies the minimum cost.
Graphically, we can graph the cost function over this feasible region, with different cost levels. When the line cost intersects one of the four points with the lowest level of cost, this is the optimum combination.
(NOTE: it is best to start with a low guessing of the cost and going up until it reaches one point in the feasible region).
The solution is for the point (M1=1, M2=3), with a cost of C=$680.
The cost function graph is attached.
Answer:
The cost recorded for the equipment=$66,500
Explanation:
When dealing with the total cost of an equipment we take the purchase cost and other additional associated costs that come with the equipment. This can be expressed as;
T=P+A
where;
T=total cost
P=purchase cost
A=additional costs(transportation cost+sales tax+installation cost)
In our case;
T=unknown
P=$60,000
A=(1,000+3,000+2,500)=$6,500
replacing;
T=60,000+6,500=66,500
The total cost=$66,500
The cost recorded for the equipment=$66,500
Answer:
The correct answer for option (a) is 7.17% and for option (b) is $48,546.69.
Explanation:
According to the scenario, the given data are as follows:
(a) Present value = $3,000
Future value = $6,000
Time period = 10 years
So, we can calculate the annual rate of return by using following formula:
Rate of return = (( FV ÷ PV)^1/t -1)
= (( $6,000 ÷ $3,000)^1/10 -1)
= (2)^0.1 - 1
= 1.07177346254 - 1
= .07177 or 7.17%
(b) Present value = $12,000
Rate of interest (r) = 15%
Time period = 10 year
So, we can calculate the Future value by using following formula:
FV = PV × ( 1+r)^t
= $12,000 × ( 1 + 15%)^10
= $12,000 × 4.04555773571
= $48,546.69