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diamong [38]
3 years ago
12

Ivan Knobel holds a well-diversified portfolio that has an expected return of 11.0% and a beta of 1.20. He is in the process of

buying 1,000 shares of Syngine Corp at $10 a share and adding it to his portfolio. Syngine has an expected return of 13.0% and a beta of 1.50. The total value of Ivan's current portfolio is $90,000. What will the expected return and beta on the portfolio be after the purchase of the Syngine stock?
Business
1 answer:
-Dominant- [34]3 years ago
8 0

Answer:

Exptected return = 11.2%

Beta = 1.23  

Explanation:

The post-purchase expected return of the portfolio is the weighted average return of Syngine stock and pre-purchase return of the portfolio, calculated as below:

Post-purchase portfolio return = (Market value of Synhine stock purchase/Total market value of post-purchase portfolio)x Syngine stock return + (Market value of pre-purchase porfolio/Total market value of post-purchase portfolio) x Pre-purchase return

= [(1,000 x 10)/(1,000 x 10 + 90,000)] x 13% +  [(90,000)/(1,000 x 10 + 90,000)] x 11% = 11.2%

Using the same concept, beta of the post-purchase is calculated as below:

Post-purchase portfolio beta = [(1,000 x 10)/(1,000 x 10 + 90,000)] x 1.5 +  [(90,000)/(1,000 x 10 + 90,000)] x 1.2 = 1.23

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An analysis of stockholders' equity of Hahn Corporation as of January 1, 2010, is as follows:
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Answer:

Additional paid-in capital is $904,200

Explanation:

Number of shares, issued and outstanding = 93,000 shares

Acquired 2,460 shares of its stock for $75,000.

Sold 2,000 treasury shares at $35 per share.

Sold the remaining 460 treasury shares at $20 per share.

i) Acquired 2,460 shares of its stock for $75,000.

= Treasury Stock Dr $75,000

ii) Sold 2,000 treasury shares at $35 per share.

Treasury Stock (2,000 × $35) = Dr $70,000

iii) Sold the remaining treasury shares at $20 per share.

Treasury Stock (460 × $20) = Dr $9,200

Total Treasury Stock = $75,000 - $70,000 - $9,200

= ($4,200)

Paid in Cap-tresury stock= 10,000-5000=5000

Additional Paid in capital = Paid in Capital - treasury stock

= 900,000 + 4,200 = $904,200

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3 years ago
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Mashcka [7]
The right answer for the question that is being asked and shown above is that: "d. AD and AS curves are more horizontal to the multiplier is more effective." Fiscal policy is limited when the slope of the <span>AD and AS curves are more horizontal to the multiplier is more effective.</span>
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The ability to attract and manage _________may be the most important skill a strategic leader must have.
Mars2501 [29]

Answer:

Human Capital

Explanation:

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6 0
1 year ago
Suggest and critically discuss a model that helps managers to decide to adopt an offensive or defensive route. Support your answ
Artist 52 [7]

Answer:

Please find the complete question in the attached file.

Explanation:

In order to study the impact on five forces and thereby decipher an offensive or defense strategy to stay competitive and maintained, management must employ a prototyping approach as Porters 5 Headed framework. Samsung, for example, should adopt a great combination because of its subsequent globalization.

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The threat of new entries – The risk is significantly greater because new entries from low-cost China carriers can eat Samsung share since they are tax- and licensing-friendly.

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6 0
2 years ago
Consider the following linear program: Min s.t. 8X + 12Y 1X + 3Y &gt;= 9 2X + 2Y &gt;= 10 6X + 2Y &gt;= 18 A, B &gt;= 0 a. Use t
mihalych1998 [28]

Answer: Graph of (A) (B) and {D) are attached accordingly.

Explanation:

A)

The critical region of the constraints can be seen in the following diagram -

(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white

The intersection points are found by using these equations -

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 48

(9,0) x+3y = 9; y = 0 72

(2,3) 2x+2y = 10; 6x+2y = 18 52

(0,9) 6x+2y = 18; x = 0 108

So, we can see the minimum value of the objective function occurs at point (3,2) and the minimum value of the objective function is = 48.

------------------------------------------------------------------------------------------------------------------------------------------------------------------

B)

When we change the coefficients of the variables in the objective function, the optimal solution may or may not change as the weights (coefficient) are different for each constraints for both the variabls. So, it all depends on the coefficient of the variables in the constraints.

In this case, the optimal solution does not change on changing the coefficient of X from 8 to 6 in the objective function.

The critical region would remain same (as shown below) as it is defined by the constraints and not the objective function.

(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white

However, the optimal value of the objective function would change as shown below-

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 42

(9,0) x+3y = 9; y = 0 54

(2,3) 2x+2y = 10; 6x+2y = 18 48

(0,9) 6x+2y = 18; x = 0 108

So, we can see that the minimum value now has become 42 (which had to change obviously).

-------------------------------------------------------------------------------------------------------------------------------------------------------

C)

Now, when we change the coefficient of the variable Y from 12 to 6, again the critical region would remain same as earlier. But in this case, the optimal solution changes as shown below -

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 36

(9,0) x+3y = 9; y = 0 72

(2,3) 2x+2y = 10; 6x+2y = 18 34

(0,9) 6x+2y = 18; x = 0 54

We can see that the minimum value now occurs at (2,3) which is 34, so both the optimal solution and optimal value have changed in this case.

----------------------------------------------------------------------------------------------------------------------------------------------------------

D)

When we limit the range of the variables as -

4 \leq X \leq 8 \:\: and\:\: 12\leq Y \leq 24,

the critical region now becomes -

So, the new critical points are (4,12), (4,24), (8,24) and (8,12).

So, the values of the objective function at these points can be calculated as -

Vertex Value of Objective

(4,12) 8*4+12*12 = 176

(4,24) 8*4+12*24 = 320

(8,24) 8*8+12*24 = 352

(8,12) 8*8+12*12 = 208

So, the new optimal solution is (4,12) and the optimal value is 176.

if we knew the range of the variables in the part B and C earlier, we could have just said that the optimal solution will not change as the value would have been no longer depended on the coefficients of variables in the constraints.

7 0
3 years ago
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