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Sophie [7]
3 years ago
11

Chris purchased a 10 year 100 par value bond where 6% coupons are paid semiannually. Cheryl purchased a 100 par value bond where

6% coupons are paid semiannually. There is no maturity date or redemption value for Cheryl’s bond. Cheryl paid $100 for her bond. The yield for Chris’s bond is 80% of the yield for Cheryl’s bond. How much did Chris pay for his bond?
Business
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

The amount Chris pay for his bond = $109.44

Explanation:

Given that:

Chris purchased a 10 year 100 par value bond where 6% coupons are paid semiannually. Cheryl purchased a 100 par value bond where 6% coupons are paid semiannually.

The Price of the Cheryl's bond is 6% given that it is purchased at at par value where 6% coupons are paid.

Suppose The yield for Chris’s bond is 80% of the yield for Cheryl’s bond.

Then:

Price of the Cheryl's bond = Present Value of the coupon in perpetuity

∴

100=\dfrac{3}{Yield}

Yield=\dfrac{100}{3}

Yield =0.03

Yield =  3%

The Yield of Chris = 0.8 × 3

The Yield of Chris =  2.4% semiannual  

However;

Present Value of the coupons is:  PV= \dfrac{A*[ (1+r)^n -1]}{[(1+r)^n * r] }

PV= \dfrac{3*[ (1+0.024)^{20} -1]}{[(1+0.024)^{20} *0.024 ] }

PV= \dfrac{3*[ (1.024)^{20} -1]}{[(1.024)^{20} *0.024 ] }

PV= \dfrac{3*[1.606938044 -1]}{[1.606938044 *0.024 ] }

PV= \dfrac{3*[0.606938044]}{[0.03856651306 ] }

PV= \dfrac{1.820814132}{0.03856651306  }

PV = 47.21

The PV of the face value = \dfrac{100}{(1+r)^n}

The PV of the face value =  \dfrac{100}{(1+0.024)^{20}}

The PV of the face value = \dfrac{100}{(1.024)^{20}}

The PV of the face value = \dfrac{100}{1.606938044}

The PV of the face value = 62.230

Finally:

The amount Chris pay for his bond =  PV of the coupons + PV of the face value

The amount Chris pay for his bond = 47.21 + 62.230

The amount Chris pay for his bond = $109.44

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Explanation:

Solution

From the example given, we solve for which demand location will have an unmet demand

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The Maximum quantity that can be shipped from Harrisburg to Cleveland is 175

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The Maximum quantity that can be shipped from Harrisburg to Dayton is 175

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By applying a least cost method called greedy, we need to remove our least costing node and then provide minimum of demand and supply unit a present to each cell.

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From Harrisburg to Erie 25 units. Next least cost is Harrisburg to Dayton.

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From Harrisburg to Cleveland 75 units.

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Answer:

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