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jojo · 2025年04月14日

这题默认AUD和USD的本金都是1million吗?

NO.PZ2019052801000050

问题如下:

A US company entered into a one-year currency swap with quarterly reset six months ago. The notional principle is $1,000,000, At the swap’s initiation, the US company receives the notional amount in Australian dollars and pays to the counterparty the notional amount in US dollars. At the swap’s expiration, the US company pays the notional amount in Australian dollars and receives from the counterparty the notional amount in US dollars.The annual fixed swap rates for Australian dollars is 4% and for US dollars is 3.6%.The current spot exchange rate is A$1.2 / $ .

The US term structure is:

  • r(90)=3.58%

  • r(180)= 3.74%

The Australian term structure is:

  • r(90)=3.82%
  • r(180)= 4.1%

What is the value of the currency swap to US company?

选项:

A.

$-142,145million.

B.

$142,145million.

C.

$166 ,385.

D.

$-166 ,385.

解释:

C is correct.

考点:货币互换估值.

解析:

美国公司收美元本金和利息的价值:

lB  $  =0.009e0.0358×0.25+1.009e0.0374×0.5=0.999227{l}B_{\;\$}\;=0.009e^{-0.0358\times0.25}+1.009e^{-0.0374\times0.5}\\=0.999227

美国公司支澳大利亚元本金和利息的价值:

lB  A$  =0.01e0.0382×0.25+1.01e0.041×0.5=0.999411{l}B_{\;A\$}\;=0.01e^{-0.0382\times0.25}+1.01e^{-0.041\times0.5}\\=0.999411

lV=(0.9992270.999411÷1.2)×1,0000,000=166,385{l}V=(0.999227-0.999411\div1.2)\times1,0000,000=166,385

  1. 我看了一下 Example & Other Currency Swaps的例题,题干是有说不同currenc所对应的NP。这里没说就是默认NP都是1mm咯?
  2. 货币互换是不是期初交换(1mm的AUD和1mm的USD),期末再换回来,名义本金的金额不变?
  3. 我的图是ok的吧?最后的答案差了一点点,应该是小数点之后少取了几位。



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NO.PZ2019052801000050问题如下A US company entereinto a one-yecurrenswwith quarterly reset six months ago. The notionprinciple is $1,000,000, the swap’s initiation, the US company receives the notionamount in Australillars anpays to the counterparty the notionamount in US llars. the swap’s expiration, the US company pays the notionamount in Australillars anreceives from the counterparty the notionamount in US llars.The annufixeswrates for Australillars is 4% anfor US llars is 3.6%.The current spot exchange rate is A$1.2 / $ . The US term structure is: r(90)=3.58% r(180)= 3.74% The Australiterm structure is: r(90)=3.82% r(180)= 4.1%Whis the value of the currenswto US company? A.$-142,145million.B.$142,145million.C.$166 ,385.$-166 ,385. C is correct. 考点货币互换估值.解析美国公司收美元本金和利息的价值lB  $  =0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227{l}B_{\;\$}\;=0.009e^{-0.0358\times0.25}+1.009e^{-0.0374\times0.5}\\=0.999227lB$​=0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227美国公司支澳大利亚元本金和利息的价值lB  A$  =0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411{l}B_{\;A\$}\;=0.01e^{-0.0382\times0.25}+1.01e^{-0.041\times0.5}\\=0.999411lBA$​=0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385{l}V=(0.999227-0.999411\v1.2)\times1,0000,000=166,385lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385 1、“本题是1年期的swap,每季度交换一次,已经过了半年了,求value的问题”求的不是0时刻签约的value?那是求1年到期,现在的value?2、FRM考试只要给了利率没说怎么计息,就用连续复利吗?

2024-07-08 21:16 3 · 回答

NO.PZ2019052801000050 问题如下 A US company entereinto a one-yecurrenswwith quarterly reset six months ago. The notionprinciple is $1,000,000, the swap’s initiation, the US company receives the notionamount in Australillars anpays to the counterparty the notionamount in US llars. the swap’s expiration, the US company pays the notionamount in Australillars anreceives from the counterparty the notionamount in US llars.The annufixeswrates for Australillars is 4% anfor US llars is 3.6%.The current spot exchange rate is A$1.2 / $ . The US term structure is: r(90)=3.58% r(180)= 3.74% The Australiterm structure is: r(90)=3.82% r(180)= 4.1%Whis the value of the currenswto US company? A.$-142,145million. B.$142,145million. C.$166 ,385. $-166 ,385. C is correct. 考点货币互换估值.解析美国公司收美元本金和利息的价值lB  $  =0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227{l}B_{\;\$}\;=0.009e^{-0.0358\times0.25}+1.009e^{-0.0374\times0.5}\\=0.999227lB$​=0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227美国公司支澳大利亚元本金和利息的价值lB  A$  =0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411{l}B_{\;A\$}\;=0.01e^{-0.0382\times0.25}+1.01e^{-0.041\times0.5}\\=0.999411lBA$​=0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385{l}V=(0.999227-0.999411\v1.2)\times1,0000,000=166,385lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385 老师按照基础班讲义的238和239页的原理来说,美国公司在期初收A,付US在期末就要付A,收美元;老师上课的时候说本金的交换方向和收利息的方向是反过来的嘛,但是这里为啥本金和利息的折现都是同一个方向呢?

2023-09-21 20:39 2 · 回答

NO.PZ2019052801000050问题如下 A US company entereinto a one-yecurrenswwith quarterly reset six months ago. The notionprinciple is $1,000,000, the swap’s initiation, the US company receives the notionamount in Australillars anpays to the counterparty the notionamount in US llars. the swap’s expiration, the US company pays the notionamount in Australillars anreceives from the counterparty the notionamount in US llars.The annufixeswrates for Australillars is 4% anfor US llars is 3.6%.The current spot exchange rate is A$1.2 / $ . The US term structure is: r(90)=3.58% r(180)= 3.74% The Australiterm structure is: r(90)=3.82% r(180)= 4.1%Whis the value of the currenswto US company? A.$-142,145million.B.$142,145million.C.$166 ,385.$-166 ,385.C is correct. 考点货币互换估值.解析期初和期末美国公司收美元本金和利息的价值lB  $  =0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227{l}B_{\;\$}\;=0.009e^{-0.0358\times0.25}+1.009e^{-0.0374\times0.5}\\=0.999227lB$​=0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227期初和期末美国公司支澳大利亚元本金和利息的价值lB  A$  =0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411{l}B_{\;A\$}\;=0.01e^{-0.0382\times0.25}+1.01e^{-0.041\times0.5}\\=0.999411lBA$​=0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385{l}V=(0.999227-0.999411\v1.2)\times1,0000,000=166,385lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385 请问这个是基础课程哪里的知识点,这个答案公式有什么含义

2023-04-03 14:16 1 · 回答

NO.PZ2019052801000050 问题如下 A US company entereinto a one-yecurrenswwith quarterly reset six months ago. The notionprinciple is $1,000,000, the swap’s initiation, the US company receives the notionamount in Australillars anpays to the counterparty the notionamount in US llars. the swap’s expiration, the US company pays the notionamount in Australillars anreceives from the counterparty the notionamount in US llars.The annufixeswrates for Australillars is 4% anfor US llars is 3.6%.The current spot exchange rate is A$1.2 / $ . The US term structure is: r(90)=3.58% r(180)= 3.74% The Australiterm structure is: r(90)=3.82% r(180)= 4.1%Whis the value of the currenswto US company? A.$-142,145million. B.$142,145million. C.$166 ,385. $-166 ,385. C is correct. 考点货币互换估值.解析期初和期末美国公司收美元本金和利息的价值lB  $  =0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227{l}B_{\;\$}\;=0.009e^{-0.0358\times0.25}+1.009e^{-0.0374\times0.5}\\=0.999227lB$​=0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227期初和期末美国公司支澳大利亚元本金和利息的价值lB  A$  =0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411{l}B_{\;A\$}\;=0.01e^{-0.0382\times0.25}+1.01e^{-0.041\times0.5}\\=0.999411lBA$​=0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385{l}V=(0.999227-0.999411\v1.2)\times1,0000,000=166,385lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385 (1)0.009,1.009,0.01,1.01怎么来的?(2)年末本金不回收吗,如果回收了在公式哪里表现了呢(3)4%和3.6%没用到吗

2023-01-26 13:33 2 · 回答

NO.PZ2019052801000050 问题如下 A US company entereinto a one-yecurrenswwith quarterly reset six months ago. The notionprinciple is $1,000,000, the swap’s initiation, the US company receives the notionamount in Australillars anpays to the counterparty the notionamount in US llars. the swap’s expiration, the US company pays the notionamount in Australillars anreceives from the counterparty the notionamount in US llars.The annufixeswrates for Australillars is 4% anfor US llars is 3.6%.The current spot exchange rate is A$1.2 / $ . The US term structure is: r(90)=3.58% r(180)= 3.74% The Australiterm structure is: r(90)=3.82% r(180)= 4.1%Whis the value of the currenswto US company? A.$-142,145million. B.$142,145million. C.$166 ,385. $-166 ,385. C is correct. 考点货币互换估值.解析期初和期末美国公司收美元本金和利息的价值lB  $  =0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227{l}B_{\;\$}\;=0.009e^{-0.0358\times0.25}+1.009e^{-0.0374\times0.5}\\=0.999227lB$​=0.009e−0.0358×0.25+1.009e−0.0374×0.5=0.999227期初和期末美国公司支澳大利亚元本金和利息的价值lB  A$  =0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411{l}B_{\;A\$}\;=0.01e^{-0.0382\times0.25}+1.01e^{-0.041\times0.5}\\=0.999411lBA$​=0.01e−0.0382×0.25+1.01e−0.041×0.5=0.999411lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385{l}V=(0.999227-0.999411\v1.2)\times1,0000,000=166,385lV=(0.999227−0.999411÷1.2)×1,0000,000=166,385 请问为什么这里用连续复利求解呢,谢谢

2022-09-21 16:40 3 · 回答