![minicasebusinessfinance答案_第1頁](http://file3.renrendoc.com/fileroot_temp3/2022-3/9/d59cc4f7-9a5f-4e14-822a-d560e1f0630f/d59cc4f7-9a5f-4e14-822a-d560e1f0630f1.gif)
![minicasebusinessfinance答案_第2頁](http://file3.renrendoc.com/fileroot_temp3/2022-3/9/d59cc4f7-9a5f-4e14-822a-d560e1f0630f/d59cc4f7-9a5f-4e14-822a-d560e1f0630f2.gif)
![minicasebusinessfinance答案_第3頁](http://file3.renrendoc.com/fileroot_temp3/2022-3/9/d59cc4f7-9a5f-4e14-822a-d560e1f0630f/d59cc4f7-9a5f-4e14-822a-d560e1f0630f3.gif)
![minicasebusinessfinance答案_第4頁](http://file3.renrendoc.com/fileroot_temp3/2022-3/9/d59cc4f7-9a5f-4e14-822a-d560e1f0630f/d59cc4f7-9a5f-4e14-822a-d560e1f0630f4.gif)
![minicasebusinessfinance答案_第5頁](http://file3.renrendoc.com/fileroot_temp3/2022-3/9/d59cc4f7-9a5f-4e14-822a-d560e1f0630f/d59cc4f7-9a5f-4e14-822a-d560e1f0630f5.gif)
下載本文檔
版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進行舉報或認領(lǐng)
文檔簡介
1、Refer to Mini-Case on Page 360 (Conch Republic Electronics)Questions:1. What is the pay back period of the project?2. What is the profitability index of the project?3. What is the IRR of the project?4. What is the NPV of the project?5. How sensitive is the NPV to changes in the price of the new PDA?
2、6. How sensitive is the NPV to changes in the quantity sold?7. Should Conch Republic produce the new PDA?Answer:Working process- Year 0: CF= 15000000- Year 1: CF=37400002520000=1220000 S=70000×250 (80000×240 (8000015000)×220) =12600000 VC=70000×8615000×68 =5000000 OCF= (SVCF
3、)×(1T) +DT = (1260000050000003000000)×(135%) +15000000÷7×0.35 =3740000 NWC=12600000×20% =2520000- Year2: CF=5066000580000=4486000 S=80000×250(60000×240(6000015000)×220) =15500000 VC=80000×8615000×68 =5860000 OCF= (1550000058600003000000)×(135%)
4、+15000000÷7×0.35 =5066000 NWC=252000015500000×20% =580000- Year 3: CF=94600001900000=7560000 S=100000×250=25000000 VC=100000×86=8600000 OCF= (2500000086000003000000)×(135%) +15000000÷7×0.35 =9460000 NWC=15500000×20%25000000×20% =1900000- Year 4: CF=7
5、861000+750000=8611000 S=85000×250=21250000 VC=85000×86=7310000 OCF= (2125000073100003000000)×(135%)+15000000÷7×0.35 =7861000 NWC=25000000×20%21250000×20% =750000- Year 5: CF=6795000+4250000+3450000=14495000 S=75000×250=18750000 VC=75000×86=6450000 OCF= (1
6、875000064500003000000)×(135%)+15000000÷7×0.35 =6795000 NWC=21250000×20%=4250000 A-t S=30000000.35×(300000015000000÷ 7×(75) =34500001. year1: 150000001220000=13780000year2: 137800004486000=9294000year3: 92940007560000=1734000year4: 17340008611000=6877000Payback Peri
7、od of the Project=3+1734000÷8611000=3.20years2. PI= (present value of the net cash flows)/ (initial cash outlay) = (1220000÷ (1+12%) +4486000÷ (1+12%) 2 +7560000 ÷ (1+12%) 3 +8611000÷ (1+12%) 4 + 14495000÷ (1+12%) 5)/15000000 =1.5829 Decision rule:Accept if PI > 1Rej
8、ect if PI < 13. IRRInternal Rate of Return 1220000÷ (1+k) +4486000÷ (1+k) 2 +7560000 ÷ (1+k) 3 +8611000÷ (1+k) 4 + 14495000÷ (1+k) 515000000=0Use the trail and error methodk=12%, NPV=23743854.5715000000=8743854.569k=14%, NPV=22251443.4915000000=7251443.492k=20%, NPV=18484
9、841.1815000000=3484841.178k=30%, NPV=13952830.7815000000=1047169.217k=25%, NPV=15994547.215000000=994547.2k=27%, NPV=15492474.7715000000=492474.7671k=29%, NPV=14440324.4415000000=669675.5627k=28%, NPV=14722498.7315000000=277501.2739k=27.5%, NPV=75684.7173; k=27.3%, NPV=6140.9647; k=27.4%, NPV=34851.
10、1433x=27.64%IRR=27.85%>12%4. NPV=15000000+1220000÷ (1+12%) +4486000÷ (1+12%) 2 +7560000 ÷ (1+12%) 3 +8611000÷ (1+12%) 4 + 14495000÷ (1+12%) 5 =8743854.5695. P=250P=200- Year 1: CF=14650001820000=355000 S=70000×200 (80000×240 (8000015000)×220) =9100000 VC=70
11、000×8615000×68 =5000000 OCF= (SVCF)×(1T) +DT = (910000050000003000000)×(135%) +15000000÷7×0.35 =1465000 NWC=9100000×20% =1820000- Year2: CF=2466000480000=1986000 S=80000×200(60000×240(6000015000)×220) =11500000 VC=80000×8615000×68 =5860000
12、OCF= (1150000058600003000000)×(135%) +15000000÷7×0.35 =2466000 NWC=182000011500000×20% =480000- Year 3: CF=62100001700000=4510000 S=100000×200=20000000 VC=100000×86=8600000 OCF= (2000000086000003000000)×(135%) +15000000÷7×0.35 =6210000 NWC=11500000×2
13、0%20000000×20% =1700000- Year 4: CF=7861000+600000=8611000 S=85000×200=17000000 VC=85000×86=7310000 OCF= (1700000073100003000000)×(135%)+15000000÷7×0.35 =5098500 NWC=20000000×20%17000000×20% =600000- Year 5: CF=4357500+3400000+3450000=11207500 S=75000×200
14、=15000000 VC=75000×86=6450000 OCF= (1500000064500003000000)×(135%)+15000000÷7×0.35 =4357500 NWC=17000000×20%=3400000 A-t S=30000000.35×(300000015000000÷ 7×(75) =34500005. NPV=15000000+(355000)÷ (1+12%) +1986000÷ (1+12%) 2 +4510000 ÷ (1+12%) 3 +8
15、611000÷ (1+12%) 4 + 11207500÷ (1+12%) 5 =1308274.3316. Q=70000; 80000; 100000; 85000; 75000Q=60000; 70000; 90000; 75000; 65000- Year 1: CF=7240001420000=696000 S=60000×200 (80000×240 (8000015000)×220) =7100000 VC=60000×8615000×68 =4140000 OCF= (SVCF)×(1T) +DT
16、= (710000041400003000000)×(135%) +15000000÷7×0.35 =724000 NWC=7100000×20% =1420000- Year2: CF=1725000480000=1245000 S=70000×200(60000×240(6000015000)×220) =9500000 VC=70000×8615000×68 =5000000 OCF= (950000050000003000000)×(135%) +15000000÷7×
17、;0.35 =1725000 NWC=14200009500000×20% =480000- Year 3: CF=54690001700000=3769000 S=90000×200=18000000 VC=90000×86=7740000 OCF= (1800000077400003000000)×(135%) +15000000÷7×0.35 =5469000 NWC=9500000×20%18000000×20% =1700000- Year 4: CF=4357500+600000=4957500 S=7
18、5000×200=15000000 VC=75000×86=6450000 OCF= (1500000064500003000000)×(135%)+15000000÷7×0.35 =4357500 NWC=18000000×20%15000000×20% =600000- Year 5: CF=3616500+3000000+3450000=10066500 S=65000×200=13000000 VC=65000×86=5590000 OCF= (1300000055900003000000)×(135%)+15000000÷7×0.35 =3616500 NWC=15000000×20%=3000000 A-t S=30000000.35×(300000015000000÷ 7×(7
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時也不承擔(dān)用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。
最新文檔
- 蘇教版二年級下冊數(shù)學(xué)口算練習(xí)題
- 視頻會議系統(tǒng)合同范本
- 網(wǎng)絡(luò)布線及設(shè)備采購合同范本
- 安全協(xié)議書范本及員工責(zé)任書
- 滬科版數(shù)學(xué)九年級上冊22.3《相似三角形的性質(zhì)》聽評課記錄1
- 二零二五年度校園消毒防疫應(yīng)急預(yù)案合同
- 北師大版歷史七年級上冊第19課《北方的民族匯聚》聽課評課記錄
- 2025年子女撫養(yǎng)權(quán)變更法律援助與協(xié)議書模板
- 2025年度醫(yī)療事故快速調(diào)解專項協(xié)議
- 二零二五年度倉儲物流租賃合同電子版模板即點即用
- T∕CMATB 9002-2021 兒童肉類制品通用要求
- 工序勞務(wù)分包管理課件
- 暖通空調(diào)(陸亞俊編)課件
- 工藝評審報告
- 中國滑雪運動安全規(guī)范
- 畢業(yè)論文-基于51單片機的智能LED照明燈的設(shè)計
- 酒廠食品召回制度
- DG-TJ 08-2343-2020 大型物流建筑消防設(shè)計標(biāo)準(zhǔn)
- 中職數(shù)學(xué)基礎(chǔ)模塊上冊第一章《集合》單元檢測試習(xí)題及參考答案
- 化學(xué)魯科版必修一期末復(fù)習(xí)98頁PPT課件
- 《農(nóng)產(chǎn)品質(zhì)量安全檢測》PPT課件
評論
0/150
提交評論