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Unformatted text preview: First Test; IME 3100; Wednesday May 27, 2009 m 9.63
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Problem 1
The total relevant cost of holding inventory in a plant for purchased materials is given, W ' glq TRC = {6000!Q) (60) + 0.30( 12) (Qf2} + 600002), where 'Q = economic order quantity.
Determine the optimum order quantity and the Corresponding total relevant cost (TRC).
Solution: 'I'RC : (600010) (60) + 0.30(12)(Q‘;2)+ 6000(12‘) Diﬁ‘erentialing TRC with respect to Q, we get 0 (“racde = {360,000} (3: + 1.3 Q = (436000003) =44? TRC = (6000104?) (60) + 03002) (44712) + 6000(12) = $73,610 Problem 2 A highqech company in the US can. have one of the popular home made offshore at half the price of making them in
the US. However. about 90% of the items made oﬁ'shore will he returned within the warranty period of 1 year for
repairs. The following additional data are available. Cost of making an item in the US = $250 Cost of repairing an item = $50 Number 30101r year = 500,000 Conmute the cost savings that the company makes per year by making the item oﬁithore. Solution: Cost savings per item = $250015 0)  0.90 (50): $80 Cost savings per year = $500,000(80) = $40,000,000 Problem 3 m [f the time required for the ﬁrst and eighth units are 7.29 hours and 10' hours respectively in a production operation
determine the learning curve percentage for this operation. A. 72.9%13. 95% C. 81% D. 90% Solution: T3: T1 (8)1; 7.29 = 10(3):.13 = 0.152 Log (xvlog 2 = 0.152 x= 90%
The answer is “D“. Problem 4 Pace Tech Engineering Inc manufactures an electronic toy. The company has a building with a ﬂoor area of 180,000 gguare feet for both the manufacturingand warehousing operations. The company produces all of the
components that gointo the toy. Variable costs are estimated to be $_20_pe_r unit, and ﬁxed costs per year are $108 'STheselﬁzugew
l} The bre even quantity for the item assembled in the plant is 5 22 . 2) The proﬁt made by'the company, if it sold 1,00'01mits in a year is g 1L5 .3’ [32
3) Determine the number of unitsz’year required to be sold to make a proﬁt of$25.000.. Solution: 95' l 9
Let “X“ be the breakeven volume. ' ' ‘—'/ 39X = 10,375 + 2015((39  20) X = 10,375 X = [email protected] ‘_
[loco572) 7W4 "'
at? '0 'l Proﬁt for 1000 units = 390,000) — {10,375 + 20( 1000)} = $3,125 Let “Y" be the number of units required to be sold to make a proﬁt of $25,000. q .—
3931410375 3 21m = 25,000 19v = 25,000 + 10,375 25. °° ° I ‘ " Problem 5
An entrepreneur is considering opening; a‘cofl‘ee shop in downtown Cookeville. The building that he is considering will have a monthly lease payment of $3200 and basic utility costs of $600 per month. 'I‘wo employees will be hired at $10.00fhourz'emloyge (including am and beneﬁts). Each employee will work an average of _170 hours per
month. The average revenue per custorner is estimated at $7. 00. The variable cost of serving each customer is estimated at $2.
(a) Calculate ow many customers per month it will take for the coﬂ‘ee shop owner to breakeven. (b) How many customers would need to be served per month to achieves monthly proﬁt of$5,000?
Solution: (a)AtbrealteVen, rc=m 3200* Tc=3,2oo +500+2(10) (170)+2X'I'R=?X 6°91 Setting TC=TR. o o... _1qoo 3.200+600+21   — oo+2x=7x 2.1V? “La—3" 7.200=5xx 1,440custornersfmonth \/ 1.1" 77¢ \Hw
(h)TRTC=5, “ vest“i as 2° 1.; 7x  (3.200 + see + 2(10).(170)+ 2X) = 5.000 7  77,9 0 7x  2): = 5,000 5x = 12,200 5 +~ ’ 1.0%
= 2.440 C I ’ 1' MW mersfmonth \/ 5000" 5 . al 1 Problem 6
How much would you need to invest at 6% interest compounded quarterly on Jarurary l, 2008 in order to accumulate $10,000 by December 31, 2010?
Solution: F= $10000. =1f2008 tol
P= F (PIP. i. n) =10, 000 (0. 8364) — $3 3'64. 00_ '
Problem 7 Gene Milton borrowed a sum of $5,000 fromhis uncle Ben and after three years paid a sum of $5,000 and paid
another $1000 after 4 years to pay oﬂ' the loan. Determine the interest rate Gene paid if the payments were based on yearly compounding \g a 0
Solution: ==' ‘5,” 5.000= 5,000(PfF, t, 3) +1,000(PIF,1, 4) 0 Try i = 6% 5,000 = 5.000(08396) + 1,000(0.?921) = $4,990.10? N0 '5 ‘i Try i = 5% _ 5,000 = 5,000(0.8633)&1.8227} = $5,141.?0? N0 {000 htterpolating. We get i 5.93%. , Problem 8 You are repaying a debt of $10 000 with four equal end of the year payments Ifthe interest rate is 10% per year.
how match of the original $10 000 principal will be paid 1n the second payment? 1w] aoD Solution: ‘ ”/
Principal = $10,000 Number of payments, n= 4 % per period A =P(AfP.i,n)=10,000(AfP,10%.4)=10 000 (o [email protected] % E AI : .E
First payment: interest on unpaid balance— — 10 000 x 0 10 — 000 Principal repaid = A  1,000 = $2,155.00 A11 is“? Second payment: Principal due = 10,000  2,155 = $?,845
Interest on unpaid balance = 7,84S(0.10) = $784.50
Principal paid in. the second payment = 3,155  784.50 $2,370.50 1
4' = 5 "r/ 17 Problem 9 u: l t 0
Assume that a piece ofproperty is purchased for $200 000. A 20% down payment is made and the rest is ﬁnanced through a 15year mortgage loan with a S ‘xl % annual interest rate, cOmpounded monthly. The loan “rim
in equal monthly payments. Calculate the monthly payments.
What is the total interest paid through the life of the loan? hot ‘7 Solution: A a ?
Down payment— _ 0. 20(200, 000)= $40, 000 Amount ofloan= 200, 000 40 ,000— — ' : E l I I Monthly payments A— — 160 000 (JVP, 0.4 0) — 160, 000 _.(0 0008039 = $1, 236. 22
interest paid = tam lessen—mono = 37152121 Problem 10 A young "scientist has just'stnrted her career with a federal government agency. She has go'odjob security and
expects consistent 5% annual salary increases. Her starting annual salary is $30,000; She plans to save 10% of her
monthly salary at a bank, Whitih pays 5.4%. internist, compounded monthly. At the end of each year, she will put the
accumulated savings into a money market fund that is expected to pay. 8% interest at the end of each year. Calculate
the amount of money he expects to have alter ten years. = '5"
Solution: Monthly salary during the ﬁrst year = 30,000! 1'2 = $2,500
A = 0.10 (2,500) = 3250(during the ﬁrst year) interest ratefmonth = 54%! 12 = 0.45% 4: g S .t,‘ I. In
A: = 250 (FZA, 0.45%, 12) = 250(12.3015) = $1075.38. Interestper year = 8% g = 5% P 2 m [(1  {(1 + g)..(1+i)..}] x (i  g) = 3,075.38 . .03) no}! (nos  0.05)] = $25,167.54
F = 25,167.54(Fm, an, 10) = 22151540159 _= $54,334.33. __________...' Problem 1]
Twelve deposits of 5 1,000 are made at the end of every quarter at an interest rate of 10% compounded quarterly and
another deposit of $2,000 is {nude every six months at an interest rate of 12% for three years. ‘ \ 1 i. What is the total accumulation a person would have after three years?
ii. What is the effective interest rate in'both the cases? Solution: _ _ _ m ‘3: 33;
i. F = 1.000(FR'A, 2 5’1 “/13, 12) + 2 000(FKA, 6 Va, 6) = 1,000(13.796)a1 5' 27,746 ii.i(1+0.92)+1=09g4 @ i.(1+ 0.0632—1=o.123 k... A 5““; «t::r’
Problem 12 it
U“ “'1‘," = t hwy.
An adjustable interest loan was taken by Raleigh Waters for a sum of $20,000. The monthly interest rate was With a.‘ 5 £7_
for the ﬁrsttiﬂ months. For the remaining 60 months, the interest rate was reduced to 'é%. A a e 1:0“
Determine the monthly payments for the ﬁrst 60 months and the last 60 months. Solution:
Monthly payment for theftrst 60 months = 20.0mm. 324%, 120) = zuocowmzv) on; 2W Loan balance inunediately alter the 60opayment = 254(PHA, 3/4013, 60) = 254(.48.l74) : 2 12,2 6:2]
Monthly payment for the last 60 months =12,236.20(NP, V: “/0, 60) =12,236.20 (0.0193. 35 236.16
Problem [3 Determine the effective interest rate for the following cases. i. 10% compounded weekly ii. 12% compounded daily
iii. 9% compounded continuously Solution: i. Eﬁ‘eetive interest rate, ia'= (1 + 0.1053524 = 0.1051 1.0.5 ° ii. Effective interest rate, ia= (I + 0.12865)er .127 2 12.71%
iii. Eﬁ'ective interest rate, in = ears 1 = 0.094 = 9.42% ' Problem 14 You have just deposited $1,000 in an investment that promises to pay you $100 per year for the next 20 years
(starting at the end of this year). What is the effective annual interest rate earned for this investment, assuming
annual compounding? As always, assume your account balance will be zero at the end of 20 years (i.'e., your investment will he ﬁdly amortized). Show all work! A 7 " 9 °
Solution: _ it
P =$t,000 A = $100 n=20 years
1,000 =100(P!A, i, 20) (PIA, i, 20) = LODGING = 10 1 ’0
i (PIA, i, 20}
2% 10.594 loco
8% 9.818
On interpolation, we [email protected]
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Problem 15
For the cash ﬂows shown in table below, if the interest rate is 9%, what is the value of “X"? 1
Year: 0 I 2 3 4 5
Cash Flows: 5' X 1,000 300 600 400 1,000
Solution: “a“ X: l,000(.P'/A,9%, 5)200(P}’G, 9%, 5) 1,2' , , g... 6»
= 1,0000390) 4000.11 1)  t,200(0.64‘99) =€ $1,687.92 3 to" ...
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