Q1
(a)
Twenty enty (20) (20) year yearss ago, ago, your your com company pany has has purc purcha hassed a smal smalll fact factor ory y building costing RM 300,000. owe!er, boo" !alue of the building remains RM 230,000 only due to the decrease in mar"et !alue. The factory is sold at the price of RM #$0,000. %etermine the !alue from each of the following cost&
(i)
'ash 'ost (2 mar"s)
(ii)
oo" 'ost (2 mar"s)
(iii)
un" 'ost (2 mar"s)
(i!)
*pportunity 'o 'ost (2 mar"s)
(!)
tandard 'o 'ost (2 mar"s) Solution:
i.
ii.
iii.
i!.
!.
'ash 'ost uildin ding - 'ost of the factory buil
+
RM 300,000 000 ..... .......... 2 mar" ar"s
+ +
RM 300, 000 230,000 -0,000 . .............. 2 mar"s
+ +
230,000 / #$0,000 RM 0,000 .......... 2 mar"s
*pportunity 'ost - 1alue of building sold
+
RM #$0,000 ........ 2 mar"s
tandard 'ost - ie ied d %epr %eprec eciiatio ation n
+
RM -,00 -,000 0 .... ...... .... .. .. 2 mar"s ar"s
oo" 'ost - Total decrease !alue un" 'ost - oo" !alue less old prices
45 30602
(b)
7 group of 88 team is ma"ing analysis in a decision to produce a new product with two alternati!e plants. %etails are as follows9 Details :abour 'ost (RM per unit) Raw Material 'osts (RM per unit) actory Manager(RM per month) Rental (RM per month) elling 4rice (RM per unit)
(i)
Plant A Plant B 3;.00 3$.00 -3.00 $2.00 ;,;00 ,$00 -,$00 ;,300 #0.00 #0.00
'alculate the Total 1ariable 'ost (1')< ied 'ost (') (= mar"s)
(ii)
%etermine the rea"e!en 4ointunits per month (= mar"s)
(iii)
uggest the most profitable plant for production if the consumer demand is 00 units per month. (= mar"s) (i!)
The selling price has reduced by RM.00 for site chosen in (iii). 'alculate how many units to be sold if the company is to maintain a profit le!el of RM2,$$0.00 (3 mar"s)
Solution: Details
#.
Plant A
Plant B
Variable Cost (VC) :abour 'ost (RM per unit) Raw Material 'osts (RM per unit) Fixed Cost (FC) Rental (RM per month) actory Manager(RM per month)
2.
elling 4rice (RM per unit) rea"e!en 84 + '> (4 1')
3;.00 -3.00
3$.00 $2.00
109.00
120.00
;,;00 -,$00
,$00 ;,300
14,400
12,100
#0.00 84 + '> (4 1') + #=,=00>(#0 #06) + 351.2195 units
2
#0.00 84 + '> (4 1') + #2,#00>(#0 #20) + 403.3333 units
45 30602
3.
4rofitable site Re!enue RM#0 00 + RM-,000
=
4 #0 to #= 2,$$0>(#= /#06) + $0 units =$0 #= + ;6,;00
Production Cost (109 500! " 14,400 # $%,900 Pro&it # ' ) 'C # *5,000 ) $%,900 # $,100 Selected +lant due to i-er +ro&it lo/er BP
Production Cost (120 500! " 12,100 # *2,100 Pro&it # ' ) 'C # *5,000 ) *2,100 # 2,900
84 + '> (4 1')
84 + '> (4 1')
+ #=,=00>(#= #06) + #2,#00>(#= #20) + =00units + =$= units 84 ? 4rofit :e!el 84 ? 4rofit :e!el + =00 ? $0 + =$= ? $0 # 4%0 units # 5$4 units (=$0 #06) ? #=,=00 ;;,-20 4rofit :e!el + TR T'
+ ;6,;00 ;;,-20 + 2,$$0 Q2(a)
The table @2 (a) below shows the past price of tandard Malaysia Rubber (MR) since 20#2, whereby 20#3 is the reference year ha!ing 2=; as an inde !alue. The weight place on MR '1 is one (#) time, MR : is one and half (#.) times and MR is two (2) time 'ale Q2 (a!
MR
4rice (sen > "g ) in Aear 20#2
20#3
20#=
MR '1
#0$$
6#6
-3
MR :
#0=;
$32
;6;
MR
6-=
-6=
-6
#. 'alculate a weighted inde for the price of a "g of MR in 20#=. (= mar"s) 2. 'alculate the corresponding 20# prices of MR from 20#= if 2#$ is the inde !alue in 20#. (; mar"s)
Solution:
3
45 30602
B# ('n#>'"2) ? B2 ('n2>'"2) ? B3 ('n3>'"3) ////////////////////////////////////////////////////////// D C B# ? B2 ? B3
Cn20#= +
n20##
#(-3>6#6) ? #.(;6;>$32) ? 2(-6>-6=) ///////////////////////////////////////////////////////// D 2=; + #63.##;3 # ? #. ? 2 0.$#63 ? #.2=$ ? #.=$= /////////////////////////////// D 2=; + #63.##00 =. #63.##;3 *R #63.##00 . (= mar"s)
+ +
+
'n20#=MR '1 +
'"20## (Cn20#3>C"20##) + -3(2#$>#63.##;3) + $0.02;; sen>"g
'n20#MR :
+ + +
'n20#3MR +
'"20## (Cn20#3>C"20##) + -6(2#$>#63.##;3) + ;3.;0;# sen>"g
(b)
...... (2 mar"s)
'"20## (Cn20#3>C"20##) ;6; (2#$>#63.##;3) -$.;$20 sen>"g ...... (2 mar"s)
... (2mar"s)
The structural engineering design section of 7gile 8/4ower, a multinational electrical utility corporation has de!eloped se!eral standard designs for a group of similar transmission line towers. The detailed design for each tower is based on one of the standard designs. 7 transmission line proEect in!ol!ing ;0 towers has been appro!ed. The estimated number of engineering hours needed to accomplish the first detailed tower design is #23. %etermine9 (i)
The number of engineering hours needed to design theeight and siteenth tower using a 95% learning curve. ( mar"s)
(ii)
The reduction percentage when the production is doubled. ( mar"s)
(iii)
The estimated cumulati!e a!erage hours reFuired to produce the first fi!etower designs. ( mar"s)
S'6:
=
45 30602
(i)
5 + #23 hours s + 0.6 (6G learning cur!e) n + (log 0.6) > (log 2) + /0.0-= H$ + #23($)/0.0-= + #0.=; hours H#; + #23(#;)/0.0-= + #00.#$ hours
(ii)
........................................... (# mar"s) ......................................... (2 mar"s) ......................................... (2 mar"s)
(H$ / H#;) > H$ + (#0.=;/#00.#$)>#0.=; + 0.0 7ssumption > conclusion9 The 6G learning cur!e results in a G reduction in number of engineering hours each time the Fuantity of the tower needed to be designed is doubled. ( mar"s)
(iii)
T
+ #23 I#0 J+#u/0.0-= + #23 K#/0.0-= ? 2/0.0-=? 3/0.0-=? =/0.0-=? /0.0-=L + #23K# ? 0.600 ? 0.62#6 ? 0.3$ ? 0.3036L + #23 K3.3=3L + =3=.-2 hours ........................................... (3 mar"s)
'
+ T > + T> + =3=.-2 > + $;.6= hours
................... (2 mar"s)
