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Unformatted text preview: UNIVERSITY EXAMINATIONS UNIVERSITEITSEKSAMENS
Q
U N I SA Mes“
DSC3703 Maleune 2013
SIMULATION Duratlon 2 Hours 90 Marks
EXAMINERS ' FIRST DR S MUKERU SECOND MS J LE ROUX EXTERNAL DR A DE VILLIERS Programmable pocket calculator Is permismble. Closed book examination. This examination questnon paper remains the property of the Unwersuty of South Africa and may not be
removed from the examination venue. This paper consists of 4 pages plus a list of formulae (pp. i  ii). ANSWER ALL THE QUESTIONS Question 1 Suppose you need to generate ﬁve random numbers from U [0, 1) (a) Prepare to do the by usmg the muted congruentlal method X0 = 14, Xn+1 = (13Xn + 15) (mod 32) for n = 0,1,2, to generate a sequence of ﬁve random mteger numbers between 0 and 31 [5]
(b) Convert these random mteger numbers to obtaln random numbers from U[0, 1) [1]
(c) Determlne the perlod of thlS generator Substantlate your answer [1] [TURN OVER] DSC3703
2 May/ J un 2013 Question 2 @ Janet Dawes rs a purchaser for a factory that produces clothes To ensure its supply for the
upcommg year, her company must s1gn a contract wrth one of several textile supphers After a
few enqurrres1 she has narrowed the ch01ce down to two spec1ﬁc supphers The ﬁrst w1ll supply
as much fabrlc as she needs durlng the upcoming year for R20 per metre and W111 guarantee
supply throughout the year at the prroe The second supplier has a prlce schedule that depends
on how much Janet orders over the next year It IS as follows Orders in meters Prlce per metre 1n rand Fast 20 000 metres 21
Next 10 000 metres 19 Next 10 000 metres 17
Thereafter 15 After carefully cons1der1ng the uncertamty about sales from the new garment hue, Janet decides
to model her uncertalnty about the amount of fabric requlred over the next year as a lognormal
distribution LN (a, 02) where ,u = 10 and a = 0,5 Both supphers are w11hng to supply the
amount of fabrlc that Janet Will need Suppose you are asked to adv15e Janet on wh1ch suppher rs the most costeffectrve Opthn (that
13, wrth the mmunum cost) If we denote by D the amount of fabrrc (1n meters) that Janet
Will need next year, then D ~ LN (p, 02) Denote by C1 the total cost 1f the ﬁrst suppher rs
consrdered and by 02 the total cost if the second suppher 1s considered Answer the followmg quest1ons (a) Which type of srmulatron 1s su1table to model th1s problem (Monte Carlo snnulatron or
dynanuc sunulation)? Justify your answer [2] (b) Is 1t sultable to model this problem by a determlnlstlc simulatlon model or by a stochastic srmulatlon model‘7 Justify your answer [2]
(c) Prove that 01 = 20D and
2113 If 0 g D < 20000 420 000 + 19(D — 20000) 1f 20 000 5 D < 30 000
610000 + 17(D — 30 000) 1f 30 000 S D < 40 000
780000 + 15(1) —— 40 000) If D 2 40 000 02 = {21 (d) Usmg the random numbers U1, U2, , from Appendix A, generate three values of D and
determine the correspondmg costs for both supphers Ind1cate, for each of these values of
D, which suppher you would adv1se Janet to choose Show all your worklngs Are you
satlsﬁed Wlth the result? J ustrfy your answer [15} (e) In order to dec1de which suppller should be chosen, a s1mulatron model for th1s problem IS
run for 10 000 consecutive Iterations The values of the dlfference 02 w Cl are collected for
each Iteration The correspondrng mean 15 1 019,68 and the staude dewatlon is 36 760,76
Determine a. 95% conﬁdence 1nterval for that dlfference Whrch suppher would you adVISe
Janet to choose? Substantiate your answer [4] [TURN OVER] DSC3703
3 May/ Jun 2013 Question 3 A small reparr shop spec1ahses 1n repalrmg German cars The shop has one techn1c1an The
tune requlred to repan a car has an exponent1al dlstrlbutlon w1th mean 0,2 days The shop’s
busmess has been steadlly moreasmg The shop’s management pI'OJeCtS that, by next year,
German cars Wlll arr1ve randomly to be reparred at a mean rate of 4 cars per day, so the tlme
between arrlvals w111 have an exponential dlstr1butlon w1th mean 0,25 days The management
would hke the expected the total tune 1n the shop unt11 completlon of a repau‘ Job to be no more
than 0,5 days (a) (b) (C) Formulate a Slmulatron model for this system Clearly deﬁne any variable; that you may
use and glve corrospondmg lnltlal values where necessary You should present your model
In the form of ﬂowchart(s) [14] Use the random numbers U1, U2, U3, U4 from Appendm A to generate the ﬁrst ﬁve arnval
tlmes (the ﬁrst arnval takes place at tune D) and use U5, U6, U7, U3 and U9 to generate the
correspondmg servrce tlmes Present your answers In the form of a table as shown below Car Arrwal Semce Begmmng End /
number tlme tlme serv1ce Departure Usmg only these ﬁve arrlvals, deternune an estrmate of the average tlme a car spends 1n
the shop
Is the objectwe of the management attalned'? Substant1ate your answer [10] The management 1s oons1der1ng hmng a second techmman who specmllsee 1n German cars
so that two such cars can be repau‘ed srmultaneously (only one techn1c1an works on any
car) Usmg the mural tlmes and servrces tlmes of Questmn (b), determme a new estlmate
of the average tame a car spends 1n the shop [6] {TURN OVER] DSC3703
4 May / J un 2013 Question 4 A certaln pro_]ect has elght dlfferent tasks denoted A, B, C, D, E, F, G and H. The prOJect
manager has the followmg data ava1lable Completlon tune
4 days Wlth probablhty 0,36
5 days w1th probabﬂlty 0,44
6 days w1th probabﬂrty 0,20 The manager wants to determlne whlch of the three paths ACFG, BDFG and BEH rs the
crrtrcal path (that 1s, the path whlch takes the largest tune to complete) To determme the
length of a given path, we snnply add the completlon times of each of 1ts act1v1t1es For example
the length of ACFG lS glven by addmg the completlon tune for the tasks A, C, F and G Answer the followmg questrons (a) Usmg random numbers U1, U2, U3,U4, U5, U5, U7, U3 generate completlon tlmes for tasks
A, B, C, D, E, F, G and H respectwely [20] (b) Usmg the results of Quwmon (a), determme the correspondmg cntlcal path [6] (c) Can you adwse the project manager to conSIder the obtamed path 1n Questlon (b) as the
erltlcal path of the pr0Ject7 Justlfy your answer [2] TOTAL: 90 ©
UNISA 2012 DSC3703
1 May] Jun 2013 APPENDIX A
Formulae and numbers Random numbers The follounng sequence of 20 random numbers was generated sequentzally from a U[0, 1) desira
button. U1 I 0,45 U9 = 0,35 U17 = 0,81 U25 2 0,69 U33 = 0,78
U2 = 0,71 U10 = 0,19 U13 = 0,63 U25 = 0,68 U34 2 0,92
U3 = 0,97 U11 = 0,70 U19 = 0,87 U27 = 0,60 U35 = 0,19
U4 = 0,80 U12 = 0,12 U20 = 0,50 U23 = 0,20 U35 = 0,56
U5 = 0,57 U13 = 0,85 U21 = 0,96 U29 = 0,26 U37 = 0,86
Us = 0,73 U14 = 0,56 U22 = 0,12 U30 = 0,19 U33 = 0,73
U7 = 0,05 U15 = 0,42 U23 = 0.80 U31 = 0,83 U39 = 0,42
U3 = 0,43 U15 = 0,73 U24 = 0 33 U32 = 0,08 U40 = 0,54 The chisquare test value Percentile points of the chisquare distribution Degrees of freedom (d f ) a
1/ 0,95 0,90 0,10 0,05 3
4
5
6
7
8
9
10 N)
a; The KolmogorovSmirnov test values 13+ = max [1 — Exam] 0* = max [Fxmw — i] D = max(D+, 13) lﬁzgn Tl. 131.51: T = ENE + 0,12 + 0,11/ﬁ) Percentile points for T: 2,5
1,133 DSCB703 11 May/ Jun 2013
Cumulative distribution functions 1 — "(m/5) f > 0
Exponentlal dlstrlbutlon F(:c) = e 1 a: 0 otherw1se 1 — (3/13)“ f > 0
Welbull dlstributlon F(:n) = e l a: 0 othermse 0 1f 2: < a
Umform distrlbutlon F (2:) = f}: 1f (1 S 17 S b 1 If m > b 0 1f 2: < a x—a 2
f < < Triangular dlstnbutlon F(:c) = 5—“ CE“ 2 1 a — m C 1—47LHb—a—mb—c 1fc<m<b 1 1f :1: > 0 Conﬁdence intervals and the t—distribution A 100(1 — 00% conﬁdence interval for E(X) 18 gwen by _ {5'2
X 3': t(af2,n—1) T";— Y=Z£ 32:2”:(Xl_X)2 71—1 where
7:] 1:1 Percentile points of the t—distribution Degrees of ireedom (d f ) or
7/ 0,1 0,05 0,025 0,01
1 3,078 6,314 12,706 31,821
2 1,886 2,920 4,303 6,965
3 1,638 2,353 3,182 4,541
4 1,533 2,132 2,776 3,747
5 1,476 2,015 2,571 3,365
6 1,440 1,943 2,447 3,143
7 1,415 1,895 2,365 2,998
8 1,397 1,860 2,306 2,896
9 1,383 1,833 2,262 2,821
10 1,372 1,812 2,228 2,764
00 1,282 1,645 1,960 2,326 Central limit theorem The sum of 71 Independent and 1dentrcally d16tr1buted random variables, each with mean ,u and
ﬁnite vanance 02, IS approxmlately normally dlstrlbuted w1th mean 71.11 and varlance 7102 ...
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