MP 4

4. Informasi pendukung
• Materi tiap pertemuan, download dari: slideshare
• Reference, download melalui : gen.lib.rus.ec
• Jounral atau hasil penelitian, download dari: Google scholar.com

6. References
Main Readings
1. Dennis Lock. 2007. Project Management. Ninth Edition. Gower Publishing Company, USA.
2. R.C. Mishra & Tarun Soota. 2005. Modern Project Management. New Age International (P) Limited,
Publisher.
3. John C. Goodpasture, PMP. 2004. Quantitative Methods In Project Management. J. Ross Publishing,
Inc., USA.
Supporting Readings
4. Harvey Maylor.2003. Project Management. Third Edition. Pearson Education, Ltd Gary R.
5. Heerkens, PMP. 2002. Project Management. MvGrow-Hill, USA.
6. Sidney M. Levi. 2005. Project Management in Construction. Fifth Edition. McGrow-Hill.

7. (1) (2) (3)
(4) (5)
(6)

8. RPS Manajemen Proyek
1. Introduction to project management, factors for project success or
failure, defining the project task
2. Estimating the project cost
3. First steps in planning the timescale
4. Financial appraisal and the business plan (2x)
5. Uncertainty and Risk
6. Project organization
7. Work breakdown structure
8. Detailed Planning
9. Scheduling (2x)
10.Managing project strat-up and commercial
11.Managing procurement, progress, changes, and project costs
12.Earned-value analysis and reporting

9. PROJECT MANAGEMENT

10. FIRST STEPS IN PLANNING THE
TIMESCALE
 Introduction to project palnning
 What makes an ideal project plan?
 Distinction between planning and time scheduling

52. Cost effectivenes and profitability
Contoh: Proyek manufaktur memperhitungkan cost effectiveness dan
optimum profitabilitas:
 Implementasi keputusan proyek, memperhatikan contribution margin
dan profitabilitas
 Keutusan penentuan kapastias produksi manufaktur yang optimal
 Keputusan penentuan lokasi proyek manufaktur

53. Contribution margin & BEP
• CM = P- VCQ
• BEP = FC/(P – VCQ)

54. FIX COST
1000 2000
Q
Q
Rp
Rp
FC
AFC
1000 2000
50.000
50
25
0
0

55. VARIABLE COST
VC
AVC
Q
Q
Rp
Rp
0
0
1000 2000
1000 2000
100
100.000
200.000
A
B
A’ B’

56. TOTAL COST (TC)
FC
TC
Q
Q
Rp
Rp
0
0
1000 2000
1000 2000
50.000
150.000
250.000
AC
150
125

57. 6. Luas dan pola produksi
 Hubungan luas produksi dengan biaya :
BEP
Q
$ $ $TR
FC
TC
AC
AC
Q
Q
100.000
500
1000
50.000
2000
125
150
Kejenuhan
.

58. Break Even Point (BEP) - 1
FC
TC
Q
Rp
0
400 600
50.000
100.000
500
TR
BEP90.000
120.000
80.000
110.000
Laba
Rugi

59. Break Even Point (BEP) - 2
FC
TC1
Q
Rp
0
400
50.000
100.000
500
TR
BEP1
Catatan :
BEP = FC / (P – VC)
BEP = 50.000/ (200-75) =400 unit
TR = 400 x Rp 200 = Rp 80.000
TC = 50.000 + (400 x Rp 100) = Rp 80.000
TC2
BEP2
80.000
Catatan :
- Pangsa Pasar < 500 unit
- Solusinya , VC ditekan
dari Rp Rp 100 perunit
menjadi Rp Rp 75 per unit

60. Break Even Point (BEP) - 3
FC1
TC1
Q
Rp
0
400
50.000
100.000
500
TR
BEP1
Catatan :
BEP = FC / (P – VC)
BEP = 40.000/ (200-100) =400 unit
TR = 400 x Rp 200 = Rp 80.000
TC = 40.000 + (400 x Rp 100) = Rp 80.000
TC2
BEP2
80.000
Catatan :
- Pangsa Pasar < 500 unit
- Solusinya , FC ditekan
dari Rp Rp 50.000 pertahun
menjadi Rp Rp 40.000 pertahun
FC240.000

61. Break Even Point (BEP) - 4
TC
Q
Rp
0
400
50.000
100.000
500
TR1
BEP1
Catatan :
BEP = FC / (P – VC)
BEP = 50.000/ (225-100) =400 unit
TR = 400 x Rp 225= Rp 90.000
TC = 50.000 + (400 x Rp 100) = Rp 90.000
TR2
BEP2
90.000
Catatan :
- Pangsa Pasar < 500 unit
- Solusinya , P dinaikkan
dari Rp Rp 200 perunit
menjadi Rp 225 perunit
FC

62. Programasi Linear
• Makximum profit
• Minimum cost

63. Optimitation models - Linear programming
• Linear programs
• Metode grafik
• Metode simplex Tableau
• Solusi permasalahan
• Maksimum
• Minimum
• Contoh 1 :
Max Z = 3 A + 4 B
Subject to :
(1) 10 A + 6 B ≤ 1.200
(2) 6 A + 10 B ≤ 900
(3) 7 A + 5 B ≤ 700
A, B ≥ 0
• ZX = 3(0) + 4(9) = 360
ZY = 3(62.5 ) + 4(52.5 ) = 397.5 (max)
ZZ= 3(100 ) + 4(0 ) = 300 A
B
200
120
140
100 150
90
10A+6B=1200
7A+5B=700
6A+10B=900
X (0,90) Z (100,0)
Y (62.5, 52.5)
7A+5B=700
6A+10B=900
14A+10B=1400
6A+10B=900
8A =500
A=62,5
B =52.5

64. MiaxZ = 40X1+ 30X2
Sobject to
2X1 + 3X2 ≤ 60
2X2 ≤ 30
2X1 + X2 ≤ 40
X1, X2 ≥ 0
Optimitation models - Linear programming
X1
X2
20
30
2X1+3X2=60
2X1+X2 =40
2X2=20
X2=10
2x1+3(10)=60
2X1=30
X1=15
10
15 20
40
15
2X1+3X2=60
2X2=30
4X1+6X2=120
6X2=90
4X1=30
X1=7.5
7.5
B(7.5 , 15)
C(15,10)
D(20,0)
A(0,15)
2X2=30
2X1+X2=40
2X1+X2=60
ZA=40(0)+30(15)=450
ZB=40(7.5)+30(15)=750
ZC=40(15)+30(10)=900 (Max)
ZD=40(20)+30(0)=800
Contoh 2 :

65. Catatan tabel 1:
a) Baris Cj = angka pada fungsi obyektif
b) Baris Bj = variabel yang digunakan & variabel tambahan
c) Kolom CB = angka fungsi obyektif masing2 variabel baris ybs
d) Kolom Cj = jumlah kendala pada masing2 fungsi kendala
e) Kolom X1, X2, Si, S2, S3 = dari koefisien fungsi kendala
MiaxZ = 40X1+ 30X2 Pertidaksamaan menjadi persamaan :
Sobject to 2X1 + 3X2 +S1 = 60
2X1 + 3X2 ≤ 60 2X2 + S2 = 30
2X2 ≤ 30 2X1 + X2 + S3 = 40
2X1 + X2 ≤ 40
X1, X2 ≥ 0
Tabel -1
Varb dlm Cj 40 30 0 0 0
CB Basic Bj X1 X2 S1 S2 S3 Index
0 S1 60 2 3 1 0 0 60/2=30
0 S2 30 0 2 0 1 0 30/0=∞
0 S3 40 2 1 0 0 1 40/2=20
Zj 0 0 0 0 0 0
Zj-Cj 0 -40 -30 0 0 0
Optimitation models - Linear programming
Metode Tablo

66. Optimitation models - Linear programming
Metode Tablo
Catatan tabel-2:
1. Kolom kunci = negatif terbesar dari Cj-Zj = X1 (pada Tabel 1)
2. Baris kunci = index terkecil = (kolom kendala) dibagi (kolom kunci) = S3 (Pada Tabel 1)
3. Nomor kunci = 2 (Pada Tabel 1)
4. Tabel-2
a) Baris S3 diganti oleh X1
- Baris S3 lama = 40 2 1 0 0 1
- Baris X1 baru (mengganti S3)
40/2= 2/2= 1/2= 0/2= 0/2= 1/2=
20 1 0.5 0 0 0.5
Angka kolom CB untuk X1 = angka X1 pada baris Cj =40
b) Baris S1
- S1 lama = 60 2 3 1 0 0
- S1 baru=angka (baris S1 lama) dikurangi (angka X1 x angka kolom kunci S1 atau 2 )
60-20(2)= 2-1(2)= 3-0.5(2)= 1-0(2)= 0-1(2)= 0-0.5(2)=
20 0 2 1 0 -1
c) Baris S2
- S2 lama 30 0 2 0 1 0
- Baris S2 baru=angka (baris S2 lama) dikurangi (angka X1 x angka kolom kunci S2 atau 0 )
30-30(0)= 0 -1(0)= 2-1.5(0)= 0-0.5(0)= 1-0(0)= 0-0(0)=
30 0 2 0 1 0
d) Baris Zj = (angka kolom CB) x (angka kolom ybs)
Tabel -2
Varb dlm Cj 40 30 0 0 0
CB Basic Bj X1 X2 S1 S2 S3 Index
0 S1 20 0 2 1 0 -1 20/2=10
0 S2 30 0 2 0 1 0 30/2=15
40 X1 20 1 0.5 0 0 0.5 20/0.5=40
Zj 800 40 20 0 0 20
Zj-Cj 800 0 -10 0 0 20

67. Optimitation models - Linear programming
Metode Tablo
Catatan tabel-3:
1. Kolom kunci = negatif terbesar dari Cj-Zj = X2 (Pada Tabel 2)
2. Baris kunci = index terkecil = (kolom kendala) dibagi (kolom kunci) = S1 (Pada Tabel 2)
3. Nomor kunci = 2 (Pada Tabel 2)
4. Tabel-2
a) Baris S1 diganti oleh X2
- Baris S1 lama = 20 0 2 1 0 -1
- Baris X2 baru (mengganti S1)
20/2= 0/2= 2/2= 1/2= 0/2= -1/2 =
10 0 1 0.5 0 -0.5
Angka kolom CB untuk X2 = angka X1 pada baris Cj = 30
b) Baris S2
- S2 lama = 30 0 2 0 1 0
- S2 baru=angka (baris S1 lama) dikurangi (angka X1 x angka kolom kunci S1 atau 2 )
30-10(2)= 0-0(2)= 2-1(2)= 0-0.5(2)= 1-0(2)=
10 0 0 -1 1
c) Baris X1
- X1 lama 20 1 0.5 0 0 0.5
- Baris S2 baru=angka (baris S2 lama) dikurangi (angka X1 x angka kolom kunci S2 atau 0 )
20-10(0.5)= 1-0(0.5)= 0.5-1(0.5)= 0-0.5(0.5)= 0-0(0.5)=
15 1 0 -0.25 0
d) Baris Zj = (angka kolom CB) x (angka kolom ybs) = 30
Tabel -3
Varb dlm Cj 40 30 0 0 0
CB Basic Bj X1 X2 S1 S2 S3 Index
30 X2 10 0 1 0.5 0 -0.5
0 S2 10 0 0 -1 1 1
40 X1 15 1 0 -0.25 0 0.75
Zj 900 40 30 5 0 15
Zj-Cj 900 0 0 5 0 15
Keterangan : Karena (Zj - Cj) tidak ada lagi negatif, maka dinyatakan optimal
0.5-(-0.5x0.5)=
0.75
0-(-0.5x2)=
1

68. • Contoh (Primal) :
Min Z = 30 X1 + 50 X2
Subject to :
(a) 2X1 + 4X2 ≤ 80
(b) X1 + X2 ≤ 25
X1, X2 ≥ 0
Optimitation models - Linear programming
X1
X2
40
20
25
2515
2X1+4X2 =80
X1+X2 =25
A(0,25)
B(10,15)
C(40,0)
Titik B:
2X1+4X2=80
X1 + X2 =25
2X1+4X2 =80
2X1 +2X2=50
2X2= 30
X2=15
X1 = 25-15 =10
10
15
ZA = 30(0) + 50(25)=1250 (Min)
ZB = 30(10)+ 50(15)=1050 (Min)
ZC = 30(40) + 50(0) =1200

69. • Contoh Dual :
Max Z = 80A + 25 B
Subject to :
(1) 2A + B ≥ 30
(2) 4A + B ≥ 50
A,B ≥ 0
Optimitation models - Linear programming
A
B
15
30
12.5
50
10
10
X (0,30)
Y(10,10)
Z(12.5 , 0)
ZX = 80(0) + 25(30)=750
ZY =80(10)+25(10)=1050 (Max)
ZZ = 80(12.5) +25(0) = 1000

70. Penentuan lokasi proyek manufaktur
A= 5.000 + 12 Q
B= 10.000 + 10 Q
C= 15.000 + 5 Q
D= 17.000 + 6 Q
Lokasi Q Q Q
1000 2000 3000
A 17,000 29,000 41,000
B 20,000 30,000 40,000
C 20,000 25,000 30,000
D 23,000 29,000 35,000
Cost (Rp)
50,000
A A=C
40,000 B 5000+12Q=15000+5Q
D 7Q=10000
30,000 C Q = 1429
20,000
10,000 A C C
1000 1429 2000 3000 Q
Pendekatan FC dan VC