Description
Concept of Crashing in Operations
Crashing
Problem - 1
Normal & crash times and cost are given below for an expansion project Activity Pre. Activity Normal Dur Crash Dur Normal Cost Crash Cost
A
B C D
A C
3
6 3 2
2
4 2 1
40
200 20 20
50
300 35 32
E
F G H
A
D D,E B,F,G
1
5 7 4
1
3 6 3
20
150 120 160
20
190 150 195
All Durations are in months and cost are in ‘000. If the company has Rs 7,76,000/- available for this project, how should it allocate funds to minimise overall completion time, to the nearest 0.1 month? What is the minimum completion time?
Problem – 1 . .contd
Activity P Activity A B A C D C E A F D G D,E H B,F,G
Normal Dur
3
6
3
2
1
5
7
4
2 A 3 1 C 3 3 D 2 D0 1 E
B 6 G 7 5 F 5
6
H 4
7
1-2-6-7 = A-B-H = 13 - IV 1-2-5-6-7 = A-E-G-H = 15 - II
4
1-3-4-6-7 = C-D-F-H = 14 - III
1-3-4-5-6-7 = C-D-D0-G-H = 16 - I
Problem – 1 . .contd
Normal Dur Crash Dur Normal Cost Crash Cost Max crash Cost Slope
Activity
A
B C D E F G H
3
6 3 2 1 5 7 4
2
4 2 1 1 3 6 3
40
200 20 20 20 150 120 160
50
300 35 32 20 190 150 195
1
2 1 1 0 2 1 1
10
50 15 12 0 20 30 35
Problem – 1 . .contd
• Normal Project Cost = Rs 7,30,000/• Total available funds = Rs 7,76,000/• Excess funds available for crashing = Rs 46,000/-
I Critical Activity C
1-3-4-5-6-7 Max crash 1
C-D-D0-G-H Cost Slope 15
16 /15 Rank 2 Remarks (i)
D
D0 G H
1
1 1
12
30 35
1
3 4
II A E G H
1-2-5-6-7 1 1 1
A-E-G-H 10 30 35
15 /15 1 2 3
III C D F H
1-3-4-6-7 1 1 2 1
C-D-F-H 15 12 20 35
14 /13 2 1 3 4 (i)
IV A B H
1-2-6-7 1 2 1
A-B-H 10 50 35
13 /13 1 3 2
Problem – 1 . .contd
• Step (i)
? ?
Crashed activity D by 1 month Now new project durations are:
o o
o
o
I = 15 II = 15 III = 13 IV = 13
? ?
Increase in cost = Rs 12000 x 1 = 12000 Funds available = 46000 – 12000 = Rs 34000
• Now path II also becomes critical
I Critical Activity C
1-3-4-5-6-7 Max crash 1
C-D-D0-G-H Cost Slope 15
16 /15 /14 Rank 2 Remarks ( ii ) (i)
D
D0 G H
1
1 1
12
30 35
1
3 4
II A E G H
1-2-5-6-7 1 1 1
A-E-G-H 10 30 35
15 /15 /14 1 2 3 ( ii )
III C D F H
1-3-4-6-7 1 1 2 1
C-D-F-H 15 12 20 35
14 /13 /12 2 1 3 4 ( ii ) (i)
IV A B H
1-2-6-7 1 2 1
A-B-H 10 50 35
13 /13 /12 1 3 2 ( ii )
Problem – 1 . .contd
• Step (ii)
? ?
Crashed activity C by 1 month & A by 1 month Now new project durations are:
o o
o
o
I = 14 II = 14 III = 12 IV = 12
? ?
Increase in cost = Rs 15000 x 1 + 10000 x 1 = 25000 Funds available = 34000 - 25000 = Rs 9000
• No new path becomes critical
I Critical Activity C
1-3-4-5-6-7 Max crash 1
C-D-D0-G-H Cost Slope 15
16 /15 /14 /13.7 Rank 2 Remarks ( ii ) (i)
D
D0 G H
1
1 1
12
-
1
3 4
0.7
30 35
II A E G H
1-2-5-6-7 1 1 1 0.7
A-E-G-H 10 30 35
15 /15 /14 /13.7 1 2 3 ( ii )
III C D F H
1-3-4-6-7 1 1 2 1
C-D-F-H 15 12 20 35
14 /13 /12 /12 2 1 3 4 ( ii ) (i)
IV A B H
1-2-6-7 1 2 1
A-B-H 10 50 35
13 /13 /12 /12 1 3 2 ( ii )
Problem – 1 . .contd
• Step (iii)
? ?
Crashed activity G by 0.3 months Now new project durations are:
o o
o
o
I = 13.7 II = 13.7 III = 12 IV = 12
? ?
Increase in cost = Rs 30000 x 0.3 = 9000 Funds available = 9000 - 9000 = Rs 0
• Since funds are exhausted crashing is over
Problem – 1 . .contd
• • • • Normal Project Duration = 16 months Crashed Project Duration = 13.7 months Normal Project Cost = Rs 7,30,000 Crashed Project Duration = Rs 7,76,000
Problem - 2
The time-cost estimates for various activities of a project are given below: Activity A B C D E F G Pre. Activity Normal Dur A B A D,E C 8 7 5 4 3 5 4 Crash Dur 6 5 4 3 2 3 3 Normal Cost 8000 6000 7000 3000 2000 5000 6000 Crash Cost 10000 8400 8500 3800 2600 6600 7000
The project manager wishes to complete the project in minimum possible time. However he is not authorised to spend more than Rs 5000 on crashing.
Suggest the least-cost schedule for achieving the objective of the project manager. Assume that there are no indirect or utility cost.
Problem – 3
The duration below gives duration, cost of activities of a project Activity Normal Dur Crash Dur Normal Cost Crash Cost
1-2
1-3 2-4 3-4
2
8 4 1
1
5 3 1
100
150 200 70
150
210 280 70
3-5
4-6 5-6
2
5 6
1
3 2
80
100 120
150
190 360
Crash the activities of the project and find the minimum cost and corresponding duration given that an indirect cost per day is Rs 40.
Problem – 3 . . contd
1-2-4-6 = 11 - II 2 2 1 4 3 2 5 6 1 4 4 5 6 1-3-4-6 = 10 - III
1-3-5-6 = 12 - I
Problem – 3 . . contd
Normal Dur 2 8 4 1 2 5 6 Crash Dur 1 5 3 1 1 3 2 Normal Cost 100 150 200 70 80 100 120 Crash Cost 150 210 280 70 150 190 360 Max crash 1 3 1 0 1 2 4 Cost Slope 50 20 80 0 70 45 60
Activity 1-2 1-3 2-4 3-4 3-5 4-6 5-6
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 Cost Slope 20 70 60 12 /11 Rank 1 3 2 Remarks (i)
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11
2 3 1
III 1-3 3-4 4-6
1-3-4-6 3 2 2 20 45
10 /9 1 2 (i)
Problem – 3 . . contd
• Step (i):
? ?
Crashed activity 1-3 by 1 day Now new project durations are:
o o
o
I = 11 II = 11 III = 9
?
Increase in cost = Rs 20 x 1 = 20
• Now path II also becomes critical
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 Cost Slope 20 70 60 12 /11 /9 Rank 1 3 2 Remarks ( ii ) ( i )
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11 /9
2 3 1 ( ii )
III 1-3 3-4 4-6
1-3-4-6 3 2 2 20 45
10 /9 /5 1 2 ( ii ) ( ii ) ( i )
Problem – 3 . . contd
• Step (ii)
? ?
Crashed activity 1-3 by 2 days & 4-6 by 2 days Now new project durations are:
o o
o
I=9 II = 9 III = 5
?
Increase in cost = 20 x 2 + 45 x 2 = 130
• No new path becomes critical
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 3 Cost Slope 20 70 60 12 /11 /9 /8 Rank 1 3 2 ( iii ) Remarks ( ii ) ( i )
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11 /9 /8
2 3 1 ( ii )
( iii )
III 1-3 3-4 4-6
1-3-4-6 3 2 20 45
10 /9 /5 /5 1 2 ( ii ) ( ii ) ( i )
Problem – 3 . . contd
• Step (iii)
? ?
Crashed activity 5-6 by 1 day & 1-2 by 1 day Now new project durations are:
o o
o
I=8 II = 8 III = 5
?
Increase in cost = 60 x 1 + 50 x 1 = 110
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 32 Cost Slope 20 70 60 12 /11 /9 /8 /7 Rank 1 3 2 ( iii ) ( iv ) Remarks ( ii ) ( i )
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11 /9 /8 /7
2 3 1
( iii )
( iv ) ( ii )
III 1-3 3-4 4-6
1-3-4-6 3 2 20 45
10 /9 /5 /5 /5 1 2 ( ii ) ( ii ) ( i )
Problem – 3 . . contd
• Step (iv)
? ?
Crashed activity 5-6 by 1 day & 2-4 by 1 day Now new project durations are:
o o
o
I=7 II = 7 III = 5
?
Increase in cost = 60 x 1 + 80 x 1 = 140
• Crashing is over since CP II is fully crashed
Problem 3 . . contd
• • • • Normal Project duration = 12 days Crashed Project duration = 7 days Normal Project Cost = Rs 820 Crashed Project cost:
? ? ? ? ?
Normal Project cost = Rs 820 Increase in cost at step (i) = Rs 20 Increase in cost at step (ii) = Rs 130 Increase in cost at step (iii) = Rs 110 Increase in cost at step (iv) = Rs 140 Total = Rs 1220
Problem 3 . . contd
Step
Project Duration
Direct cost
Indirect Cost
Total Cost
12
i ii ii iii iv 11 10 9 8 7
820
820 + 20 = 840 840 + 65 = 905 905 + 65 = 970 970 + 110 = 1080 1080 + 140 = 1220
480
440 400 360 320 280
1300
1280 1305 1330 1400 1500
Problem 3 . . contd
cost 1500 1400 1300 1200 1100 1000 900 800
12
11
10
9
8
7
Days
Optimum cost = Rs 1280 and optimum duration = 11 days
Problem - 4
The detailed activities in a building project are given below Activity A B C D E F G H Pre. Activity Normal Dur A A C E B,C F,G 9 14 4 6 14 6 5 2 Crash Dur 6 4 3 4 13 6 3 1 Normal Cost 12000 14000 2000 44000 1600 4000 4000 12000 Crash Cost 18000 24000 2400 56000 1800 4000 4800 14000
Indirect cost per day is Rs 5000. Find the minimum time schedule and optimum cost.
Problem 5
A small marketing project in the table given below. With each job is listed its normal time and crash time. The cost in Rs / day of crashing each job is also given. Cost of completing the project in normal situation is Rs 5000.
Activity
1–2 1–3 1–4
Normal Duration
9 8 15
Crash Duration
6 5 10
Cost of crashing (Rs/day) 20 25 30
2–4
3–4 4-5
5
10 2
3
6 1
10
15 40
(1) What is Normal project length & minimum project length (2) Overhead costs total Rs 35 per day, what is the optimum length schedule
Problem 5 . .contd
1-2-4-5 = 16 - III
1-4-5 = 17 - II
2
1-3-4-5 = 20 - I
9
5
1 15
4
2 10
5
8 3
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 Rank 2 1 3 (i) Remarks
II 1-4 4-5
1-4-5 5 1 30 40
17 / 17 1 2
III 1-2 2-4 4-5
1-2-4-5 3 2 1 20 10 40
16 / 16 2 1 3
Problem – 5 . . contd
• Step (i):
? ?
Crashed activity 3-4 by 3 days Now new project durations are:
o o
o
I = 17 II = 17 III = 16
?
Increase in cost = Rs 15 x 3 = 45
• Now path II also becomes critical
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 / 16 Rank 2 1 3 (i) (ii) Remarks
II 1-4 4-5
1-4-5 5 1 30 40
17 / 17 / 16 1 2 (ii)
III 1-2 2-4 4-5
1-2-4-5 3 2 1 20 10 40
16 / 16 / 15 2 1 3 (ii)
Problem – 5 . . contd
• Step (ii):
? ?
Crashed activity 4 - 5 by 1 day Now new project durations are:
o o
o
I = 16 II = 16 III = 15
?
Increase in cost = Rs 40 x 1 = 40
• No new path becomes critical
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 / 16 / 15 Rank 2 1 3 (i) (ii) (iii) Remarks
II 1-4 4-5
1-4-5 5 1 4 30 40
17 / 17 / 16 / 15 (iii) 1 2 (ii)
III 1-2 2-4 4-5
1-2-4-5 3 2 1 20 10 40
16 / 16 / 15 / 15 2 1 3 (ii)
Problem – 5 . . contd
• Step (iii):
? ?
Crashed activity 3 - 4 by 1 day & 1 – 4 by 1 day Now new project durations are:
o o
o
I = 15 II = 15 III = 15
?
Increase in cost = 15 x 1 + 30 x 1 = 45
• Now all 3 paths are critical
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 / 16 / 15 / 12 Rank 2 1 3 Remarks (iv) (i) (ii) (iii)
II 1-4 4-5
1-4-5 5 1 4 1 30 40
17 / 17 / 16 / 15 / 12 (iii) (iv) 1 2 (ii)
III 1-2 2-4 4-5
1-2-4-5 3 2 1 2 20 10 40
16 / 16 / 15 / 15 / 12 (iv) 2 1 3 (iv) (ii)
Problem – 5 . . contd
• Step (iv):
?
?
Crashed activity 1 - 3 by 3 days, 1 – 4 by 3 days, 2 – 4 by 2 days and 1 – 2 by 1 day. Now new project durations are:
o
o
o
I = 12 II = 12 III = 12
?
Increase in cost = 25 x 3 + 30 x 3 + 10 x 2 + 20 x 1 = 205
• Crashing is over since all activities in CP I are crashed
Problem 5 . . contd
• • • • Normal Project duration = 20 days Crashed Project duration = 12 days Normal Project Cost = Rs 5000 Crashed Project cost:
? ? ? ? ?
Normal Project cost = Rs 5000 Increase in cost at step (i) = Rs 45 Increase in cost at step (ii) = Rs 40 Increase in cost at step (iii) = Rs 45 Increase in cost at step (iv) = Rs 205 Total = Rs 5335
Problem 5 . . contd
Step Project Dur 20 i i i ii iii iv iv 19 18 17 16 15 14 13 Direct cost 5000 5000+15 = 5015 5015+15 = 5030 5030+15 = 5045 5045+45 = 5090 5090+45 = 6035 6035+65 = 7000 7000+65 = 7065 Indirect cost Total Cost 700 665 630 595 560 525 490 455 5700 5680 5660 5640 5650 6560 7490 7520
iv
12
7065+75 = 7140
420
7560
Problem 6
Consider the foll activities with normal time and cost together with extra cost of saving a day on selected activity Activity 1–2 1–3 1–4 2–3 2–5 2–6 3–4 3–6 4–6 5-6 Normal Dur 4 6 2 5 2 7 4 3 7 12 Crash Dur 3 5 2 3 1 5 2 2 4 9 Cost of crashing (Rs/day) 10 25 35 15 5 15 20 30 5 10
In addition overhead costs are Rs 35 per day. Calculate optimum time
doc_974177801.ppt
Concept of Crashing in Operations
Crashing
Problem - 1
Normal & crash times and cost are given below for an expansion project Activity Pre. Activity Normal Dur Crash Dur Normal Cost Crash Cost
A
B C D
A C
3
6 3 2
2
4 2 1
40
200 20 20
50
300 35 32
E
F G H
A
D D,E B,F,G
1
5 7 4
1
3 6 3
20
150 120 160
20
190 150 195
All Durations are in months and cost are in ‘000. If the company has Rs 7,76,000/- available for this project, how should it allocate funds to minimise overall completion time, to the nearest 0.1 month? What is the minimum completion time?
Problem – 1 . .contd
Activity P Activity A B A C D C E A F D G D,E H B,F,G
Normal Dur
3
6
3
2
1
5
7
4
2 A 3 1 C 3 3 D 2 D0 1 E
B 6 G 7 5 F 5
6
H 4
7
1-2-6-7 = A-B-H = 13 - IV 1-2-5-6-7 = A-E-G-H = 15 - II
4
1-3-4-6-7 = C-D-F-H = 14 - III
1-3-4-5-6-7 = C-D-D0-G-H = 16 - I
Problem – 1 . .contd
Normal Dur Crash Dur Normal Cost Crash Cost Max crash Cost Slope
Activity
A
B C D E F G H
3
6 3 2 1 5 7 4
2
4 2 1 1 3 6 3
40
200 20 20 20 150 120 160
50
300 35 32 20 190 150 195
1
2 1 1 0 2 1 1
10
50 15 12 0 20 30 35
Problem – 1 . .contd
• Normal Project Cost = Rs 7,30,000/• Total available funds = Rs 7,76,000/• Excess funds available for crashing = Rs 46,000/-
I Critical Activity C
1-3-4-5-6-7 Max crash 1
C-D-D0-G-H Cost Slope 15
16 /15 Rank 2 Remarks (i)
D
D0 G H
1
1 1
12
30 35
1
3 4
II A E G H
1-2-5-6-7 1 1 1
A-E-G-H 10 30 35
15 /15 1 2 3
III C D F H
1-3-4-6-7 1 1 2 1
C-D-F-H 15 12 20 35
14 /13 2 1 3 4 (i)
IV A B H
1-2-6-7 1 2 1
A-B-H 10 50 35
13 /13 1 3 2
Problem – 1 . .contd
• Step (i)
? ?
Crashed activity D by 1 month Now new project durations are:
o o
o
o
I = 15 II = 15 III = 13 IV = 13
? ?
Increase in cost = Rs 12000 x 1 = 12000 Funds available = 46000 – 12000 = Rs 34000
• Now path II also becomes critical
I Critical Activity C
1-3-4-5-6-7 Max crash 1
C-D-D0-G-H Cost Slope 15
16 /15 /14 Rank 2 Remarks ( ii ) (i)
D
D0 G H
1
1 1
12
30 35
1
3 4
II A E G H
1-2-5-6-7 1 1 1
A-E-G-H 10 30 35
15 /15 /14 1 2 3 ( ii )
III C D F H
1-3-4-6-7 1 1 2 1
C-D-F-H 15 12 20 35
14 /13 /12 2 1 3 4 ( ii ) (i)
IV A B H
1-2-6-7 1 2 1
A-B-H 10 50 35
13 /13 /12 1 3 2 ( ii )
Problem – 1 . .contd
• Step (ii)
? ?
Crashed activity C by 1 month & A by 1 month Now new project durations are:
o o
o
o
I = 14 II = 14 III = 12 IV = 12
? ?
Increase in cost = Rs 15000 x 1 + 10000 x 1 = 25000 Funds available = 34000 - 25000 = Rs 9000
• No new path becomes critical
I Critical Activity C
1-3-4-5-6-7 Max crash 1
C-D-D0-G-H Cost Slope 15
16 /15 /14 /13.7 Rank 2 Remarks ( ii ) (i)
D
D0 G H
1
1 1
12
-
1
3 4
0.7
30 35
II A E G H
1-2-5-6-7 1 1 1 0.7
A-E-G-H 10 30 35
15 /15 /14 /13.7 1 2 3 ( ii )
III C D F H
1-3-4-6-7 1 1 2 1
C-D-F-H 15 12 20 35
14 /13 /12 /12 2 1 3 4 ( ii ) (i)
IV A B H
1-2-6-7 1 2 1
A-B-H 10 50 35
13 /13 /12 /12 1 3 2 ( ii )
Problem – 1 . .contd
• Step (iii)
? ?
Crashed activity G by 0.3 months Now new project durations are:
o o
o
o
I = 13.7 II = 13.7 III = 12 IV = 12
? ?
Increase in cost = Rs 30000 x 0.3 = 9000 Funds available = 9000 - 9000 = Rs 0
• Since funds are exhausted crashing is over
Problem – 1 . .contd
• • • • Normal Project Duration = 16 months Crashed Project Duration = 13.7 months Normal Project Cost = Rs 7,30,000 Crashed Project Duration = Rs 7,76,000
Problem - 2
The time-cost estimates for various activities of a project are given below: Activity A B C D E F G Pre. Activity Normal Dur A B A D,E C 8 7 5 4 3 5 4 Crash Dur 6 5 4 3 2 3 3 Normal Cost 8000 6000 7000 3000 2000 5000 6000 Crash Cost 10000 8400 8500 3800 2600 6600 7000
The project manager wishes to complete the project in minimum possible time. However he is not authorised to spend more than Rs 5000 on crashing.
Suggest the least-cost schedule for achieving the objective of the project manager. Assume that there are no indirect or utility cost.
Problem – 3
The duration below gives duration, cost of activities of a project Activity Normal Dur Crash Dur Normal Cost Crash Cost
1-2
1-3 2-4 3-4
2
8 4 1
1
5 3 1
100
150 200 70
150
210 280 70
3-5
4-6 5-6
2
5 6
1
3 2
80
100 120
150
190 360
Crash the activities of the project and find the minimum cost and corresponding duration given that an indirect cost per day is Rs 40.
Problem – 3 . . contd
1-2-4-6 = 11 - II 2 2 1 4 3 2 5 6 1 4 4 5 6 1-3-4-6 = 10 - III
1-3-5-6 = 12 - I
Problem – 3 . . contd
Normal Dur 2 8 4 1 2 5 6 Crash Dur 1 5 3 1 1 3 2 Normal Cost 100 150 200 70 80 100 120 Crash Cost 150 210 280 70 150 190 360 Max crash 1 3 1 0 1 2 4 Cost Slope 50 20 80 0 70 45 60
Activity 1-2 1-3 2-4 3-4 3-5 4-6 5-6
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 Cost Slope 20 70 60 12 /11 Rank 1 3 2 Remarks (i)
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11
2 3 1
III 1-3 3-4 4-6
1-3-4-6 3 2 2 20 45
10 /9 1 2 (i)
Problem – 3 . . contd
• Step (i):
? ?
Crashed activity 1-3 by 1 day Now new project durations are:
o o
o
I = 11 II = 11 III = 9
?
Increase in cost = Rs 20 x 1 = 20
• Now path II also becomes critical
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 Cost Slope 20 70 60 12 /11 /9 Rank 1 3 2 Remarks ( ii ) ( i )
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11 /9
2 3 1 ( ii )
III 1-3 3-4 4-6
1-3-4-6 3 2 2 20 45
10 /9 /5 1 2 ( ii ) ( ii ) ( i )
Problem – 3 . . contd
• Step (ii)
? ?
Crashed activity 1-3 by 2 days & 4-6 by 2 days Now new project durations are:
o o
o
I=9 II = 9 III = 5
?
Increase in cost = 20 x 2 + 45 x 2 = 130
• No new path becomes critical
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 3 Cost Slope 20 70 60 12 /11 /9 /8 Rank 1 3 2 ( iii ) Remarks ( ii ) ( i )
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11 /9 /8
2 3 1 ( ii )
( iii )
III 1-3 3-4 4-6
1-3-4-6 3 2 20 45
10 /9 /5 /5 1 2 ( ii ) ( ii ) ( i )
Problem – 3 . . contd
• Step (iii)
? ?
Crashed activity 5-6 by 1 day & 1-2 by 1 day Now new project durations are:
o o
o
I=8 II = 8 III = 5
?
Increase in cost = 60 x 1 + 50 x 1 = 110
Problem – 3 . . contd
I Critical Activity 1-3 3-5 5-6 1-3-5-6 Max crash 3 2 1 4 32 Cost Slope 20 70 60 12 /11 /9 /8 /7 Rank 1 3 2 ( iii ) ( iv ) Remarks ( ii ) ( i )
II
1-2 2-4 4-6
1-2-4-6
1 1 2 50 80 45
11 /11 /9 /8 /7
2 3 1
( iii )
( iv ) ( ii )
III 1-3 3-4 4-6
1-3-4-6 3 2 20 45
10 /9 /5 /5 /5 1 2 ( ii ) ( ii ) ( i )
Problem – 3 . . contd
• Step (iv)
? ?
Crashed activity 5-6 by 1 day & 2-4 by 1 day Now new project durations are:
o o
o
I=7 II = 7 III = 5
?
Increase in cost = 60 x 1 + 80 x 1 = 140
• Crashing is over since CP II is fully crashed
Problem 3 . . contd
• • • • Normal Project duration = 12 days Crashed Project duration = 7 days Normal Project Cost = Rs 820 Crashed Project cost:
? ? ? ? ?
Normal Project cost = Rs 820 Increase in cost at step (i) = Rs 20 Increase in cost at step (ii) = Rs 130 Increase in cost at step (iii) = Rs 110 Increase in cost at step (iv) = Rs 140 Total = Rs 1220
Problem 3 . . contd
Step
Project Duration
Direct cost
Indirect Cost
Total Cost
12
i ii ii iii iv 11 10 9 8 7
820
820 + 20 = 840 840 + 65 = 905 905 + 65 = 970 970 + 110 = 1080 1080 + 140 = 1220
480
440 400 360 320 280
1300
1280 1305 1330 1400 1500
Problem 3 . . contd
cost 1500 1400 1300 1200 1100 1000 900 800
12
11
10
9
8
7
Days
Optimum cost = Rs 1280 and optimum duration = 11 days
Problem - 4
The detailed activities in a building project are given below Activity A B C D E F G H Pre. Activity Normal Dur A A C E B,C F,G 9 14 4 6 14 6 5 2 Crash Dur 6 4 3 4 13 6 3 1 Normal Cost 12000 14000 2000 44000 1600 4000 4000 12000 Crash Cost 18000 24000 2400 56000 1800 4000 4800 14000
Indirect cost per day is Rs 5000. Find the minimum time schedule and optimum cost.
Problem 5
A small marketing project in the table given below. With each job is listed its normal time and crash time. The cost in Rs / day of crashing each job is also given. Cost of completing the project in normal situation is Rs 5000.
Activity
1–2 1–3 1–4
Normal Duration
9 8 15
Crash Duration
6 5 10
Cost of crashing (Rs/day) 20 25 30
2–4
3–4 4-5
5
10 2
3
6 1
10
15 40
(1) What is Normal project length & minimum project length (2) Overhead costs total Rs 35 per day, what is the optimum length schedule
Problem 5 . .contd
1-2-4-5 = 16 - III
1-4-5 = 17 - II
2
1-3-4-5 = 20 - I
9
5
1 15
4
2 10
5
8 3
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 Rank 2 1 3 (i) Remarks
II 1-4 4-5
1-4-5 5 1 30 40
17 / 17 1 2
III 1-2 2-4 4-5
1-2-4-5 3 2 1 20 10 40
16 / 16 2 1 3
Problem – 5 . . contd
• Step (i):
? ?
Crashed activity 3-4 by 3 days Now new project durations are:
o o
o
I = 17 II = 17 III = 16
?
Increase in cost = Rs 15 x 3 = 45
• Now path II also becomes critical
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 / 16 Rank 2 1 3 (i) (ii) Remarks
II 1-4 4-5
1-4-5 5 1 30 40
17 / 17 / 16 1 2 (ii)
III 1-2 2-4 4-5
1-2-4-5 3 2 1 20 10 40
16 / 16 / 15 2 1 3 (ii)
Problem – 5 . . contd
• Step (ii):
? ?
Crashed activity 4 - 5 by 1 day Now new project durations are:
o o
o
I = 16 II = 16 III = 15
?
Increase in cost = Rs 40 x 1 = 40
• No new path becomes critical
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 / 16 / 15 Rank 2 1 3 (i) (ii) (iii) Remarks
II 1-4 4-5
1-4-5 5 1 4 30 40
17 / 17 / 16 / 15 (iii) 1 2 (ii)
III 1-2 2-4 4-5
1-2-4-5 3 2 1 20 10 40
16 / 16 / 15 / 15 2 1 3 (ii)
Problem – 5 . . contd
• Step (iii):
? ?
Crashed activity 3 - 4 by 1 day & 1 – 4 by 1 day Now new project durations are:
o o
o
I = 15 II = 15 III = 15
?
Increase in cost = 15 x 1 + 30 x 1 = 45
• Now all 3 paths are critical
Problem 5 . .contd
I Critical Activity 1-3 3-4 4-5 1-3-4-5 Max crash 3 4 1 1 Cost Slope 25 15 40 20 / 17 / 16 / 15 / 12 Rank 2 1 3 Remarks (iv) (i) (ii) (iii)
II 1-4 4-5
1-4-5 5 1 4 1 30 40
17 / 17 / 16 / 15 / 12 (iii) (iv) 1 2 (ii)
III 1-2 2-4 4-5
1-2-4-5 3 2 1 2 20 10 40
16 / 16 / 15 / 15 / 12 (iv) 2 1 3 (iv) (ii)
Problem – 5 . . contd
• Step (iv):
?
?
Crashed activity 1 - 3 by 3 days, 1 – 4 by 3 days, 2 – 4 by 2 days and 1 – 2 by 1 day. Now new project durations are:
o
o
o
I = 12 II = 12 III = 12
?
Increase in cost = 25 x 3 + 30 x 3 + 10 x 2 + 20 x 1 = 205
• Crashing is over since all activities in CP I are crashed
Problem 5 . . contd
• • • • Normal Project duration = 20 days Crashed Project duration = 12 days Normal Project Cost = Rs 5000 Crashed Project cost:
? ? ? ? ?
Normal Project cost = Rs 5000 Increase in cost at step (i) = Rs 45 Increase in cost at step (ii) = Rs 40 Increase in cost at step (iii) = Rs 45 Increase in cost at step (iv) = Rs 205 Total = Rs 5335
Problem 5 . . contd
Step Project Dur 20 i i i ii iii iv iv 19 18 17 16 15 14 13 Direct cost 5000 5000+15 = 5015 5015+15 = 5030 5030+15 = 5045 5045+45 = 5090 5090+45 = 6035 6035+65 = 7000 7000+65 = 7065 Indirect cost Total Cost 700 665 630 595 560 525 490 455 5700 5680 5660 5640 5650 6560 7490 7520
iv
12
7065+75 = 7140
420
7560
Problem 6
Consider the foll activities with normal time and cost together with extra cost of saving a day on selected activity Activity 1–2 1–3 1–4 2–3 2–5 2–6 3–4 3–6 4–6 5-6 Normal Dur 4 6 2 5 2 7 4 3 7 12 Crash Dur 3 5 2 3 1 5 2 2 4 9 Cost of crashing (Rs/day) 10 25 35 15 5 15 20 30 5 10
In addition overhead costs are Rs 35 per day. Calculate optimum time
doc_974177801.ppt