Question

Problem No. 10.2-3. Consider the following project network (as described in Sec. 9.8), where the number over each node is the time required for the corresponding activity. Consider the problem of finding the longest path (the largest total time) through this network from start to finish, since the longest path is the critical path. FINISH START (a) What are the stages and states for the dynamic programming formulation of this problem? (b) Use dynamic programming to solve this problem. However, instead of using the usual tables, show your work graphically. In particular, fill in the values of the various fi (5) under the corresponding nodes, and show the resulting optimal are to traverse out of each node by drawing an arrowhead near the beginning of the are. Then identify the optimal path (the longest path) by following these arrowheads from the Start node to the Finish node. If there is more than one optimal path, identify them all. (e) Use dynamic programming to solve this problem by constructing the usual tables for all n.

Answer 1

Consider the following network problem and the associated figure from the textbook, the objective is to find the longest distance from start to finish.

(a)

There are four stages that have been taken from the origin to the destination in the given problem. The first stage is the origin or the Start. The second stage is the next step from origin towards the destination. It can take any of the two states or
В
.

Now at stage-2 contains states and
В
.

At third stage are the next possible nodes in the route. So the stage-3 has three states
C,D
and
E
. As move further, the possible nodes in the route are
F,G,H
and
1
.

At states form the stage-4. The next stage will then have just one more step before the finish. This is stage-5 and it has states
JAK
and
7
. Thus finally reach the Finish of the network through these stages and states in the given dynamic programming formulation

(9)

In the dynamic programming problem, start from the destination node and compute the longest distance at each stage. Eliminate the shorter routes at each stage in the process. In the given figure, it is given that

f: ()=5 fs *(K)=4 f (L)= 7

To obtain
fi(F)
compute the distance of path
F-J
– Finish.

f. (F)= f4(F,J)+f; (1) = 1+5 = 6

To obtain
f. (6)
compute the distance of path
G-K
– Finish.

f. (G)= f,(G,K)+f, (K) = 4+4 = 8

To obtain
f. (H)
compute the distance of path
H-K
– Finish.

f. (H)= f:(H,K)+f, (K) = 6+4 = 10

To obtain
1:1)
compute the distance of path
7-1
– Finish.

f. (1)=f4(1,1)+f, (L) = 2 +7 = 9

Now at stage-3 computations,

To obtain
f(C)
compute the largest distance from paths
C-F-
– Finish and
C-G-K
– Finish.

$ (C)= max{";(C,F)+ fi(F)] (C,G)+f: (G) $ (C)= max(4+67 4+8) (10) f: (C)= max 12 = 12

To obtain
3 (D)
compute the largest distance from paths
D-H-K
– Finish and
D-1-L
– Finish.

S1;(D,H)+f: (H) f (D)=max 1 (0,1)+1: (1) 2+10 $ (D)=max (2+9 12 L (D)=max 11 = 12

To obtain
f3(E)
compute the largest distance from paths
E-H-K
– Finish and
E-1-L
– Finish.

1. (E)= max {(E,H)+ fi (H) (6 (1,1)+1: (1) (3+10 f; (E)= max 13+9) 13 f; (E)= max 12) = 13

Now at stage-2, to obtain the value of

(4)

f2 (A)=max $13(4,0)+f; (C)] 13(A,D)+f; (D) 5+12 fi (A) = max 5+12 171 f. (A)= max (17) = 17

Similarly find,

fi (B) = f(BE)+f; (E) = 3+13 = 16

Finally the stage-1 computations,

fi (S)= max (S, A)+fi (A (S,B)+f? (B) ti (S)= max {0+17) 10+16 (171 f (S)= max (16) = 17

So the graphical solution of the dynamic programming problem is,

Stages: 2 3 States: 17 А 12 C 17 G 4 K S X 12 D 10 Н. 16 13 L B E 7 1

So the longest path from start to finish is
Start - A-C-G-K - Finish
or
Start - A-D-H-K - Finish
, both giving the longest time as 17.

(c)

The finish node can be reached through either state. Therefore at stage-5 the solution is,

J,K or L

n= 5	s	f* 5 (s)	x 5 *
	J	5	J
	K	4	K
	L	7	L

As we proceed to stage-4, we compute the table for. So,

n=4

n= 4	s	f 4 (s)	f* 4 (s)	x 4 *
		J	K	L
	F	6	-	-	6	J
	G	-	8	-	8	K
	H	-	10	-	10	K
	I	-	-	9	9	L

Further for stage-3 we have table for as,

n=3

n= 3

s

f 3 (s)

f* 3 (s)

x 3 *

F

G

H

I

C

10

12

-

-

12

G

D

-

-

12

11

12

H

E

-

-

13

12

13

H

The next step is to compute the table for n = 2 at stage-2,

n= 2	s	f 2 (s)	f* 2 (s)	x 2 *
		C	D	E
	A	17	17	-	17	C or D
	B	-	-	16	16	E

To proceed further, node can go to states A and B so,

n= 1	s	f 1 (s)	f* 1 (s)	x 1 *
		A	B
	Start	17	16	17	A

Thus the longest path is
Start - A-C-G-K - Finish
or
Start - A-D-H-K - Finish
with longest duration as
[17]
.