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MP CHAPTER 8 SOLUTIONS
SECTION 8.2
1. First label node 1 with a permanent label: [0* 7 12 21 31
44]
Now node 2 receives a permanent label [0* 7* 12 21 31
44].
Node
Temporary Label (* denotes next assigned permanent
label)
3
min{12,7+7} = 12*
4
min{21,7+12} = 19
5
min{31,7+21} = 28
6
min{44,7+31} = 38
Now labels are [0* 7* 12* 19 28 38]
Node
Temporary Label (* denotes next assigned permanent
label)
4
min{19,12+7} = 19*
5
min{28,12+12} = 24
6
min{38,12+21} = 33
Now labels are [0* 7* 12* 19* 24 33]
Node
Temporary Label (* denotes next assigned permanent
label)
5
min{24,19+7} = 24*
6
min{33, 19+12} = 31
Now labels are [0* 7* 12* 19* 24* 31]
Node
Temporary Label (* denotes next assigned permanent
label)
6
min{31,24+7} = 31
Now labels are [0* 7* 12* 19* 24* 31*]
31 24 = c56, 24  12 = c35, 12  0 = c13. Thus 136 is the
shortest path (of length 31) from node 1 to node 6.
2. We begin by permanently labeling node 1 and assigning
temporary labels to other nodes: [0* 2 8 ]. Then we give
node 2 a permanent label: [0* 2* 8 ]
Node Temporary Label (* denotes next assigned permanent 2
label)
3
min{8,2+5} = 7
4
min{, 2+4} = 6*
5
min{,2+12} = 14
Now labels are [0* 2* 7 6* 14]
Node
Temporary Label (* denotes next assigned permanent
label)
3
min{7} = 7*
5
min{14,6+10} = 14
Now labels are [0* 2* 7* 6* 14]. Since there is no node
joining the newest permanently labelled node (node 3) to node
5, we may give make node 5 a permanent label. We obtain [0*
2* 7* 6* 14*]. Since c25 = 14  2 and c12 = 2  0 we find the
shortest path from node 1 to 5 to be 125(length 14). 3.
Node 1
Node 2
Node 3
Node 4 Node 2
2
0 Node 3
8
5 M Node 4 Node 5
M
4 M
12 Supply
1
1 0 5 M 1 M M 0 10 1 1 1 1 1 Demand
4. We begin by giving node 1 a permanent label [0* 2 1 ].
Next node 3 obtains a permanent label: [0* 2 1*]. Node 4 now
obtains a new temporary label of min{,1+1} = 2, and node 2
obtains a permanent label yielding [0* 2* 1* 2]. Finally node
4's temporary label becomes permanent and we obtain [0* 2* 1*
2*] which yields the shortest" path 134. Of course, 1234
with length 1 is a shorter path, but we fail to find this
because Dijkstra's method assumes that because node 3 is not 3
connected to node 1 by an arc, node 3 cannot be the closest
node to node 1.
5. Node 1 = beginning of year 1, Node 7 = end of year 6 or
beginning of year 7
c12 = 3300, c13 = 4800, c14 = 7600, c15 = 9800, c16 = 12,400, c17
= 15,600, c23 = 3300, c24 = 4800, c25 = 7600, c26 = 9800, c27 =
12,400, c34 = 3300, c35 = 4800, c36 = 7600, c37 = 9800, c45 =
3300, c46 = 4800, c47 = 7600, c56 = 3300, c57 = 4800, c67 = 3300
We begin by giving node 1 a permanent label [0* 3300 4800
7600 9800 12,400 15,600]. Next we give node 2 a permanent
label: [0* 3300* 4800 7600 9800 12,400 15,600]
Node
Temporary Label (* denotes next assigned permanent
label)
3
min{4800,3300+3300} = 4800*
4
min{7600,4800+3300} = 7600
5
min{9800,7600+3300} = 9800
6
min{12,400,9800+3300} = 12,400
7
min{15,600,12,400+3300} = 15,600
We now make node 3's label permanent and obtain
[0* 3300* 4800* 7600 9800 12,400 15,600].
Node
Temporary Label (* denotes next assigned permanent
label)
4
min{7600,4800+3300} = 7600*
5
min{9800,4800+4800} = 9600
6
min{12,400,4800+7600} = 12,400
7
min{15,600,4800+ 9800} = 14,600
We now make node 4's label permanent and obtain [0* 3300*
4800* 7600* 9600 12,400 14,600].
Node
Temporary Label (* denotes next assigned permanent
label)
5
min{9600,7600+3300} = 9600*
6
min{12,400,7600+4800} = 12,400
7
min{14,600,7600+7600} = 14,600
We next give node 5 a permanent label and obtain [0* 3300*
4800* 7600* 9600* 12,400 14,600].
Node
Temporary Label (* denotes next assigned permanent
label) 4
6
min{12,400,9600+3300} = 12,400*
7
min{14,600,9600+4800} = 14,400
We now give node 6 a permanent label and obtain [0* 3300*
4800* 7600* 9600* 12,400* 14,400]
Node
Temporary Label (* denotes next assigned permanent
label)
7
min{14,400,12,400+3300} =14,400
We now make node 7's label permanent obtaining [0* 3300*
4800* 7600* 9600* 12,400* 14,400*]. Since 14,400  9600 = c57,
9600  4800 = c35, and 4800  0 = c13, we find the shortest
path from node 1 to node 7 to be 1357. Thus car should be
kept for a two year period and then sold.
6. Node 1 = beginning of year 1, Node 7 = end of year 6 or
beginning of year 7
c12 = 60, c13 = 90, c14 = 130, c15 = 190, c16 = 260 c23 = 60, c24 =
90, c25 = 130, c26 = 190, c27 = 260, c34 = 60, c35 = 90, c36 =
130, c37 = 190, c45 = 60, c46 = 90, c47 = 130, c56 = 60, c57 = 90,
c67 = 60
We begin by giving node 1 a permanent label [0* 60 90 130 190
260 ]. Next we give node 2 a permanent label: [0* 60* 90 130
190 260 ] .
Node
Temporary Label (* denotes next assigned permanent
label)
3
min{90,60+60} = 90
4
min{130,60+90} = 130
5
min{190,60+130}=190
6
min{260,60+190} = 250
7
min{,60+260} = 320
We now make node 3's label permanent and obtain
[0* 60* 90* 130 190 250 320].
Node
Temporary Label (* denotes next assigned permanent
label)
4
min{130,90+60} = 130*
5
min{190,90+90} = 180
6
min{250,90+130} = 220
7
min{320,90+190} = 280
We now make node 4's label permanent and obtain [0* 60* 90*
130* 180 220 280] 5
Node
Temporary Label (* denotes next assigned permanent
label)
5
min{180,130+60} = 180*
6
min{220,130+90} = 220
7
min{280,130+130} = 260
We next give node 5 a permanent label and obtain [0* 60* 90*
130* 180* 220 260]
Node
Temporary Label (* denotes next assigned permanent
label)
6
min{220,180+60} = 220*
7
min{260,180+90} = 260
We now give node 6 a permanent label and obtain [0* 60* 90*
130* 180* 220* 260]
Node
Temporary Label (* denotes next assigned permanent
label)
7
min{260,220+60} =260
We now make node 7's label permanent obtaining [0* 60*
90* 130* 180* 220* 260*]. Since 260  130 = c47 and 130  0 =
c14 we find that 147 yields the shortest path. Thus the
phone should always be kept for three years.
7. Let cij = cost incurred if a machine is purchased at
beginning of year i and is kept until beginning of year j.
Then we may formulate the problem of minimizing total cost
incurred over 5 years as the following transshipment problem: 2 3 4 5 6 1
2 4
5 6 6
208 258 355 537 841 1 0 258 278 375 557 1 M 0 248 298 395 1 1 3
4 1 5 1
M M 0 288 388 1 M 0 388 1 1 1 1 1 M M 1 1 1 The optimal tableau above indicates that we should first keep
the machine for two years and then trade it in. Then we
should keep the new machine for three more years (until
problem is over).
8. Let cij = cost of storing books >i inches and j inches on
a single shelf.c04 = 2300 + 200(.5)(4)(5) = $4300
c08 = 2300 + 300(.5)(8)(5) = $8300
c0,12 = 2300 + 380(.5)(12)(5) = $13,700
c4,8 = 2300 + 100(.5)(5)(8) = $4300
c4,12 = 2300 + .5(180)(5)(12) = $7700
c8,12 = 2300 + .5(5)(12)(80) = $4700
It is easily found that the shortest path from 0 to 12
is 0 412 (cost of $12,000). Thus a 4" shelf should be built
for 4" books, and a 12" shelf is built for 8" and 12" books.
9. For j>i let xij = 1 if a type i box is used to meet demand
for boxes of types i, i+1, ..., j1 . Then we have a
transshipment problem which may be formulated as the
following balanced transportation problem (optimal solution
is also given); all supplies and demands=1: Note: xii = 1
means that a type i box is not used. 7 24,100
1 5
1
2 70,300 83,500 90,100 25,000 46,000 52,000 64,000 70,000 14,200 0 14,000
1 32,200 37,400 47,800 53,000 0 4
8 63,700 10,000 3
7 40,600 M 0 17,800
1 22,600 32,200 37,000 M M M 0 4800 12,400 16,200
1 M M M M 0 8200 11,800 0 4400 1 1
M 3 2
6 M M M M
1 4
5
6
7 For example,c36 = 1000 + 26(500+700+200) = $37,400
We find minimum total cost of $72,100 is attained by using a
size 33 box for size 33 and 30 demand, a size 26 box for size
26 demand, a size 24 box for size 24 demand, and a size 19
box for the remaining demand.
10. To illustrate we consider Example 2 and set up a
transshipment problem whose solution will yield the shortest
path from node 1 to all other nodes. We create a node 0 with
a net supply of 6  1 = 5 units and create an arc (0, 1). We
give node 1 a net supply of 0 and nodes 2, 3, 4, 5, 6 a net
demand of 1. The optimal solution to this problem will ship a
single unit from node 1 to each of nodes 16. The single unit
shipped from node 1 to node i will be shipped along the
shortest path from node 1 to node i; if the single unit
shipped from node 1 to node i in the "optimal solution" was
not shipped along the shortest path from node 1 to node i, 8
then by transferring this unit to the shortest path from node
1 to node i we would find a better solution to the
transshipment problem, thereby contradicting the assumed
optimality of our current solution.
SECTION 8.3 SOLUTIONS
1. max z = x0
s.t. xso,16, xso,22 x121, x323, x133, x3,si2, x247,
x4,si7
x0 = xso,1 + xso,2 (Node so)
xso,1 = x13 + x12 (Node 1)
x12 + x32 + xs0,2 = x24 (Node 2)
x13 = x32 + x3,si (Node 3)
x24 = x4,si (Node 4)
x3,si + x4,si = x0 (Node si)
All variables 0
Initial flow of 0 in each arc. Begin by labeling sink
via path of forward arcs (so, 1)  (1, 3)  (3, 2)  (2, 4) (4, si). Increase flow in each of these feasible arcs by 3,
yielding the following feasible flow: Arc
Flow
so1
3
so2
0
13
3
12
0
24
3
3si
0
32
3
4si
3
Flow to sink 3
Now label sink by (so2)  (24), (4, si). Each arc is
a forward arc and we can increase flow in each arc by 2.
This yields the following feasible flow: 9
Arc
Flow
so1
3
so2
2
13
3
12
0
24
5
3si
0
32
3
4si
5
Flow to sink 5
Now label sink by (so1)  (1, 2)  (3, 2)  (3, si).
All arcs on this path are forward arcs except for (3, 2),
which is a backwards arc. We can increase the flow on each
forward arc by 1 and decrease the flow on each backward arc
by 1. This yields the following feasible flow:
Arc
Flow
so1
4
so2
2
13
3
12
1
24
5
3si
1
32
2
4si
5
Flow to sink 6
The sink cannot be labeled, so we found the maximum flow
of 6 units. The minimum cut is obtained from V' = {3, 2, 4,
si}. This cut consists of arcs (1, 3), (1, 2), (so, 2) and
has capacity of 3 + 1 + 2 = 6 = maximum flow.
2. max z = x0
s. t.
xso,12, x124, x1,si3,x2,si2, x231,x3,si2,xso,31
x0 = xso,1 + xso,3 (Node so)
xso,1 = x1,si + x12 (Node 1)
x12 = x23 + x2,si (Node 2)
x23 + xso,3 = x3,si (Node 3)
x1,si + x2,si + x3,si = x0 (Node si)
All variables 0 We begin with a flow of 0 through each arc. Label the
sink by (so, 1)(1, 2)(2, 3)(3, si). We can increase the
flow on each of these arcs by one unit, obtaining the 10
following feasible flow: 11
Arc Flow (so,1) 1 (so,3) 0 (1,2) 1 (1,si) 0 (2,3) 1 (2,si) 0 (3,si) 1 Flow to
Sink 1 We next label the sink by (so, 1)(1, 2)(2, si) and
increase the flow in each of these arcs by 1 unit.
We
obtain the following feasible flow: Arc Flow (so,1) 2 (so,3) 0 (1,2) 2 (1,si) 0 (2,3) 1 (2,si) 1 (3,si) 1 Flow to
Sink 2 Now we label the sink by (so, 3)(3, si), and increase 12
the flow in each of these arcs by 1 unit.
following feasible flow: We obtain the 13
Arc Flow (so,1) 2 (so,3) 1 (1,2) 2 (1,si) 0 (2,3) 1 (2,si) 1 (3,si) 2 Flow to
sink 3 Now the sink cannot be labeled, so we have obtained a
maximal flow.
V' = {1, 2, 3, si} yields the minimal cut
(arcs (so, 1) and (so, 3)) with a capacity of 2+1=3 =
maximal flow.
3. max z = x0
s.t. xso,11, xso,23, x134, x123, x3,
x0 = xso,1 + xso,2 (Node so)
xso,1 = x13 + x12 (Node 1)
x12 + xso,2 = x2,si (Node 2)
x13 = x3,si (Node 3)
x3,si + x2,si = x0 (Node si)
All variables 0 si 1, x2,si 2 We begin with a flow of 0 through each arc and label
the sink by (so2)(2si). We increase the flow in each of
these arcs by 2 units. This yields the following feasible
flow.
so1 0 so2 2 12 0 13 0 14
2si 2 3si 0 Flow to
Sink 2 Now label the sink by the chain (so, 1)(1, 3)(3, si)
and increase the flow in each of these arcs by 1 unit. This
yields the following feasible flow: so1 1 so2 2 12 0 13 1 2si 2 3si 1 Flow to
Sink 3 The sink cannot be labeled, so we have obtained a
maximal flow.
V' = {si} yields the minimal cut (3,
si), (2, si) with a capacity of 1+2 = 3 = maximum flow.
4. Maximum flow is 45. Min Cut Set = {1, 3, and si}.
Capacity of Cut Set = 20 + 15 + 10 = 45. See Figure. 15 16 5. Maximum flow = 9. Min Cut Set = {2, 4, si}. Capacity of
Cut = 3 + 1 + 3 + 2 = 9. See Figure. 17 18 19 6. See figure. If the maximum flow = 21, then all 20
packages can be loaded. 7. See figure. 21 8. See figure. 9. Step 1Add 4 units of flow along so13si
Step 2Add 2 units of flow along so24si
Step 3Add 1 unit of flow along so243si
Step 4Add 1 unit along so213si
Cannot label si. Maximum flow = 8 Arcs in cut are 3si
and 4si with capacity of 8. Flow along each arc is as
follows:
Arc
Flow
so,1
4
so,2
4
2,1
1
1,3
5 22
1,4
2,4
3,si
4,3
4,si 0
3
6
1
2 10. Step 1Add 7 units along so2si
Step 2Add 6 units along so13si
Step 3 Add 6 units along so4si
Cannot label si. Maximal flow = 19. Cut set is derived from
V' = {si} and consists of arcs (3, si), (2, si), (4, si)
with capacity 6+7+6=19
Arc
so,1
so,2
so,4
1,3
2,si
3,si
4,si Flow
6
7
6
6
7
6
6 11. Suppose maximum flow = k. At each step the flow to the
sink increases by an integer amount of at least one
unit. Thus at most k iterations of the FordFulkerson
method will yield the maximum flow. Also, after each
iteration the flow to the sink is an integer, so the
optimal flow will be an integer. 12. Construct a "supersource" that has arcs of infinite
capacity leading to each real source. Also construct a
"supersink' that has infinite capacity arcs leading
from each real sink to the supersink. 13. Suppose we know the flow into node i cannot exceed 10
units.
Add a new node, node i', to the network.
Replace each arc of the form (j,i) in the original
network with an arc (j,i') and add an arc (with
capacity 10 units) (i',i).
This will ensure that at
most 10 units flow into node i! 14. See figure. 23 15. Let F = a set of flights and CUTF = cut
corresponding to the sink, nodes associated with
flights not in F, and nodes associated with airports
not used by F. Then CUTF consist of arcs from so to
nodes for flights not in F and arcs from nodes for
airports used by F to si.
Then
Capacity of Cut F = (revenue from flights not in F) +
(costs associated with airports used by F).
Therefore
(Profit from F) = (total revenue from all flights) (capacity of CUTF).
Thus set of flights F' corresponding to CUTF' of
minimum capacity will maximize profit. Note that the
network contains cuts other than those CUTF cuts
corresponding to a set of flights, but all these other
cuts have an infinite capacity. Thus the minimal cut
must be CUTF' for some set of flights F'.
Once the
minimal cut is found, the airline knows what set of
flights will maximize profit. 16. All arcs from month i workers to Project j have a
capacity of 6. All projects can be completed if and
only if the maximum flow from source to sink equals 24
30. SECTION 8.4
1. 2. We could not determine ET(2) without knowing ET(4).
Similarly, we could not determine ET(4) without knowing
ET(3). To find ET(3), however, we need to know ET(2).
Thus we are in trouble! The reason for this is the arc
from a higher numbered node (4) to a lower numbered
node (2).
Activity Predecessors Duration (wks.) A = Design  5 B = Make Part A A 4 C = Make Part B A 5 D = Make Part C A 3 E = Test Part A B 2 F = Assemble A and
B C,E 2 G = Attach C D,F 1 25
H = Test Final
Product G 1 From the project diagram we find that
ET(1) = 0 LT(1) = 0 TF(1,2) = 0 FF(1,2) = 0 ET(2) = 5 LT(2) = 5 TF(2,3) = 0 FF(2,3) = 0 ET(3) = 9 LT(3) = 9 TF(3,4) = 0 FF(3,4) = 0 ET(4) = 11 LT(4) = 11 TF(2,4) = 1 FF(2,4) = 1 ET(5) = 13 LT(5) = 13 TF(4,5) = 0 FF(4,5) = 0 ET(6) = 14 LT(6) = 14 TF(2,5) = 5 FF(2,5) = 5 ET(7) = 15 LT(7) = 15 TF(5,6) = 0 FF(5,6) = 0 TF(6,7) = 0 FF(6,7) = 0 Looking at the activities with TF of 0, we find that
the critical path is 1234567 (length 15 days).
The appropriate LP is
min z = x7  x1
s.t.
x2x1 + 5
x3x2 + 4
x4x3 + 2
x4x2 + 5
x5x2 + 3
x5x4 + 2
x6x5 + 1
x7x6 + 1
all variables urs 26 3.In the project diagram we use the mean time for each
activity. For each activity's duration we find Activity Mean Variance (1,2) 6 0.44 (1,3) 4.33 1 (2,4) 3.33 1 (3,4) 9 1 (3,5) 10 2.78 (3,6) 12.17 3.36 (4,7) 8.83 1.36 27
(5,7) 2 0.11 (6,8) 3.33 0.44 (7,9) 15 2.78 (8,9) 8.83 0.69 From the project diagram we find that
ET(1) = 0 LT(1) = 0 TF(1,2) = 4 FF(1,2) = 0 ET(2) = 6 LT(2) = 10 TF(1,3) = 0 FF(1,3) = 0 ET(3) = 4.33 LT(3) = 4.33 TF(2,4) = 4 FF(2,4) = 4 ET(4) = 13.33 LT(4) = 13.33 TF(3,4) = 0 FF(3,4) = 0 ET(5) = 14.33 LT(5) = 20.16 TF(3,5) = 5.83 FF(3,5) = 0 ET(6) = 16.5 LT(6) = 25 TF(3,6) = 8.5 FF(3,6) = 0 ET(7) = 22.16 LT(7) = 22.16 TF(4,7) = 0 FF(4,7) = 0 ET(8) = 19.83 LT(8) = 28.33 TF(5,7) = 5.83 FF(5,7) = 5.83 ET(9) = 37.16 LT(9) = 37.16 TF(6,8) = 8.5 FF(6,8) = 0 TF(7,9) = 0 FF(7,9) = 0 28
TF(8,9) = 8.5 FF(8,9) = 8.5 We find the critical path to be 13479 with expected
length of the project being 4.33 + 9 + 8.83 + 15 = 37.16 and
the variance of the project length being given by 1 + 1 +
1.36 + 2.78 = 6.14.
Since the standard deviation of the length of the
critical path is given by 6.141/2 = 2.48 we find that
CP  37.16
40  37.16
P(CP40) = P( ) =
2.48
2.48
= P(Z1.15)
= .875 . Alternatively we could find answer in EXCEL with
the formula =NORMDIST(40,37.16,2.48,1)= .875.
We can find the critical path from the following LP:
min z = x9  x1
s.t.
x2x1 + 6
x3x1 + 4.33
x4x2 + 3.33
x4x3 + 9
x5x3 + 10
x6x3 + 12.17
x7x4 + 8.83
x7x5 + 2
x8x6 + 3.33
x9x7 + 15
x9x8 + 8.83
all variables urs
4a. See figure 4b. For the duration of each activity we find that 29 Activity Mean Variance (1,2) 3 0.11 (2,3) 2 0.11 (2,4) 6 1.78 (4,7) 2 0.11 (2,8) 3 0.44 (3,5) 3 0.11 (4,8) 5 0.44 30
(4,5) 1 0.03 (5,6) 1.5 0.03 (6,8) 2 0.11 (7,8) 0 0.00 ()
From the project diagram we find that
ET(1) = 0 LT(1) = 0 TF(1,2) = 0 FF(1,2) = 0 ET(2) = 3 LT(2) = 3 TF(2,3) = 2.5 FF(2,4) = 0 ET(3) = 5 LT(3) = 7.5 TF(2,4) = 0 FF(2,4) = 0 ET(4) = 9 LT(4) = 9 TF(4,7) = 3 FF(4,7) = 0 ET(5) = 10 LT(5) = 10.5 TF(2,8) = 8 FF(4,8) = 0 ET(6) = 11.5 LT(6) = 12 TF(3,5) = 2.5 FF(3,5) = 2.33 ET(7) = 11 LT(7) = 14 TF(4,8) = 0 FF(4,8) = 0 ET(8) = 14 LT(8) = 14 TF(4,5) = 0.5 FF(4,5) = 0 TF(5,6) = 0.5 FF(5,6) = 0 TF(6,8) = 0.5 FF(6,8) = 0 TF(7,8) = 2.5 FF(7,8) = 2.5 The critical path is 1248 with expected length 3 + 6 + 5
= 14.
4c. The variance of the critical path is 0.11 + 1.78 + 0.44
= 2.33. Then the standard deviation of the critical path is
2.331/2 = 1.53.
Let D = random variable representing
duration of the project and d = number such that 99% of the
time the project duration is at most d days. Then
P(Dd) = .99 or
D14 d14 P(
1.53 ) = .99.
1.53 31
We
find
that
F(2.33)
=
NORMINV(.99,14,1.53)= 17.56
d14
Thus
1.53 .99. Also we may use = 2.33 or d = 17.56 Now if we begin project on June 13 we have 18 days to
complete the project. Thus if we begin the project on June
13, there is a 99% chance of completing the project by the
end of June 30.
4d.
min z = x8  x1
s.t.
x2x1 + 3
x3x2 + 2
x4x2 + 6
x8x7
x5x4 + 1
x5x3 + 3
x6x5 + 1.5
x7x4 + 2
x8x2 + 3
x8x4 +5
x8x6 +2
all variables urs 32
5a.From the project diagram we find that
ET(1) = 0 LT(1) = 0 TF(1,2) = 0 FF(1,2) = 0 ET(2) = 5 LT(2) = 5 TF(2,3) = 0 FF(2,3) = 0 ET(3) = 13 LT(3) = 13 TF(3,5) = 0 FF(3,5) = 0 ET(4) = 17 LT(4) = 17 TF(3,6) = 8 FF(3,6) = 8 ET(5) = 23 LT(5) = 23 TF(3,4) = 0 FF(3,4) = 0 ET(6) = 26 LT(6) = 26 TF(4,5) = 0 FF()4,5 = 0 Both are 12356 and 123456 are critical paths
having length 26 days. 5b. Let A = number of days by which we reduce duration of
activity A, etc. and xj = time that event at node j occurs
min z = 30A + 15B + 20C + 40D + 20E + 30F + 40G
A2, B3, C1, D2, E2, F3, G1 33
x2x1 + 5  A
x3x2 + 8  B
x4x3 + 4  E
x5x3 + 10  C
x5x4 + 6  F
x6x5 + 3  G
x6x3 + 5  D
x6  x120
A,B,C,D,E,F,G0, xj urs 6a. From the project network we find that
ET(1) = 0
LT(1) = 0
TF(1,2) = 0
ET(2) = 2
LT(2) = 2
TF(2,3) = 0
ET(3) = 6
LT(3) = 6
TF(3,5) = 0
ET(4) = 8
LT(4) = 9
TF(3,4) = 1
ET(5) = 9
LT(5) = 9
TF(3,6) = 9
ET(6) = 19
LT(6) = 19
TF(5,6) = 0
TF(4,5) = 1 FF(1,2)
FF(2,3)
FF(3,5)
FF(3,4)
FF(3,6)
FF(5,6)
FF(4,5) =
=
=
=
=
=
= 0
0
0
0
9
0
1 The critical path is 12356. The length of the critical
path is 19 days. 34
6b. min z = x6  x1
st
x2x1 +
x3x2 +
x4x3 +
x5x3 +
x5x4
x6x5 +
x6x3 +
xj urs 2
4
2
3
10
4 7a. From the project network we find
ET(1) = 0
LT(1) = 0
TF(1,2)
ET(2) = 3
LT(2) = 3
TF(2,3)
ET(3) = 17
LT(3) = 17
TF(3,4)
ET(4) = 25
LT(4) = 25
TF(4,5)
ET(5) = 29
LT(5) = 29
TF(2,5)
ET(6) = 37
LT(6) = 37
TF(5,6)
ET(7) = 46
LT(7) = 46
TF(6,7) =
=
=
=
=
=
= 0
0
0
0
20
0
0 FF(1,2)
FF(2,3)
FF(3,4)
FF(4,5)
FF(2,5)
FF(5,6)
FF(6,7) =
=
=
=
=
=
= 0
0
0
0
20
0
0 The critical path is 1234567 with a duration of 46
days. The critical path may also be found from the following
LP: min z = x7 st
x2x1
x3x2
x4x3
x5x4 x1
+
+
+
+ 3
14
8
4 35
x5x2 + 6
x6x5 + 8
x7x6 +9
xj urs
7b. min z = 100A + 80B + 60C + 70D + 30E + 20F + 50G
st A3, B4, C5, D2, E4, F4, G4
x2x1 + 3  A
x3x2 + 14  C
x4x3 + 8  D
x5x4 + 4  E
x5x2 + 6  B
x6x5 + 8  F
x7x6 + 9  G
x7  x130
A, B, C, D, E, F, G, _0, xj urs
8a. Using the fact that each constraint in the LP represents
an arc and the rhs of any constraint is the duration of the
activity associated with that arc, we obtain the project
network given in the answer to problem 5.
8b. Looking at the rows with dual prices of 1 we find that
12 3456 is a critical path, the length of the project
is 26 days, and the activities corresponding to arcs (1,2),
(2,3), (3,4), (4,5) and (5,6) are critical activities.
9. Since ET(j)LT(j) we find that FF(i, j)TF(i, j) will
hold for any activity.
10. See Figure. Note that without the dummy arc Rule 4 would
have been violated. 36 11. Let CD = Duration of path 134 AB = Duration of path
124.
For CD there are 9 equally likely outcomes; C may last a
time units and D a time units, etc Each of these outcomes
has a probability of 1/9.. Thus
P(CD = 12) = 1/9, P(CD = 14) = 3/9, P(CD = 13) = 2/9 P(CD =
15) = 2/9, P(CD = 16) = 1/9.
Similarly, P(AB = 7) = 1/9, P(AB = 11) = 2/9, P(AB = 15)
= 3/9, P(AB = 19) = 2/9, P(AB = 23) = 1/9
Now Probability that only 124 is critical path
is 37
given by
P(CD = 12)P(AB>12) + P(CD = 14)P(AB>14) + P(CD = 13)P(AB>13)
+P(CD = 15)P(AB>15) +P(CD = 16)P(AB>16) = (1/9)(6/9) + (3/9)
(6/9)
+(2/9)(6/9) + (2/9)(3/9) +(1/9)(3/9) = 15/27.
Probability that both 124 and 134 are critical paths is
P(AB = 15)P(CD = 15) = 2/27.
Then the probability that 134 is the only critical path is
1  2/27  15/27 = 10/27
12a. See Figure 12b. Critical paths are ADG, AEG, BDG, and BEG, all
having length 10.
13. See figure. 38 14. MODEL:
1]SETS:
2]NODES/1..6/:TIME;
3]ARCS(NODES,NODES)/
4]1,2 1,3 2,3 3,4 3,5 4,5 5,6/:DUR,CR,LIM,COST;
5]ENDSETS
6]MIN=@SUM(ARCS(I,J):COST(I,J)*CR(I,J));
7]@FOR(ARCS(I,J):TIME(J)>TIME(I)+DUR(I,J)CR(I,J););
8]TIME(6)TIME(1)<25;
9]@FOR(ARCS(I,J):CR(I,J)<LIM(I,J););
10]DATA:
11]LIM=5,5,0,5,5,5,5; 39
12]COST=20,10,0,30,3,40,50;
13]DUR=9,6,0,7,8,10,12;
14]END DATA
END
15. min z = 300A+200B+350C+260D+320E
st A5, B5, C5, D5, E5,
x3x190,
x2x1 + 20  A
x3x1 + 25  B
x4x2 + 50  C
x4x3 + 40  D
x5x4 + 30  E
All variables 0 16. ET(1) = 0, ET(2) = 2, ET(3) = 8, ET(4) = 6,
ET(5) = Max{ET(3)+4, ET(4)+2} = 12
ET(6) = Max{ET(3)+1, ET(5)+2, ET(4)+1} = 14
LT(6)=14, LT(5) = 12,
LT(4) = Min{LT(5)2,LT(6)1} = 10
LT(3) = Min{LT(5)4,LT(6)1} = 8
LT(2) = Min{LT(3)6,LT(4)4} = 2
LT(1) = LT(2)  2 = 0
12356 is critical path(length 14)
Arc
TF
FF
1,2
202=0
202=0
2,3
826=0
826=0
2,4
1024=4
624=0 TF not equal to FF!
3,5
1284=0
1284=0
3,6
1481=5
1481=5
4,5
1262=4
1262=4
4,6
1461=7
1461=7
5,6
14122=0
14122=0
17. ET(1) = 0, ET(2) = 3, ET(3) = 3+3=6,
ET(5) = Max{ET(3)+4 ,ET(2)+4, ET(4)+3} =
ET(6) = Max{ET(3)+5 , ET(5)+5} = 15
ET(7) = Max{ET(6)+6, ET(5)+4, ET(4) + 2}
ET(8) = ET(7) + 6 = 27
LT(8) = 27, LT(7) = 27  6 = 21, LT(6) =
LT(5) = Min{LT(7)4, LT(6)5} = 10
LT(4) = 10  3 = 7 ET(4) = 3+3=6
10
= 21
21  6 = 15 40
LT(3) = Min{LT(6)5,LT(5)4} = 6
LT(2) = Min{LT(5)4, LT(4)3, LT(3)3} = 3
LT(1) = 3  3 = 0
Arc
1,2
2,3
2,4
2,5
3,5
3,6
4,5
4,7
5,6
5,7
6,7
7,8 TF
303=0
633=0
733=1
1034=3
1064=0
1565=4
1063=1
2162=13
15105=0
21104=7
21156=0
27216=0 FF
303=0
633=0
633=0
TF not equal to FF!
1034=3
1064=0
1565=4
1063=1
2162=13
15105=0
21104=7
21156=0
27216=0 Critical Path is 1235678; length 27.
SECTION 8.5
1. min z = 4x12 + 3x24 + 2x46 +3x13 + 3x35 + 2x25 + 2x56
st
x12 + x13 = 1 (node 1 constraint)
x12 = x24 + x25 (node 2 constraint)
x13 = x35 (node 3 constraint)
x24 = x46 (node 4 constraint)
x25 = x56 (node 5 constraint)
x46 + x56 = 1 (node 6 constraint)
all xij0
If xij = 1 the shortest path from node 1 to node 6 will
contain arc (i, j) while if xij = 0 the shortest path from
node 1 to node 6 does not contain arc (i, j).
2. Let yij be the dual variable corresponding to the arc (i,
j) constraint. Then the dual to the CPM LP is
max w = 6y13 + 9y12 + 8y35 + 7y34 + 10y45 + 12y56
st
y13 y12 = 1
y12  y23 = 0
y13 + y23  y35  y34 = 0
y34  y45 = 0
y35 + y45  y56 = 0 41
y56 = 1
all yij0
This LP represents the problem of sending one unit
of flow from node 1 to node 6 through a path of maximum
length in the project diagram. It is a MCNFP because each
variable has a +1 coefficient in a single constraint and a
1 coefficient in a single constraint. By the Dual Theorem,
the optimal zvalue for the above LP = Length of the
critical path. Thus the length of the critical path in a
project network is equal to the length of the longest path
in the network. 3. Node
Net Outflow
Detroit
6500
Dallas
6000
City 1
5000
City 2
4000
City 3
3000
Dummy
500
All arcs from Detroit or Dallas to City 1, 2, or 3 have a
capacity of 2200. Other arcs have infinite capacity
Arc
Shipping Cost
DetroitCity 1
$2800
DetroitCity 2
$2600
DetroitCity 3
$2300
DetroitDummy
$0
DallasCity 1
$2300
DallasCity 2
$2000
DallasCity 3
$2000
DallasDummy
$0
4a. All arcs have infinite capacity
Arc
Shipping Cost
BosChic
800 + 80 = $880
BosAustin 800 + 220 = $1,020
BosLA
800 + 280 = $1,080
RalChic
900 + 100 = $1,000
RalAus
900 + 140 = $1,040 42
RalLA
900 + 170 = $1,070
ChicAus
$40
ChicLA
$50
Problem is balanced so no Dummy Point is needed
City
Net Outflow
Boston
400
Raleigh
300
Chicago
0
LA
400
Austin
300 4b. If at most 200 units can be shipped through Chicago, we
must add a node Chicago'. Delete the RaleighChicago and
Boston Chicago arcs and insert the following arcs
Arc
BostonChicago'
RaleighChicago'
ChicagoChicago' Shipping Cost
$880
$1000
0 Arc Capacity 200 5a. Net Outflow (in Hundred Thousand Barrels/Day) 43
SD
LA
Dallas
Houston
Chic NY
5
4
0
0
4
3
Unit Cost For Each Arc
SDDallas:$420
SDHouston:$100
LADallas:$300
LAHouston:$110
SDDummy:$0
LADummy:$0
Dallas  Chic.: 700 + 550 = $1,250
DallNY:700 + 450 = $1,150
Houston  Chic.: 900 + 530 = $1,430
Hous.  N.Y.:900 + 470 = $1,370 Dummy
2 5b. Delete arcs leading into Dallas and Houston. Add nodes
Dallas' and Houston' to the network. Also add the following
arcs:
Arc
Shipping Cost
LADallas'
$300
SDDallas'
$420
LAHouston'
$110
SDHouston'
$100
Dallas'Dallas
0
Houston'Houston
0 Arc Capacity 500
500 6. The objective function represents firing costs + hiring
costs + salary costs. (1) Ensures that at least 20 workers
are working during month 1. (2) Ensures that at least 16
workers are working during month 2. (3) Ensures that at
least 25 workers are working during month 3.
An LP with constraints (i)(iv) has the same feasible
region as the LP with constraints (1)(3). To see this
observe that any point in the feasible region for (1)(3)
clearly satisfies (i) (iv). If a point satisfies (i)(iv),
then the point satisfies (1), satisfies, (i) + (ii) = (2),
and (iv) = 3. Thus the two LP's have the same feasible
region. The LP with constraints (i) (iv) has the following
constraint matrix.
x12 x13 x14 x23 x24 x34
e1
e2
e3 RHS
 Node
Node
Node
Node 1'
2'
3'
4' 1
1
0
0 1
0
1
0 1
0
0
1 0
1
1
0 44
0
1
0
1 0
0
1
1 1
1
0
0 0
1
1
0 0
0
1
1 20
4
9
25 Each variable has a +1 coefficient in a single
constraint and a 1 coefficient in a single constraint, so
these constraints are the flow balance equations for the
network in the figure. The variable e 1 represents "flow"
from node 2' to node 1', the
variable e2 represents flow
from node 3' to node 2' and the variable e 3 represents flow
from node 4' to node 3'. All arcs have infinite capacity.
Shipping costs are as follows:
Arc
Shipping Cost
Arc
Shipping Cost
1'2'
50 + 100 + 140
4'3'
0
1'3'
50 + 100 + 280
3'2'
0
1'4'
100 + 420
2'1'
0
2'3'
50 + 100 + 140
2'4'
100 + 280
3'4'
100 + 140 7. See Figure 45
8. The network requires the following nodes:
Monday
Tuesday
Wednesday Thursday
Phil
Phil
Phil
Phil
NY
NY
NY
NY
Wash
Wash
Wash
Wash Friday
Phil
NY
Wash Six arcs emanate from each node, indicating whether the
truck is loaded or empty and where the truck will be
tomorrow. For example, from the Monday Philadelphia node we
have the following six arcs: P = Philadelphia N = New York W
= Washington M = Monday, T = Tuesday, etc.
Cost of Arc
MPTP(loaded)
$700
MPTP(empty)
$400
MPTW(loaded)
$1000
MPTW(empty)
$800
MPTN(loaded)
$1000
MPTN(empty)
$800
Arcs which correspond to fulfilling a shipping
requirement are handled by making the arc have lower bound =
upper bound = requirement. Thus MPTN(loaded) has upper
bound = lower bound = 2. All other arcs have a lower bound
of 0 and upper bound of . Arcs are needed from Friday
nodes to Monday nodes(six arcs leave each Friday node). Each
node has a net outflow of 0.
SECTION 8.6
1. We begin at Gary and include the GarySouth Bend arc.
Then we add the South BendFort Wayne arc. Next we add the
GaryTerre
Haute
arc.
Finally
we
add
the
Terre
HauteEvansville arc. This minimum spanning tree has a total
length of 58 + 79 + 164 + 113 = 414 miles.
2. We begin at node 1 with arc (1,3). Then we add (in turn)
arcs (3,5), arc (3,4), and arc (3,2). Total length of the
tree is 15.
3. If you replace at' with at you still have a spanning tree
(this follows because at connects Ct1 to Ct1' so all nodes
are 46
still connected). The spanning tree with at has a shorter
total
length than the spanning tree with at'. Thus the tree with
at' is
not a minimum spanning tree. This contradiction proves that
the algorithm yields a minimum spanning tree.
4a. Clearly MST has length 1+1 = 2.
4b. See Figure. Length of MST is AD+DC+DB=3DC. CDE is a 3060 right triangle with angle DCE = 30 degrees and angle CDE
= 60 degrees. Length CE = .5. Let length DE = x. Then length
DC = (.25+x2)1/2. Since sine 30 degrees = .5, x/(.25+x2)1/2 = .
5 which yields x = .5*(3)1/2. Then length DC = 1/31/2 and
length of MST = 3DC = 31/2, which is 13% smaller than 2! Section 8.7
Problems 47 1a. 1b. One possible bfs is
2 1 4 1 1
(1) 0 1c) 6 1
3 5 0 Find the Y values.
Y1=0, Y1Y2=4, Y1Y3=3, Y2Y4=3, Y5Y6=2, Y4Y6=2
This yields (1) 48
Y1=0, Y2=4, Y3=3, Y4=7, Y5=7, Y6=9
Find the row 0 coefficients for each nonbasic
variable.
_
C25=4(7)2=1 (violates optimality condition)
_
C35=3(7)3=1 (violates optimality condition)
Ties are broken arbitrarily, so choose to enter X 25
into the basis.
2 1 4 1 1
(1)
(1) 6 1 0 3 0+ 5 =1
Either X24 or X46 leaves the basis. Arbitrarily choose X46. 2 0 4 1
(1) 1 6 1
0 3 5 1 (1) 49
Find the Y values.
Y1=0, Y1Y2=4, Y1Y3=3, Y2Y4=3, Y2Y5=2, Y5Y6=2
This yields
Y1=0, Y2=4, Y3=3, Y4=7, Y5=6, Y6=8
Find the row 0 coefficients for each nonbasic variable
__
C35 = 3 – (6) 3 = 0 (satisfies optimality condition)
__
C46 = 7 – (8) 2 = 1 (satisfies optimality condition)
Thus, an optimal solution to this MCNFD is
(Basic Variables) X12 = 1, X25 = 1, X56 = 1
(Basic Variables at lower bound) X13 = 0, X24 = 0
(Nonbasic Variables at lower bound) X35 = 0, X46 = 0
The optimal zvalue is
Z = 1(4) + 1(2) + 1(2) = $8
(Note that there are alternate optimal solutions.)
2.) Many answers may be correct as long as the arcs
corresponding to
the basic variables form a spanning tree. A bfs which
happens to be optimal is shown. 50 3.) Find the Y values.
Y1 = 0, Y1 – Y2 = 15, Y2 – Y4 = 5, Y2 – Y5 = 10, Y3 – Y4 =
4
This yields
Y1 = 0, Y2 = 15, Y3 = 16, Y4 = 20, Y5 = 25
Find the row 0 coefficients for each nonbasic
variable.
_
C13 = 0(16)11 = 5 (satisfies optimality condition) 51
_
C23
=
15(16)5
=
4
(satisfies
optimality
condition)
_
C35 = 16(25)5 = 4 (violates optimality condition)
_
C45 = 20(25)14 = 9 (satisfies optimality
condition)
Enter X35 into the basis. 10+ 2 4 (22) 20
10(32) 1
1212
3 5 (10) 52
= 10, X25 leaves the basis 20 2 4 (22) 20
(32) 1
2
12 3 5 (10) 10
Find the Y values.
Y1 = 0, Y1 – Y2 = 15, Y2 – Y4 = 5, Y3 – Y4 = 4, Y3 – Y5 = 5
This yields
Y1 = 0, Y2 = 15, Y3 = 16, Y4 = 20, Y5 = 21
Find
_
C13 =
_
C23 =
_
C25 =
_
C45 = the row 0 coefficients for each nonbasic variable.
0(16)11 = 5 (satisfies optimality condition)
15(16)5 = 4 (satisfies optimality condition)
15(21)10 = 4 (satisfies optimality condition)
20(21)14 = 13 (satisfies optimality condition) Thus, an optimal solution to this MCNFP is
(Basic Variables) X12 = 20, X24 = 20, X34 = 2, X35 = 10
(Nonbasic Variables at upper bound) X13 = 12
(Nonbasic Variables at lower bound) X23 = 0, X25 = 0, X45 = 0
The optimal zvalue is
Z = 20(15) + 12(11) + 20(5) + 2(4) + 10(5) = $590
4.) Many answers may be correct as long as the arcs 53
corresponding to the basic variables form a spanning
tree. A bfs which happens to be optimal is shown. 5.) Find the Y values.
Y1 = 0, Y1Y2 = 15, Y2Y4 =10, Y2Y6 = 5, Y3Y6 = 8, Y4Y5 = 30, Y5Y7 = 10, Y6Y8 = 15
This yields
Y1 = 0, Y2 = 15, Y3 = 12, Y4 = 25, Y5 = 55, Y6 =
20, Y7 = 65, Y8 = 35 Find the row 0 coefficients for each
variable.
_
C13 = 0(12) 10 = 2 (satisfies
condition)
_
C23
=
15(12)5
=
8
(satisfies
condition)
_
C25 = 15(55) 10 = 30 (violates
condition)
_
C35
=
12(55)25
=
18
(violates
condition) nonbasic
optimality
optimality
optimality
optimality 54
_
C47
condition)
_
C58
condition)
_
C67
condition)
_
C78
condition) = 25(65)45 = 15 (satisfies optimality = 35(35)10 = 10 (satisfies optimality 15 (violates optimality 75 (satisfies optimality =
= 20(65)30
65(35)45 Enter X25 into the basis =
= 4 2525 2
50 7 (25) 25
5 (100) 1 50 25
3
50 6 8
75 = 25, Choose X45 to leave the basis (75) 55 0 4 2
50 25 1 (100) (25) 7 25
5 50 25
3
50 6 8 (75) 75 Find the Y values.
Y1=0, Y1Y2=15, Y2Y4=10, Y2Y5=10, Y2Y6=5, Y3Y6=8, Y5Y7=10,
Y6Y8=15
This yields
Y1=0, Y2=15, Y3=12, Y4=25, Y5=25, Y6=20, Y7=35, Y8=35
Find
_
C13 =
_
C23 =
_
C35 =
_
C45 =
_
C47 =
_
C58 =
_
C67 =
_ the row 0 coefficient for each nonbasic variable
0(12)10 = 2 (satisfies optimality condition)
15(12)5 = 8 (satisfies optimality condition)
12(25)25 = 12 (satisfies optimality condition)
25(25)30 = 30 (satisfies optimality condition)
25(35)45 = 35 (satisfies optimality condition)
25(35)10 = 0 (satisfies optimality condition)
20(35)30 = 15 (satisfies optimality condition) 56
C78 = 35(35)45 = 45 (satisfies optimality condition)
Thus, an optimal solution to this MCNFP is
(Basic variables) X12 = 50, X25 = 25, X26 = 25, X36 = 50, X57 =
25, X68 = 75
(Basic variables at lower bound) X24 = 0
(Nonbasic variables at upper bound) X13 = 50
(Nonbasic variables at lower bound) X23 = 0, X35 = 0, X45 =
0, X47 = 0, X58 = 0, X67 = 0, X78 = 0
The optimal zvalue is
50(15)+50(10)+0(10)+25(10)+25(5)+50(8)+25(10)+75(15) = $3400 SOLUTION TO CHAPTER 8 REVIEW PROBLEMS
1a. The closest node to New York is Cleveland (via NYClev
path of length 400 miles.)
Second closest node to New York is Nashville (via
NYNash path of length 800 miles).
Third closest node to New York is St. Louis (via NYSt.
Louis path of length 950 miles).
Fourth closest node to New York is Dallas (via NYClevDallas path of length 1300 miles)
Fifth closest node to New York is Salt Lake City (via
NY NashSLC path of length 2000 miles)
Sixth closest node to New York is Phoenix (via NYSt.
Louis Phoenix path of length 2050 miles).
Seventh closest node to New York is LA (via NYSt.
Louis PhoenixLA) path of length 2450 miles.)
1b. CLEV NY 1 CLEV 1 STL 1 NASH 1 PHO 1 STL NASH PHO DALL SLC LA 400 950 800 M M M M 0 M M 1800 900 M M M 0 M 1100 600 M M M M 0 M 600 1200 M M M M 0 M M 400 M M M M 0 M 1300 M M M M M 600 M 57
DALL 1 SLC 1
0+1 0+1 0+1 0+1 0+1 0+1 1+1 1c. NY has a net outflow of +1 and LA a net outflow of 1.
All other nodes have a net outflow of 0. Each arc has a
shipping cost equal to its length, and each arc has an
infinite arc capacity.
2a. City 1 = NY,... City 6 = LA (see figure)
max z = x0
s.t. x12 + x13 = x0 (Node 1)
x12 = x24 + x25 (Node 2)
x13 = x34 + x35 (Node 3)
x24 + x34 = x46 (Node 4)
x25 + x35 = x56 (Node 5)
x46 + x56 = x0 (Node 6)
x12500, x13400, x24300, x25250, x34200, x35150,
x46400, x56350,All variables 0
2b. Begin by labeling sink via chain (1,2)(2,4)(4,6) and
add a flow of 300 to each of these arcs. Then label sink by
path (1,3) (3,5)(5,6) and add a flow of 150 to each of
these arcs. Then label sink by chain (1,2)(2,5)(5,6) and
add a flow of 200 to each arc on this path. Finally label
the sink by (1,3)(3,4) (4,6) adding a flow of 100 to each
arc on this path. This yields the following (maximum) flow
Arc
Flow
NYCHIC
500
NYMEMP
250
CHICDENV 300
CHICDALL 200
MEMPDENV 100
MEMPDALL 150
DENVLA
400
DALLLA
350
INTO SINK 750
We can tell this is a maximum flow because the cut
defined by node 6 (LA) with arcs (4, 6) and (5, 6)
has 58
capacity 400 + 350 = 750 = flow. 3a. See figure (we use expected duration of each activity in
figure). 59 60 3b. and 3c.
= 0 ET(1) = 0 LT(1) = 0 TF(1,2) = 0 FF(1,2) ET(2) = 6 LT(2) = 6 TF(2,3) = 0 FF(2,3) ET(3) = 9 LT(3) = 9 TF(3,5) = 0 FF(5,6) ET(4) = 5 LT(4) = 16 TF(5,6) = 0 FF(5,6) ET(5) = 11 LT(5) = 11 TF(6,7) = 0 FF(6,7) ET(6) = 14 LT(6) = 14 TF(7,8) = 0 FF(7,8) = 0
= 0
= 0
= 0
= 0
ET(7) = 18 LT(7) = 18 TF(1,4) = 11
FF(1,4)
=
0
ET(8) = 20
LT(8) = 20
TF(4,7) = 11
FF(4,7) = 11
The critical path is 1235678 with a duration of 20
days.
3d. min z = x8  x1
st
x2x1 + 6
x3x2 + 3
x5x3 + 2
x6x5 + 3
x7x6 + 4
x8x7 + 2
x4x1 + 5
x7x4 + 2
xj urs 3e. Node 1 has an outflow of +1, node 8 has an outflow of
1, and all other nodes have an outflow of 0. For each arc
the shipping cost equals the arc length, and the goal is to
maximize total cost. Each arc has infinite capacity. If the
flow through an arc equals one, then the arc is on a
critical path.
3f. Mean length of Critical Path = 6 + 3 + 2 + 3 + 4 + 2 =
20 61
Variance of length of Critical path =
64 + 4 + 4 + 16 + 16 + 16
36
= 3.33
Standard deviation of length of critical path = 3.33 1/2 =
1.82.
12  20
We must determine P(CP12) = P(Z  ) = 0! Or
1.82
from EXCEL =NORMDIST(12,20,1.82,1) = 0.
3g. min z = 80A + 60B + 30C + 60D + 40E + 30F + 20G
st
x2x1 + 6  A
x3x2 + 3  C
x5x3 + 2  D
x6x5 + 3  E
x7x6 + 4  G
x8x7 + 2  H
x4x1 + 5  B
x7x4 + 2  F
x8  x112
A,B,C,D,E,F,G2
xj urs A...G0
4. Dummy (which receives Total SupplyTotal Demand) must be
added so we know how much will be `shipped' out of each
production point. Node Rt represents month t regulartime
production and Node Ot represents month t overtime
production, Node
R1
O1
R2
O2 Net Outflow
1,000
400
1,200
400 62
R3
1,200
O3
400
Dummy
Inflow of (3400+1200)4300=300 so bDummy= 300
Demand 1
1000
Demand 2
1500
Demand 3 has bi of 1800. All arcs have infinite capacity.
Arc
Shipping Cost
R1Dem 1
$4
R1Dem 2
$5.50
R1Dem 3
$7.00
O1Dem 1
$6
O1Dem 2
$7.50
O1Dem 3
$9
R2Dem 2
$4
R2Dem 3
$5.50
O2Dem 2
$6
O2Dem 3
$7.50
Arc
Shipping Cost
R3Dem 3
$4
O3Dem 3
$6
All arcs to dummy have 0 shipping cost.
5. Begin at NY and include NYCleveland arc (length 400).
Next include NYNashville arc (length 800). Next include
NashvilleDallas(length
600).
Next
include
DallasSt.
Louis(length 600). Next include DallasPhoenix(length 900).
Next include PhoenixLA (length 400. Finally include Salt
Lake CityLos Angeles (length 600). Total length of minimum
spanning tree is 4300 miles. 6. ARC
P11C1 SHIPPING COST
33 + 51 ARC CAPACITY
 63
P11I'
33 + 51
P21C1
30 + 42
P21I'
30 + 42
P12C2
43 + 60
P22C2
41 + 71
I'I
0
6
IC2
13
P11D
0
P21D
0
P12D
0
P22D
0
Pij = Plant i production during period t
I = Inventory at end of period 1
C1 = Period 1 demand
C2 = Period 2 demand
D = Dummy demand point
NODE
NET OUTFLOW
P11
7
P12
4
P21
9
P22
9
C1
9
C2
11
I'
0
I
0
D
(29  20)
7a. See Figure 64 65
7b. BEF is a critical path of length 15 (follow DUAL
PRICES of 1).
8. See Figure. Each "Professor" node has a net outflow of 4.
Each "Course" node has a net outflow of 4. All other nodes
have a net outflow of 0. Arcs from, say, Fall Finance to
Finance have a lower bound of 1 to ensure that the minimum
number of courses are taught each semester. The cost for
each arc from a "professor" to a "seasoncourse" node is
(reward associated with that teaching assignment). Thus
Professor 1 to Fall Marketing has a cost of (3+6) = 9. 66 9. The key is to think of demands in terms of minutes needed
on a machine! Let node Dij represent month i demand for
product j(in terms of minutes of machine time required). Let
node Mij represent available minutes during month i on
machine j. Node
M11
M12
M21
M22
D11
D12
D13
D21
D22
D23
DUMMY Net Outflow
2400
2400
2400
2400
30(50)
20(70)
15(80)
30(60)
20(90)
15(120)
100 Arc
Shipping Cost/Minute
M11D11 40/30
M11D12 45/20
M11D21 (40 + 15)/30
M11D22 (45 + 10)/20
M12D12 60/20
M12D13 55/15
M12D22 (60 + 10)/20
M12D23 (55 + 5)/15
M21D21
40/30
M21D22
45/20
M22D22
60/20
M22D23
55/15
M11DUMMY 0
M12DUMMY 0
M21DUMMY 0
M22DUMMY 0
All arcs have an infinite capacity. ...
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This homework help was uploaded on 04/02/2008 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at University of California, Berkeley.
 Fall '07
 HOCHBAUM

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