Ch08 - 1 MP CHAPTER 8 SOLUTIONS SECTION 8.2 1. First label...

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Unformatted text preview: 1 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 1-3-6 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 1-2-5(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 1-3-4. Of course, 1-2-3-4 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 1-3-5-7. 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 1-4-7 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- 4-12 (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, ..., j-1 . 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 1-6. 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 so-1 3 so-2 0 1-3 3 1-2 0 2-4 3 3-si 0 3-2 3 4-si 3 Flow to sink 3 Now label sink by (so-2) - (2-4), (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 so-1 3 so-2 2 1-3 3 1-2 0 2-4 5 3-si 0 3-2 3 4-si 5 Flow to sink 5 Now label sink by (so-1) - (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 so-1 4 so-2 2 1-3 3 1-2 1 2-4 5 3-si 1 3-2 2 4-si 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 (so-2)-(2-si). We increase the flow in each of these arcs by 2 units. This yields the following feasible flow. so-1 0 so-2 2 1-2 0 1-3 0 14 2-si 2 3-si 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: so-1 1 so-2 2 1-2 0 1-3 1 2-si 2 3-si 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 1-Add 4 units of flow along so-1-3-si Step 2-Add 2 units of flow along so-2-4-si Step 3-Add 1 unit of flow along so-2-4-3-si Step 4-Add 1 unit along so-2-1-3-si Cannot label si. Maximum flow = 8 Arcs in cut are 3-si and 4-si 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 1-Add 7 units along so-2-si Step 2-Add 6 units along so-1-3-si Step 3 Add 6 units along so-4-si 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 Ford-Fulkerson 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 1-2-3-4-5-6-7 (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 1-3-4-7-9 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 1-2-4-8 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 D-14 d-14 P( 1.53 ) = .99. 1.53 31 We find that F(2.33) = NORMINV(.99,14,1.53)= 17.56 d-14 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 1-2-3-5-6 and 1-2-3-4-5-6 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 1-2-3-5-6. 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 1-2-3-4-5-6-7 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 1-2- 3-4-5-6 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 1-3-4 AB = Duration of path 1-2-4. 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 1-2-4 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 1-2-4 and 1-3-4 are critical paths is P(AB = 15)P(CD = 15) = 2/27. Then the probability that 1-3-4 is the only critical path is 1 - 2/27 - 15/27 = 10/27 12a. See Figure 12b. Critical paths are A-D-G, A-E-G, B-D-G, and B-E-G, 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, x3-x190, 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 1-2-3-5-6 is critical path(length 14) Arc TF FF 1,2 2-0-2=0 2-0-2=0 2,3 8-2-6=0 8-2-6=0 2,4 10-2-4=4 6-2-4=0 TF not equal to FF! 3,5 12-8-4=0 12-8-4=0 3,6 14-8-1=5 14-8-1=5 4,5 12-6-2=4 12-6-2=4 4,6 14-6-1=7 14-6-1=7 5,6 14-12-2=0 14-12-2=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 3-0-3=0 6-3-3=0 7-3-3=1 10-3-4=3 10-6-4=0 15-6-5=4 10-6-3=1 21-6-2=13 15-10-5=0 21-10-4=7 21-15-6=0 27-21-6=0 FF 3-0-3=0 6-3-3=0 6-3-3=0 TF not equal to FF! 10-3-4=3 10-6-4=0 15-6-5=4 10-6-3=1 21-6-2=13 15-10-5=0 21-10-4=7 21-15-6=0 27-21-6=0 Critical Path is 1-2-3-5-6-7-8; 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 z-value 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 Detroit-City 1 $2800 Detroit-City 2 $2600 Detroit-City 3 $2300 Detroit-Dummy $0 Dallas-City 1 $2300 Dallas-City 2 $2000 Dallas-City 3 $2000 Dallas-Dummy $0 4a. All arcs have infinite capacity Arc Shipping Cost Bos-Chic 800 + 80 = $880 Bos-Austin 800 + 220 = $1,020 Bos-LA 800 + 280 = $1,080 Ral-Chic 900 + 100 = $1,000 Ral-Aus 900 + 140 = $1,040 42 Ral-LA 900 + 170 = $1,070 Chic-Aus $40 Chic-LA $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 Raleigh-Chicago and Boston- Chicago arcs and insert the following arcs Arc Boston-Chicago' Raleigh-Chicago' Chicago-Chicago' 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 SD-Dallas:$420 SD-Houston:$100 LA-Dallas:$300 LA-Houston:$110 SD-Dummy:$0 LA-Dummy:$0 Dallas - Chic.: 700 + 550 = $1,250 Dall-NY: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 LA-Dallas' $300 SD-Dallas' $420 LA-Houston' $110 SD-Houston' $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 MP-TP(loaded) $700 MP-TP(empty) $400 MP-TW(loaded) $1000 MP-TW(empty) $800 MP-TN(loaded) $1000 MP-TN(empty) $800 Arcs which correspond to fulfilling a shipping requirement are handled by making the arc have lower bound = upper bound = requirement. Thus MP-TN(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 Gary-South Bend arc. Then we add the South Bend-Fort Wayne arc. Next we add the Gary-Terre Haute arc. Finally we add the Terre Haute-Evansville 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 Ct-1 to Ct-1' 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, Y1-Y2=4, Y1-Y3=3, Y2-Y4=3, Y5-Y6=2, Y4-Y6=2 This yields (-1) 48 Y1=0, Y2=-4, Y3=-3, Y4=-7, Y5=-7, Y6=-9 Find the row 0 coefficients for each non-basic 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, Y1-Y2=4, Y1-Y3=3, Y2-Y4=3, Y2-Y5=2, Y5-Y6=2 This yields Y1=0, Y2=-4, Y3=-3, Y4=-7, Y5=-6, Y6=-8 Find the row 0 coefficients for each non-basic 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 (Non-basic Variables at lower bound) X35 = 0, X46 = 0 The optimal z-value 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 non-basic 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 non-basic 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 (Non-basic Variables at upper bound) X13 = 12 (Non-basic Variables at lower bound) X23 = 0, X25 = 0, X45 = 0 The optimal z-value 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, Y1-Y2 = 15, Y2-Y4 =10, Y2-Y6 = 5, Y3-Y6 = 8, Y4Y5 = 30, Y5-Y7 = 10, Y6-Y8 = 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) non-basic 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, Y1-Y2=15, Y2-Y4=10, Y2-Y5=10, Y2-Y6=5, Y3-Y6=8, Y5-Y7=10, Y6-Y8=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 non-basic 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 (Non-basic variables at upper bound) X13 = 50 (Non-basic variables at lower bound) X23 = 0, X35 = 0, X45 = 0, X47 = 0, X58 = 0, X67 = 0, X78 = 0 The optimal z-value 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 NY-Clev path of length 400 miles.) Second closest node to New York is Nashville (via NY-Nash path of length 800 miles). Third closest node to New York is St. Louis (via NY-St. Louis path of length 950 miles). Fourth closest node to New York is Dallas (via NY-ClevDallas path of length 1300 miles) Fifth closest node to New York is Salt Lake City (via NY- Nash-SLC path of length 2000 miles) Sixth closest node to New York is Phoenix (via NY-St. Louis- Phoenix path of length 2050 miles). Seventh closest node to New York is LA (via NY-St. Louis- Phoenix-LA) 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 NY-CHIC 500 NY-MEMP 250 CHIC-DENV 300 CHIC-DALL 200 MEMP-DENV 100 MEMP-DALL 150 DENV-LA 400 DALL-LA 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 1-2-3-5-6-7-8 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 Supply-Total Demand) must be added so we know how much will be `shipped' out of each production point. Node Rt represents month t regular-time 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 R1-Dem 1 $4 R1-Dem 2 $5.50 R1-Dem 3 $7.00 O1-Dem 1 $6 O1-Dem 2 $7.50 O1-Dem 3 $9 R2-Dem 2 $4 R2-Dem 3 $5.50 O2-Dem 2 $6 O2-Dem 3 $7.50 Arc Shipping Cost R3-Dem 3 $4 O3-Dem 3 $6 All arcs to dummy have 0 shipping cost. 5. Begin at NY and include NY-Cleveland arc (length 400). Next include NY-Nashville arc (length 800). Next include Nashville-Dallas(length 600). Next include Dallas-St. Louis(length 600). Next include Dallas-Phoenix(length 900). Next include Phoenix-LA (length 400. Finally include Salt Lake City-Los Angeles (length 600). Total length of minimum spanning tree is 4300 miles. 6. ARC P11-C1 SHIPPING COST 33 + 51 ARC CAPACITY - 63 P11-I' 33 + 51 P21-C1 30 + 42 P21-I' 30 + 42 P12-C2 43 + 60 P22-C2 41 + 71 I'-I 0 6 I-C2 13 P11-D 0 P21-D 0 P12-D 0 P22-D 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. B-E-F 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 "season-course" 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 M11-D11 40/30 M11-D12 45/20 M11-D21 (40 + 15)/30 M11-D22 (45 + 10)/20 M12-D12 60/20 M12-D13 55/15 M12-D22 (60 + 10)/20 M12-D23 (55 + 5)/15 M21-D21 40/30 M21-D22 45/20 M22-D22 60/20 M22-D23 55/15 M11-DUMMY 0 M12-DUMMY 0 M21-DUMMY 0 M22-DUMMY 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.

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