OA 4202, Homework 5. Nedialko B. Dimitrov

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1 OA 4202, Homework 5 Nedialko B. Dimirov 1. (AMO) Formulae his problem as a minimum cos flow problem. We creae a biparie graph. On he lef, we have one node for each Serviceman. On he righ, we have one node for each Posing. Service man i is conneced o posing j if he is elligable for he posing. If he is elligable, edge (i, j) has coss/lower/upper bounds equal o: (moving cos, 0, 1). We give a supply of 1 o each serviceman, and a demand of 1 o each posing. Tha makes he b(i) for serviceman i be 1 and he b(j) for posing j be (Adaped from AMO) We are in charge of a food disribuion warehouse. Our warehouse holds and ships ou wo kinds of foods, apples and bananas. Apples are ypically sored in a cube of boxes on a palle. 1

2 The palle can be picked up by a fork-lif o be moved around he warehouse or o he loading dock o ship hings ou. Bananas are sored in a similar way. A palae eiher holds boxes of bananas or boxes of apples, bu no boh. The warehouse is divided ino hree zones, he Norh, Souh, and Eas zone. Each zone, i, can only sore so many palles B i. Also, each zone i is d i yards away from he loading dock. Whenever a palle is shipped ou of he warehouse, a fork-lif has o go and pick up he palle from he zone conaining he palle, and bring i o he loading dock. Every day, our warehouse ships a apples palles of apples, and a bananas palles of bananas. We would like o make he warehouse as efficien as possible, minimizing he movemen of fork-lifs. Formulae he problem of deciding which sorage locaions we should pick for apples and bananas o minimize fork-lif movemen as a minimum cos flow problem. Norh Apples Bananas Souh Eas We have one node for apples, one node for bannanas, each wih supply corresponding o he number of palles of each frui ha we ship every day. We have one node for each zone of he warehouse. The frui nodes are conneced o he zone nodes by arcs wih (0, 0, ) cos/lower/upper bounds. The zone nodes are conneced o a super-sink node,, by arcs wih (d i, 0, B i ) cos/lower/upper. 3. (AMO) Imagine ha here are wo classes of library iems, say books and CDs. Also, imagine ha here are hree possible sorage locaions. Formulae he problem finding a minimum cos sorage plan for our iems as a minimum cos flow problem. 2

3 Sorage 0 CDs, iem 0 Books, iem 1 Sorage 1 Sorage 2 We have one node for each iem, one node for each sorage faciliy, and one super-sink. We give each iem node supply equal o a i, he oal number of iems of ha ype ha have o be sored. We give he super-sink node demand equal o i a i. We connec each iem node i o sorage node j using an arc wih daa (u i v j, 0, 0). The cos of his arc comes from he fac ha we will, on average, rerieve an iem of ype i u i imes a cos v j each ime. We connec each sorage node o he super-sink using an arc wih daa (0, 0, b j ). This arc capures he capaciy of sorage faciliy j. 4. (AMO) s p0 p1 p2 p3 We creae a node for each ime period. We also creae one super-source node, s, and one super-sink node,. We add edges from s o every period. These edges have daa (p i, 0, α i ) and flow on hese edges represens purchasing an iem. We add edges from every period o he following period. These edges have daa (w i, 0, β i ) and flow on hese edges represens soring iems. We also 3

4 add edges from every period o. These edges have daa ( s i, γ i, ) and flow on hese edges represens selling an iem. Finally, we creae one edge from o s o creae a feasible problem, where he opimizaion decides he number of iems are being bough/sold. 5. (AMO) The book suggess you look a applicaion 9.6, bu ha may no be very useful. One hing ha likely is useful, is looking a Figure 9.6 in he book as you are rying o creae a graph for his problem. Jan 1 Feb 1 Mar 1 Apr 1 May The picure above represens he graph for jus he firs four monhs. We creae a node for each monh. The flow from January o February represens he cars ha we have during he monh of January. Since we are required o have a leas 430 cars during he monh of January, he lower bound on his flow is 430. We se up similar arcs for he res of he monhs. We have o have one more node han he number of monhs we are modeling. So, if we model everyhing up unil April, we also have o have a node for May. Tha is because we need he flow from he node for April o he node for May o signify he number of cars we have in April. Similarly, if we are modeling everyhing up unil June, we have o have one node for July. 4

5 We have 200 cars on Jan 1 and we have o give hose cars back a he end of Feb. We represen his as a supply of 200 on he node for Jan, and as a demand of 200 on he node for March. Tha s because we have hose 200 cars for boh he monh of January and he monh of February. Finally, he represen he possibiliy of enering a leasing conrac, we add edges o creae cycles. A 3 monh leasing conrac ($1700) for leasing cars for he monhs of Jan, Feb, and Mar shows up as a backward arc from Apr o Jan wih daa (1700, 0, ). Oher possible leasing conracs show up in a similar way. For example, a 4 monh leasing conrac for he monhs of Jan, Feb, Mar, Apr shows up as an arc from May o Jan wih a daa (2200, 0, ). 6. (AMO) Exp, D0 Disric 0 Exp Con Reg, D0 Exp, D1 s Reg, D1 Disric 1 Reg Con Exp, D2 Disric 2 Reg, D2 We creae a node for each conracor, a node for each disric, a node for each (experienced, disric), (regular, disric) pair, and wo nodes s and. We add edges from each disric j o wih daa (0, r j, ). These represen he fac ha we need o hire r j conracors for he disric. We add edges from he (experienced, disric j) 5

6 node o he disric j node wih daa (0, 1, ). These represen he fac ha each disric requires one experienced conracor. We add edges from (regular, disric j( o disric j wih daa (0, 0, ), represening ha here are no resricions on he number of inexperienced conracors we hire. We add edges from experienced conracor i o all he nodes (experienced, disric j) wih daa (c ij, 0, ). These represen ha hiring a eam from his conracor in disric j coss c ij and ha here are no resricions on how many of hem we hire from his paricular provider i. Similarly, we add edges from inexperienced conracor i o all he nodes (regular, disric j) wih daa (c i j, 0, ). We add edges from s o each conracor i wih daa (0, 0, u i ), represening he fac ha he conracor only has u i eams available. Finally, we add an edge from o s wih daa (0, 0, ) o make he sysem feasible. 7. (AMO) We ll assume ha h k > r k, because oherwise you would hire an infinie number of ships o ge an infinie amoun of revenue. Also, assume ha he shipping company sars ou wih n ships. Ren Y1 Ren Y2 Ren Y3 Year 1 Year 2 Year 3 Year 4 -n n We creae a node for each year, and one addiional node for he p + 1s year. Tha is because he flow from node labeled Year 1 o he node labeled Year 2 represens he number of ships 6

7 we have in year 1. We also creae a renal node for each year we are modeling and a sink node. We add edges from each year i o year i + 1 wih daa ( r i, d(i), ). These edges represen ha we are required o have d(i) ships in year i, and each ship gives us a revenue of r i. We add edges from each year i o wih daa ( s i, 0, ). The flow on hese edges is he number of ships we sell in year i. Finally, we add edges from year i + 1 o he renal node for year i and from he renal node for year i o year i. These edges have daa (0, 0, ) and (h i, 0, ) respecively. The flow on hese edges represens he number of ships we ren in year i. To represen he n ships we have o sell, we add a supply of n on Year 1 and a demand of n on. This problem is very similar o he car renal problem. We could have dropped he renal nodes enirely, and jus had backward arcs from year i + 1 o year i wih daa (h i, 0, ). 8. (AMO) We creae he same graph as in he anker problem, bu he wih following daa changes: (a) The arcs from s o he sar node of each shipmen have daa (5, 0, ), represening he fac ha i coss $5 o bring a ship ino service. (b) The arcs from he sar node of each shipmen, o he end node of he same shipmen have daa ( shipmen profi, 0, 1), meaning ha we ge a profi by compleing he shipmen. The upper bound of 1 prevens he model from sending muliple ships o complee he same shipmen and hinking ha is geing more profi for i. How would you formulae he original anker problem, where we have o complee all shipmens bu we wan o use he minimum number of ships as a min-cos flow problem? Is i faser o solve he original anker problem in he min-cos-flow way or he max-flow way? 9. (AMO) 7

8 Formulae his problem as a minimum cos flow problem. A mehod ha does no work for solving his problem is o have a node for each run, and a node for each bus driver. Then, he flow beween a run and a bus driver signify he number of hours assigned o he bus driver. This allows us o pu nodes afer he bus-driver node ha spli his hours ino regular pay and overime pay. Bu, his doesn work because i may assign par of a run o a bus driver. For example, consider having wo morning runs wih 7, 2 hours each and wo afernoon runs wih 12, 0 hours each, and overime happens afer 10 hours. Such a formulaion would wan o spli up he 12 hour run amongs he wo bus drivers. Tha is why his problem is essenially jus abou maching morning runs and afernoon runs. The bus drivers don show up as nodes a all. We creae a graph where on he lef we have morning runs, and on he righ we have afernoon runs. We connec each morning run wih an edge o each afernoon run. The edge has daa ( duraion morning + duraion afernoon - D, 0, ). This signifies ha a maching of his morning run and his afernoon run gives duraion morning + duraion afernoon - D over ime hours. To finish he formulaion, we give each morning run a supply of 1 and each afernoon run a demand of 1. 8