The Transportation and Assignment Problems. Hillier &Lieberman Chapter 8

Transcription

1 The Transportation and Assignment Problems Hillier &Lieberman Chapter 8

2 The Transportation and Assignment Problems Two important special types of linear programming problems The transportation problem How to optimally transport goods The assignment problem Assigning people to tasks

3 The Transportation and Assignment Problems A very large number of constraints and variables most of the a ij coefficients in the constraints are zeros the relatively few nonzero coefficients appear in a distinctive pattern A special streamlined algorithms Dramatic computational savings by exploiting the special structure of the problem.

4 The reason to consider transportation problem Transportation problem is too restricted for general production optimization The reason to consider are: Additional flexibility achieved through slacks/dummy variables It might be a good idea to determine the subproblem that fits to the transportation formulation, resolve it and keep updated through the iterations of the global optimization

5 The transportation problem chapter 8.1

6 The Transportation Problem The main products of the P&Tcompany canned peas The peas are prepared at three canneries Bellingham Eugene Albert Lea The products are then shipped by truck to four distributing warehouses Sacramento Salt Lake City Rapid City Albuquerque

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8 P&T The shipping costs are a major expense, the mgt is studying to reduce them as much as possible. which plan for assigning these shipments to the various cannery-warehouse combinations would minimize the total shipping cost.

9 The table describes units of truckloads along with the shipping cost per truckload for each cannery-warehouse combination. Cannery Sacramento Sal Lake City Rapid City Albuquerque output Bellingham $464 $513 $654 $ Eugene $352 $416 $690 $ Albert Lea $995 $682 $388 $ allocation

10 network representation [75] cannery T1 warehouse V1 [-80] [125] [100] T2 T3 V2 V3 [-65] [-70] number out of location V4 [-85]

11 Minimize Z: where Z Mathematical formulation c x Z ij ij x n j1 m i1 ij x x ij ij m i1 s i d 0,all n j1 j c ij denotes totalshipping cost, x cannery i to warehouse j ij, is unit shipping cost from cannery i to warehouse j is the number of truckloads to be shipped from i 1,2,..., m s is supply j 1,2,..., nd is demand i, j

12 x x x x x x x x x x x x Z 85 x x x 70 x x x 65 x x x 80 x x x 100 x x x x 125 x x x x 75 x x x x ,2,3,4) j 1,2,3; (i 0 x ij objective function and the constraints

13 Constraint coefficients for P&T Co. x11 x12 x13 x14x21 x22x23 x24 x31 x32 x33 x A = Supply constraints Demand constraints The special structure in the pattern of the coefficients distinguishes this problem as a transportation problem

14 Terminology for the transportation problem truckloads of canned peas three canneries four warehouses output from cannery i allocation to warehouse j shipping cost per truckload from cannery i to warehouse j units of a commodity m sources n destinations supply s i from source i demand d i at destination j cost c ij per unit distributed from source i to destination j

15 The feasible solution property all the basic variables have integer values; m i1 n s i d j1 j

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19 The Northern airplane company; an example with a dummy destination

20 Introduction Company builds commercial airplanes for various airline companies around the world. The aim is a schedule developed for the number of engines to be produced in each of the 4 months so that the total of the production and storage costs will be minimized.

21 Production scheduling data month scheduled installations maximum production unit cost of production unit cost of storage

22 Formulation source i= production of jet engines in month i (i=1,2,3,4) destination j= installation of jet engines in month j (j=1,2,3,4) x ij = number of engines produced in month i for installation in month j c ij =cost associated with each unit of x ij si=? dj= number of scheduled installations in month j

23 Cost per unit distributed Source/ Destination Supply ? 2? ? 3?? ? 4??? 1.130? Demand ?= it is impossible to produce engines in one month for installation in an earlier month

24 Constraints x11+x12+x13+x14 25 x21+x22+x23+x24 35 x31+x32+x33+x34 30 x41+x42+x43+x44 10 Constraints on the amount that can be supplied constraints are in the form of inequalities instead of equalities To convert these inequalities to equalities in order to fit the transportation problem model, slack variables are introduced. In this context the slack variables are allocations to a single dummy destination that represent the unused production capacity in the respective month.

25 solution inequalities>equations> slack variables= a dummy destination= unused production capasity Source/ Destination (D) Supply M M M M M M Demand ? It is impossible to produce engines in an earlier months. 5. month demand = Max Production Demand = 30 ( )-( )=30

26 Final solution Ratkaistaan rajoitusten mukaisesti kokonaiskustannusfunktion minimi ja lasketaan yhteen kuukausittaiset valmistuserät = SUPPLY x 11 =10 ; x 12 =15; x 22 =0; x 23 =5; x 25 =30 x 33 =20; x 34 =10; x 44 =10 Z=$77,30

27 Metro water district An example with a Dummy Source An agency that administers water distribution in a large geographical area The area must purchase and bring in water from outside the region. The sources of water Colombo Sacron Calorie river Customers Berdoo Los devils San go hollyglas

28 Table 8.10 Water resources data for Metro Water District cost (tens of dollars) per acre foot Berdoo Los devils San Go Hollyglass Supply Colombo River Sacron River Calorie River Minimum neede Requested Allocate all the available water from the rivers to the four cities in such a way as to meet the essential needs of each city while minimizing the total cost to the district

29 Metro water district It is not clear what the demands at the destinations are. The amount to be received at each destination is a decision variable, with both a lower bound and an upper bound. The upper bound is the amount requested unless the request exceeds the total supply remaining after the minimum needs of the other cities are met, in which case this remining supply becomes the upper bound Hollyglass ( )-( )=60.

30 Metro Water district The demand quantities must be constants The requested allocations are viewed first as the demand quantities Excess demand capacity Introduction of a dummy source to send the unused demand capacity The imaginary supply source is the amount by which the sum of the demands exceeds the sum of the real supplies: ( )-( )=50

31 Table 8.11 Parameter table without minimum needs for Metro Water District cost (tens of dollars) per acre foor Berdoo Los devils San Go Hollyglass Supply Colombo River Sacron River Calorie River M 50 Dummy Requested Calorie river water Cannot be used to supply ( )-( )=60 Hollyglass M min ( )-( )=50 excess demand capacity

32 Metro Water District Next each city s minimum needs are taken into account San Go no minimum need, it is all set Hollyglass ok Lod Devils minimum need equals its requested allocation, Its entire demnad of 70 must be filled from the real sources rather than dummy source. A huge cost of M to the allocation from the Dummy source ensures this allocation will be zero. Berdoo s minimum need is 30 Adjustments must be made to prevent the dummy source from contributing more than 20 to Berdoo s total demand of 50. Splitting Berdoo into two destinations Demand of 30 with a unit cost of M for any allocation from the dummy source The other having a demand of 20 with a unit cost of zero for the dummy source allocation.

33 Table 8.12 Parameter table for Metro Water District Berdoo Berdoo 2 Los devils San Go Hollyglass Supply Colombo River Sacron River Calorie River M 50 Dummy M 0M Requested min 30 min 70

34 table forming form the standard simplex table including supply column and demand row (Table 8.12)

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36 8.2 A streamlined simplex method for the transportation problem Let us first review how the general simplex method would set up a transportation problem in a tabular form.

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39 Transportation simplex method obtain the same info in much simpler way Besides the input data (c,s,d), the only info needed is the current BF solution, the current values of u and v and the resulting values of c-u-v for nonbasic varaibles x

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41 Initialization General procedure for constructing an initial BF solution 1. From the rows and columns still under consideration, select the next basic variable (allocation) according to some criterion. 2. make that allocation large enough to exactly use up the remaining supply in its row or the remaining demand in its column (whichever is smaller<)

42 8.2 General procedure for constructing an initial BF solution 3. eliminate that row or column (whichever had the smaller remaining supply or demand) from further consideration. 4. If only one row or only one column remain under consideration, then the procedure is comleted by selecting every remaining variable associated with that row or column to be basic with the only feasible allocation. Otherwise return to step 1.

43 format of a transportation simplex tableau

44 Northwest corner -rule / criteria for step 1 begin selecting x 11 ( start in the northwest corner of..)(x 11 =30). select x i,j+1 if source i has any supply remaining (move one column to the right) otherwise select next x i+1,j (move one row down)

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46 An Initial BF Solution Käyväksi aloitusratkaisuksi saadaan nyt x 11 =30 x 12 =20 x 22 =0 x 23 =60 x 33 =10 x 34 =30 x 35 =10 x 45 =50 Z=16* x * x * x * x * x * x * x 35 +0* x 45 Z=16*30+16*20+14*0+13*60+20*10+23*30+10*M+0 *50 = M

47 An Initial BF Solution criteria for step 1 Northwest corner rule Vogel s approximation method Russell s approximation method

48 Vogel s approximation method for each row and column remaining under consideration, calculate its difference, which is defined as the aritmetic difference between the smallest and next to the smallest unit cost c ij still remaining in that row or column, in that row or column having the largest difference, select the variable having the smallest remaining unit cost. Eliminate that row or column whichever had the smaller remaining supply or demand

49 row supply difference source M (D) M 0 M demand select x44=30 colum difference eliminate column 4

50 row supply difference source M (D) M 0 M demand select x45=20 colum difference eliminate row 4(D)

51 row supply difference source M 50 0 demand select x13=50 colum difference eliminate row 1

52 row supply difference source M 50 0 demand select x25=40 colum difference M-15 eliminate column 5

53 row supply difference source demand select x23=20 colum difference eliminate row 2

54 1 2 3 supply source demand select x31=30 x32=20 Z=2460 x33=0

55 Russell s approximation method for each source row i remaining under consideration, determine its u i, which is the largest unit cost c ij still remaining in that row. for each destination column j remaining under consideration, determine its v j, which is the largest unit cost c ij still remining in that column. for each variable xij not previously selected in these rows and columns, calculate ij c ij u i v j Select the variable having the largest negative value of. ij

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59 Initialization transportation simplexmethod An Initial BF solution using one of the three methods described earlier Optimality test Iterations

60 Optimality test A BF solution is optimal if and only if c ij u i v j 0 For every (i,j) such that x ij is nonbasic. c ij =u i +v j for each (i,j) such that x ij is basic. m+n-1 basic variables, m+n-1 of these equations Since number of unknowns (u and v) is m+n. one of these variables can be assigned a value arbitrarily without violating the equations. Select the u i that has the largest number of allocations in its row and assign to it the value zero.

61 Because two of these values negartive, the current BF solution is not optimal.

62 Setting u 3 =0 ( since row 3 of table 8.19 has the largest number of allocations 3)

63 An iteration Determine an entering basic variable (step 1) A leaving basic variable (step 2) Identify the resulting new BF solution (step 3)

64 Step 1: Find the entering basic variable c ij -u i -v j represents the rate at which the objective function will change as the nonbasic variable x ij is increased, the entering variable must have a negative value. Select the one having the larger (in absolute terms) negative value of c ij -u i -v j to be the entering variable x 25

65 Step 2: find the leaving basic variable Increasing the entering basic variable from zero sets off a chain reaction of compensating changes in other basic variables in order to continue satisfying the supply and demand constraints. The first basic variable to be decreased to zero becomes the leaving basic variable

66 X 15 leaving variable Reach a zero allocation first (2,5 ) and ( 1,3) recipient cells the donor cells (1,5) and ( 2,3)

67 Step 3: find the new BF solution Adding the value of the leaving basic variable to the allocation for each recipient cell and subtracting this same amount from the allocation for each donor cell. X 15 =10 Table 8.22

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72 Summary of the Transportation Simplex Method Initialization Construct an initial BF solution by the procedure outlined.go to the optimality test Optimality test: Derive u i and v j by selecting the row having the largest number of allocations, setting its u i =0, and then solving the set of equations c ij =u i +v j for each (i,j) such that x ij is basic. If c ij -u i -v j >=0 for every (i,j) such that x ij is nonbasic, then the current solution is optimal and stop. Otherwise iterate.

73 Summary of the Transportation Simplex Method Iteration: 1. Determine the entering basic variable: select the nonbasic variable x ij having the largest (in absolute terms) negative value of c ij u i -v j 2. Determine the leaving basic variable: Identify the chain reaction required to retain feasibility when the entering basic variable is increased. From the donor cells, select the basic variable having the smallest value. 3. Determine the new BF solution: Add the value of the leaving basic variable to the allocation for each recipient cell. Subtract this value from the allocation for each donor cell.

74 Assignment problem A special kind of transporation problem The number of assignees (sources) and the number of tasks (destinations) are the same Each source is assigned to one task Each task is performed by one assignee The costs c ij are associated with assignee i performing task j The objective is to minimize the total costs

75 Assigning products to plants Two options are available: Permit product splitting (the same product is produced by more than one plant) Prohibit product splitting Unit cost for product Capacity Plant

76 Permit product splitting Results in a transportation problem: A dummy destination is introduced to balance the source and destination capacity Unit cost for product (D) Plant M Capacity

77 Prohibit product splitting Results in an assignment problem Two assignees are used for each of plant 1 and 2 Unit cost for product (D) Plant 1a b a M a M M Capacity

78 An example: the cost bounds with varying demands and supply The transportation problem with varying demand and supply is considered = min = 0

79 An example: the cost bounds with varying demands and supply The cost bound under varying demands and supply needs to be estimated: max = max, min and the constraints on variables

80 An example: the cost bounds with varying demands and supply The costs bound problem is not a linear programming problem, because the objective is to maximize with respect to s i and d j and minimize with respect to The dual LP to the transportation problem should be considered to replace the minimization with respect to to maximization with respect to,

81 An example: the cost bounds with varying demands and supply The objective: max,,, The constraints are: + +, 0, unconstrained in sign

82 An example: the cost bounds with varying demands and supply Literature: The total cost bounds of the transportation problem with varying demand and supply, Omega, 31, pp