Missing S Supply and Missing D Demand Values

The NETFLOW Procedure

Missing S Supply and Missing D Demand Values

In some models, you may want a node to be either a supply or demand node but you want the node to supply or demand the optimal number of flow units. To indicate that a node is such a supply node, use a missing S value in the SUPPLY list variable in the ARCDATA= data set or the SUPDEM list variable in the NODEDATA= data set. To indicate that a node is such a demand node, use a missing D value in the DEMAND list variable in the ARCDATA= data set or the SUPDEM list variable in the NODEDATA= data set. Once a missing S or missing D value is found in the input data sets, the THRUNET option is automatically made active. Suppose the Oil example in the "Introductory Example" section is changed so that crude oil can be obtained from either the Middle East or U.S.A. in any amounts. You should specify that the node "middle east" is a supply node, but you do not want to stipulate that it supplies 100 units, as before. The node "u.s.a." should also remain a supply node, but you do not want to stipulate that it supplies 80 units. You must specify that these nodes have missing S supply capabilities.

title  'Oil Industry Example';
title3 'Crude Oil can come from anywhere';
data miss_s;
   missing S;
   input   _node_&$15. _sd_;
   datalines;
middle east         S
u.s.a.              S
servstn1 gas      -95
servstn1 diesel   -30
servstn2 gas      -40
servstn2 diesel   -15
;

The following PROC NETFLOW run uses the same ARCDATA= and CONDATA= data sets used in the "Introductory Example" section.

   proc netflow
      nodedata=miss_s       /* the supply (missing S) and */
                            /* demand data                */
      arcdata=arcd1         /* the arc descriptions       */
      condata=cond1         /* the side constraints       */
      conout=solution;      /* the solution data set      */
   run;
   print some_arcs('middle east' 'u.s.a.',_all_)/short;
proc print;sum _fcost_;run;

The following messages appear on the SAS log:

NOTE: Number of nodes= 14 .
NOTE: Number of supply nodes= 2 .
NOTE: Number of demand nodes= 4 .
NOTE: Total supply= 0 , total demand= 180 .
NOTE: Number of arcs= 18 .
NOTE: Number of iterations performed (neglecting any 
      constraints)= 9 .
NOTE: Of these, 0 were degenerate.
NOTE: Optimum (neglecting any constraints) found.
NOTE: Minimal total cost= 50040 .
NOTE: Number of <= side constraints= 0 .
NOTE: Number of == side constraints= 2 .
NOTE: Number of >= side constraints= 2 .
NOTE: Number of arc and nonarc variable side 
      constraint coefficients= 8 .
NOTE: Number of iterations, optimizing with 
      constraints= 3 .
NOTE: Of these, 0 were degenerate.
NOTE: Optimum reached.
NOTE: Minimal total cost= 50075 .

The PRINT statement reports the arcs directed away from the supply nodes, shown in Figure 4.14. The amount of crude obtained from the Middle East and U.S.A. is 145 and 35 units.

Oil Industry Example

Crude Oil can come from anywhere

NETFLOW PROCEDURE

_N_	_from_	_to_	_cost_	_capac_	_lo_	_name_	_FLOW_
1	middle east	refinery 1	63	95	20	m_e_ref1	20
2	u.s.a.	refinery 1	55	99999999	0		125
3	middle east	refinery 2	81	80	10	m_e_ref2	10
4	u.s.a.	refinery 2	49	99999999	0		25

Figure 4.14: Print statement, Oil example, missing S supplies.

The CONOUT= data set is shown in Figure 4.15.

Obs	_from_	_to_	_cost_	_capac_	_lo_	_name_	_SUPPLY_	_DEMAND_	_FLOW_	_FCOST_	_RCOST_	_ANUMB_	_TNUMB_	_STATUS_
1	refinery 1	r1	200	175	50	thruput1	.	.	145.00	29000.00	.	7	2	KEY_ARC BASIC
2	refinery 2	r2	220	100	35	thruput2	.	.	35.00	7700.00	17	8	3	LOWERBD NONBASIC
3	r1	ref1 diesel	0	75	0		.	.	36.25	0.00	.	10	5	KEY_ARC BASIC
4	r1	ref1 gas	0	140	0	r1_gas	.	.	108.75	0.00	.	9	5	KEY_ARC BASIC
5	r2	ref2 diesel	0	75	0		.	.	8.75	0.00	.	12	6	KEY_ARC BASIC
6	r2	ref2 gas	0	100	0	r2_gas	.	.	26.25	0.00	.	11	6	KEY_ARC BASIC
7	middle east	refinery 1	63	95	20	m_e_ref1	S	.	20.00	1260.00	8	2	1	LOWERBD NONBASIC
8	u.s.a.	refinery 1	55	99999999	0		S	.	125.00	6875.00	.	3	4	KEY_ARC BASIC
9	middle east	refinery 2	81	80	10	m_e_ref2	S	.	10.00	810.00	32	4	1	LOWERBD NONBASIC
10	u.s.a.	refinery 2	49	99999999	0		S	.	25.00	1225.00	.	5	4	KEY_ARC BASIC
11	ref1 diesel	servstn1 diesel	18	99999999	0		.	30	30.00	540.00	.	17	8	KEY_ARC BASIC
12	ref2 diesel	servstn1 diesel	36	99999999	0		.	30	0.00	0.00	12	18	10	LOWERBD NONBASIC
13	ref1 gas	servstn1 gas	15	70	0		.	95	68.75	1031.25	.	13	7	KEY_ARC BASIC
14	ref2 gas	servstn1 gas	17	35	5		.	95	26.25	446.25	.	14	9	NONKEY ARC BASIC
15	ref1 diesel	servstn2 diesel	17	99999999	0		.	15	6.25	106.25	.	19	8	KEY_ARC BASIC
16	ref2 diesel	servstn2 diesel	23	99999999	0		.	15	8.75	201.25	.	20	10	KEY_ARC BASIC
17	ref1 gas	servstn2 gas	22	60	0		.	40	40.00	880.00	.	15	7	KEY_ARC BASIC
18	ref2 gas	servstn2 gas	31	99999999	0		.	40	0.00	0.00	7	16	9	LOWERBD NONBASIC
										50075.00

Figure 4.15: Missing S SUPDEM values in NODEDATA

The optimal supplies of nodes "middle east" and "u.s.a." are 145 and 35 units, respectively. For this example, the same optimal solution is obtained if these nodes had supplies less than these values (each supplies 1 unit, for example) and the THRUNET option was specified in the PROC NETFLOW statement. With the THRUNET option active, when total supply exceeds total demand, the specified nonmissing demand values are the lowest number of flow units that must be absorbed by the corresponding node. This is demonstrated in the following PROC NETFLOW run. The missing S is most useful when nodes are to supply optimal numbers of flow units and it turns out that for some nodes, the optimal supply is zero.

data miss_s_x;
   missing S;
   input   _node_&$15. _sd_;
   datalines;
middle east         1
u.s.a.              1
servstn1 gas      -95
servstn1 diesel   -30
servstn2 gas      -40
servstn2 diesel   -15
;
proc netflow
      thrunet
      nodedata=miss_s_x     /* No supply (missing S)      */
      arcdata=arcd1         /* the arc descriptions       */
      condata=cond1         /* the side constraints       */
      conout=solution;      /* the solution data set      */
   run;
   print some_arcs('middle east' 'u.s.a.',_all_)/short;
proc print;sum _fcost_;run;

The following messages appear on the SAS log. Note that the Total supply= 2, not zero as in the last run.

NOTE: Number of nodes= 14 .
NOTE: Number of supply nodes= 2 .
NOTE: Number of demand nodes= 4 .
NOTE: Total supply= 2 , total demand= 180 .
NOTE: Number of arcs= 18 .
NOTE: Number of iterations performed (neglecting any 
      constraints)= 13 .
NOTE: Of these, 0 were degenerate.
NOTE: Optimum (neglecting any constraints) found.
NOTE: Minimal total cost= 50040 .
NOTE: Number of <= side constraints= 0 .
NOTE: Number of == side constraints= 2 .
NOTE: Number of >= side constraints= 2 .
NOTE: Number of arc and nonarc variable side 
      constraint coefficients= 8 .
NOTE: Number of iterations, optimizing with 
      constraints= 3 .
NOTE: Of these, 0 were degenerate.
NOTE: Optimum reached.
NOTE: Minimal total cost= 50075 .

The PRINT statement and the CONDATA= data set are very similar; the supplies of the supply nodes are 1, not missing S. Otherwise, the solutions are identical.

If total supply exceeds total demand, any missing S values are ignored. If total demand exceeds total supply, any missing D values are ignored.

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