4Initial Solution by North West Corner Rule SourcesDestinationCapacityD1D2D3S12Xij4315S25725S3830Demand10204070S1D1 box is the most north column so we fill this first. See its column total which is 10 and row total which is 15. we place the less no in the box so is 10. Now the column total has been exhausted/completely filled and we mark zero to below box and the row total decrease to 5.
5Sources Destination Capacity Demand 2104Xij35S2725S3830Demand204070S1D2 box is the most adjacent west column so we fill this first now. See its column total which is 20 and row total which is 5. we place the less no in the box so is 5. Now the row total has been completely filled and we give zero to respective row total and the column total decreases to 15.
6Sources Destination Capacity Demand S2D2 box is the most adjacent north column to the previous cell just filled. So we fill this first. See its column total which is 15 and row total which is 25. We place the less no in the box so is 15. Now the column total has been exhausted and we assign zero to relevant column total and the row total decreases to 10.SourcesDestinationCapacityD1D2D3S1210453S2Xij725S3830Demand154070
7Sources Destination Capacityy Demand 210453S2157XijS3830Demand4070S2D3 box is the most adjacent west column so we fill this first. See its column total which is 40 and row total which is 10. we place the less no in the box so is 10. Now the row total has been totally filled and the column total decrease to 30.
8Sources Destination Capacity Demand 210453S2157S38Xij30Demand70S3D3 box is the last column so we fill this first. See its column and row total which is 30 . So we place 30 in the box and both column and row balance will be utilized fully.Every time while finding the initial solution the column and row total of last box will be equal as in the given case 30.
13Total cost Total cost = Xij * Cij Total is calculated on the basis of all the basic variables because non basic variables are zero and their cost will also becomes zero when multiplied with zero units.= 10 x 2 +5 x x x 7+30 x 3=245
14Criteria for Optimality All the shadow cost, which must be non negativeShadow cost, for basic variables is always zero so there is no need to find their shadow costShadow cost:Vij = (Ui + Vj) –CijSo we need to find the U and V.
16If we find the values of U1,U2,U3, and V1,V2 and V3 than we can easily find the shadow cost Write the equations for the basic variablesU1+V1=2U1+V2=4U2+V2=3U2+V3=7U3+V3=3
17There are 6 basic variables and 5 equations so we will give one variable arbitrary value which is equal to zeroLet U2=0By putting this U2 zero we can easily find the value of all other variables which will be as followsU1=1 V1=1U2=0 V2=3U3=-4 V3=7
18Sources Destination Capacity Demand 104515U1=1S23725U2=0S3830U3=-4DemandV1=120V2=340V3=770Shadow cost of S1D1Vij = (Ui + Vj) –CijV11 = (U1 + V1) –C11=(1+1) -2=0So save time and do not calculate the shadow cost for basic variables. Write the shadow cost on opposite side of cost in the columnShadow cost of S1D3Vij = (Ui + Vj) –CijV13 = (U1 + V3) –C13=(1+7) -3=5
19Sources Destination Capacity Demand 104515U1=1S23725U2=0S330U3=-4DemandV1=120V2=340V3=770Our criteria of optimality is non positivity of shadow cost which is not satisfied as we have S1D3 box with positive shadow cost so we use θ
20Sources Destination Capacity Demand 1045-θ0+θ15U1=1S2315+θ710-θ25U2=0S330U3=-4DemandV1=120V2=340V3=770Add θ in cell having largest positive shadow cost this will disturb the row and column balances for which you need to add and subtract θ from some other cells
21Max θ = Min of (10,5)=5The boxes from which the θ has been subtractedReplace the θ with 5 and make new tableau.‘
22Sources Destination Capacity Demand 21043515U1S220725U2S3830U3DemandV1V240V370Again find the shadow cost by finding the same rules as discussed above
23=220 (cost decreased as compared to previous tableau) Equations: Total Cost:=10*2+5*3+20*3+5*7+30*3=220 (cost decreased as compared to previous tableau)Equations:U1+V1=2U1+V3=3U2+V2=3U2+V3=7U3+V3=3
24Let U2=0By putting this U2 zero we can easily find the value of all other variables which will be as followsU1=-4 V1=6U2=0 V2=3U3=-4 V3=7
25Sources Destination Capacity Demand 2103515U1= - 4S220725U2=0S330U3= - 4DemandV1= 6V2=340V3=770Our criteria of optimality is non positivity of shadow cost which is not satisfied as we have S2D1 box with positive shadow cost so we use θ
26Sources Destination Capacity Demand 210-θ35+θ15U1= - 4S20+θ2075-θ25U2=0S330U3= - 4Demand10V1= 6V2=340V3=770We can no add θ in S3D2 because its above and below boxes have zero units and we can not balance in zero .
27The boxes from which the θ has been subtracted Max θ = Min of (10,5)=5The boxes from which the θ has been subtractedReplace the θ with 5 and make new tableau.