Ques :

The Big Bazaar at netaji subhash place, pitampura dailyneeds between 32 and 40 workers depending on the timeof the day. The rush hours are between noon nnd ·2 p. m.

,-=--

The table below indicates the number of workers neededat various hours when the bazaar is open.

Time Period Number of workers needed

9 a.m. - 11 a.m. 3211 a.m. - 1 p.m. 401 p.m. - 3 p.m. 353 p.m. - 5 p.m. 33

Big Bazaar now employ's 34 full-time workers, but needsa few part-time workers also. A part-time worker mustput in exactly 4 hours per day, but can start any timebetween 9 a.m. and 1 p.m. Full time workers work from 9a. m. to 5 p. m. but are allowed an hour for lunch (half ofthe full-timers eat at 12 noon, the other half at 1 p.m.).The management of Big Bazaar limits part-time hours toa maximum of 50 per cent of the day's total requirement.Part timers earn Rs. 48 per day on the average, whilefull-timers earn Rs. 140 per day in salary and benefits onthe average. The management wants to set a schedulethat would minimize total manpower costs.Formulate this problem as a Linear Programming Model tominimize total daily manpower cost.

Solution:

MATHEMATICAL FORMULATION :

The key decision is to determine the number of full-time andpart-time workers needed in Big Bazaar.Decision variables: Let y represents the number of full-timeworkers and xj represent the part-time workers starting at9 a.m., 11a.m., and 1 p.m. respectively. ( j = 1,2,3 ).Making use of given information, the appropriate LPPis:

Minimize (total daily manpower cost)z= 140y + 48 (Xl + X2+ X3)

subject to constraints:

y + Xl ~ 32 (9 a.m. - 11a.m. need)~ y + Xl + X2~ 40 (11a.m. - 1 p.m. need)t y + X2+ X3~ 35 (1 p.m. - 3 p.m. need)y + X3~.33 (3 p.m. - 5 p.m. need)y s 34 (Full timers available)4(Xl + X2+ X3) s 0.50 (32 + 40 + 35 + 33)

(Part timer's hours cannot exceed 501'0 of total hoursrequired each day which is the sum of the workers neededeach hour)

Also,y ~ 0 and Xl', X2 , X3z 0 (Non-negative restrictions)

We see that the constraints are of greater than type and

also of less than type so I

So we add artificial and slack variables and modify the

constrai nts.

And Artificial variables be All A21 A31 A4 ~ O.

Now the problem becomes:

Max Z* = - { Min ( Z )}

Subject to constraints

y + Xl - Sl + Al = 32

t y + Xl + X2 - S2 + A2= 40

t y + X2 + X3 - S3+ A3 = 35

y + X3 - S4 + A4 = 33

Y + S5 = 34

4(Xl + X2 + X3) + S6 = 0.50 ( 32 + 40 + 35 + 33 )

Initial Simplex table is as follow's:

Cj -140 -48 -48 -48 0 0 0 0 0 0 -M -M -M -MCb Xb B y Xl X2 X3 51 52 53 54 55 56 Al A2 A3 A4

-~-M A1 32 1* 1 0 0 -1 0 0 0 0 0 1 0 0 0

-M A2 40 1/2 1 1 0 0 -1 0 0 0 0 0 1 0 0

-M A3 35 1/2 0 1 1 0 0 -1 0 0 0 0 0 1 0

-M A4 33 1 0 0 1 0 0 0 -1 0 0 0 0 0 1

0 55 34 1 0 0 0 0 0 0 0 1 0 0 0 0 0

0 56 70 0 4 4 4 0 0 0 0 0 1 0 0 0 0

Zj -3M -2M -2M -2M -M -M -M -M 0 0 -M -M -M -M

- - - - -M -M -M -M 0 0 0 0 0 0Zj - Cj 3M+l 2M+ 2M+ 2M+

40 48 48 48_.-

iHere we see that the key element is 1 from the 'y' column,The key row is Al and the key column is y.

Therefore,

y enters the basis and Al departs from the basis.Since Al is an artificial variable we eliminate its column from thenext simplex table.

Table 2 :

Cj -140 -48 -48 -48 0 0 0 0 0 0 -M -M -MCb Xb B Y Xl Xz X3 SI Sz S3 S4 S5 S6 Az A3 A4

-140 Y 32 1 1 0 0 -1 0 0 0 0 0 0 0 0

-M Az 24 0 1/2 1 0 1/2 -1 0 0 0 0 1 0 0•-M A3 19 0 -1/2 1 1* 1/2 0 -1 0 0 0 0 1 o -f-+

-M A4 1 0 -1 0 1 1 0 0 -1 0 0 0 0 10 55 2 0 -1 0 0 1 0 0 0 1 0 0 0 0

0 56 70 0 4 4 4 0 0 0 0 0 1 0 0 0

Zj -140 -140 -M -2M 140 M M M 0 0 -M -M -M

0 -92 -M+ - 2M 140 M M M M M 0 0 0Zj - Cj 48 +48

iHere we see that the key element is 1 from 'X3' column,The key row is A3 and the key column is X3 .

Therefore,

X3 enters the basis and A3 departs from the basis.Since A3 is an artificial variable we eliminate its column from thenext simplex table.

Table 3:

Cj -140 -48 -48 -48 0 0 0 0 0 0 -M -M

Cb Xb B Y Xl X2 X3 Sl S2 S3 S4 S5 S6 A2 A4

-140 Y 32 1 1 0 0 -1 0 0 0 0 0 0 0-~

-M Az 24 0 1/2 1 0 1/2* -1 0 0 0 0 1 0

-48 X3 19 0 -1/2 1 1/2 0 -1 0 0 0 0 0 0

-M A4 18 0 -1/2 -1 0 1/2 0 1 0 0 0 -1 10 S5 2 0 -1 0 0 1 0 0 0 1 0 0 0

0 S6 -6 0 6 0 0 -2 0 -4 0 0 1 0 0

Zj -140 -116 -48 -48 -140 M+ -M 0 0 0 0 -M- M 48

0 68 0 0 -140 M+ -M 0 0 0 M 0Zj - Cj -M 48

i

Here we see that the key element is 1/2 from '51' column,The key row is A2 and the key column is S1.

Therefore,

S1 enters the basis and A2 departs from the basis.Since A2 is an artificial variable we eliminate its column from thenext simplex table.

Table 4:

Cj -140 -48 -48 -48 0 0 0 0 0 0 -M

Cb Xb B y Xl X2 I X3 SI S2 S3 S4 S5 S6 A4

-140 Y 80 1 2 2 0 1 -2 0 0 0 0 0

0 51 48 0 1 2 0 0 -2 0 0 0 0 0

-48 X3 19 0 -1/2 1 1/2 0 -1 0 0 0 0 0

-M A4 42 0 -1 -2 0 0 1* 1 0 0 0 1 -f-+

0 Ss 46 0 -2 -2 0 0 0 0 0 1 0 0

0 S6 18 0 8 4 0 0 -4 -4 0 0 1 0

Zj -140 -256+ -328 -24 0 -232 -M 0 0 0 -MM +M -M

0 M- 2M- 24 0 -232 -M 0 0 0 0Zj - Cj 208 280 -M

i

Here we see that the key element is 1 from 'S2'column,The key row is A4 and the key column is S2.

Therefore,

S2 enters the basis and A4 departs from the basis.Since A4 is an artificial variable we eliminate its column from thenext simplex table.

Table 5:

Cj -140 -48 -48 -48 0 0 0 0 0 0

Cb Xb B Y Xl Xl X3 SI Sl S3 S4 S5 S6

-140 Y -4 1 0 -2 0 0 0 2 0 0 0

0 SI -36 0 1 0 0 1 0 2 0 0 0

-48 Xg -23 0 -3/2 -3/2 1/2 0 0 1 0 0 0

0 Sl -42 0 -2 -2 0 0 1 1* 0 0 0 ---+

0 55 -46 0 4 -2 0 0 0 0 0 1 0

0 56 -150 0 8 -4 0 0 0 0 0 0 1Zj -140 72 352 -24 0 0 -328 0 0 0

0 120 400 24 0 0 -328 0 0 0Zj - Cj

i

Here we see that the key element is 1 from 's3'column,The key row is Sl and the key column is S3.

Therefore,

S3enters the basis and S2departs from the basis.

Table 6:

Cj -140 -48 -48 -48 0 0 0 0 0 0

Cb Xb B Y Xl X2 X3 Sl S2 S3 S4 S5 S6

-140 Y 80 1 2 1 0 0 -2 0 0 0 0

0 Sl 48 0 1 4 0 1 -2 0 0 0 0

-48 X3 19 0 1/2 1/2 1/2 0 -1 0 0 0 0

0 S3 -42 0 -1 -2 0 0 1 1 0 0 0

0 55 -46 0 -2 -2 0 0 0 0 0 1 0

0 56 -150 0 4* -4 0 0 0 0 0 0 1 f----.

Zj -140 -304 -164 -24 0 -232 0 0 0 0

0 -266 116 24 0 -232 0 0 0 0Zj - Cj

iHere we see that the key element is 4 from 'Xl'column,The key row is 56 and the key column is Xl.

Therefore,

Xl enters the basis and S6 departs from the basis.

Table 7:

Cj -140 -48 -48 -48 0 0 0 0 0 0

Cb Xb B y Xl X2 X3 SI S2 S3 S4 S5 S6

-140 Y 155 1 0 3 0 0 -2 0 0 0 -1/2

0 51 271/2 0 0 5 0 1 -2 0 0 0 -114

-48 X3 151/4 0 0 1 1/2 0 -1 0 0 0 -118

0 S3 -159/2 0 0 -3* 0 0 1 1 0 0 1/4 ~

0 S5 -121 0 0 -4 0 0 0 0 0 1 1/2

-48 Xl -150/4 0 1 -1 0 0 0 0 0 0 1/4

Zj -140 0 -468 -24 0 328 0 0 0 0

0 48 -420 24 0 328 0 0 0 76,Zj - Cj

i

Here we see that the key element is 4 from 'X2 column,The key row is 53 and the key column is X2 .

Therefore,

X2 enters the basis and S3 departs from the basis.

Table 8 :

Cj -140 -48 -48 -48 0 0 0 0 0 0

Cb Xb B y Xl X2 X3 Sl S2 S3 S4 S5 S6

-140 Y 453/6 1 0 0 0 0 -1 1 0 0 -1/20 Sl 3 0 0 0 0 1 -1/3 5/3 0 0 -1/4

-48 X3 135/12 0 0 0 1/2 0 -2/3 1/3 0 0 -1/8-48 X2 159/6 0 0 1 0 0 -1/3 -1/3 0 0 -1/120 S5 -15 0 0 0 0 0 -4/3 -4/3 0 1 1/6

-48 Xl 132/4 0 1 0 0 0 -1/3 -1/3 0 0 1/2Zj -140 -48 -48 -24 0 204 468/3 0 0 17

0 0 0 24 0 204 468/3 0 0 17Zj - Cj

Here we see that all Zj - Cj's are positive or zero hencethis is the optimal solution.Therefore, the minimum value of the objective functionoccurs when,y = 453/6 = 76 ,Sl = 3,X3 = 15/12 = 1 ,X2 = 159/6 = 27 ,S5 = -15 I

Xl = 132/4 = 33 .

Hence,

Min Z= 140 (453/6) + 48 ( 132/4 + 135/12 + 159/6)Min Z = Rs. 13958

Thus I total minimum daily manpower cost is Rs.13958.

, ~~--------------------------------J

CONCLUSION

Minimum total daily manpower cost is Rs .13958 whennumber of full time workers is 76 and total part-timeworkers is 61 cit Big Bazaar, netaji subhash place .

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