Weight Moving Average = n

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2 Forecastig Simple Movig Average- F = 1 N (D + D 1 + D 2 + D 3 + ) Movig Weight Average- Weight Movig Average = W id i i=1 W i Sigle (Simple) Expoetial Smoothig- F t = F t 1 + α(d t 1 F t 1 ) or F t = (1 α)f t 1 + αd t 1 if previous forecastig is ot give F t = αd t + α(1 α)d t 1 + α(1 α) 2 D t 2. Where F t = Smoothed average forecast for period t F t 1 =Previous period forecast α=smoothig costat Liear Regressio- Y = a + bx Forecastig Error- Bias- Mea Absolute Deviatio- Mea Square Error- Mea Absolute Percetage Error- y = a + b x xy = x + b x 2 e t = (D t F t ) Bias = 1 N (D t F t ) MAD = 1 N D t F t MSE = 1 N (D t F t ) 2 MAPE = 1 N D t F t 100 Ivetory If D= demad/year, C o =Order cost, C c = Carryig cost, P= Purchase price/uit Q =Ecoomic Order Quatity, K= Productio Rate ad C S =Shortage Cost/uit/period Case-1 Purchase Model With Istataeous Repleishmet ad Without Shortage- 1. EOQ Q = 2DC o C c 2. No. of order = D Q D t at EOQ Ivertory cost = Order cost 3. Time Take Per Order = Q D 4. Total Cost = Uit cost + Ivertory cost + Order cost = (D P) + ( Q 2 C c) + ( D Q C o)= (D P) + 2DC c C o Case-2 Maufacturig Model Without Shortage- 2

3 1. EOQ Q = 2C od 2. t 1 = Q K 3. t 2 = Q (1 D K ) D C c (1 D K ) 4. Total optimum cost = 2DC c C o (1 D K ) Case 3 Purchase Model With Shortage- 1. Q = EOQ = 2DC o ( C s+c c ) C c C s 2. Q 1 = 2DC o ( C s ), Q C c C s +C 2 = Q Q 1 c 3. t = Q, t D 1 = Q 1 ad t D 2 = Q 2 D 4. Total optimum cost = 2DC c C o ( C s C s +C c ) Case 4 Maufacturig Model With Shortfall 1. Q = EOQ = 2DC o C c (1 D K ) (C s+c c C s ) 2. Q 1 = 2DC o ( C s ) (1 D ) C c C s +C c K 3

4 3. Q 1 = (1 D K ) Q Q 2 4. t = Q, t D 1 = Q 1 ad t K D 2 = Q 1 D 1. t 3 = Q 3 D ad t 4 = Q 2 K D Lead Time Demad + Safety Stock = Reorder Poit PERT ad CPM EFT = EST + activity time LFT = LST + Duratio of activity Total Float- Free Float- FFo= (Ej-Ei)-Tij Idepedet Float - Example- 1. Total float = L2 (E1 + t12) = 57 ( ) = Free float = E2 E1 t12 = 0 3. Idepedet float = E2 (L1 + t12) = -18 PERT Expected time- t e = t 0+4t m +t p 6 1. t0 = Optimistic time i.e., shortest possible time to complete the activity if all goes well. Stadard deviatio- 2. tp = Pessimistic time i.e., logest time that a activity could take if everythig goes wrog. 3. tm = Most likely time i.e., ormal time of a activity would take. Variace - Crashig- Stadard Normal Variatio (SNV)- 4

5 Liear Programmig Simplex Method Case 1. Maximizatio Problem Max Z = 3x1 + 5x2 s / t 3x1 + 2x2 18 ( I) x1 4 ( II) x2 6 ( III ) x1, x2 0 Stadard Form: Max Z = 3x1 + 5x2 + 0w1 + 0w2 + 0w3 3x1 + 2x2 + w1 + 0w2 + 0w3 = 18 x1 + 0x2 + 0w1 + w2 + 0w3 = 4 0x1 + x2 + 0w1 + 0w2 + w3 = 6 To prepare iitial Table: Table - I Ij = (Zj-cj) = ( aij.ci)-cj Iterpretatio of Simplex Table Table - I Key Colum Mi Ij [ Most Negative ] Key Row Mi positive ratio. How to get ext table? Leavig variable : w3 Eterig variable : x2 Key o. = 1 For old key row : New No.= Old No./key No. For other rows: ( Correspodig Key Row No.). ( Correspodig Key Colum No.) New No. = Old No. Key No (6*2)/1 = 6 I(w3) = [1*(-5)]/1 = 5 5

6 Table - II Key Colum Mi Ij Key Row Mi positive ratio Table - III This is the fial Table The Optimal Solutio is x1 = 2, x2 = 6 givig Z = 36 Type of Solutios : Basic, Feasible/Ifeasible, Optimal/ No-Optimal, Uique/Alterative Optimal, Bouded/Ubouded, Degeerate/No-Degeerate Aalysis of Solutio 1. This is a Basic solutio, as values of basic variables are Positive 2. This is a feasible solutio, as values of basic variables, ot cotaiig Artificial Variable, are Positive ad all costraits are satisfied 3. This Feasible solutio is a Optimal, as all values i Idex Row are positive. 4. If there is a Artificial Variable, as Basic variable i fial table, it is called as Ifeasible solutio 5. This solutio is uique Optimal, as the umber of zeroes are equal to umber of basic variables i Idex Row i fial Table. 6. If the umber of zeroes are more tha umber of basic variables i Idex Row i fial Table, it is a case of more tha oe optimal solutios. 7. This is a Bouded Solutio, as the values of all Basic variables i fial table, are fiite positive. 8. This is a No-degeerate Solutio, as value of oe of the basic variables is Zero, i fial table. 9. If value of at least oe of the basic variables is Zero i Idex Row i fial Table, it is a Degeerate Solutio. Duality With Example 1. Case-1 Max.Z = x1 - x2 + 3x3 s/t x1 + x2 + x x1 - x2 - x3 2 2x1-2x2-3x3 6 x1, x2, x3 0 Dual of this would be Mi Z = 10y1 +2y2 + 6y3 s/t y1 + 2 y2 + 2y3 1 y1 - y2-2y3-1 y1 - y2-3y3 3 y1, y3, y3 0, 6

7 2. Case -2 Mi Z = 20 x x2 s/t - 4 x1 - x2-8 5 x1-3 x2-4 x1, x2 0 Dual of this would be Max Z = 8 y1 + 4 y2 S/t 4 y1-5 y2 20 y1 + 3 y2 23 y1, y2 0, 3. Case-3 Max.Zp = x1+2x2-3x3 s/t 2 x1 + x2 +x x1 - x2 + 2 x3 110(this eed to be coverted i less tha form) x1 + 2x2 - x3 = 4 x1, x3 0, x2 urestricted Dual of this would be Mi. Zd = 10 y1-110 y2 +4 y3 s/t 2 y1-3 y2 + y3 1 y1 + y2 +2 y3 = 2 y1-2 y2 - y3-3 y1, y2 0, y3 urestricted Assigmet Steps of Solvig Assigmet Problem- 1. Idetify the miimum elemet i each row ad subtract it from every elemet of that row. 2. Idetify the miimum elemet i each colum ad subtract it from every elemet of that colum 3. Make the assigmets for the reduced matrix obtaied from steps 1 ad 2 i the followig way: I. For each row or colum with a sigle zero value cell that has ot be assiged or elimiated, box that zero value as a assiged cell. II. For every zero that becomes assiged, cross out (X) all other zeros i the same row ad the same colum. III. If for a row ad a colum, there are two or more zeros ad oe caot be chose by ispectio, choose the cell arbitrarily for assigmet. IV. The above process may be cotiued util every zero cell is either assiged or crossed (X) 4. A optimal assigmet is foud, if the umber of assiged cells equals the umber of rows (ad colums). I case you have chose a zero cell arbitrarily, there may be alterate optimal solutios. If o optimal solutio is foud, go to step Draw the miimum umber of vertical ad horizotal lies ecessary to cover all the zeros i the reduced matrix obtaied from step 3 by adoptig the followig procedure: I. Mark all the rows that do ot have assigmets. II. Mark all the colums (ot already marked) which have zeros i the marked rows. III. Mark all the rows (ot already marked) that have assigmets i marked colums. IV. Repeat steps 5 (ii) ad (iii) util o more rows or colums ca be marked. V. Draw straight lies through all umarked rows ad marked colums. 7

8 1. Select the smallest elemet (i.e., 1) from all the ucovered elemets. Subtract this smallest elemet from all the ucovered elemets ad add it to the elemets, which lie at the itersectio of two lies. Thus, we obtai aother reduced matrix for fresh assigmet. Note- For maximizatio problem the matrix is coverted ito miimizatio problem by subtractig each elemet by the elemet havig maximum value i the matrix. Queuig Theory Arrival rate is λ ad mea service rate is deoted by µ. Kedall s otatio:- (a/b/c) : (d/e) a = Probability law for the arrival time b = Probability law accordig to which the customers are beig served. c = umber of chaels d = capacity of the system e = queue disciplie. Traffic Itesity( ρ) = Mea arrival rate/ Mea service rate = λ / µ Formulas of Queuig Theory 1. Expected umber of customers i the system (Ls) = λ / (µ - λ) 2. Expected umber of customers i the queue (Lq) = λ 2 / µ (µ - λ) 3. Expected waitig time for a customer i the queue(wq) = λ / µ (µ - λ) 4. Expected waitig time for a customer i the system (Ws) = 1 / (µ - λ) 5. Probability that the queue is o-empty P(>1) =( λ / µ) 2 6. Probability that the umber of customers, i the system exceeds a give umber, P(>=k) = =( λ / µ)k 7. Expected legth of o-empty queue = µ / (µ - λ). Sequecig ad Schedulig Sequecig 1. Johso s Problem 1. 2 machies ad jobs Step 1: Fid the miimum amog various ti1 ad ti2. Step 2a : If the miimum processig time requires machie 1, place the associated job i the first available positio i sequece. Go to Step 3. Step 2b : If the miimum processig time requires machie 2, place the associated job i the last available positio i sequece. Go to Step 3. Step 3: Remove the assiged job from cosideratio ad retur to Step 1 util all positios i sequece are filled machies ad jobs two coditios of this approach The smallest processig time o machie A is greater tha or equal to the greatest processig time o machie B, i.e., Mi. (Ai) Max. (Bi) The smallest processig time o machie C is greater tha or equal to the greatest processig time o machie B, i.e., Max. (Bi) Mi. (Ci) If either or both of the above coditios are satisfied, the we replace the three machies by two fictitious machies G & H with correspodig processig times give by Gi = Ai + Bi Hi = Bi + Ci Where Gi & Hi are the processig times for ith job o machie G ad H respectively Time Study- Normal time = Observed time x Ratig factor Stadard time = Normal time + allowaces 8

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