Last Minute Revision For Operations Research April 2014 Exams

2mrw 0.R. at 3.00pm

Carry
★hall ticket
★I.d. card
★2 pens n 1 black pen
★scale pencil eraser
★calc(dont forget)

All d best… u will rock 🙂

Last day revision.
OPERATIONS REAEARCH

# Simplex – If all replacement ratios are negative, then the solution is Unbounded.

# Simplex – If ▲j value of non basic variable is 0, then there is Alternate solution.

# Simplex – If solution value (bi value) of basic variable is 0, then Degeneracy exist.
# Transportation – In penalty method always select maximum penalty and then lowest cost with maximum demand/supply.

# Transportation – 
Cij = Ui + Vj
▲ij = Cij – (Ui + Vj)
Start Loop from most negative (-) value of ▲ij.

# Transportation – 
RIM Condition (m + n – 1) = no of occupied cells.

# Transportation – When RIM Condition does not satisfy Degeneracy occurs and Epsillon will come.

# Transportation – Epsillon should be added to an unoccupied cell with least cost where either Ui or Vj one value is known and other value is unknown.

# Transportation – When the value of an unoccupied cell is 0 in final answer, Alternate solution exist.

# Transportation – When in question Profit is given instead of Cost start it by making Regret table and then in last in transportation schedule multiply Quantity with Profit.
# CPM – 
ES = Tail □
EF = ES + Duration
LF = Head △ 
LS = LF – Duration

# CPM – Floats
TF = LS – ES
FF = TF – HS
INDF = FF – TS
INTF = HS

CRASHING REVISION: punch line crash more and more days with minimum increase in cost. Rule no 1 u can crash only ur critical activities . Rule no 2  first crash the critical activity wit minimum cost slope. Rule no 3 (applicable wen there r more than 1 critical path) either we crash common act or two uncommon act wit same no of days. Key point 1.  while crashing make sure ur hero always remains hero . Side hero can become hero but hero can never become side hero 2. Direct cost increses and indirect cost decreases while crashing.

TEST OF OPTIMALITY 
1. Only for Occupied Cells
find Ui, Vj using Cij =Ui+Vj
(Ui pata hai ya Vj? Jo pata hai usko cost m se minus karo)
2. Only for unoccupied cell
Find Opportunity cost.
Δij= cij- (ui+vj)
( ye unoccupied hai? Iska Ui kitna hai? iska Vj kitna hai? Total of Ui + Vj?
Cost – Total = Δij)
If all Δij > or equal to 0, solution is optimum. If not, go to nxt step 
3. MODI
a. start loop from most negative value of Δij ( Khali dabba)
b. Loop can go up or down
c. Ensure that all corner points are Occupied cells.
d. Khali dabbe me + sign. then alternatively – + – on corners.
e. Least allocation amongst cells to whom – sign is assigned will b added or deducted from allocations
f. For nxt table, sabse pehle un Occupied cell ko chap jisko loop ne affect nhi kiya. thn mak changes in allocation due to loop. Check if no. of occupied cells is same

Steps in simplex 1 write standard form (to convert constraints into=by a adding  slack , surplus and artificial var) 2 make initial simplex table ( wit help of standard form) 3 cal zj ( our formula is this × this + this × this …) 4 cal delta^j ( cj _ zj 5 test of  optimality (all delta j shld b <=0) if nt optimum 6 find key column (select col wit most +^j ) 7 find RR (bi/□) 8 find key row (least+RR) 9 replace out goin var wit incoming 10 find new row ( KR/KE) 11 make remaining element of key col 0 by making working note and copy same if its already 0 . Formula for working note is old val_ (key col val×new row)   after this new table is ready so again start same procedure from  zj ……

Source : TYBMS students